[8]Demidovich B Problems in Mathematical Analysis Mir 1970

MIRPUBL1SHF..RS r. c. C. M. c. n OeMU008U'l" B. A Et/JuMeHf(O, Jl JlyHU,. E cP nOpwHeea, E Cbllieea, B. et>PO/lOB, P. fl. lllocmaK, A. P. f/nnoAbcKuiJ. napaHeHK08.. B. KOcUH, r n. 3AJlAlIH 11 ynPA)I(HEHI1S1 no MATEMATHLJECKOMY AHAJll13Y nOG penaKt4lJl't1 B. n. A E M 1-1 A 0 B M q A rocyiJapcmseHHoe U3aameAbCmpo AU mepDlIlYPbl t/JU3UKO-ManteMamU'l.eCKOil M OCK.6Q G. Baranenkov, B. Drm;dovich V. Epmenko, S. Kogan, G. Lunt~) E. Por~hnlva, E. ~!/(:h(ta, S. fl0lCJv, R. ~ho~takt A. Yanpolsky PROBLEMS IN MATHEMATICAL ANALYSIS Under the editorship of B. DEMI DOVICH Translated /rorn the Russian. by G. YANKOVSKY MI~ PUBLISHERS Moscow TO THE READER M/R Publishers would be glad to have yo"r opinion of the translatton and the design of this book. P lease send your suggest tons to 2. Pervy Rlzh~kg Pereulok, Mo~(.ow, u. S. S. R. Second PrintiRR Printed In the Union of Soviet Socialist Republics CONTENTS Preface • • . • • • • • • • 9 Cizapter 1. INTRODUCTION TO ANALYSIS Sec. Sec. Sec. Sec. Sec. 1. Pu nctions 2 3 4 5. . . . . . . Graphs of Elementary Functions Li mits . Infinitely Small and Large Quantities. Continuity of Functions . . . . . . . . . . . 11 16 22 33 36 Chapter II DIFFERENTIATION Of ·FUNCTIONS Sec 1. Calculating Derivatives Directly . . . • • . . . . . . • Sec 2 Tabular Differentiation Sec. 3 The Dprivat'ves of Functions Not Represented Explicitly Sec. 4. Geometrical and Mechanical Applications of the Derivative . Sec 5 DeflvatlveC) of Hig ler Orders ..... Sec 6 DifTerE:'ntials of First and Higher Orders Sec 7 Mean Value Theorems . Sec. 8 Taylor'\) Formula Sec 9 The L'Hospital-Bernoulli Rule for Evaluating Indeterminate Forms ,. . ...•.. 42 46 56 60 66 11 15 17 78 Chapter III THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF A DER IVATIVE Sec. Sec. Sec Sec Sec. 1. The Extrema of a Function of One Argument 2 3 4. 5. Chapter IV Sec. 1 Sec 2 Sec 3 Sec. 4 Sec. 5. 83 The Dirfct ion of Concavity Points of Inflection 91 AsymptoteC) . . • • • • 93 Graph Ing Functions by Characteristic Points • • • • • 96 DilTerential of an Arc Curvature • • • • • . • . • • • . 101 I~DEFINI"E INTEGRALS Direct lntellration . . by Substitution • • Integrat 1 0n by Part~ . . . . . Standard Integral'\ ContaininR a Quadratic Trinornlal Integration of Rational Functions • • • • • • • • • • Inte~ration • . . . • 107 113 116 118 121 Contents Sec. 6. Init'grating Certain IrratIonal Functions •• 125 ~€c 7. lr;t£~rating 1 rifonrrretric Functions . . • •• 128 ~ec. 8 Integrat ion of H Yferbolic Functions . . . . . 133 Sec 9. USIng I ngonometric and Hyperbolic Substitutions for Finding Integrals of the Form ~ R (x. Vaxl+bx+cl dx. Where R tional Sec 10 Sec 11 Sec. 12. FunctIon . . . . . . . . . . . . . . . . . . . . I ntf,:rat ion of Varlouc; Transcendental Functions Using Reduction Formulas. ..... MIscellaneous Examples on Integration . • • • • IS a Ra133 135 • • 135 136 Chapter V DEFINITE INTEGRALS Sec. 1. The Denn ite I ntegral as the Limit of a Sunl . • • • • • • • 138 ~€c 2 Evaluatll";~ [(tirite Intfgrals by N.eans of lndeflniteIntegrals 140 Sec. 3 Improper Integrals . . . . . . . . . . . • • 143 Sec 4 Charge ~f Vari2ble in a Dffinite Integral • 146 ~ec. 5. In1f~ration by Parts . • • • • • . . . • 149 Sec 6 Mean- Value Theort'm . . • • • • 150 Sec. 7. The Areas of Plant' Fi~ures • • 153 Sec 8. The Arc Length of a Curve • 158 Sec 9 Volumes of Sol ids . . . . • • • • 161 5ec 10 The Area of a ~urface of RE'volution • 166 Sec 11 Norrfnts Centres of Cravity CJuldln's Thforems • • 168 ~ec 12. Applyir.g [ef!r.ite Int€grals to the Solution of Physical Prcblems •• • • • . • • • • • . . . • • • . • . • . • 173 Chapter VI. FUNCTIONS OF SEVERAL VARIABLES Sec. Sec. Sec Sec Sec Sec. Stc. Sec Sec Sec Sec. Sec Sec. Sec Sec Sec Sec. 1. Basic Notions • • • 2. Contlnulty 3 4 5 6. • 180 P.artlal Derivatives • Total DlfJerential of a Function ...•• Dlffprpntiation of Comroslte F-unctions •• Dt-f1Vatlve In a GIven DirtAct10n and the Gradient of a Function 7 H I~lfel -Crder Cenvat lVfS and Differentials • • • • • 8 I ntegration of 10tal D;f;erentials • • 9 Dif.erentiation of I mpllclt functions. • • • • 10 Chanf!e of Variables . . • . . . . . . . •• 11. The Tangent Plant' and the Normal to a Surface • 12 1 aylor'~ Formula for a F-un('tion of 5pveral Variables 13 The Ex tremum of a FunctIon of Several Variables • 14 Fir dlf:~ 1ht- Createst and c. n~allt"st Values of Functions •• 15 Slnl'ular POInts ot Plane Curves • 16 Envelope • • • • . • 17. Arc Length of a Space Curve •• , •. 184 185 181 190 193 197 202 205 211 217 220 222 227 230 2.32 234 Contents S~c. 18. The Vector Function of a Scalar Argument • Sec. 19 The Natural Trihedrnn of a Space Curve Sec. 20. Curvature and Torsion of a Space Curve • 7 · 2JS • 238 · 242 Chapter VII. MULTIPLE AND LINE INTEGRALS Sec. 1 The Double Integral in Rectangular Coordinates . · . . 246 Sec. 2 Chan~e of Variables in a Double Integral • •• 252 Sec. 3. Com pu ti ng Areas • . • • • • • • • • 256 Sec. 4. Computing Volumes . • • . . • • • • • • 258 Sec. 5. Computing the Areas of Surfac~s . • . . . · 259 Sec. 6 Applications of the Double Integral in Mechanics • 230 Sec. 7. Triple Integrals . 262 Sec. 8. Improper Integrals Dependent on a Paralneter. Improper Multi lIe Integrals. . • • • •• 269 Sec. 9 Line Integrals . • • • • . . . . · 273 Sec. 10. Surface Integrals • . . . . . . • · 284 Sec. 11. The Ostrogradsky-Gauss Formula • • 286 Sec. 12. Fundamentals of Field Theory · 288 Chapter VIII. SERIES Sec. Sec. Sec. Sec. 1. Number Series 2. Functional Series 3. Taylor's Series • 4. Fourier's Series • •• 293 · 304 . . . . • . 311 · 318 Chapter IX DIFFERENTIAL EQUATIONS Sec. 1. Verifying Solutions. Forming Differential Equations of FaOli· lies of Curves. Initial Conet itions . . . • • • • . . . . . . . . . Sec. 2 First-Order Differential Equations • • • • • . • • . • . . . Sec. 3. First-Order Diflerential Equations with Variables Separable. Orthogonal Trajer.tories • • . . . . . . . . . . . . . • • . . . . Sec. 4 First-Order Homo~eneous Differential Equ:Itlons • • • . . • Sec. 5. First-Order Linear Differential Equations. Bernoulli's Equation . . . . . . . • • . • • . • • . . . . . . . . • • • . • Sec. 6 Exact Differential Equations. Integrating Factor • . • • .• Sec 7 First-Ordpr DIfferentIal Equations not Solved for the Derivative Sec. 8. The Lagrange and Clairaut Equations .• Sec. 9. Miscellaneous Exercises on First-Order Differential Equations Sec. 10. I-ligher-Order Differential Equations • • • • • • • • • • • . Sec. 11. Linear Differential Equations • . • • • • • • • . . • • • • Sec. 12. LInear Differential Equations of Second Order with Constant Coefficients • • • • • • • • • • • • • • • • • • • • • • • • • • • 322 324 327 330 3::12 335 337 339 340 345 349 351 Contents 8 Sec. 13. Linear Differential Equations of Order Higher than Two with Constant CoefficIents .....• . . Sec 14. Euler's Equations . . . . . . . . • • • • • • • • • • • • • Sec 15. Systems of Diff'l'rentlal Equa tions . . . . • . . . . . . . . Sec. 16. Integration of Dit'ien'ntial Equations by Means of Power Series . . . . . • • . . • . . . . Sec 17. Problenls on Fourier's Method • • • • • • • • • • • • • • 356 357 359 361 363 Chapter X. APPROXIMATE CALCULATIONS Sec. 1 Operations on Approxarnatc NUlllbers • • • • • Sec. 2. Interpolation of Functions . . . . . Sec. 3. ConlputtT1f.! thc,Rcnl Roots of EquCltions . • • ~ec. 4 NUlllprital, In tet!ration of FUllct ions . . . . . . Sec. 5. f\un encal Inte~'r(ltion of Crdlrary Dtf'1t:rtntial Sec. 6. ApprO}dITlating P<. tlfllr'S Ccefficlents • . . • • • • . . " • Equations •• . 367 372 • • APPENDIX . . . . . . • · I. Grrcl< A1ph a bpt . · 11. SOlne COl1~tants . · III. Inverse Quantities, Powers, Roots, Lo~arithlns . · IV Trigollollletric Funct ton~ • • . . • • • • • • • • . V. EXrorel~ tIaI, 11 y ~erbolic and Trigon01l1etric Functions • • • • • VI. Some Curves . • . . • • . • . . • . • . . . . • . • · 396 ANSWERS .....• • • • • 376 382 384 3Q3 475 475 475 476 478 479 480 PREFACE This collection of problems and exercises in mathematical analysis covers the maximum requirements of general courses in higher mathematics for higher technical schools. It contains over 3,000 problems sequentially arranged in (~hapters I to X covering all branches of higher mathematics (with the exception of analytical geometry) given in college courses. Particular attention is given to the most important sections of the course that require established skills (the finding of limits, differentiation techniques, the graphing of functions, integration techniques, the applications of definite integrals, series, the sol ut ion of differential equations). Since some institutes have extended courses of mathematics, the authors have included problems on field theory, the Fourier method, and approxima1e calculations. Experience sho\vs that the nunlber of problclns given in this book not only fully satisfies the requirclT:€n s of the student, as far as practical 11las~ering of the various sections of the course goes, but also enables the instructor to supply a varied choIce of problems in each section and to select problenls for tests and examinations. Each chap.er begins with a brief theoretical introduction that covers the basic definitions and formulas of that section of the course. Here the rnost inlportant typical problems are worked out in full. We beli('ve that this will greatly Silllplify the work of the student. Answers are given to all computational problems; one asterisk indicate~ that hints to the solution are given in the answers, two asterisks, that the solution is given. The problems are frequently illustrated by drawings. This collection of problems is the result of many years of teaching higher mathelnatics in the technical schools of the Soviet Union. It includes, in addition to original problems and exam· pies, a large nunlber of commonly used problelTIs. . \). Determine the donlaln of definItion of the funchon I Y= V x2-1 . Inverse functions. x I> 1. 8S real. *) Hencelorth all values will be conSIdered stated. b). and la . Functions 1°. Rational and irrational numbers are collectively known 1 he ab\olutf value of a real nurnber a IS undt'rstood to be the nonnegative numb~r I a. defined by the conditions· I a I = a if a ~ 0. 2°. AI~o pOSSible IS a ITIOre complex structure of the dOllJ3111 01 defInition of 3 function (see./e-vQlupd functIon. bl. Fronl now on we shall u~e the \\ford "functIon" only in the meaning of a ~ln[!. Solution. In the slnlplest cases. If the equation y = f (x) may be solved unIquely for the variable x. 1hen y is said to be a funcllon (~lnRle-valued) of x or a dependent 1 artable def! ned on the ~et E. for Instance. x is the argument or tndependent vanable The fact that II IS a functIon of x IS expressed In brief form by the notation y = f (x) or y == F (. Probleln 21) Exanlple 1. wh Ich :s the set of real nuolhers that satisfy the In('quallLes a < x < b. the' domaIn of a function IS either a clofied Interval [a. and the I'ke If to every value of x belonging to somt» set E there corresponds one or several values of the vanabl~ !i. that is. The function is defined if x 2 -1 > 0.I ~nd I < x < + 00 4°. if not otherwise .00 < x < . the domaIn of the function is a set of two intervals: . Thus. = . Real nurrl:ers. lhe following in(qual1ty holds for all real numbers CJ as real numbers and b: 'a+bl~lal+'bl. then y is called a multIple-valued functton of x defined on E.Chapter I INTRODUCTION TO ANALYSIS Sec.a if a < o. 1. Definition of a function. If to every val ue *) of a variable x. if there is a function x =g (y) such that y == f [g (y». if . which is the set of n~al nUlnbprs x that satisfy the inequal itJes a ~ ~ ~ b. The collection of values of x for which the gIven functlon IS dt. that is. wh ich belongs to SOllie collect Ion (set) E. or an open tnterval (a.\Hned IS called the dOTnaln 01 de{tTulton (or the domain) of thiS function. 11 not other\vl~e stated 3° The domain of deltnition of a function. there corresponds one and onl y one finite value of the quantity y. Corrros it e and ireplicit functicns. (1) Solution.co < Y < 1. etc.:o an ImplIcit function.~)./( ~).1(1). b) Ix--t-1/>2. =. 1**. {(-x). A set of poi nts (x. b) + laI =a 2 2 . c) I~ I= ::: d) V£i2=-. . Solve the inequalities: a) Ix-II<3.. Prove the following equalities: a) Iab I = I a 1·1 b I. The function {(x) is linear.x =1-y and x=- IO~~~. if f(x)=x ' -6xl + Ilx-6. Find this function. 4. that is. Prove that if a and b are real numbers then Ilal-l b II ~ la-b I ~ lal+lb (. Solving equation (1) for x. Find 1(0). is the tnverse of Y=f(x). A function Y of x defined by a ~e­ ries of equalit1t~sy=f(u). Obviou~lv. {(4). Find f (~ ). A function d€fined by an fqu~tion not solved for the derencent variable is calh.3. 1(3). y =g (x). Find /(-1). 2. I( . Cetennine the inverse of the funchon y= 1-2. (b =1= 0).J (y) ~uch that y == f If-) (y)J for all Y that are values of the function f (x) I:xhn Jjle 2.Y) *).:\al. 7. {(2). For €xam~le. f~X)' if l(x)=V1+x 2 • 6. tre rquation y = f (x) refines a multiple-valued Inver~e fupct Ion x = f. 3. g(f(x))=:x. is called a comoosite function. we have 2. in standard notation. is caIJed the graph of the given funct~on. or. if /(-1)=2 and f (2) = . 1(10). whose coordinates are connectEd by the equation y (x).x .(x) is the Inverse of g (x) (a nd v ice versa). • ) Log x is the logarithm of the number x 10 the base 10. Obviou~ly. (2) tl-'e dotr'ain of cefinition of the function (2) Is . f (x) = arc cos (log x). d) Ix-II<lx+ll. [(0). c) 12x+ 11< I. whereu=cp(x). f (1).. The graph of a function. or a function of a fun~tiol. the function . 5°. 1 then the function x = g (y). 6°. the equation x' + yl= 1 defines y as an implicIt function of x.Introduction to Analysis 12 (Ch. 5. In 11 r J ereJ al ca~e. y) in an xy-plane. y=V2+x-x . Find 4 q> (x) 1 = 2" (I (x) -t. Prove that any function f (x) defined in the interval . x-l2 2x 19.x) J. 2 b) y~x x -2. • • ( X) 20. 14**. Given that f (4) == . y==log 13.xl . Y= log ~+~ V -V ~ 18. x. Y= b) Y= ~ x+ 1.X ). 9.l < x < 1 may be represented in the fOfln of the SUln of an even function and an odd function. Y-=4--2' 17. V(x+ 1)1+ V(x-I)I. if f(O)==1. f (5) = 6. Determine the domains of definition of the following functions: 11. 3) if we consider the function J (x) on the interval 4 ~ x ~ 5 linear (linear interpolation 01 a function). 16. f(x)=2x -3x ' -5x 2 +6x-l0. 3 +2 x .2.f ( .x) = I (x) and odd if f (-x) == . f (x) defined in a symmetric region . A function is called even if Determine which of the following functions are even and which are odd: a) f (x) = ~ (a~ +a. a) Y= 12. /(1)=0 and f(3)=5.x) ) and 1 '1'( x) = 2" [/ (x) . y=arccos l+x' l 1 Y= V -x+ Y2+x' 15. V X-Xl. Determine the domain of definition of the function Y= Vsin2x. 22. Find the rational integral function f (x) 01 degree two. Vx+ 1. Write the function f (x) = { 0. Y= arc SIn log 10 • 21. Approximate the value of 1(4. e) l(x)=log(x+Vl+x J ). a) Y = x 2 -2. b) f (x) = VI c) I(x)= d) + x +x'. 23. l+x f (x) = log -1-x.l < x < I f (. If x>O as a single formula using the absolute-value sign. . 10.Vl-x +x 2 .I (x).I(. ~f x ~ 0. 24.Functions Sec 1) 13 8. and for tLe perlcdic ft:ncticns flnd tt. = sin (VX).(Ch. Construct the graphs of these functions. Xl' XI' J (Xl)' x. d) f (x) = e) / (x) sin l x. Ce1ermine \\hirh of the rol1ewing functions are perIodic. The linear density (that is. and that the product of an even function by an odd function is an odd function. 26. Show that if f (x) =kx+ b and the numbers the nurn ters gression. if f(x-l)=x l • 32.'W(x)) and ¢. Express the mass m of a variable segn. f (x) = Vtan x.. D C 28. Construct the graph of this function.1)-1 (n) = O. 27. 29. 33.ent AM =X of this rod as a function of x. mass per unit length) of a rod AB = I b (Fig. 2) on the segments AC = I" CD == 12 and DB == la (L \ + I" + La ==. I Introduction to Analysis 14 25. 2 Fig. Express the length of the segment y.l) A is equal to q J' q 2 and q a' respec/3 c ~--a--~ Fig.(x)). f (x) = a sin AX + b cos AX. Find f(x+ 1). Let f (n) be the sum of n terms of an arithmetic progression.q.= M N and the area S of the figure AA1N as a fun~tl0n of x= AM (Fig 1). Show that / (n + 3)-3/ (n + 2) + 3/ (n.eir least period T: a) b) c) f (x) = 10 sin 3 x. form an arithmetic progression. then / (XI) and f (XI) likewise form such a pro- . Prove that the product of two even functions or of two odd functions is an even fl'nctlon. 1 tively. A function f (x) IS called periodic if there exists a positive numter T (the period of the tunction) such that f (x + T) = f (x) for all valres of x within the de main of definition of f (x). 30. Find f tf If (xlI}. if <p(X)=K and 'I'(x)=2 x . Find q. if f (x) = 1 ~x • 31. i-x Show that f (x) + f (y) = f ( . b) y = 2 -t. Let 34.x. l+x f(x)= log .Xl . and the numters x" XI' XI form an arithmetic progression. d) y = log ~ . trIgonometric. d) y = arc sin (3 -X2 ). f(1) if f (x) = { arc sin x for-t ~ t" ~ 0. then the numbers f (Xl)' f (x 2 ) and {(x. Let cp (x) = 2" (aX + a-X) and Show that 'P (x) = 21 (aX _a-X).cp (y) 'I' (x). .) form a geometric progression. Prove that if f (x) = aX (a > 0). if x~O. J] Functions 15 f (x) is an exponential function.3 . b) y=x 2 -1. Find the inverse of the function I X. 1 +x - · 39. d) y = x· . Find /(-1). {(O). and the like): a) y = (2x-5)10. rp (x + y) = cp (x) <p (y) + 'P (x) '" (y) and '" (x + y) == cp (x) 'I' (y) +. that is. F:nd the inverse of the function y if: a) y = 2x + 3. 37. b) y = 2cos x. . 41. Write the given functions as a series of equalities each member of which contains a simple elementary function (po\ver. 35.3x. c ) Y=v .Iog-c) y = I-x -t- Xl.Sec. y =) Xl. In what regions will these inverse functions be defined? 40. exponential.i/-l. arctanxforO<x ':+00. 38. 2x e) y. x c) y = log tan 2 .:~) · 1 36. Determine the roots (zeros) of the region of positivity and of the region of negativity of the function y if: a) y == I -t. if x > o. 2 c) y=arctan3x.x .X . u=Vv.. explicitly. if u>O... Calculations are best done by a ~ltde rule. .2. U= sin x. Graphs of Elementary Functions Graphs of functions Y= f (x) are l11ajnly constructed by marking a sufficiently dtln~e net of points Nli(xit I/i). 43. y -.. Proceeding from the graph of Y=f (x). functions of y defined by the equations: a) arc cos y =---1(. 2) !I.f (x) is the mirror image of the graph r about the x-axis... 2. y 0. Write.. . v=logx..y. Find the domains of definition of the given implicit functions.. . ... . if u ~ 0.s 16 . . b) y=arctanu.:[::.I_n_t.. Write as a single equation the composite functions represented as a series of equalities: 8) y=U'. (f) we get the graphs of the following functions by means of simple geometric construct ions: 1) YI = . tt=x'-l. ..._od_u_c_t..._I 42.=f(xj) (i==-O._on_t_o_A_n_a~ly:-. = f (-I') is the mirror image of the graph r about the y-axis.s_.. Sec. c) = { 2u.. 3 Graphs of the basic eleJnentary functions (see Ap pendix VI) are readily learned through their construction.. r- c) x+1YI=2y. ) and by connecting the pOInts with a line that takes account of intermediate pOints. 1. . b) lOx + 1()Y = 10. where Y. x Fig....-C_h_...... Construct the graph of the function y = sin ( x .2. -1. -I. 46. 51*. 1) 2. -1. y = 1. 55. = f (x-a) Is the r graph displaced along th? x-axis by an amount a. y=(X-x o )2. 2) 17 3) Y.2. -2. Example. if: I) a=l. b=-2. 1. Y = x 2 + C. ii xo=O. 2.1)2.5'x -}. Y = x 4 • 57.2. The desired line Is a sine curve g = sin x displaced along the x-axis to the: right by an amount ~ (Fig. 48. y = 2 t.. 50. Y = x 3 (cub ic parabola). I) 2. Construct the graphs of the following rational integral tunction~ of degree above two: 53*. y=ax 2 . 45. + 4) Y4 = b I (x) is the r graph displaced along the y-axis by an amount" (Fig.: ) . Ij=x+ b. 3). y = x'. -2.3x -t. 4 Construct the graphs of the following linear functions (straight I inps): 44. if Yo = 0.x-x2 • Fi nd the points ot intersect ion of th is parabola with the x-axis. y=kx. . y=+. . 56. 1. c=3.1/2. 52. Construct the graphs of the following linear fractional functions (hyperbolas): 68*. Y = Yo + (x . 4) x Fig. 1/2.I. if b=O. Construct the graphs of rat ional integral fun. Solution.tions of degree two (parabolas).Graphs of Elementary P'J. if Q= 1.nct. if c = 0. O. 54.2. y=2+-(x-I)'. -1. if k=O. y=2x l -x4. 49. c=O. b=6. 47.oni Sec.2. 2) a=-2. y=axl-~-bx+c. -2. 1.1. . if n= 1.cot x. [eh.. 84*. y= x+2. ~. y= f V25-x· ± V xZ-l (elltpse). x 67*. (hyperbola). x~. 1/2. 82*. y = 5 sin (2x-3). 1 77. y= x+ 2x . r 79. y=y o -+.on to Analysis 18 1 59. 86. 3. 3. 88.oid of Diodes). y=-tanx. y= J/72. y=sec x. 1/2.2~ 1 (Newton's serpentine). y= V 1. y= ± x (c. 85*. 73*. x 65*. if A = 1. ~. y= A sin x. -2. 87*. x2 + 1· !I = . y=sin(x-q». y= I-x.<. 1 . yo::::a-l.: .duct. 2. 1 69. y= ~ x Vb (cis. . 74.. (trident of Newton). 10. X 2x-3 62 * · Y=3x+2" o Construct the graphs of the fractional rational functions: 1 63. 89*. 61*. y=cosx. '1= v x. y= • x . 83*. 68. · 64. y= x 1 66.Xl · 78*. y= ViZ (Niele's parabola). y. y = x.l~ 1 (W itch of Agnesi). m=6. Vi VX- ± 75*. y= sin nx. 81*. if xo=l. y=cosec x.. y = Xl +!.emicubical parabola). x-2 60. n. y=x+-. 70. 76. y= ± xV 25-x 2 • Conslruct the graphs of the trigonometric functions: 80*. Y = sin x. if q>=O. y =.Intr. Y = Construct the graphs of the irrational functions: 71*. . 92*. Y = logl x. Y= cos x+ 2 cos 2x.x sinx. e(e=2. y=co~hx. functions~ . y=logx 2 • 110. 97. 108... x 120*. y=arc sinx. y = 2 (x x I)· +. y=2- 1 x 2 • 109. if Q= 6. . b) y=logv7Ixl. 95. 127. g= sinx-{ sin3x.. 128. n 99*.. . y=arctanx. 123. y=log(logx). 124. y=log(cosx). 93*. . y=x+arccotx. Y= lOx. y = tan! x. y=log-. 3-X2 \vhen Ixf~ I. where co~hx=1/2(ex+e-X). y=log(-x). V lOt.. Y = e. when Ixl> 1. 21 Graphs of Elementary ftlnctions 19 90*. y = 1-2 cos x. I 98. Y = x sin x. Y= a sin x+ b COS X. Y= x+ sin x. 718. sinh x 105*. b -= -8. x 1 106. Y = sin x -1. e.x2 (probab ility curve).l j = -og l. Y = cos -x . x I 119*. g=a". 129. y=arcsin. . if a= 10. 1 113. 91. 115. y = arc cot x. Y = tanh x. . Construct the graphs of the functions: 125. 104*. 117. x Construct the graphs of the inverse trigonometric lt8*. Y = log! (l + x). y=2. y=cos1x. lIt. a) y=sinx+lsinxl. 1 112. where sinhx=lj2(eX-e-~. 103*. )*).cos x. J07*. a) y=xlxl. 96.Sec. 100. 122. Y=lxl. Y = arc cos x. 22 for more details. *) About the number e see p. y=arccos-. 1 126. x 114. 102*.. g= lo&z x. Construct the graphs of the exponential and logarithmic runetions: 94*.:. 2. y= { .. y = ± sin x. where tan h x = -cos h. b) y=slnx-lsinxl. if a=2. y=sinhx. 116.!. 121 *. Xl = co~ y. . 13S*.(straight line). . y=a(slnt-tcost) (involute of a circle ). q> 135. (hyperbol ic sptral). 1 130. y = (' 142*. X=l~tal Y=l~:a (folium of Descartes).Xl + yl = 25 (c i r (l e) . y= sin t (ellipse). X= 10 cos t. 133*. Constrt:ct the graphs of the following functions in the polar coordina.(branch of a hyperbola).1 = a l eos 2cp (a > 0) (lemniscate). r = 2 cos 'P (cirlle). the grea~est in:eger less than or equal to x. 136. 149. y=t 2 __ t a• 150. x==t-t 2 . 1 t 147. r = _1. 158. <p) (r ~ 0): 131. where [xl is the In~egral part of the nurnLer x. 100+64=1 (elllPse). that is. CJnstruct the graphs of the following functions defined implicitly: t 51 *. 145*. b) y=x-[x]. y=2s1n 2 t (segntent of a straight line).. t slnt). x~~t +2-'. . 143*.. r = 10 sin 3cp (three-leafed rose) 139*. 143*. 155. 134*. 15l. x = t a. x+y=-lOlogy. a) y=[x].e system (r. r=sec! ~ (parabola). 149.X Z ). y=2 _2. r=a(l +-coscp) (a>O) (cardioid). x=2co~Zt. Y = 10 sin a t (astroid). x==-~J2cost-co~2t). 151*. CJns~ruct the graphs of the functions represented parametri- cally: 141*. r (spiral of Archimedes).. x=a(cost+ (sernicubical parabola). I 2 2 156*. . =! 132*. y=a(2sint-sin2t) (cardIoid). y! = Xl ( 100 . + y"3 = a3" (astroid). 146 • X = rot a Ir . Y= r 1 + /2 y2 I • at ( I + tl 0 154. 153*0 yJ=2x (parabola). I) se"LtCtrc e . x .Introduction to Anall/sis 20 (Ch. x = 10 cos' t. 144*. xy= 12 (hyperbGla). r =!:. r·":= e<P (Iog'lrithmic spiral). Sin q> 137. r = 1. 165. Construct the graph of the function obtained. 162. A '--'-----~ 8 Fig. y=cosx. d) x" y= 10. d) lO-x=x. c) x!-x+y=4. Solve the systems of equations graphically: a) xy = 10. 164. Give a graphic solution of the equations: a) 2x"-5x-r2=O. altitude h = 6) is a rectangle (Fig. b) xy = 6. Plot the grapn of this function and find its greatest value. Inscribed in a triangle (base b == 10.!. c) logx=O. 159*.lx. . 5). Derive the conversion fornlula trom the Celsius scale (C) to the Fahrenheit scale (F) if it is known that O°C corresponds to 32°P and 1000e corresponds to 212°F. e) x = I -+ 0 5 sin x. + + e) Y= sinx. 6). ABC as a function of x. x+ y"= 6. Express y = area 6. 160*. x -t y = 7. 161. y!-2x=O. Express the area of the rectangle y as a func-tion of the base x.!. (O<x<2n). 163.y" = earc tan . b) XS + x-I = 0. Given a triangle ACB with BC=a. 5 FIg 6 Construct the graph of this function and find its greatest value.-x Z •• (logarlthmtc spiral).Sec. XZ y" = 13. X 8 +y3_3x!J=0 (folium of Descartes). AC=b and a variable angle ?J ACB = x (Fig. 2) 2t Graphs of Elementary Functions VX -t. f) cot x=x (O<x<:t). l)/(n + 1). for every positive nUlnber e there will be a number N =. a IS f (x) = I x I > tv (8). where E i~ an arbitrary positive number 3°. or tJ sequence lim xn=a. The numbers f (a-O) = lim f (x) and J (a + 0) = lim f (x) -P x ~ a -. The limit of a function. the ltmlt on the left of the function f (x) at the point a and thf limIt on the f.!-l such e that for n > N we will have irequality (2) Consequently. The number a is the limit of xJ ' • • . henc~t formula (I) is true. respectively. or lim f (x):= A. We say that a function f (x) .. then we write conventionally x -+ a-O. . The limit of a sequence.. which means that I f (x) I > E for 0 < . Similarly: x if I f (x) - linl f (x) = A. s!ml1arly.. Thus. 3. If x < a and x -+ a. (I) n+l n-+ct) Solution. . A as x ~ a (A and a are numbers). . x -+a if tor every e > 0 we have ~ = ~ (e) > 0 such that If (x). x". also used: CO. x-a I < f) (E)..Introd'lction to Analysis (Ch. 2°.. Limits XI' to. Form the difference 2ft + 1_ 2 = __ 1_ n+1 n+l' Evaluating the absolute value of this difference..0 -? x ~ a+0 . .. Show that Ihn 2n + 1=2. -+ 00 A I < e for The following conventional notation lim x .. One-sided limits. 1 Sec. \ve have: I + 2n 1 1 j n + 1 .A I < e for 0 < I x-a I < ~. the number 2 is the limit of the sequence x n = (2n-t. ..are catted. (8) such that Example 1. . n + C1J If for any e > 0 there is a number N = N Ixn-al<e when n>N. then we Wrtte x a +0. if I > -e n -1 =N (2) (e).ght of the function f lX) at the point a (if these nuolbers eXist). if x > a and x a.2 = n + 1 < e. I) +.)X = X Jim (1 «-+0 + ala =e=2 71828 .0.1.. ••• t is equal to zero. x~a f" (x). Find the limits on the right and left of the functioD 1 f (x) =arc tanx as x -+ o.a x~a 1old: 1) lim [f.)=-~. c) 161. For which values of n will we have the inequality 1 nz < 8 (8 is an arbitrary positive number)? Calculae numerically for a) 8=0. 3) For the existence of the linlit of a function f (X) as x and sufficient to have the follcw lng equal ity: ~ a. .0 2 Obviously. x-.+ 00 the linlit of the sequence 1 1.. 166.. Prove that as n . ) 8=O.. (x)· lim x~a If.. 4 t 1 nz' .!. x~a The following two limits are frequently used: lim sin x = 1 -+ . I f the linlits lim f 1 (x) and lim f I (x) exist. then the following theorems.01... (x) x~a 3) litn + f2 (x)] = '2 (xli = lim f... Prove that the limit of the sequence n x"=n+l (n=1.2. the function f (x) in this case has no limit as x -. it Is necessary f (a-O) =:-f (a+O). f(-O)= lim x x-+ .!. b) 8=0.)=~ X 2 (aJctan.. Solution. x~a (x)! Brn x~a f 2 (x) (lim x~a f 2 (X) :I: 0). We have '(+0)= lim (arctan x ~ +0 and -!. (x)! f 2 (x) J = lim f.OOl~ . x~a x~a lim .23 Limits Sec... (x) x~a 2) lim [f.• Example 2.¥ X 0 and 1 Jim (1 X-+(. (x) + lim f2 (x). n n~oo 173.. 0. 0.t o 00 is unity. b) 8=0. + 3"176. +-21l 172. c) '" 1 1 Find the limits: · ( 171 · Iun 1 2 3 2+2-12n n n~oo.24 Introduction to Analysis (Cia.. 2 -... . .. for a given positive number positive number 6 so that the inequality 8. x -~ X +0 .... some Ix -41< 8 l should follow from Ix. I as n . Give the exact meaning of the following notations: a) liin log x = . + ~ ) .2333. 21~ 1-"3+ 9-27+ . d) 0.233.( 2 V'2. Y V2.n n-l) .I)n-l n . 1 1 2 2 178•. -"2' 3' 2 b) 4 1 (. ..00. lim n~oo 2n + 1 +3 1Z + 1 2n +. ..1. <» n •• +n l • • • . lirn (n+l)(n~2)(n+3) . liln l!. n-+oon 175.'] 177. 6 2n T' "3' '5' . Prove that How should one choose. For which values of n> N will we have the inequality Ixn-ll< e (e is an arbitrary positive number)? Find N for a) 8=0. + ... c) Iinl f (x) = 00.:t3+5+7+"'+'(2n-I)_2n+IJ n+l n-+oo 2 · 174• I1m n+(-I)n -(-1)".2 1<cS? Compute 6 for a) 8=0.2 0. x + 00 ~ 00 170..01. -( 2 Vf. 169. · [ 1 I 1 ( _ 1) n .4-' . lim 1 +2 +3 : n .001.1.23. .. b) 8=0.n • ~ }i~ ( ~ + ~ + ~ + . c) 8=0... 168.01: c) 8=0. b) lim 2x = + 00..001..... Find the limits of the sequences: 1 1 a) 1. . . . 2n=1' . 2) (x - t. .. HIn (2x 3)3 (3x-2)! ---5--- 0 185. x' -=-=.3. +5· 3_8'\: x . Example 1.2) = 1) lim x +2 c::.t -+ x 3 x -1.4c= 8.3)( x =2. 2. 11111 .t+3 188.3.Vn).+1 X2 ° x~oo + 00 +1 189 · 1lIn V --1-. -1 · . -t. 11111 t -+- rJ:: .2-1 · ~ 1 186.tl l x -4 - 3~ +2 z:: lim (z-2) (x .ft ~ 2 x-I · . 182. 1°lIn x~oo (x+ 1)2 hIll .Seco 3] L_l_om_i_t.6) 3+(2 .5 x+ 0 190. 11111 00 x + 00 x 1 2~2-x-1X ~ 187 · 1lIn • x2-5t+ 1 183. where n Is the highest de~ree of these l\oly~omia!so A si 1T1 ilar procedure is also possi ble in nlany cases for fractioos eontaiQfng irrational terms. .!ill) X 3-~-~-~3 t+7fJ a x x2 Example 2o x 1I•III x~rJ::3/~\3+10 V 181. lim P (x) t ~a Q (x) is obtained directly.¥ -+ 2 (x .s _ (V n + 1.l~7 • X · 2x~-3x-4 11In --. then it is advisable to laJllel the binonllal out of the fraction ~ ~:~ once or several times. Example 3.¥ Ai&a-Q .- If P (x) and Q (x) arp integral polynomials and P (0) '# 0 or <J then the lilnlt of the rational fraction lll) :f: 0. 179. -r-+ 1 · n 180. lIru 11 -+ • . But if P (a) = Q (a) =0.J t lOOOx I 1:11 -.V x J • x2 -~ oc 10 +x V "t . htll n -+«J When seekipg the lilnit of a ratio of two integra t polynomials In x as x -+ 00... linl (2x-3) (3t + 5) (4x-6) 3x'+x-1 5)( 4 .r =:: ~ X OCt 2 _L I- • 184. it is useful first to divide both ternlS of the ratio by x n . t Ii 111 ~ too r Y:-==:x====~ -V x + V x -f V-. lim n-fJ(J) n "ill nl .. x 3V-TO == .r -::---==4 ° x+ooX- 00 1 1--1->. . 11-+0 VX . lim (I I . y+1 2 Y~l ° 2O1. % -"14 Vx-I Jim y:. 1lIn x-. lim x %-+0 208.Va _ x-a Um x-a -.. Find lim Vf+X-I . 1m ¥-+. r 205.Vx-3 x2 . ~-+IV. I. --%-+Q(x-a)(Yx+Ya) lim 1 %-+aVx+Va r 2.t-l 206.oduction to Analysis (Cho J ° x'-3x+ 2 11m 195.. Putting I +x=y'. Example 5. -x .4 9 ' 204. froln the denominator to the numerator. lim ~~Q VX" . I1m x V" 1 + x-I -+.1= Y-+ly-I lim y!+Y+I=~.TI x-h x . lirn YX+hh. 8 x -2 203.. a · 197 • I1m h -+ 0 I) x+a a • (X+h)'_X' h • 198.(-2-1)2 Xl I X • Another way of finding the litnit of an irrational expression is to transfer the Irrational term froln the nUlnerator to the denominator.. 200.r~oVI+x_I Solution..I V V x-I V-x + 1 V . 0). X I X The exrressions containing irrational terms are In many cases rational- tzed by introducing a new variableo Example 4... X-+l~ 4-4 x + 3 ' . "We have Yf+X-l Urn 199o 1lIn 0 x -+ J .I 3 .Yl-x 207.Int. or vice versa.). 11m I' xYx -8 X-+l V-x -4 · ° 202. 10 V- Vx-I · 1m ~--. I 2 Va (a> r1m 3-Y5+x v--. x2_(a+ 196 • IlIn 3 x -a x . x -. x-+41- 5-x VI +x. X x.¥ -.CPs a • a · 225. 213.. It is taken for granted that lin] Sin x = Sin a and lim cos x ==C<JS a_ -'~" x-+a Example 6. lim ¥-.. a) lim b) 1Lll . 11 : 1 1 . eX) x -+ Cf) sin 3x - sin 5x .1-+ .. 7 a) I·lIn x sIn .. I1m ..c . 212... 228. X-+ 0 x . lim sin x-sin a x-a tan 3tX 1· 22 4· . V-x . x~eX) The formula Urn x Sitl -+ 0 x x= I i~ frequently u~E'd when solvin~ the following examples... .y x + 2x-6 1. t b) Ij'n x sin x.5) = 1.. 229... 211..¥-++ aJ 2 2 In) 4 2 JC~3 . 220. o X . lim (1 -.{n 5x . - z 222 • 11m . IS) ¥~+oo V 1- 215.0 x x~o X X -+2 sin X-CflS x i-tan x • • x · arc tcln 2t 23 5.¥ . I I~n--.. ILIl .5== 5. 2 X-+I 218 • 1lin -2. 221.. I en .. O 4)Ul x · l-x2 236 .. lLn cot x .. (V x (x + a)-x). 11 x .-t aJ (V x! -5x -+ 6-x).. v- r1m x+hh h-._~C~2 x+ 2 x2 sin (x r in] 11 ~~- • · art' sin. . liln (x -t Xl)... . IIrn . 22 J. n n-+cx> .. cos mx-cf'S nx 233...x .t --1> 1 Sll1llX . I-sin~ S(fl1(X 219.. n · 231 • I l~n 1..-I .. 1un 2 3t-x 1.. r . lim 214. ILn y x -2t+6.2 C0S x 11..3 .x) tan nx • 217.c('sx 2x cot ( . sin x -.. lim sin 5x = liIn (c. Q \ 5x JC-t> 0 216. Sill -!..o 210. IL11 x <V x! + I-x).t~o x · 232. 1Inl 230. l Sill 311X liin (n sin ~)..t -+ Q • + h)-sin x h I) -+-0 22 6.Vx). cC"s x .3x x-a u 223.Limits ~') Sec.. t a.x) .x 234 ..¥ -++ 3 x-x+ .. . si fl x .. iim (V x x-... 209. I E¥ample 7. lim .: ± ( X -+ "1>Q 00.Introduction to Analysis · 11m 237 x-c. I YI +sinxx Y I-sjnx When taking limits of the form linl (cp (x)]~ JC (X) = C (3) -+a ()ne should bear in mind that: 1) If there are final limits lim cp (x) = A and lim '" (x) = B.. We have 1+ 1 x+l 1 11m . '1' (Xl . then the problem of finding (J the linllt of (3) is 'iolved in straightforward fashion. 1 x %-+0 Example 8. then we x-+a where a (x) c= lim l' -+ x-+a 0 as x -+ '1' (X) 1 lim a (XI {[I+a(x)]a(x)~~(.. ~ +11x lim x. [Ch. X x-+o Solution. cos- liIn 240.. ~. lim (Sin 2X) =2 and Urn (1 +x) = I. 3) if linl q> (x) = 1 and lun 'i' (x) = 00..in 2x :r . 11m x-. X-+Q ~-+a then C= An. hen\.e. Find sin 2x \ I+~ Urn ( .o (Sin 2X)1 +x =2 =2. is Napier's number.. a and..~)~(X)=ex-+a lim [~(x) -IJ =~-+a +a where e = 2..= Urn --=~ -+ 2x + 1 x -+ 00 2 + l 2 x x fI:J aod put q> (x) = I + a (x).... Here. 0 X + S1Il 3. X 1- 239... ~-+o x bence. 2) if lim <p (x) = A #= 1 and lim 'l' (x)::. 0 VCOS"X x2 x -+0 1(X 238.718 . Find lim x-+ c» (X+ 1 )XJ 2x +1 • Solution.) . x+ (X_')~+J 249. it is useful to remernber that lim %-+ . (Xl +2 ).c • .o .x CJ) t . We have t--!. _X)n 250.!J)x+l. (X-I)~ . lim x-I = lim _ _ x =1 x-+CI>x+l ('~oo l+-!. QD 247 lin) (1 ¥~QI + !)J& .. a) lim (cos x) ~ : ..J(X+I\X ---) 2x + 1 2 11m x-.. lim ( 1 n.)II n b) lim (cos x)iI.... I 2 246.t+3) sln% · .. Find litn % . . % %-+0 I ~ . \\'e have I"1m x-.. CX) Example 9. · x Transforming.2} I+x = eX l:n"" .o + sin x) .= I"1m x+1 (II x -+ (YJ l + (X-J -+I x I )J" = -1 X+ 1 !X -2)1 ..t 245...x _ e\x+1 = lim {[ I-t.' I .29 Limits Sec 3) 1"here{ore. lim ( I X-+>D X x 248. =0.!. 1 244 I· --. (~i~ x+3 x 243. lirn (1 ¥-. x+l Solution.. 00 (X-I )'x .. z.. 1i~n(XI.orting to the general procedure: lim JC -+ '" (X-l)Xo=: X+1 lim (1-+ X+ '" ( 1+ ~ ' rr J[I~mct 1( 1-++f"r r x ~I~ ( 1 =e-l=e. as indicated above..... :~~ 2xl + 1 252**. (l+~)x=efl )-i+I . Itm 3 c.o \ ..~~ xZ -3x+2 · % 1 251.: . (2+X)"~ 241. + e Generally. lim ( · 242.. %-+. lim ( 1 + n • • n-+QD (xl-2./.(' %-+ct'> 0 t• In this case it is ealijer to find the lilnlt without rec..-)~ . o (see Problems 103 and 104).a x. 0 x ¥-+O 1 • (a>O). . . n-+oo (a x b) Iimcosh~-l >0).a Um [In f (x)] = In rUm f (x)]. 258*. x-..Jt I b) lim I ~tn-x . . lim .Introduction to Analysis 30 [Ch. Iim[ln(2x+l)-ln(x+2)].JI where tanh x = e-x--+e. 260*.x_I.. ¥-+-:» ¥-++QD • 266.. Prove that Urn In (1 +x) x x-+o 1. In (Cf'S x) 257 • I1m 2 262.. a) Ii. lOR ( 1 + lOx) 254.. a) lim .. We have Formula (*) is frequently used in the solution of probleuls.c-. OD x b) lim % ~ + V ~! + 1 OD Vx + 1 2 265.O 259*.t-++o .. • ~~ ( ~ In y: ~. 261. bIn .. x ~ . J 268. 253. ""'-1 1 +e" ... then .. 1 +e In (1 +e-. a) liln % -+ .. " .I-J '... lim %-+0 -=- 1 X -x SlIl %-+0 oX aX • x 263. it is useful to know that if the limit lim f (x) exists and is positive. I1m x K-+' 255. lim~.) . +GID X X-+O K. a) lim • x x-+ -00 In (I + eX) b) lirn x JC-++oo K 267...n tanh x.. (*) Solution. . g Example 10. b) lim tanh x. Find the' following limits that occur on one side: x 264. a) lim ~irh x .+o X . 1 When solving the problems that follow. I . lim x (In (x+ 1) -In xl. a) rlin x-+ - 0 x b) lim 'sin x I • ¥-. lim n (Via -1) • eaX_e fJX 256... eX _e. . dra\vn at the poi nts x = 0. 3.Limits x-I 270. y = lim V 1 +xn (x ~ 0). n-+(J) 275. a-+o 274. Find the limit of the perimeters of regular n-gons inscribed in a circle of radius R and circumscribed about it as n . . 282. ~ind the limit of the perimeter of a broken line MoM I •• • M" as on inscrIbed in a logarithmic spiral I -=. 279.00. Regard it as the limit of the corresponding finite fraction. a) lim~2 . where x=: 1. Transform the following mixed periodic fraction into a common fraction: a = 0. Y = lim (eOS Zn x)o n-+~ 272*.. 2. y=lim n-+oo 273. as n -+ 00.. 2. if the cotfficient a approaches zero while the coefficients band c are constant. ~f 1. Y = lLn (arc tan nx). . 00. '--=-11 · %-+2+0 x- Construct the graphs of the following functions: 271 **. I I. 1. Y = lim x I+ (x~O)_ n X Vx-z-+-a.. n-+~ 276.. a) 11m -. 0 I- %-+2-0 X X-+I-O 81 x-I b)X-+I+O 1m x x- b) 1im~2. Find the ltmit of the interior angle of a regular n-gon as n --.13555. Find the limit of the sum of the lengths of the ordinates of the curve y = e. 277. 2E:O. provided that n ~ 00. n. What will happen to the roots of the quadratic equation ax! + bx +c=O.x cos nx. Find the limit of the ~um of the areas of the squares constructed on the ordinates of the curve 0 y=2 1 - . . X ba~es..e-CP . 269.-z. and b#=O? 278. n. .Int . q:>1 AB = a nn n = 2"' .on to Analysis (Ch. = li'n Q~n). ... The point C1 divides a sc~nlent AB ---l in hCllf.. 285. the point C 2 divides a segnlent AC.. Deternlinc the linlit of the area of the step-like figure thus formed if n ~ (X).. Dctpfl11inc the linuting position of the pOInt e. \vhen 11-'" cu.. 8).n • Find Q. and so on. in half.C 3 in half. 7) is divided into n equal parts.::~.oduct. respectively.ll line thus fortned ciilTers frOl11 the len~~th of AB despite the fact that in the linllt the broken line "geornetrlcal1y nl{'r!~eS \Vlt~ the segrnent AB".. on each of \vhich is constructed an inscri bed rectangle (rJig. the point C~ divides a spglnent C 2 C.45°. q>tJ = 2 · (Fi~.0. 2. isoscelf's triangle \\'lth base angles u . Sho\v th:lt the linlit of the perulletcr of tl1(~ hrok(. the point C. the polar angles €PI 283. divides C:!. ench pnrt serving ns the bClse of A Q~ cHl 8 Fig 8 Fig. (i -t I) i) (i ==. 286. Find the constants k and b fronl the equation :c~r~(kX+b-. ) is proportional to the quantity of the substance available at the comlnencement of each interval and to the length of the interval. Assuming that the quantity of substance at the initial time is QQ' deternline the quantity of substance Q~n) after the elapse of time t if the increase takes place each nth part of the time interval T = !. 1. 7 284. A certain chemical process proceeds in such fashion that the increase in quantity of a substance during each interval of ti nle t' out of the in fi nite sequence of intervals (tT.. . n-. A segment = O. The side a of a right triangle is divided into n equal pnrts.::t: )=0. 1 (as 11 . if the vertices of this broken line have. in half. (1) \Vhat is the geolnetric I11eaning of (I)? 287*. (0). . If lin1 a (x) = 0. By virtue of this theoreln. Vii VX"' 2°. Example 1. ~2(~ . we can subtract fronl (or add to) the numerator or denominator infinitesitnals of higher orders chosen so that the resultant quantities should be equivalent to the original quantities. x-+a Le. Infinitely small quantities (Infinlteslmals).a: sin x-x. but if C =0. a are also infinitesimals as x -+a. {j (e). In (1 +x)-x and so forth. then the functions a (x) and P(x) are called infinitesimals of the same order. 41 Infinitely Small and Large Quantities 33 Sec.p (x). If for an arbitrarily large number N there exists a ~ (N) such that when 0 < I x-a I < ~ (N) we have the inequality If (x) I> Nt then the function 2-1900 f (x) is called an infinite as x --+ Q. Inftnltely large quantities (InOnltes). then the function a (x) is an In 51ltlilar fashion we define the infinitesImal a (x) 00. P(x) are called a (x) . 4.= %-+0 In (1 + 2x) x-+o 2x 2 • . The sum and product of a linlited number of infinitesimals as x ---. For exan1ple. 0 we have (x) and (x) - equit'alent functions as x -+. If a (x) and p (x) are infinlteslnlals as x --+ a and liJn a (x)_C x-+a P(xJ - • where C is sonle number diff~rent frolll zero.."~a + 00.Sec.+ 0 and P(x) ~ 0 as x --+ a. Infinitely Small and Large Quantities 1°. tan x-x. 1 lIn1 11m . then we say that the function a (x) is an in fin ttesitnal of higher order than ~ (x). for x --. a (x) IP (x)]n =c • lim a (x)_1 . ."-+Q ~ then the function" a . when taking the limit of a fraction lim a (x) ~-+a ~ (x) . The Unlit of a ratio of two infinitesimals remains unchanged if the tenns of the ratio are replaced by equivalent quantities. if Ia (x) 1< e when 0 < I t-a I < infinitesimal as x as x ~ ---+ a. where a (x) . The function u (x) is called an in{initesit1lal of order t1 cotnpared with the function p (x) if Ihn where 0 < J C 1< If . The sUln of t\\10 infinitesimals of different orders is equivalent to the tenn whose order is lo\\'er. 21 < 6 is the inequality I f(x) I>N fulfilled if N is an arbitrary positive number? Find 6 if a) N = 10. .. c) 8=-0.. 288.. b) 8=0. b) 8=0. . In what neighbourhoods of Ix . Determine the orders of the infinitesimals relative to the infinitesimal a: a) of the chord AB. c) 8=0. o c) N = 1000. c) of the area of d ABD. Determine the order of smallness B of: a) the surface of a sphere.. Prove that the function f (x) = is an infinitesimal as x ~ quality 00. b) N = 100. 1 The definition of an infinite f (x) as x ~ Q() is analogous.01.1.----. . What will the orders be of the radius of the sphere and the volume of the sphere with respect to its surface? 292. we introduce the concept of infinites of different orders..~ 291.001. A ".01.001. 290. 9) with radius R tend to zero. As in the case of infinitesimals. b) the volume of a sphere if the radius of the sphere r is an infinitesimal of order one.Introduction to Analysis 34 [Ch.1. 9 cular sector ABO (Fig. Prove that the function 1 f (x) = x-2 is an infinite for x ~ 2. sin x x For what values of x is the ine- If (x) 1<8 fulfilled if 8 is an arbitrary number? Calculate for: a) 8 = 0. 289. Prove that the function f (x) = is an infinitesimal for quality x~ i-Xl 1. b) of the line CD.. Let the central angle a of a cirFig. For what values of x is the ine- I f (x) 1<8 fuliilled if 8 is an arbitrary positive number? Calculate numerically for: a) 8=0. b) Vx+ Vx.93 4 . demonstrate that when I x t is small we have the approximate equality VI+x~I+. c) VI0.97..06. e) tan x. approximate the following: a) Vl. 3) 105. 6) 0. lim CO~ x-cos2x • %-. Prove that the length of an infinitesimal arc of a circle of constant radius is equivalent to the length of its chord.43429. c) (1 +x)n~1 +nx (n is a positive integer). lim (x-x1)2 arc x~0 .04'.1- Compare the values obtained with tabular data. For x---+ 0 determine the orders of smallness relative to x of the functions: 2x d) 1.cos x. d) log (1 + x) = Mx. c) Vii-Vi'.Infinitely Small and Large Quantities Sec. . 4] 35 293.97. 301. approximate: 1 1 1 1) 1. 295. d) V120 and compare the val ues obtained with tabular data. find 298 1· ~ 296 I · sin 3x·sin 5x • x-.. Prove that when x~O the quantities and VI +x~i are equivalent. Using this result. r r • • x 1m I -x X-+l · 299. Can we say that an infinit~simally small segment and an infinitesimally small semicircle constructed on this segment as a diameter are equivalent? Using the theorem of the ratio of two infinitesimals. 4) V15. Using these formulas. 2* .02.o 1m 297.x~a + 2a l+x b) V l (a > 0). a) T+X. b) VO. where M=10ge=O.. S1n . (I) Applying formula (1). 294.0 I-Xl -cos X 1n (1 -. 7) log 1. Prove that when x --+ 0 we have the following approximate equalities accurate to terms of order Xl: a) _1_ ~ I-x·' x a -t. 2) 0.sin x. 5) 1.x) -i 300. 2) there exists a finite limit Urn f (x). that is. mal increment in the function. Example 1. determine the order of growth of the functions: a) x!-lOOx-l. "2 Since L\x Silly lim .. c)Vx+VX...-+O or the function f (x) is continuous at the point ~ if (and only if) at this point to an infinitesimal increment in the argument there corresponds an infinitesi. if: 1) this function is defined at the point ~. where L\~ ---+. Prove that the function y=sinx ts continuous for every value of the argument x. lim f (x) = f (~).36 ICh.0. there exists a number f (~). 6. (1) X-+. condition (1) may be rewritten as lim L1f (~) = lim (f (~+ L\~)-f (~)) = 0. Let X--'(X).-+o d.= 1 and Icos(x+ dX-+O L\x L\2X)I~I. etc. If a function is continuous at every point of some region (interval. Continuity of Functions 1°. Solution. then it is said to be continuous in this region. 3) this lim- it is equal to the value of the function at the point x-. dX-+O Hence. the function sin x is continlJOUS when . We have ~ Ay = sin (x + Ax) - sin L1x x 2 sin = ~x cos ( x+ ~x) = Sl:x2 • cos ( x+ ~x) . (2) ~. Definition of continuity..I -~ • • • +an (ao::. d) V x-2x l • Sec. Le.).-2 . ~. Xl b) x. Putting x=~+L\~. 2 it fol lows that for any x we have lim /).OOO. Show that for x ~ 00 the rational integral function p (x) = aox rl + a1xn .y=O..oo<x< + co. 1 Introduction to Analysis 302. Ax. Taking x to bean infinite of the first order. A function f (x) is continuous when X= 6 (or ccat the point ~").bO) is an infinitely large quantity equivalent to the term of highest degree aoxn • 303. . We say that a function f (x) has a dIscontinuity lat x=xo (or at the point xo) within the domain of definition of the function or on the boundary of this domain if continuity of the function at this point. Points of discontinuity of a function. If the function f (x) has finite limits: 11m f (x) = f (xo-D) and lim f (x) = f (xo D). it is necessary and sufficient that then . X-+Xo . 10 how we choose the number f (1). This function is not defined at the point x= 1. th~re 10 a) is a break in the is discontinuous when x= 1. 5J 37 Continuity of Functions 2°. In particular. The lunct'ion f (x) = (1 1x)t (Fig. For continuity of a function f (x) at a point xo. X o is called a removable discont inuit y. the redefined function f (x) will not be continuous for x = 1.0 X-+Xo +0 + and not all three numbers f (xo)' f (xo-D). if f (xo-O)=f (xo+O). f (xo + D) are equal. Example 2. then Xo is called a discontinuity of the first kind.Sec. and no matter y y=£(x) 2 ---~ I __ ~ I I I I I 2 o 1 2 (b) (a) -I (c) FIg. e. then the cOlnposite function cp (f (x)] is continuous in (a.q < I}. 3) l (x) takes on all interrrlediate values between the two given values. there will be at least one valuex=v (a<v<~) 6uch that f (V)::::IC. b). Indeed. is discontinuous (Fig. At an other points this funct ion is. Discont inuities of a function that are not of the first kind are called discont inuities of the second kind. where E (x) denotes the integral part of the number x [i. 2) f (x) has a minimum ann a maximum value on [a. These ar.b. then E (n-O):. obviously. bear in mind the following theorems: I) the sum and product of a limited number of functions continuous in some region is a function that is continuous in this region. b). b). When testing functions for conti- nu ity. + Indeed. and all the discontinuities are of the first kind. and a function cp (x) is continuous in (A. Example 5. ~hat is. b) has the following properties: 1) f (x) is bOl'fnded on [a. . f (+ 0) = lim x-++o sin x = x +1 and '(-0)= lim sin x = . b). A function f (x) continuous in an interval [a.t .e points Xo such that at least one of the one-sided limits. Properties of continuous functions. 10c) at the point x=O has a x discontinuity of the second kind. ~). The function y=cos ~ (Fig. continuous. here. and a set of its values is cont~fned in the interval (A. B). then the equation [(x) =0 has at least one real root in the interval (a. Infinite discontinuities also belong to discontinuities of the second kind. E (x) is an integer that satisfies the equal ity x = E(x) q. + tOb) at every integral point: x=O.. Le.(CIl. 2) the quotient of two functions continuous in some region is a continuous function for all values of the argument of this region that do not make the divisor zero. ±I. where O<. f (xo-O) or f (xo+O). if l(a) f (~)<O. Show that the function y = of the argument x... . 3) if a function f (x) is continuous in an interval (a. The ~unction y = E (x).= n-I and E (n 0) = n. is equal to Q() (see Example 2). The function f (x) = ~I: ~ has a discontinuity of the first kind at x=O.1-+-0 -x Example 4. if f (a) = A and f (~) = B (a ~ a < ~ ~ b).. if n is an integer. then no matter what the number C between A and B. since both one-sided limits are nonexistent here: lim cos ~ %-+-0 X and lim cos ~ • x-++o X 3°. 304. . there is some number M such that If(x) f<M when a~x<. 1 38 Example 3. In particular. B). is continuous for any value Xl . ±2. 312. Prove that if the function f (x) is continuous and nonnegative in the interval (a. Prove that the function y == cos x is continuous for any x.x is meaningless for x=O. .. Prove that the function y= Vi is continuous for x ~O. b). Show that the function y = I x I is continuous. + an boxm + b. Plot the graph of this function. Prove that the absolute value of a continuous function is a continuous function.. The right side of the equation f (x) = I-x sin-!.x m.a n is continuous for any value of x. Is it possi ble to define the val ue of f(2) in such a way that the redefined function should be continuous for x = 2? . . How should one choose the val ue of the function A = f (2) so that the thus redefined function f (x) is continuous for x = 2? Plot the graph of the function y = f (x). 5) Continuity of Functions 39 305.1 + . 309*. then the function F (x) = V! (x) is likewise continuous in this interval. 306. A for x=2. 308. How should one choose the value {(O) so that f (x) is continuous for x == O? 315.. -t. Prove that the rational integral tunction p (x) = aox n + a1xn . 313..Sec. A function is defined by the formulas f (x) == " t x2-4 --2 x- for X=F 2. 310. For what values of x are the functions a) tan x and b) cot x continuous? 311 *. is cont in uous for all val ues of x except those that make the denominator zero.x n-l+ . 307*. The function 1 f (x) = arc tan x-2 is meaningless for x = 2. Prove that the rational fractional function R( )X - aox n+ a.1 + b rn + . 314. y = e X+l 322. 331.stn x X 323. 1 316. • 1 321. d) f (x) = eX _e.-x . is discontinuous for every value of x.Introduction to Analysis 40 50 (Ch. Y= .J- J +e 1 - x . · 325. 330. The function f (x) is not defined for x = O. y=lnltanil. 2 e) Investigate the following functions for continuity: 1 317. f) f (x) = x cot x. a) y= sin -. Prove that the Dirichlet function X(x). which is zero for irrational x and unity for rational x. 318.X x . Plot the graph of this function. y= lim 1+1 n-+<» x n (x ~ 0). f (x) = x sin 1.3 x 326. . c) f (x) = In (I +X)-. Y=XX 2 . Investigate the following functions for continuity and construct their graphs: 332. y == lim (x arc tan nx). III (I-x) _ . b) y= x1 · -i2 328. x 329. y = In (cos x). Y= 1 +x 319. Y= 320. Y= -:-. Define f (0) that f (x) is continuous for x= 0. 333. b) f(x)= I-XC1osx . Y= e x sin ~. if: a) f(x) = (1 +x)n-l x (n is a positive integer). for x>3.. y= { X2. y=(1 +x)arctan I I x•• . y= iXl 1 +x' 324. y=arc tan V~.. n~CD . 327. 2x+l for x ~ 3. 336. If x<o. b) y=xE(x). 5] Continuity of Functions 41 334. { -I. Prove that the equation tan x=x has an infinite number of real roots. 337*. c) y == sgn (sin x). Give an example to show that the sum of two discontinuous functions may be a continuous function. b) y = x sgn x. a) y = sgn x. . a) y=x-E(x).S~c. if x >0.~fx=O. Let a be a regular positive fraction tending to zero (O<a< I). which is true for all values of a? 338. Approximate this root. where E (x) is the integral part of the number x. 339. 335.+ 1 =0 has a real root in the interval (1. Prove that any pol ynomial P (x) of odd power has at least one real root. Can we put the limit of a into the equality E (I + a) = E (l-Q) + 1. where the function sgn x is defined by the formulas: + 1. 340. Show that the equation x'-3x. sgnx= O.2). and Y=f (x) and Yl =f (Xl) are corresponding values of the function y = f (x). If x and Xl are values of the argurnent x. 11) and is called the mean rate of change -of the function y over the interval (x.1A and Ay=AN). For the function y=x l -5x+6 . Example 1. then Ax=xl-x is called the increment of the argument X in the interval (x. x+ Ax). where l\x=l. Xl)' and /1Y=YI-Y or AY=f (x 1 )-f (x) = f (x + Ax) -f (x) ( \) y Fig. 1t. 11 Is called the increment of the function y in the same interval (x. The ratio :~=tana js the slope of the secant MN of the graph of the function Y= f (x) (Fig. 1. Increment of the argument and increment of the function. xJ) (Fig.Chapter II DI FFERENTIATION OF FUNCTIONS Sec. Calculating Derivatives Directly 1°. y'= 11m liy= lim (2x+Ax)=2. that is. From formula (I) we have fig = (x + fiX)t-xl =-2xAx4.(Ax)' and fiy ~x = 2x+ fix. We have Ax= 1. 2°.1 + 6) = . Hence. 1] 43 calculate ~x and ~y. b} Ax=2-3 = .1..Calculating Derivatives Directly Sec.1 + 6)-(1 2 -5.1. ~x -+ 0 Ax 6.x -+ 0 ae• One-sided derivatives.(x) = lim f (x+ Ax)-f (x) 6.1-1=0.29. Solution. 10 secant passing through the points M ( 3.(31 -5.1. corresponding to a change in the argument: a} from x= 1 to x= 1. The expressions t'. find the slope of the x · "3. In the case of the hyperbola y= . fiy 1 k= lix=. tan~ent MT fo the Finding the derivative y' is usually called diOerentiallon of the functlon... t. = (1.0.. 1) Solution.3 + 6) :::::0. The derivative y' = ~ of a function y = f (x) with reo spect to the argument JC is the limit of the ratio ~~ when Ax approaches zero.30· fix = 10-3=7 / t ) . Tbe derivative. Example 3. 11): y' =tan lp. Here. The deriv at ive y' = f' (x) is the rate of change of the functton at the point x. and N (10. a} Ay 1 Example 2.y = (2:1-5· 2 + 6) .. Find the derivative of the function y=x l • Solution.%-+-0 fix and I' + (x) = 11m AX ~+Q f (x+ fix)-f (x) ~x .1 2 -5. The magnitude of the derivative yields the slope of the graph of the function y = f (x) at the point x (Fig.1. and i I 7 fiY=10-a=-M. b) fromx=3 to x=2. Hence. OOOI. Example 4 Find f~ (0) and f~ (0) of the function f (x) =. the left-hand or rtght-hand derivative of the function x. and ~x==O. We have Vi: Vf:j.y if Ax = 5. Find the increment of the function y == x 2 that corresponds to a change in argument: a) from x = I to Xl = 2. Why can increment 6. In this case. 343. Find the tions: a) 1 Y=(X t _2)2 b) y== Vi c) y = log x we. 2 are called. . c) x=a.44 · Differentiation of Function"s [Ch. Find ~y of the function y ~ Vx if: a) x=O.y and the ratio ~y for the funcx forx=l forx=O for x= 100. b) x=8. By the definition we have f~(O)= lim lL\xl=_l.1. ~x~O. ~x AX-+.x = .000. L\x 2 341. Example 5. Ilx=h. lim L\x -+ +0 I:!x Infinite derivative. determine the all we know is the corresponding increlnent the function y = Xl this cannot be done? increment tJ. c) from x = I to Xl == I + h. 342. and t!. respectively. while for 344. Ax L\x -+ -0 f~ 4 0 • (0) = Il:!xl= 1. Solution. For f' (x) to exist. Pi nd I' (0) of the funct ion y= Solution. b) from x = I to Xl = 1.000 and~x=0. x I. the tangent to the graph of the function y = f (x) is perpendicular to the x-axis.(x) = f~ (x).90. for the function y = 2x + 3.OOl. If at some point we have lim f (x+ I:!x)-f (x) = ClO. L\x==-9.4. it is necessary and sufficient that f (x) at the point t'.0 then we say that the continuous function f (x) has an infinite derivative at x.x f' (0)= lim -A-= lim L\x -+ 0 uX L\x -+ 0 1 V-=oo. x 2 =0.l. e) y = 2x . What is the mean rate of change of the function y = Xl in the interval 1 ~ x ~ 4? 348. 1] 45 Calculating Derivatives Directly 345. I'he law of motion of a point is s = 2t 2 + 3t + 5.~I.Ol. b) the instantaneous rate of rotation. Let In = f (x) be the mass of a non-homogeneous rod over the interval (0. What is to be understood by the rise of the curve y = f (x) at a given point x? 352.9. Find the slope of the secant to the parabola y=2x-x 2 . Find the ratio ~~ of the function y = at the point x=2. 351. x-t-l\x]. What is to be understood by the rate of reaction of a su bstance in a chemical reaction? 355.Whatisthederivative y' when x~: 2? + . jf the abscissas of the points of intersection are equal: a)x. b) xl==l. b) the rate of cooling at a given instant? 354. x t = I + h. Find the mean rise of the curve y = f (x) in the interval lx. What is to be understood by: a) the Inean linear density of the rod on the interval [x. Xl ~ 1. where the distance s is given in centimetres and the time t is in seconds. if: a) l\x=l.Sec. c) l\x=O. Find Ay and ~~ which correspond to a change in argument from x to x-~ ~x for the functions: a)y=ax+b. c) To what limit does the slope of the secant tend in the latter case if h ~O? 347. d)y=Vx.xl=2. Find the mean rise of the curve y = 2x in the interval 1~x~5. b) y === x' . 350. Xl. What is to be understood by: a) the mean rate of cooling. b) l\x=O. Define: a) the tnean rate of rotation. A hot body placed in a medium of lower temperature cools ott. b) the linear density of the rod at a point x? 356. 346. 1 c) Y == ~2 . 353. x + l\x]. f) y = In x. What is the average velocity of the point over the interval of time fron] t=1 to t=5? 349. if f (x) = x. if t(x)= x(x-I)I(x-2)1. V- 359. What are the slopes of the tangents to the curves y =. 6 ) (!!.Ix' drawn at a point with abscissa x = 2. .. Find the slope of the tangent to the curve y = sin x at the point (n. Find f' (0). 364.. Show that the following functions do not have finite derivatives at the indicated points: a) y= V Xl b)y= x-I c) Y = I cos xl V- at x=O. ± I. 366*. 2) (x)' = 1. 'i' (x) are functions that have derivatives. The law of motion of a point is s = 5t l . 1 b) Y= z.• Sec. Basic rules for finding a derivative. 361. Calculate f' (8). If c is a constant and u = q> (x). f (x) = I' (x)? 362. Find y' = lim ~ of the functions: ~X-+ 0 a) y = Xl. 5) (uv)' = u'v+ v'u. 358.2 DitJerentiation of Functions 46 357**. I 3) (u ± v)'=u' ± v'. ±2. x and y = Xl at the point of their intersection? Find the angle between these tangents. that is. Find the speed at t = 3. 2. 360.[Ch. 7) ( ~ )' = - ~~' (v ~ 0). Find the derivative of the function y = tan x. 363.t: 0). where the dis. d) y = cot x.!. at x = I. x c) y=Vx. 4) (cu)' =cu'. Find the slope of the tangent to the curve Y= O. Tabular Differentiation v= 1°. 365. 0)..)' V VU' -v'U Vi (v . tance s is in metres and the time t is in seconds. Find the value of the derivative of the function f (x) = ~ x at the point x = X o (x o =/= 0). At what points does the derivative of the function f (x) = x' coincide numerically with the value of the function itself. 2k+l at x=--rn. f' (2). 367**. then 1) (c)' =0. f' (1). . k=O. (tanh X)'=-!. 2] Tabular DltJerentiation 47 2°. V. 1 = 1+x l • -I X. If y=f(u) and u=q>(x). (cos x)' = . then Y~=Yu U x (1) or in other notations dy_dy du dx. (x n )' = nxn . (arc coth x)' = xl-I (I x r > 1).I (x 2 fr x III. (aX)' = aX In a. . I). 3°. VII. y = f [q> (x)]. a > 0). (cot x)' = sin -. sIn x XIX. (1" x)' =-!. where the functions y and u have derivatives. (arc cos x)' IX. (Mcsinhx)'= ~.-1-x . x (x > 0).du dx· This rule extends to a series of any finite number of differentiable functions. (sinh x)' = cosh x. XVI. (tan x)' = _11_ > 0). 1 XX I. (arc tanh x)' == 1--1 (I x I < I). (cosh x)' = sinh x. • cos x -1 VI. that is. (cue lijl x) . 1 X IV. XIII. Table of derivatives of basic functions I. Rule for differentiating a composite function. -1 (cothx)'=~hl . (sin x)' = cos x. (arc sin x)' = y -1 (I x I < I-Xl VIII. 1 +x· XX. XVII. (arc cot x)' = x2+ 1• XI. > 0.= loga e (x x "a x XV. (loga x)' = -1-.sin x. (ex)' = eX. (arc cosh x)' y 1 x2 -1 ( I x. IV.-h • cos 2 x XVIII. -x -1 XXII.I • II. XII. (yX)' = _ .Sec. > 1). y I-xl 1 (I x 1< 1). y=-2t sin t-W-2) cost. y=2x-l-x. y=tanx-cotx. 2 ) ar. 386. _I + Vi 381 · Y. /I 2 I 380. 384. y=i-- + 2x-3. Example 2. Find the derivative of the function Y= (xl - 2x+ 3)5. IR n + • +1J2' 374. tan x-x. 389. Algebraic Functions 368. y= arc tan x + arc cot x.r1. y=x 2 -Sx+S . Putting y=u'. x 2x+3 379. v=4x. y= (I +x sin x--cos x · · X arc sin x. 383.2 Example l. y=3x3 -2x t l Xl 2 .V-· x 376*. 370.ax'+b 373.f Z B.2) = 10 (x-I) (x 2 -2x + 3)'. Y = 3 372 Y = at m 5 375. Find the derivatives of the following functions (the rule for differentiating a composite function is not used in problems 368-408). y=xcotx. Y = sin x+ cos x 388 Y = 385. Y=c+dx. we find y' = 3u 2 ·cos tJ·4 = 12 sin:! 4x cos 4x. where u=(xl -2x+3). ~ x+r-O. + bx+c. y=ax 2 -5x 371. 387. A. X a+bx 378. y=~~ln2.D'Uerentiation of Functions 48 [Ch. Y= Va 2 -t-X-'. y = 5 sin x+ 3 cos x.5x·. by formula (1) we will have y' = (U5)~ (x 2 _ 2x + 3)~ = 5u 4 (2x . I 377.. + bt . Solution. Vx a b y= V-x . Putting y=u'. y=x' -4x' 369. y =a. . u=sinv. Inverse Circular and Trigonometric Functions 382. Solution. Find the derivative of the function y=sin' 4x. Y= (x l --2x + 2)e".(3-10x)=30(1 +3x-5x 2 )18. Xl 402. use the rule for differentiating a C0111posite function with one intennediate argument. 413. y= Vf=XI.+21nx_ . 39ii. Find the derivatives of the following [unctions: 409**. y=. Y = eX arc sin x.eX • 396. 410. x x 400. y=x Inx. y=(x-l)e". 397. 401.!. Denote 1 + 3x-5x! = u. y = arc sin x arc 51 nh x. y= Va+bx' . Y=56(2:-l)7 24(2~-l)' 414. 393. 411. 416. Y=-x.Sec. y=-h-.(3-IOx). 40 (2x -1)1 • . Hyperbolic and Inverse Hyperbolic Functions 405. Y= tanh x-x. 392.3 . 415. 391. x Xl 1nx 399. f (x) == eX cos x. Y=-l nx · x2 eX • Xl 398. 403. D. y=lnxlogx-lna log a x. Exponential and Logarithmic Functions 390. y=x51nhx. y=(a2/1_x3/1)'/a. Solution. Y = arc tan x. y==x 7 . 412. y=(3+2xl )4. e 394. cos x 406. 2] 49 Tabular DiOerentlation c. f(y)=(2a+3by)l. y=(1+3x--5x2 )'o. Composite Functions In problems 409 to 466. u~=3-10x. Y= (UX/b)'. Y=i.arc tanh x. 407 404 408 y =3cothx lnx _ arc cosh x · y· Y x • = arc coth x I-x!' E. lI~ ~=30u28. then y = ulO• We have: y~=30ul8. Solution.10 cos x (3-2 sin x)·. y=(3-2sinx)'.2 cos x) == .. f (x) = a cot!-. YI _(2X)1 y= arc sin ~ ._1 . • (2x)' = 2 . !(x)=-6(1-3Cosx)l. Y= VI +arc sinx. Y= . 427. y=tanx.~ tan1x+ ~ tan'x. 438. Y = arc cot l l +x • -x .. f (x) = arc cos Vi. Y= 2x + 5 cos' x. Y= Vcotx- Vcota. y=sin3x+cos ~ +tanVx.4 cos Xl.----r---'1-r. 419.I-a . Y= sin (x l -5x+ 1) + tan ~ • x 433.lyx(YX)'=3COS3X- 1 ..50 [Ch. {(Xl = cos (ax + P). x 440. Y= = cosec· t + sec· t. X 432.· 2 . x 442. x + 1 . • 425. Y = Varc tan x.(arc sin x)'.. y' = 5 (3-2 sin x)'· (3-2 sin x)' = 5 (3-2 sin x)' (. 430. ~ ~~~: ~ . Y= V2e"-2" + I + In l x. X 424. 429.. X cost . y = 436. y'=cos3x·(3x)'-sin ~(i)' +cos.!. y' 1 439. 13 sin x-2 cos x V 5 • 1 _. 426.. Solution. Y= Vsin! x+_ cos x 1 422. f (t) = sin t sin (t + q». 418. a 1 1 Y = . 428.. Vl-4x 2 441. 434. 420. -5 sIn -5 .20 cos (5xl ) . 437. Y = arc sin 2x. Solution. 423. . Y = arc tan... 421 *.. 3 cos x cos X . 2 DifJerenttation of Functions 417. 1 y arc tan x • Y= V xe" +x. 431.. 435. 445. 447.~ V (l +X·)I. f(t)=(2t+l)(3t+2)V3t+2. . ~ VXi+~xVX+~ x V?+l~X. y = V(x+ a) (x-t. a -1.Vi· Y=~· V (l 4 + XS)I. 468. = (a + x) Vli=X. y=Inlx-ln(Inx).4 arc sin x).2X S)1 • a +bXn)". y=x"IO"x. Y = log sin x. y=In(l-xl ). 454. 15 10 3 (X-3)1 457. 466. 1 evx+ F. Y= 5e. y=In(2x+7).4 (X-3)4 1 2{x-3)1- x· 458. Y= 463. Y= 462. sX 449. y = In (eX + 5 sin x. Y= arc cos eX. Y = x (a . y = arc tan (In x) + In (arc tan x).¥2x l -2x+ 1 459 · yx • 460.xa • 444. 452. Y = 5 {x+2)S. 446. VX-l 464. Y= • . y=. 450.(x + 2)· 2 I + (x+ 2)' -2" {X+2)1· 469. Y= al x y-_. Miscellaneous Functions · • 5x cos " '3. · . y= Vln x+ I + In I). Y=3 x+2· 4 465. z=Vy+Vy. y= ( a-bx n • 9 3 467.Tabular DitJerentiation Sec. y 471. 2] 51 443. 448. 470.b) (x+ c). 3 Y (1 -}- A 2)1 . f(t)=tsin2 t • 451. x 4u115** • Y = sin 11 4 456. 453. y=-2{x-2)I-x-2. y = 8 (1 -Xl).x 2 2 Xl 461. x2 -1 485. y =r. 478. y=3b'arctan - 3 Yb 498. Y = arc sin Xl + arc cos x. y= • 500. _ (tan! x-l)(tan 4 x + 10 tan 2 x +]) 475 · Y 3 tan' x • + I). 4 cos x 481. 503 = (a sin ~x . y = 1~ cos' x (3 cos' x-5).yr~' 2ay. Y = tanS 5x. x sin a 495. y= Va sin I x+ Pcos l 483.V2 arc cot Ve x x-(3b +2x) Vbx-x'. y=arc sln~. y=arcsin(1-x)+V2x-r. y=}(arcsinx)'arccosx. y = ~ arc tan x 5tan 2 +4 497. y="3 tan 3 x-tanx+x. 502. 1 480. Y = ~ sin (x'). 476. F(t)=eCltcos~t.:: arc tan 1 -xcosa • 2 496. x·. 492. 479.21) arcsin 1/-- 493. Y = 3 sin x cos· x + sin' x. 477. y = arc sin (In x). y = sin l (t 3 ).y l 473. . SlIl X 482. y=. 484. 491.. y= v x+ 21V x-x'. y = .62 D"Oerentiation of Functions [Chi 2 1 472.~2 • . 494. F(x)=(2ma mx +b)p.3 a +-3 cotx.x . 501. y = es1n2 x.p cos ~x) e'J. x= . (. ax 499. tanx V2.x. y = 1n (arc sin 5x).. y = In (VI +ex-I)-ln (VI +ex 474.x •y a2 -1. . y = arc tan In -. 509.. Y = 3cOS b~ + . y === In (x -1. 1 ~ ~~:.. 1 512. x 532. Y=3 1n . 517. Y= In In (3-2x'). y x -i-a! -x 2 52 I.. + n anx. 2 1 531. y=In(ax!+bx-J-c). 520. 2a 2 522.Y3 528. x 1 1 cos x 523. Y=~l nI x • 514* y=In(x-2)S • 515 • y --I n x-I 5 13. Y = 2arc sin IX -1.sin(lnx-~). +. Y = 3cot xi 505. y=ln y~+x. Y = arc tan In x. 518. x x 2 -2x -+-1 I 525. sin ax 527.\I-+-x+l · 526.~ . y=xna.Val + Xl). Y = 110 r X 53 (3 sin 3x-cos 3x). x+a y=x. . y===~ln __2 _ .• 506.~ In ~ .- Va tan~+ 2 2-1- Y3 529. tan ~+ 2. y = Y2 -3.- In(x+ Vxl-a l ). 530. Y = ~ In (x 2 . y = V cos xaYC"Oi'X. y= In arc sinx+~ In x-t arc sin Inx. 2] 504. Y = I n cos -x-· . _ · Y. 510.(1. Y = -2 In tan -2 --2 sin -·-2-· x 524. f (x) = V7tl -In 1+ YX2+1 . 508. y=x-2Vx+2In(1-l-VX)· 511.at) -l.arc tan x 1 x-I y'2' + (fIn X'D · + (x 1)' · (X-I)3(X-2) -~----. y=In(a-t-x-1-V2ax+x l ). Y = 51n s (ax + b).arc cos 3X)I. x-3 1 2 sin! x I t . 507. 519. y= ~ Vxl-a l _ 516 a .Tabular DiOerentiation Sec. .!. 553. f (x) = In (l +x) + arc sin ~. + ~ ) arc sinh x.549.X (-3 sin 3x)-e. y = erJx cosh px. 545. f (x) = VIC sin x + In V I-xl. Y = (~ J • 1 546. · aa -x +x f (x) = arc sin c) f(x) =_x. Y = tan' '. f (0) = 0 .. f (x)=2"ln(l +x)-6 n(xI -x+ I )+ (:rare · 536. (x) = e.54 DiOerentiation of Functions 533. Y = sinh' 2x.· Find f' (I). Find t+(O) and ''-(0) of the functions: a) b) f (x) = V sin (Xl) .X cos 3x.x for x>O. y = arc tanh I ~XI 541. Solution.. 538. I 1 I I tan 2xY3 -I 535. 543. e) f (x) = x sin x=t=O. y = tanh' 2x. 550. 552. Y= 2 (xl-I) arc tanh al' 1 x+ 2 x. Construct the graphs of the functions y and y' .. y = arc cosh In x. • 1 x: x~O. e. Y= In [Ch. % =F 0. 542. 1(0)=0. y = In sinh 2x. 539.x for x ~ 0. Find (:~ t=. l' (x) =e. y = arc coth (sec x). Y=4 1n xl _1 +41nx+l+"2arctanx. f (0) = 0. b) y=xlxl. 554.x cos 3x. f (x) = Xl sin. 540. l' (0) = eO (-3 sin O)-eO cos 0 =-1 . Calculate I' (0) if .x . Find y' if y=lnlxl (x*O). 544.+2 are tan sm x.~ x V I + Xl • Xl 048. y = arc tanh (tan x). 1 \- 1 +e ~ d) 2 l l 2 .2 1+ V' VSinx V-. Find y' x 2 if: a) Y= Ix I. 1- 3 sin x Xl +I I x-I 1 534. 551. i-Xl 037. Fin d f' (x) if f (x) = { 1.. . y = arc sinh 547. Solution. Given the functions f (x) = tanx and cp (x) = In (I-x). Example. Find f (3) (x-3) I' (3) of the function f (x) = VI -+ x. y u whence y'=y (V'lnu+: u').x satisfies the equation xy' = (I-x) y.Tabular DifJerentiation Sec. Find the deri vative of the exponential function y=u'D. 561. 2] 55 + + 555. f (x) = I-x and cp (x) = 1. where u = cp (x) and v = W(x). . Prove that the derivative of an even function is an odd function. DifTerentiate both sides of this equation with respect to x: (Iny)'=v' In u+v (In u)'. (In y)' ==!L = f' (x) y f(x) Finding the derivative is sometimes simplified by first taking logs of the function. Given the functions . Find f (0) xl' (0) of the function f (x) = e. . Show that the funct ion y = xe-. that is. Taking logarithms we get In y == v In u.x • 556. Show that the function y = xe. and the derivative of an odd function is an even function. l' (0) find cp' (0) • 558. Prove that the derivative of a periodic function is also a periodic function.: sat isfies the equat ion xy' == (I-xl) y.sin ~x . cp' (I) find f' (1) • 559. 557. 1 563. Logarithmic Derivative A logarithmic derifJatit1e of a function Y=f (x) is the derivative of the logarithm of this function. G. or 1 1 -y'=v'lnu+v -u'. Show that the function y = 1+x+ln x satisfies the equation xy' = y (y In x-I). x2 562. 560. Y= (X+2)2 Vx. y=xx3 • (1 +~ r.!-y y' = In sin x + x cot x. 568. Y = x 570. Y = (cos x)Sin x. II a function Y =f (x) has a derivative Y~:I= 0. 580. Sec.!- 3 x whence (-1) __~ + I-x Y'=Y(3:-1 3_1_ cosx_2s1nx . l+x 2 + sinx cosx 1 X-l~XI+3cotx-2tanx). Y = I x2 577. 567. Y= 574. Find y'.I) • V x-2 VX2+ 3 569.2 DiOerentiation of Functions 56 or y'=Uf1(V'lnu+~ u') 564.[Ch.I-x V x2 1 + x2 sin' x cost x. Solution.In the following problems find y' after first taking logs of the funet ion y = f (x): 566. if y=(sinx)". 2 Solution. . -. Y = ' / x (x . 565. Yx . Y= x +2)1 V(x+3)' · 572. 3. if y= . In y =31n x + In (l-x)-ln (1 +x2 ) + 3 In sin x +2 In cos x. then the derivative of the inverse function X=/-1 (y) is 1 XII =-. y=x". Find y'. • Y(x-I)S (X-3)1I vr Yx-l 579. The derivative of an inverse function. Y= 571. ! y y '=~-. y=x (x+l)8(x+3)4. The Derivatives of Functions Not Represented Explicitly 1°. v- %.¥. Y= (x+ 1) (2x+ 1) (3x+ 1). In y = x In sin x. 573. 575. y=xs1nx • I · (X-2)9 578. y' = (sin x)" (In sin x + x cot x). y= (are tan x). If a function y Is reI ated to an argu ment x by means of a parameter t. The derivative of an implicit function. taking y as a function of x.= _ dx t.3a (y XI/') = 0. Example 3. We have Yx = 1+-=--. (3) Solution. then or. 2) to equate this derivative to zero. Whence a cos t =-cot t. y) = 0. Example 2. { y= 'I' (t). We find :~ = -a sin t t!J!. a sin t 3°. 3] The Derivatives of Functions Not Represented ExplIcitly 51 or dx 1 dy= dye dx Example 1. dy dy di dx= dx . in other notation. (1) then to find the derivative y~ == y' in the simplest cases it is sufficient: 1) tocalculate the derivative. The derivatives of functions represented parametrically. if y=x+ In x. Find :~. dt if x == a cos y=a sin Solution. x y =-+ . I I x==. 1 x+ 1 x Solution. If the relationship between x and y is gJven in implicit form. to zero. of the left side of equation (I). (2) and 3) to solve the resulting equation for y'.Sec. } t and ~ = a cos t. we get 3x· 3y·y' . Find the derivative x~.q> (t). y) =0. x x x 1 2°. to put d dx F (x. that is. + + . with respect to x. Find the derivative y~ if x·+y·-3axy=0. hence. Forming the derivative of the left side of (3) and equating it. F (x. 594. r . 586.lx+ea . \ y= b sin' t. y=a(slnt+cost)..DiOerentiation of Functions 58 [Ch.sin t). Find the derivative x~ if a) y=3x+x l b) . t t+ 1 y X 584. (_t_)' { ( 1+ t y = X = 1+ tS . x c) y=O. 2 whence 581. y=x-~ sinx. find the derivative functions y represented parametrically: 89 f x=a cos· i. . dy Solution. t 693. 5 . r 2 • ~= Vi . y=t. '\ .. = { 585. 1 + t Z {X =" ~T' y \ 3at 2 + t' • 1 t. = t~I y' = :~ of the a sin t sin t a (l-cos t) = I-cos t . VI. { Y= a (sin t-t cos t).. • a (1-12) { y=b J x = a cos' t ' = I ~Itl . X 583. :~ = 1 \ Yt:l+l· 595. 5 8{2 . x=a(cost+tsint).sin t ) . Calculate I ( X = arc cos . . { y=a(l-cost). 1 +t y=elt • {x = a( . In the following problems.. when t = ~ if x= a (t. 2 588.In tan {+ cos t.r 1 . \ Y = 587. if 3at Y= Sin 590. cos't X 591. dx Yeos 2t ' sinS t V cos 2t · Y = arc sin . . Y= t { x = e. 592.• x=2t-l. x+y 608. 2 = 1 if { In t Y=-t' t w hen t __ . x·+y·=a'. Find ddYx when t · d dy 7 F In 59. I' f { X = et co. 609. Y 616. 611. ::+~:= 1. • I l O Y 64 o. d-x 1. 613. acosl(x+y)=b. Vx-t-VY=Vli. 3] The Derivatives of Functions Not Represented Explicitly 59 and • ( dx 1t sin "2 d Y) n: = t=-.::. tany=xy. 602. Let y= Val-xl. 11 l • 614. Vx -f-V y 2=Va l x-y 607. Does it follow from this that (Xl)' = (2x)' when x == 2? 600. lny+~=c. Prove that a function y represented pararrletricaJly by the equations satisfies the equation Y= (dY)' (dy)1 dx +2 dx • 599. Inx+e-X.3 siny=x. e>'=x+y. 610. When x = 2 the following equation is true: x' = 2x. Is it possible to perform term-by-term different iat ion of Xl + yl == al? In the examples that follow it is required to find the derivative y' =:~ of the implicit functions y. 598. x -}-xy+y = · 612. 5 t. 615. 606. arc tan ~ = ~ In (x' + y').:- n l-cos- = x=tlnt. 596. arctan (x+y)=x.=c. 2x-5y+lO=O. y'=-. 605. y-O. xy=arctan-..Sec. . 4 Y =e sIn t. x 603. 601. x'=yx. 4. if 2y:=. Yo) will be . 2 618. The angle between the curves Y=/l (x) and o x Y=f2 (x) at their common point M o (x o' Yo) (Fig. Putting x = 1 and y=l. Yo)' The straight line passing through the point of tangency perpendicularly to the tang~nt is called the normal to the curve.60 DiUerentiation of Functions 617. Using a familiar formula of analytic geometry. 620. we obtain 2y'=1+3y'. The tangent and the normal determine the following four . Find the derivatives y' of specified functions y at the indicated points: a) (x + y)' = 27 (x-y) b) ye' = eX+ J c) yl == X + In JLx for x=2 and Y= I. \vhere y~ is the value of the derivative y' at the point M (x o. lCJ1. 2°. Differentiating. Vx 1+y2=carctan lLx .1). 12) is the angle ro between the tan~ents MoA and MoB to these curves at the point Mo. Geometrical and Mechanical Applications of the Derivative 1°. + Solution. whence y'=-l. Find y' at the point M (1. 3°. Sec. 1 + xy·. we get 2y' = y' 3xy 2 y'. for x=1 and y= 1. From the geometric significance of a denvative it follows that the equation of the tangent to a curve :1=/ (x) or F (x. 619. Equations of the tangent and the normal. The angle between curves. y) =0 at a point M (x o. for x=o and Y= I. we get F' 12 Ig. For the normal we have the equation 8 r x-xo+Y: (y-Yo)=O. Segments associated with the tangent and the normal in a rectangu lar coordinate system. 13 Since KM = . n = N M = Yo ~:I. .Sec 4] Geometrical and Mechanical Applications of the Deritatit'e segnlents (Fig. n = MN is the segment of the polar nornzal. y o If Sn N St X Fig. 14): x T Fig. 14 t = MT is the segment of t he polar tangent. St=TK=/ 1 I VI + (Y~)! I. St=OT is the polar subtangent. If a curve is given In polar coordinates by the equation r=f(cp). is the subnormal. = OM (Fig. Sn=IYoy~l· 4°. Segments associated with the tangent and the normal in a polar system of coordinates. Sn = ON is the polar subnormal. 14). is the segment of t he normal. Yo t = TM= and tan q> = Yo. it follows that r I~o~ V + (Y~)! I. then the angle J1 fornled by the tangent MT and the radius vector . is the subtangent. 13): t = TM St=TK n == N M Sn = KN 61 is the so-called segment of the tangent. r dr r' The tangent MT and the normal MN at the point M together with the radius vector of the point of ta ngency a nd with the perpendicular to the radius vector drawn through the pole o determine the following four segments (see Fig.=r -=-. is defined by the following fonnula: dq> r tanll. c) tan q> =-1. b) x=1/2. whence the slope of the tangent is Since the .2).1). Write the equation of the tangent and the nornlal to the parabola Y= x at the point with abscissa x = 4. n=MN=Yrl+(r')I. 626. Find the points at which the tangents to the curve y==3x 4 +4x'-12xl +20 are parallel to the x-axis. q> =0°. k 1 =-4. 622. Since the slope of the normal must be perpendicular. Y = 2. rl St=OT=-1r-'I. What angles cp are formed with the x-axis by the tangents to the curve y = X-Xl r at points with abscissas: a) x=O. 15 623. Find the equation of the parabola y = Xl + bx+c that is tangent to the straight line x=y at the point (1. q> =45°. Yr +(r')I. Sn=ON=lr'l· 621. At what angle does the curve y = eO. We have y' = k = [y'l.ointof tangency has coordi nates x = 4. At what angle does the tangent curve y = tan x intersect the axis of abscissas at the origin? 624. We have y' = 1-2x. whence the equation of the normal: y-2=-4(x-4) or 4x+y-18==O. c) x=l? Solution. v- Solution.r . . ~_. b) tan q> =0. 2 t=MT=-. . At what point is the tangent to the parabola y=x l -7x+3 parallel to the straight line 5x+ y-3 = O? 627.xy-7=O at the point (1. 628.[Ch. 15). cp = 135° (Fig." = 4 = ~ . Whence a) tan q> = 1. At what angles do the sine curves y = sin x and y = sin 2x intersect the axis of abscissas at the origin? Fig. Deter4mine the slope of the tangent to the curve Xl + yl . IX intersect the straight line x=2? 625. A t what point of the curve yl = 2x' is the tangent perpendicular to the straight line 4x-3y+2=O? 630. It follows that the equation of the tangent is y-2=lj4(x-4) or x-4y+4=O. 2 Differentiation of FunctIons 62 These segments are expressed by the following fornlutas: . 629. 635. 639. d) y = In x at the point of intersection with the x-axis. Write the equations of the tangent and the normal at the point (2.0). 636. Find the equations of the tangent and the normal to the curve Y= x-I at the point (1.2). 3 1 Y=2t 2 + 2t. Form the equations of the tangent and the normal to the curves at the indicated points: a) y = tan 2x at the origin. 641.2) to the curve V- 1 +t x=-ta . 640·. 638.1).y 2+2x-6=O at the point with ordinate y=3. e) y=e1 .x2 at the points of intersection with the straight line y = 1. 4J Geometrical and Mechanical Applications of the Derivative 63 631. Show that the segment of the tangent to the hyperbola xy==a 2 (the segment lies between the coordinate axes) is divided in two at the point of tangency. 633. Show that in the case of the astroid x 2/1 + ylll = a2/1 the segment of the tangent between the coordinate axes has a constant value equal to a. 637.3) at the points of its intersection with the x-axis.5).t. 632.Sec. Write the equations of the tangent and the normal to the curve y=x'-l-2x l -4x-3 at the point (-2. Write the equations of the tangent and the normal to the curve X l . Write the equations of the tangents and the normals to the curve y = (x-I) (x-2) (x . c) y=arccos3x at the point of intersection with the y-axis. 634. Write the equations of the tangent and the normal to the curve y4=4x4+6xy at the point (1. b) y=arc sin x 2 I at the point of intersection with the x-axis. Y = t sin t at the origin and at the point t = ~ . . Write the equations of the tangent to the curve x = t cos t. Write the equation of the tangent to the curve X S + yS_ --2xy==0 at the point (1. Find the length of the segment ot the tangent.2) evaluate the lengths of the segments of the subtangent.34). Find the angle at which the parabolas y = (X-2)2 and Y= --4 +6x-x 2 intersect. 643. . Show that the curves y=4x l +2x-8 and y=xs-x+lO are tangent to each other at the point (3.DIOerentiatlon of Functions [Ch..4)? 646. Show that the length of the segment of the subnormal in the hyperbola Xl_yl = a2 at any point is equal to the abscissa of this point. mal.-_---=:. Show that the normals to the involute of the circle x=a(cost+t sint).. and normal.cos t) X { at an arbitrary point t = to. Find the angle between the tangent and the radius vector of the point of tangency in the case of the logarithmic spiral r=ae kCP • 654.. At the point (1. What procedure of construction of the tangent to the ellipse follows from this? 652. 647. and the subnormal of the cycloid = a (t . Show that the segments of the subtangents of the ellipse x2 (i2+ fj2y2 = I and the circle x 2. Given a parabola y2=4x. x 2 _y= bl intersect at a right angle. Find the angle between the tangent and the radius vector of the point of tangency in the case of the lemniscate r l = a l cos 2q>. 648. Find the length of the segment of the subtangent of the curve y =-= 2x at any point of it. Show that the hyperbolas XY= a 2 .. y=a(sint-tcost) are tangents to the circle x l +yl=a 2.. 644.sin t). subnormal. Will we have the saIne thing at (-2. At what angle do the parabolas y = Xl and y = Xl intersect? 645.ngent. the nor. ~ 64 --. tangent. 653. the subta. Show that in the equilateral hyperbola X 2_ y 2=a 2 the length of the normal at any point is equal to the radius vector of this point. 651.+ yl = at at points with the same abscissas are equal. 650. .._----=-----------=------- 642. y = a (1 .. 649. 17) is given by the formulas (air resistance is 3-1900 . 657. 656. 17 What is the rate of motion of B when A is at a distance OA = 3 m from the origin? 660*. Find the velocity of the point at to=O. The law of motion of a material point thrown up at an angle a to the horizon with initial velocity V o (in the vertical plane OXY in Fig. t 1 =1. where t~O.. subnormal. and /2=2 (x IS in centimetres and t is in seconds). The end-points of a segment AB =-~ 5 nl are sl iding along the coordinate axes OX and OY (Fig. Find the lengths of the segments of the polar subtangent. 16 X o A x Fig. subnormal. 16)..!:. and normal. y r B o J A Fig. tangent and normal. 658.Sec. With what speed are these points receding from each other at the time of encounter (x is in centinletres and t is in seconds)? 659. r = r o. Find the lengths of the segments of the polar subtangent. tangent. 4] Geometrical and Mechanical Applications of the Derivative 65 655. and also the angle between the tangent and the radius vector of the point of tangency in the case of the spiral of Archimedes r=acp at a point with polar angle <p=2n. and also the angle between the tan· gent and the radius vector in the hyperbolic spiral r = . Moving along the x-axis are two points that have the following laws of motion: x= 100+5t and x= 1/2/ 1 . The law of ITIotion of a point on the x-axis is x=3t-t'. at an cp arbitrary point cp = CPo. A is moving at 2ol/sec. [Ch. Also determine the speed of motion and its direction. Is the law of rectilinear motion of a point.2)? 662. A nonhomogeneous rod AB is 12 cm long. A derivatzve of the second order.2 Difjerentlation of Functions 66 disregarded): X= V o t coso. Derivatives of Higher Orders to.~ is the accel- . M from the end A and is 10 gm when AM =. AM. A point is in motion along a hyperbola y = 10 so that its x abscissa x increases uniformly at a rate of 1 unit per second. while the other side. If X= f (I) or d. increases with the square of the distance of the moving point. The mass of a part of it.2 em. Definition of higher derivatives. Find the mass of the entire rod AB and the linear density at any point M. or lhe second derivative. then eration of thIs motion. a = 10 em. b. What is the rate of change of its ordinate when the point passes through (5. increases at a constant rate of 4 cnl.x2' or f" (x). :. One side of a rectangle. is of constant length. At what rate are the diagonal of the rectangle and its area increasing when b = 30 cm? 664. 5. The second derivative may be denoted as 1 d2 y II. 661. of the function y= f (x) is the derivative of its derivalive. The radius of a sphere is increasing at a unifortTI rate of 5 cm/sec. A point is in motion along the spiral of Archimedes r=acp (a = 10 cm) so that the angular velocity of rotation of its radius vector is constant and equal to 60 per second. Determine the rate of elongation of the radius vector r when r = 25 cm. What is the linear density of the rod at A and B? Sec. where t is the time and g is the acceleration of gravity. y" = (y')'. that is. Find the trajectory of motion and the distance covered.'sec. 666. At what point of the parabola y2 = IBx does the ordinate increase at twice the rate of the abscissa? 663. At what rate are the area of the surface of the sphere and the volunle of the sphere increasing when the radius becomes 50 cm? 665. cot t.. . y xxx = .y x (Y:x)~ II' . ' 1 . Find the second derivative of the function y=ln(l-x). b -1 --a.XttY. v' + n (~~ I) UC- 2) v" + . Sfi1i7 -a sin t b =. x. Leibniz rule..( y'~). For the nth derivative we use the notation y (n). Y=l-x.( ')'x -.= (a cost).= :~ . 5] 8 67 Generally. F\nd y". ( . Xt For a second derivative we have the formula " Yxx= . Higher-order derivatives x=<p(t).u (n-I. if X=Q { cos t. b . dny or dx n f (n) (x). cos t b = . if 3°. + uv(n). Y= I-x Sou =(I-x)2' 2°.I )' 1 I Ii on. . . { Y='i'(t). y xx -. " x t Ytt . . We have y'= (b sln t) ~ . Solution. -a SUI t Q and ( 'l' -~ colt ): (a cos t).. then to evaluate the nth derivative of a product of these functions we can use the Leibl1iz rule (or formula): (uv) (n) = u(n)v + n. Y =.- at sin' t • ...Derivatives of Higher Orders Sec. can successively be calculated by the formulas " . then the derivatives y.b sin t. or ' Example l. If the functions u=<p (x) and v ='1' (x) have derivatives up to the nth order inclusive.and so iorth. the nth derivative of a function y=/ (x) is the derivative of derivative of order (n-l). ( x~)a Example 2. of functions represented parametrical ~j'. y= In VI +x 674. if y=x 3 -5x 2 -f-7x-2. Find the velocity and the a< celeration of motion of MI' What is the velocity and the acceleration of M 1 at the in tial time and when it passes through the origin? What are the maximum values of the absolute velocity and tt absolute acceleration of M J ? + .00lt 3 • 1. 680. Find the law of motion of i1 projection M. Show that the function y = Xl + 2x 2 + 2 sa t·IS f·les th e dl·fTer ential equation 1 +y.68 [Ch. The equation of motion of a paill along the x-axis is x = 100 + 5t-0.x cos x satisfies the differ ential equation yIV +4y==0. 686. A point M is in motion around circle x2 y2 = at with constant angul2 Fig. Find {(O). Higher-Order Derivatives of Explicit Functions In the examples that follow. an t 2 = 10. 670. Show that the function y = e.1X satisfies th equation y" +3y' -1-2y=0 for all constants C 1 and Cz.z=-=2yy". 667. Differentiation of Functions A. 673. Y = eX!. Show that the function y = eZx sin 5x satisfies the equa tion y" -4y' + 29y = O. Show that the function y = -} x'e x satisfies the differen tial equation y"-2y' +y=e x . y=(arcsinx)l. Find y"'. f' (0). 671.x -1. if f (x) = (2x-3)5. arc tan x. 678. 682.40 X Find the velocity and the acceleration ( the point for tinles to = 0. on the x-axis if at time t = the point is at M o (a. f (x) = (1 + Xl). 668. t 1 = 1. 68:3. y 684. 685.C 2 e. 676. Find yV of the function y::= In (1 + x). 18). y=ln(x+Va 2 +x Z ). 669. 672. find the second derivative of thl given function. y=x l +7x 8 -5x+4. 677. l • a 675. 681. y = sin l x. Find f"' (3). y=acosh~ . 18 velocity 00. 0) (Fig. Find yVI of the function y = sin 2x. 679. Show that the function y == C 1e. f"(O) and f'" (0: if f (x) = eX sin x. y=(l-xl)cosx. Find fln) (0). 694. e)Y=I+x. 688. Find the nth derivative of the function U= (ax + b)lI. { x=arctant. b) 1 Y=l-t' {X = e' cos t. f) y = c) y = e-' . y=ln(1-t-t 2 ). X= a cos' t. Using the Lei bniz rule. Find the nth derivative of the functions: 1 a)lJ=sinx. 695.e". b) { . Find d---a. a) y = ~ tl . If y=¥l-t l • {x=a(t-sint). d) g=ln(l+x). { x = a (sin t .2X . .atural number. 689. a) { y = sinl t. 8J 69 l). a) { • y = a sin t.'ec. a) {x=ln t. t Y ==e sin • . if: l+x a) y=x. c) x=e-at. at y=e • 696. b) {x=arctant. . x=cos2t.t cos t). y = a (1 . if f (x) = In I Ix B. d) y= b) y=x l ·e. y=t l . 690.cos t). 692. and 8) 1/=-1-x. Jt : +: . c) 691. find ycn). Find the nth derivatives of the functions: 1 1/- b) y= v x.rlvatlves of 811her Orders 687. b) y = cos 2x. d) y=a(cost+t sint). h) y=ln(ax+b). b) { y=asin't. where n is e Il. 693. t . x=acos t. c) y == x 3 In x. {x=arcsint c) { x=lnt. Iltgher-Order Derivatives of Functions Represented Parametrically and of Implicit Functions In the following problems find :~ . d1x Y . g) y = sin l x. Yx. t Vi-for any constants a and b satisfies the differential equation 2 d2 y (l-x)dx l dy - x dx=2y. Find -d n' if m x y=t. Knowing the function y = / (x). Pmd dx 2 for / =0. . x2 706. a y= an t.1). Find y" at the point (1. find the derivatives x". { y=e. a) The function y is defined implicitly by the equation Xl 2xy + yl-4x-t. 2px. Having the equation y = x + In y.Differentiation of Functions 70 . {x=ln(l+t [Ch. y2+ X 2 1 y' = III ' l' ---r=- In the following examples it IS required to determine the derivative y" of the function y = / (x) represented implicitly.Y y x y Substituting the value of y'. y=x-tarctan y. y=t .t and ::~.t . dny { X= lo t. 710.1' = . x'" of the inverse function x = /-1 (y). wh ence y ' = . 699. If Y= /2. . 704. t 702. 697.-t.1) if x 2 +5xy+y2_2x+y-6=0. By the rule for differentiating a composite function we have x an d !:J. x=e-t cost. 707. d2y . Show that y (as a function of x) defined by the equations x = sin t.y2 = 1.1) if x·-xy+y4= 1. Solution.-x )' =----2-. dly Find dx" if x 2t y2 = al. 700.t 'sin. :1. { t 701.2y-2 = O. Y = aet Vz + be. 703.1 . Q2 y2 + b2 = 1. 708.2yy ' = 0 . we finally get: . Find y" at (0.2 2 ). find ~2. if x2 . 711. b) . . 698. + Find :~ at the point (1. 709. 705. y -xy' 2x-f. y2 --. { x = e. In the following examples find y'" = X = sec i. Find y". Calculate ~y and dy of the function y=3x 2 -x for x=l and ~x=O.0503 dy=(6x-l) ~x==5·0. 19 function is equal to the product of its derivatIve by the differential of the independent variable dy== y'dx. Example t.01 =0. which part is linear relative to the increment I1x=dx of the independent variable x. First-order differential. Second method: y' =6x-l. MT is the tangent at M (x. 6. . then the increment in the ordinate of the tangent AT=dy and the segment AN = ~y. 19).01 + 3· (0. y) and PQ =~x==dx. Hence. The diOerential (first-order) of a function. Example 2.Ol. dy=y' dx= (6x-l) dx. first method: ~y = 3 (x+ ~X)I_(X+~x) -3x 2 or ~y = (6x . and ~y= (6x-I). whence 1/ •1 :::::.01)1=0.dy . The differential of a y o Fig. 6] DiOerentials of First and Higher Orders 11 Sec. dx If M N is an arc of the graph oi the function y = f (x) (Fig. Find the increment and the differential of the function y=3x 2 -x.I) ~x +x + 3 (~X)I.Sec. Solution. Y = f (x) is the principal part of it~ increment. Differentials of First and Higher Orders 1°. dy= (6x-l) Ax= (6x-l) dx.0500. Solu tion. ~x + 3 (~X)2= 5·0. If Y == f (x) and x is an independent variable. If x is the area of the sq uare and y is its si de. 2) dx == /ix. that is. 7) df (u) =.72 DiOerentiation of Functions [Ch. where c = canst.v duo 6) d ( ~ ) V = v du - v2 u do (0 #= 0).·O. If the increment Ax of the argument x is small in absolute value. where x is an independent variable.OOl. then + d 2y = y" (dU)1 y' diu. then d 2 y = y" (dX)2. f (x + fix) (x) dx. (Here the primes denote derivatives with respect to u). 2 f 9 4°. whence ~y=:::. A second-order dtOerential is the differential of a first-order differential: We similarly define the dtOerentials of the third and higher orders.1 m2 ? Solu lion.f (x) ~ f' (x) fix. Find the increment ~y and the differential dy of the function U=5x-t Xl for x=2 and L\x=O. d 3 y = y'" (dx)'. 5) d (uv)=u dv -f. ~ f (x) + l' Example 3. Applying the differential to approximate calculations. then the differential dy of the function y=f(x) and the increment fiy of the function are approxilnately equal. Higher-order differentials. where u = cp (x). It is gi ven that x = 9 and ~x = o. d 2 u y' diU + + and so forth. By how much (approximately) does the side of a square change if its a rea increases from 9 m2 to 9. 1. But if y = f (u). 4) d(u ± v)=du ± dv. then y=yx. The increment fiy in the side of the square may be calculated approximatply as follows: Ay===dy=y' Ax= ~r.f' (u) du. 3) d (cu) = c du.1=O. day == y'" (du)a 3y" du.2 2°. . Principal properties of differentials. f (x + fix) .dy. dny = y(n) (dx)n. 1) de = 0. 712.016m. 3°. t 726. Has the function Y==lxl a differential for x=O? 719. Y=--=x tn • 727. Without calculating the derivative. 721. Give a geometric interpretation of the increment and differential of the follo\ving functions: a) the area of a circle. the increnlent in the function y = 2x . corresponding to an increment ~x in x.. 723 728. Show that \vhen ~x -+0. dy=--x-y . S = nx 2 . Y = e.. 6] DiOerentlals of First and Higher Orders 73 713.X --= 3~ • 720. 730. The area of a square S with side x is given by S=x l • Find the increment and the differential of this function and explain the geonlctric significance of the latter. find S d (I-x ) 1 for x = 1 and ~x = . Calculate the diITerential of the function Y == tan x Jl n for x::=. is. Taking advantage of the invariancy of the fOrln of a differential.. equivalent to the expression 2-'. s = arc tan e • x '·Y==T-x· I I-x y=nl-f-x· + cosec <p. !J= arc sin : . 3 and ~x -== 180 • In the following problenls find the differentials of the given functions for arbitrary values of the argunlent and its increluent. y=xlnx-x.y l = a l • Solution. v=x s • 716. 715. Using the derivative. r = cot <p x 725. O? 718. v 714.-. Y = arc tan -a . find the differential of the fUllction y = cos x for x = ~ and /).OI.x2 • 731 Find dy if x 2 -}-2xy. b) the volume of a cube. for any x. For \vhat value of x is the diITerenti al of the function y == x 2 not equivalent to the incretnent in this function as L\x ---.Sec. 724. 729. we obtain 2xdx-{-2(ydx+xdy)-2ydy=O Whence X+Yd x. Find the diITerential of the function 2 Y== Yx for x==9 and L\x==-O.In 2 ~x. 1 722. 717. V640.017. c) eO. c) 1 xl 1 l+xx . 744. Approximate the functions: 8) y=x'-4x l +5x+3 for x=I.74 DiOerentiat ion of Funct ions In the following examples find the differentials of the functiollS defined implicitly. if y'-y=6xl • 736.. b) f(x)=Vl +x for x=O.05.2).9. Y = e-Ii . (X+y)2. 743. d)y =e f or x= 105 . Find dy at the point (1. (2x+y)' = 1. Approximate VI7. V70.03. Derive the approximate formula (for I ~x \ that are small compared to x) vx + ~x ~ VX + 2 ~x · lJsing it. 740.500+0. Find the approximate value of arc sin 0. 741. V70. V200. Replacing the increment of the function by the differential. 738. 734. ~3 =0. Approximate tan 45°3'20". In V Xl + y. 737. approximate Y-5. Solution.2. Find the approximate value of sin 31°. x 733. = arc tan JLx • 735. b) tan 44°. Derive the approximate formula . 2 . What will be the approximate Increase in the volume of a sphere if its radius R = 15 cm increases by 2 mm? 739. d) In 0. Putting x=arc300=i and Ax=arc 1°= 1~0' from formula (1) (see 3°) we have sin 31°:::::::: sin 30° + 1. 732. V17. f (x) = V for X= 0 • 1 . 742. e) arc tan 1.V-x + V= ax Vx+~x~ 3 Xl and find approximate values for VI0.515.0 cos 30°=0. calculate approximately: a) cos 61°.54. find dnu. 755. then f (b) -f (a) F (b)-F (a) f' (~) = F' (~)' where a < £< b. y = eX cos a sin (x sin u). may be found approximately by the formula 8/=- I R~R. 746. . d 2 y = y" (dx 2) = . xo& Z = (-). in determining the length of the radius. If the functions f (x) and F (x) are continuous on the interval a ~ x ~ b and for a < x < b have derivatives that do not vanish simultaneously. Cauchy's theorem. -x •• find d 01 z.X . In x f d d 2 751. Show that the function f (x) = x-x' on the intervals -1 ~ x ~ 0 and 0 ~ x ~ 1 satisfies the Rolle theorem. 753. 7) 745. 754. such that (~) =0. has a derivative f (a) -= f (b). 11 748. if Y = cos 5x. 7. y = sin x In x. 756. and F (b) :/:. Find the appropriate values of 6e . show that a small change in the current. 2°. y = arc cos x.< b. Compute diy. z = x2 e.. then f (b)-f (a) = (b-a) I' (6). If a function f (x) is continuous on the interval a ~ x ~ b and has a derivative at every interior point of this interval. find d 2 y. If a function f (x) is continuous on the interval f' (x) at every interior point of tillS interval. Using Ohm's law. Rolle's theorem.. find dray. z=-X' In z. 747.. 3°. Mean-Value Theorems 1°. find d 2 u. find d'z.. 752. I 15 =~. u = i-x2 . a relative error of 1% results in a relative error of approximately 2° loin calculating the area of a circle and the surface of a sphere.F (a). 749. then the argunlent x has at least one value I' ~.5).25 cos 5x (dX)2. II = 3 sin (2x -t. where a < . due to a small change in the resistance. where a < ~ < b. Lagrange's theorem. and a ~ x ~ b. 750. Solut ion. Show that. find d 2 y. Sec. Mean-Value Theorems Sec. x obviously has a root x=O. ~1 = - -V ~ . 761.2) (x + 3). Show that this equation cannot have any other real root. The function /(x)=V(x-2)' takes on equal values 1(0)=/(4)= V4 at the end-points of the interval [0.1) (x t. n]? 759. the only suitable value is inequality -2 < ~ < I holds ~=-l.1] and find the appropriate internlediate value of Solution. The function f (x) is continuous and differentiable for all values of x.~2= -V ~.2 Solution. · 1-3~2=-2 and ~=±l. Whence and 0 < ~I < l. To find £ we fornl the equation f' (x) = 1-3.1) and B (3.r=O.4]? 758. f' (s)~~2 Hence.. Hence. we h3ve f(1)-f(-2j=O-6==(I-(-2)]f'(~). by the Lagrange fornlula. Using the Lagrange theorem.DiUerenttation of Functions 76 [Ch.. s. +h) - sin x = h cos S. Given a segment of the parabola y = x! lying between two points A (1.1]. Test whether the Lagrange theorem holds for the function {(x) = x-x 3 on the interval (-2. and I' (x)= 1-3x2 Whence. that is. on the interval [-1. Test the validity of the Lagrange theorem and find the clppropriate intermediate point ~ for the function f (x) = X"/. 764. . and f (-1) =f (0) = f (1) =0. 760.4). prove the formula sin (x where x<.9). where -1 < ~1 < 0 757. Does the Rolle theorem hold for this function on [O. the Rolle theorem is applicable on the intervals -1 ~ x <: 0 and 0 ~ x < 1. 763. find a point the tangent to which is parallel to the chord AB. The function I (x) =x-x' is continuous and differentiable for nil values of x. Show that the equation has three rea i roots. for which the 762. l~he equation f' (x) =0 eX = 1 t. Does the Rolle theorem hold for the function f (x) = tan x on the interval [0.<x+h. Let f (x) = x (x . +(x+ I)..).a)2 t" (a l + (x .. a) For the functions f (x) = r -t. and there IS a finite derivative fen) (x) at each interior point of the interval. Taylor's Formula If a function f (x) is continuous and has continuous derivatives up to the (n-I)th order inclusive on the interval a ~ x ~ b (or b ~ oX ~ a).!. Hence.. fen) (x) = 0 766. f' (x) == 3x 2 .r . + (~. f (x) = In x in powers of x-I up to . Solution. eX 2) + 4 (X--2)2 -l.. 768. 8] 77 765.~~~l-I rn-I'(aH (x:~)~ fIn) (. + (x + I). then Taylor's fornzula (x . {" (2) == 8.Taylor's Formula Sec. + (~n-:)I fIn_I) (0) + :~ fin) m. f (x) = eX in powers of ten) (-1) =. Expand the function the term with (x-I)I. 1(") (x) =e X for all n.2)2 8 (x .. +--.2] and find ~.!.. Expand the function term containing (x+ 1)1..(x-2)'. + (x+ 1)2 .. for n ~ 4. a + (x-a ) t ' (a) + -2-1f (x) == f () 3t . Sol ution. :]. holds true on the interval. rherefor~. In particular. 0<8<1.4x + 3. [~xpand the polynomial f(x)=x ' -2x l +3x+5 in positive integral powers of the binonlial x-2. 8. e =. til (x) == 6x . f"' (2) = 6. 0<8<1...I f'" (a)+ .! + (x+ 1)· e e 21 x -r-l to the e 31 e 41 e~ ' where ~=-1+8(x+l).!. when a==O we have (Maclaurin's formula) f (x) =f (0) +xf' (0) where ~=Ox.2).a) . Whence f (2) = 11.4~ f'" (x) = 6.. 7 +---. where ~==a+8(x-a) and 0<8<1.!.. +fu f" (0) + . 2! 31 or x l -2x 2 +3x+5= II + 7 (x - 767. f' (2) = 7.. x• .2 and F (x) = x' -1 test whether the Cauchy theorem holds on the interval [1.. b) do the same with respect to f(x)=sinx and F(x)=cosx on the interval [0.. Sec.2)1 6 + (.2x 2 + 3x+ 5 = II (x . 773. Determine the origin of the approximate formulas: 1 1 V-1 +x~ 1 +2"x-S-x -1 1 b) V 1 + x~ 1 + "3 x·. Let the single-valued functions f (x) and cp (x) be differentiable for 0</ x-a I <h. Due to its own weight. Show that sin (a +h) differs from sin a -+h cos a 2 by not more than 1/2 h • 772.9 x 2 a) Ixl .78 [Ch. is one of the indeterminate forms lim f (x) _lim l' (x) x~a <p (x) . if the quotient ~. 1 + 3! T 4i · 774. to within approximate eq·uality x ea ~ (~ r. Show that for small Ix I the a shape of the thread is approximately expressed by the parabola x2 y=a+ 2a · 775*. the derivative of one of them does not vanish. Evaluate the error in the formula e ~ 2+ 1 2! 1. 9. that is.1 • 771. I x I< 1 2 . If f (x) and cp (x) are both infinitesimals or both inflnites as x ~ a. ~ or .%-+-a <p' (x) provided that the limit of the ratio of derivatives exists. then co ~(~:)' at x = a. The L'Hospital-Bernoulli Rule for Evaluating Indeterminate Forms 1°. 2 DiOerent iat ion of Functions f (x) = sin x in powers of x up to the term conand to the term containing Xl. and evaluate their errors. Evaluating the indeterminate forms %and : . Expand taining Xl ing x n . Expand f (x) = eX in powers of x up to the term contain769. 770. a heavy suspended thread lies in a catenary line y = a cosh -=-. a+x a-x Sec. we have the y-. Show that for Ix I~ a. < 1. 0) (x) The indeterminate forms I 0<». x %-+0 X We thus finally get lim In x =0. To evaluate an indeterminate form like 0-00. 2°. whereas the ratios of the derivatives do not tend to any Ir~~~ (see Example 809).(form -). Example 1. (x).(~1) again yields an indeterminate form. Applying the L'Hospital rule we have lim ~ = Urn (In x)' %-+0 cot x We get the indeterminafe form x~o (cot x)' ~. I n certain cases it is useful to combine the L'Hospital rule with the finding of limits by elementary techniques. of one of the two above-mentioned types and I' (x) and cp' (x) satisfy all the requirements that have been stated for' (x) and «p (x).Sec. Compute In x 00 l1m . I( -) (the 1 x 12 (x) In the case of the indeterminate form 00 - '. 'I 1 1 (x) (x) form 00). Other indeterminate forms. cot x x~o .I«X». X'~o cot x 00 (X) . == _ Urn x-+o l sin » • x however. etc. x lim if %~a I duce the expression to the form 1_'t. 00 one should transform the 00. [I-~: ~:n '. then appropriate difference fdx)-f 2 (x) into the product f'<x) first evaluate the indeterminate form '1 and we re- 1 X (the form 0-. However. If the quotient ~. into the quehent /1 (IX) (the ~a 'I 0 ¥-wJ (x) form -0 (or. 9] L' Hospital-Bernoulli Rule for Indeterminate Forms 79 The rule is also applicable when a= 00.. at the point x=a. it should be borne in mind that the limit of the ratio f (x) may exist.. transform the appropriate product (x). where lim (x)=O and '1 lim!2 (x) =~. 00 0 are evaluated by first taking loga e rithms and then finding the limi t of the logarithm of the power ['I (x» 12 (¥' (which requires evalua ting a form like O· 00).1 (x» = 1. we can then pass to the ratio of second derivatives. since l lim ~-+o sin x I sin x --= 1m -·sln x= 1-0=0. we do nof need fo use the L'Hospital rule. Solution. Seco 4) x 2 sin! x_x4 • We obtain x 2 . we find lim x-+o (_I__ .sin 2 x 11m . (1 1) x S111 2 X x-+o 0 0 X X-+O The L 'Hospital rule gives 1. . lim .-2 = Iun 4 (form -0). Compute a lim (cos 2x) Xi (fonn 100 ) x-+o Taking logarithms and applying the L'Hospital rule. tan x--sin x lim x-sinx . lim cosh x-I 1 -cos x x~o I-x 1...r--+J ot 779.x-+o 4x 8 x-+o 12x2 0 0 0 x-. X-i-O x2 X -+0 2x • lienee.1) . .t~o .sin ~ 2 780. rlin x cos x-sin x a • x -+Q 778. 2 Example 2. .s• x-+o F·"ind the indicated limits of functions in the following exatTIpleso · XS_2XI_X+2 776 · 1lin 3-7 +6 X-+l x X · Iimx'-2xl-x+2_Il°m3x2-4x-I_1 · _ __ So IU t Ion. COlTIf'ute lim x ~o (_l__ . we get · (1 I) = ---- ltm sin 2 x x-+o x2 2 x (form -) 0 ltm x -sin! x 2 sin! x 0 · 0 x~o Before applying the L'Hospital rule. we get a lim ln (tos 2x) X2 = HIn 3ln cos 2x = _ 6 lim tan 2x = _ 6 x-+o .11m 2x-sin 2x .+o • Then. X-'l x'-7x+ 6 X-+J 3x 2 -7 2· 777. lim (cos 2x)Xi:y:e.!-) (form 00-00). sin x x 6x 6x 3 2 2 2 x-+o 2 x-+o Example 3. sin x x 2 2 Reducing to a common denomi nator. we replace the denominator of the latter fraction by an equivalent infinitesinlal (eh.. in elementary fashion.80 DiOerent iat ion of Funct ions [Ch..11 m 2-2cos 2x 1Jm ( sin 2 x x 2 . I.-'!')=liIn l-cos2x=lim 2sin2x=~. Hln (x ll e.\.t=z x-?o x-?o c:: lilll X -?>O sin x -+0 Sill X SlIl .cos 4x • 785.. n>O.\-} I .\-1 Inx • 791. 795..· (794. linl x n sin!:. X~1 = Htll 1) =Itm.. hnllnxln(x-I). Hln x sin!!:.. x .\ 796. 784.L' Hospital-Bernoulli Sec.. Iny=-xlnx: x ! Inx ==lim-=lim--==O. hmn Rule for Indeterminate Forms sec 2 x-2tanx I -1.\. x?4 x-+°cotj(x · tan x 782 · 1Im nt. . lim _x_ .\-I-x(·\-1) 1 Ill' ~ 1 x-+1}nx- I X---+l 79~ = ltnl-1. 1m 1 • x---+o n sin x X-i-- 2 . lilll arc sin x cot x. (X hIll ---n cotx 2cosx 1] V d..\"-.r-+o . 787. 9] . = 0 t cos \" nx 792.= %-+0 l. x~o We Solution. 3 (1 - x) • t~- 2 798. 788. 793.cos x) cot x.--+Inx-I x I =linl x-~11I1. In (sin nlx) 1 786. n>O. eX 783. lim (1.an Sx • 2 .X ). I lIn · (x-I) X-i-l 'x) 1t) . whence limy=l. xlnx-x+1 == In \ (X ----In x .\. X+l t"~0 . that %-+0 -. I lIn x -+3 X X. hm --. .. t'---+oo X . Itnl .. liln xx.. ltnl.1 x X-x' 789. I lIn 790. . liln (I-x) tan "2 .. 11m 81 x -+0 In x Vi - X-~'7J x • Solution. 2 _ -6 . 781.\. x~o 1) \.lil11 (l-cosx)cotx=lim (I-cosx)cosx== litn (I-cosx).-I-=2 · ~71~+X2 5) [ I JI 2 (1 - ~ x -+-l · ( -:--3.. Solutaon.!X x-+o _ x2 linllny=limxlnx=x-+o 1 %-+0 IS. have xX=y. .'n x. 20 810*. 806.82 DifJerent iat ion of 1 1 799. lim x. lim Xl-X. %-+0 %-+0 rEx 802.)tan %-+0 ~-+1 % • X J 803. x • a) 1Im_. lim (I %-+0 + x fi . linl xx. is approximately 2 S~-bh 3 wi th an arbitrari Iy small reI ati ve error when a --+ O. lim (cot x)~in 2 %. 20).-=o. %-+0 809. Iim( . x-+>+ tI) 800. 804. o Fig. %-+0 %-+J 7) ". Prove that the limits of Xl stnJ. Show that the area of a circular segment with minor central angle a. lim (cot x)ii1i. 1 801. [Ch. Find these limits directly. x~o sin x b) lim x-s~n x = 1 %-+ClDx+slnx cannot be found by the L'Hospital-Bernoulli rule.2 Funcf ions C-+J a limx4 + 1n tan~ 805. lim (l_x)cos a .!-. lim(tan %. 808. which has a chord AB=b and CD=h (Fig. 807. 1. <b. 2) if 1<x< -1. 21b»). the functioll t (x) decreases in the interval (-00. Test .the following function for increase and decrease: r ' y=x 2 -2x+5. Whence y'=O for x= 1. frolll the inequality X I <X2 we ~et the Inequality I (x I )< <I (x 2 ) (Fig 21a) [{ (XI»! (. any points Xl and X 2 which belong to this Interval. then y'>O. On a nUlnber scale we get two intervals of monotonicity: (-00. These Intervals are bounded by Cl itic poi nts x [where I' (x) =~ 0 or f' (x) does not exist]. Solution. b) and f' (x»O [t' (x)<OJ for a<.00. Frolll (1) we have: I) if -oo<x<l. 22 Fig.x 2 ) (Fig. b). Thl 1u11ctlon y= f (x) is called on some Interval if. Example 1. hence. I). 22). hence.Chapter 11/ THE EXTREMA OF A FUNCTION AND THE GEOMETRIC APPLICATIONS OF A DERIVATIVE Sec.x) increases (decreases) un the Interval [a.l finite nUlllber of intervals of incrense and decrease of the function (intervals of 11l0notontcity). fo. + + . the don13ln of definition of f (x) may be subdivided Into t. 21 111 the sinlplest cases. Increa~e dna decrease of 'unctions. t he function f (x) increases in the interval (1. The Extrema of a Function of One Argument 1°. and. We find the derivative y'=2x-2=2(x-l). then y'<O. 00) (Fig. and. co). increasing (decreaslItg) y y o (b) (a) o Fig. then f (. i i f (x) is conttnuous on the interval la. I) and (I. 2 is a discontinuity = -~22<0 of the function and y' = for x:j:. hence. f (x»! (x o). y' has no discontinuities). +00) retains its sign. o x . The rTiI nimum point or maximum point of a function is its ext remal point (bending point). Similarly. -1) and the function in this interval increases Sirnilarly.x o of this neighbourhood we have the inequality X. the function under investigation is monotonic in each of these intervals. if for any point x¥=x l of some neighbourhood of the point Xl' the inequality f (x)<t (Xl) is fulfilled. y'>O 111 the interval (-00.00.00). -t. To determine what the sign of y' is in the interval (. Determine the intervals of increase and decrease of the rune· tion 1 Y=x+2 · Solution.84 Extrema and the GeometrIc Applications of a Derivative [Ch. it is sufficient to determine the sign of y' at sorne point of the interval. for this reason. Example 3. -I).oo<x<-2 and -2<x< + 00. y'<O in the interval (0.0). one has to determine the sign of the derivative In each of the intervals. then the point Xo is called the minimum potnt of the fu nct ion y =-= f (x). (-1.1) and (1. Fig 23 whtle the nUlnber I (x o) is called the mInimum of the function y= f (x). x 2 =0. for example.0. the function being tested increases in the interval (. Since y' can change sign onl y when passing through points at which it vanishes or becomes discontinuous (in the given case. Here. The converse is not true: points at which f' (x) =0. then X J is called the maximum point of the function I (x). -1). + 00). decreases in the Interval (-1. we find the points Xl = -1. then I' (x o) = 0. To determine in wh ich of the indicated intervals the function increases and In which it decreases. does not exist (critical polnts) are not necessarily extremal points of the function I (x).1) (here. Test the following function for increase or decrease: 1 y == 5 Solution xS-a1 Xl. we can take X= - y ~). and f (Xl) is the maximum of the function (F ig. (0. Here. xs == 1 at which the derivative y' vanishes.3 Example 2. (2) Sol ving the equation X4 _X 2 =:. 2°. the derivative in each of the intervals (-00. or I' (x). and the minimum or nlaximum of a function is called the extremUtn of the' funct ion. -1). we find that y'<O in the interval (-1. 1) and again increases in the interval (I. If there exists a two-sided neighbourhood of a point Xo such that for every pC'int x:j:.00. 23). or f' (xo) does not exist (necessary condition for the existence of an extremum). If Xo is an extremal point of the function I (x). Hence. the function y decreases in the intervals (x+ ) . Extremum of a function. x= . Thus. we can use x = 1/2) and y'>O in the interval (1. we get from (2) y' = 12>0. taking X= -2.-2. 0) (as a check. if k is even.h.. Con"cquently. whence \\e find the second critical point of the functIon xt=O. xt=-=O is the Il1inl111UI11 point of the functIon y. Greatest and least val ues. \VhenCL'. The least (greatest) val ue of a cont inuous function f (x) on a given interval [a. 2.= 2x i +3 V x 2• Solution. \ve get V- x~ o·. Xl::::: . From fonnula (3) \ve have: if x==. where h IS a sufficiently sl11all positive nUl1lUer. but if f' (xo) = 0. Finally. th en Xo is the 1T1 i n imum po in t. Example 4. then the point Xo is not an extremal point. Iff' (x o) =. 1] Tlte Extrelna of a Function of One Argument 85 The sufficient conditions for the existence and absence of an extremum of a continuous function f (x) are given by the following rules: I. then X o is not (n ext re111 a I poi nt . if t<k) (xo»O But if k IS odd. we find the critical point Xl =-= -1. and t" (xo)<O. then y'>O. then Xo is the minimum point of the function f (x). lienee. and ymax ::-:-. Then. one can calculate arithmetically by taking for Iz a sufl1cillntly snlall positive number. xo+6) of a critical point Xo such that f' (x»o for X o. and f'" (xo):#:O.I is the 111aXlInUm point of the function. and !/rnin==O (FIg. then Xo is the Inaximum point.6<x<x o and I' (x)<O for xo<x<xo +~. the l11axinlUl11 point. we get: V x+ 1 ==0. If there exists a neighbourhood (x o-6./ ' 3x V x Here. y" <0 for Xl = . ~I\ore generally: let the first of the derivatives (not equal to zero at the point x o) of the function f (x) be of the order k. but if x=---= -1 h. 1" (xo) -= 0. . then y' <0*). . and if I' (x)<O for xo-6<x<xo and f' (x»O for xo<x<x() + 6. iff' (x 0) == 0 and (x 0) > 0. the point Xo IS an extrenlal point. Xl = .I IS the maXlmU111 point of the fu nctton !I.I and. -=V-( Vx derivativ~ y' (3) t to zero. b] is attained either at the critical points of the function or at the end-points of the interval {a. + Eqllatln~ thr dcnolninator of the expression of y' in (3) to zero. \\There there IS no derIvatIve !I' For x== -h. for x== Iz \ve have U'>O. 24).Sec. then Xo is not an extremal point of the functIon f (x). nalnely.: .1. Clnd it is the 111ininlum point. then Xo is the maxinlum point of the function I (x). b]. *) If It is difficult to deternline the sign of the derivative y'. if t<k) (xo)<O. if there is SOlne positive number () such that f' (x) retains its sign unchanged for 0<1 x-xo /<6. Find the derivative y'=2-t- Equating the 2 2 V-x+I). hence. It is also possible to test the behaviour of the function at the point x= -1 by means of the second derivative !I=-. 3°. Find the extrenla of the function r° y:. \ve obviously have y'<O. 4)1. Fig.Y( . 24 ~\J Comparing the values of the function at these points and the values of the function at the end-points of the given interval Y(-I)=5.Hi· ) y=(x-3)Vx. Y = xZ 818. Y(2 ~)=ll ~. 25) that the function attains its least value. 817. y = Xl (x. Solution.I ~)=4 ~. Y=--!-2. at the point x= I (at the minimum point). Find the greatest and least values of the function y=x 8 -3x+3 on the interval -P/2~X~21/1.3). it follows tha t the critical points of the function yare Xl = -I and XI = I. iions: 811. 812. 815. we conclude (Fig.f(I)=I. 813.8 Example 5. Y= 1-4x-xl • Y=(X-2)1. . (at the right-hand end-point of the interval). Y = (x -t. x- 816.86 Extrema and the Geometric Applications of a Derivative [CIl. 1 Y=(x_I)2. 814. m = I. and the greatest value M = II at the point x=2 1 /. : x. i- Determine the intervals of decrease and increase of the fune . Since y'=3xl ·-3. y y x Fig. Sec. y=2x'+3xz -12x+5. Y = X In· x. Since y'<O when x<-2. 844. y-= X (4-x 2 ) 836. Solution" We find the derivative y'=6x l +6x-12=6 (x l +x-2). we get the critical points Xl =-2 and XI = 1. Y=x l +3' 840. Equating the derivative y' to zero. 841. y=x· (x-·12)2. y=x-ln(l+x). 846.arc tan x. Y= X In x. 839. • 842. y= x' Test the following functions for extrema: 826. Solution. y = x" + 4x -t.r-I)·. determine the . Equating y' to zero. l 833. x· 832. y=xe x . We get the same result by utilizing the sign of the second derivative at the crit ical point y" = 2>0. it follows thatxl =-2 is the maximum point of the function y. 848. therefore. 1] The Extrema ot a Function of One Argument V-x. y=~ . y=x Inx.4 ' 838. eX 825. Y==3- +sinx. and ymax=25. Y = 837. y=2 cos -i +3 cosi-. Y=XIe-~. Xz = 1 is the minimum point of the function y and Ymln= -2. Y = (X-2~~8-X) • 16 835. Y= V(. 847. Determine the least and greatest values of the functions on the indicated intervals (if the interval is not given. y=2+x-x l • 828. y = x . To determine the nature of the extremum. 822. y=coshx. 830. y'=2x+4. we havey" (1»0. y= x -2x+2 x-I · 834. Similarly. 824. 845 . 820. 831. y = X (x-I)1 (X-2)'. y=x'-3x' +3x+2. and y'>O when x>-2. y = arc sin (1 + x). we get the critical value of the argument X= -2. it follows that X= -2 is the minimum point of the function. We find the derivative of the given function.6. y=x 87 I 821. and Ymln = 2. Y = 2X::-a. 829. Y = 2 sin 2x + sin 4x. Sinee y"(-2)<O. 827. 843. we calculate the second derivative y"==6(2x+l). x 819. 4 yXi"+8 x + ' l x Y=V xl . Solution. Y= sin" x + cos" x. 851. ~f . It is required to build a rectangular playground so that it should have a wire net on three sides and a long stone wall on the fourth.It> 1+x when x + o. 857. What right triangle of given perimeter 2p has the great·est area? 864.3]. when x+O. cos x > 1. Show that for positive values of x we have the inequality 1 x+-~2.'2 x when x>O. Explain the result in ~eometrlcal terms. y = Xl on the interval [-1. 854. 849. x 856. x.X -(1 +x). In the usual Hence. x-"2<ln(l+x)<x when x>O. 12x + 1 a) on the interval r-1. 8 greatest and least values of the function throughout the domain definition). and so eX 8S we set out to prove.of the playground if I metres of wire netting are available? . Prove the inequa1ities: Xl 858. What is the optimum (in the sense of area) shape . Determine the coefficients p and q of the quadratic trl'nomial y=xl+px+q so that this trinomial should have a minimum y=3 when X= 1. Y = arc cos x.xi' 853.3x l 852. Y= I 850. y = 2x' -1. 855. Prove the inequal ity .6 < sinx<x Xl 859. > 1 + x when x#: 0.5]. Bend a piece of wire of length 1 into a rectangle 80 that the area of the latter is greatest. Separate a given positive number a into two summands ·such that their product is the greatest possible. 863. 861. 2 860. y = V-x(-10---x-). Consider the function f (x) :a::. b) on the interval [-10.12].88 Extrema and the Qeometric Applications of a DerivativI [Ch. wa~ we find that this function has 8 single minimum f (0) =0. f (x) > f (0) when x ¥= O. 862. Determine the least height h=OB of the door of aver·· tical tower ABeD so that this door can pass a rigid rod MN of length i. A~8 a A B M Fig. 26 875. the end of which. In a given sphere inscribe a cylinder with the greatest volume. 869. An open tank with a square base must have a capacity of u litres. Which cylinder of a given volume has the least overall surface? 868. In a given sphere inscribe a cone \vith the greatest voluIne. What should the central angle cp be so that the channel will have maxinlunl capacity? C IJ o N A / ~. 'f . What size will it be if the least amount of tin is used? 867. What will the dimensions of the vessel be if a mininlUlTI of rnaterial is used for a given capacity? 877. Out of a circular sheet cut a sector such that when madeinto a funnel it will have the greatest possible capacity. About a given cylinder CirCUITIScribe a right cone of least volunle (the planes and centres of their circular bases coincide). In a given sphere inscribe a cylinder having the greatest lateral surface. A sheet of tin of width a has to be bent into an open cylindrical channel (Fig. An open vessel consists of a cylinder with a hemisphere at the bottom. the walls are of constant thickness. 873. 26). . It is required to make an open rectangular box of greatest capacity out of a square sheet of cardboard with side a by cutting squares at each of the angles and bending up the ends of the resulting cross-like figure. Inscribe in a given sphere a right circular cone \vith the greatest lateral surface. slides along a horizontal straight line AB. 1] The Extrema of a Function of One ArgulfLCnt 89 865. The width of the tower is d < 1 (Fig. 876. 866. 27). M. 872.Sec. Which of the cones circumscribed about a given sphere has the least volume? 874. 870. 871. On the curve y = 1. which can rotate about a point A (Fig.3 878. 881. Draw a straight line through this point so tha t the triangle which it forms with the positive semi-axes is of least area. A messenger leaving A on one side of a river has to get to B on the other side.~~--:t :A---~----. 880. to bending-to the product of the width of the cross-section by the square of its height. A homogeneous rod AB. The width of the river is h and the distance between A and B along the bank is d. 879. find the point M that receives least light (the intensity of illumination is inversely proportional to the square of the distance from the light source). 882. A point Mo (x o' Yo) lies in the first quadrant of a coordinate plane. determine the angle at which the messenger has to cross the river so as to reach B in the shortest possible time. Knowing that the velocity along the bank is k times that on the water. It is required to cut a beam of rectangular cross-section 01lt of a round log of diameter d.vB Fig. Inscribe in a given ellipse a rectangle of largest area with sides parallel to the axes of the ell ipse. 884. On a straight line AB=a connecting two sources of light A (of intensity p) and B (of intensity q). 28).Extrema and the Geometric Applications of a Derivative [Ch. Determine the length of the rod x so that the force P should be least. . 883. A Iinear cent imetre of the rod weighs q kilograms. 886.~XI find a point at which the tangent forms with the x-axis the greatest (in absolute value) angle. is carrying a load Q kilograms at a distance of a em from A and is held in equilibrium by a vertical force P applied to the free end B of the rod. ~ Note. What should the width x and the height y be of th is cross-sect io n so that the beam will offer maximum resistance a) to compression and b) to p bending? . and find P min. A lamp is suspended above the centre of a round table of radius r. Inscribe a rectangle of maximum area in a segment of the parabola y" = 2px cut off by the straight line x = 2a. The resistance of a beam to compression is proportional to the area of its crosssection. At what distance should the lamp be above the table ~o that an object on the edge of the table will get the greatest !1lunlination? (The intensity of illumination is directly proportional to the cosine of the angle of incidence of the light rays and is inversely proportional to the square of the distance from the \ig ht source.) 885. where h is the depth of the lowest point of the opening (Iz and the empirical coefficient care constant). What mass should B have so that C will have the greatest possible velocity? 888. Points of Inflection 91 887*. 2. Sec. Prove that the nl0st probable value of x is the arithmetic mean of the measurements. N identical electric cells can be formed into a battery in different ways by combining n cells in series and then combining the resulting groups (the number of groups is ~) in parallel. f (x o)) at which the direction of concavity of the graph of some function changes is called a point 0/ inflectIon (Fig. The current supplied by this battery is given by the formula 1= Nne N R+n 2 .b l )] if for a < x < b the arc of the curve is below (or for a < x < bl above) the tangent drawn at any point of the interval (a. x n are the results of Illeasurernents of equal precision of a quantity x. Sphere A of mass M moving with velocity v strikes B. Thr concavity of the graph of a function. 29). . then its Inost probable value will be that for wh ich the sunl of the squares of the errors n (}= ~ (x-xi)Z i=l is of least value (the principle of least squares). and R is its external resistance.b/)] (Fig. A sufficient condition for the concavity downwards (upwards) 0 a graph Y=f (x) is that the following inequality befulfilled in the appropriate interval: f" (x) < 0 If" (x) > 0]. is its internal resistance. 2°.Sec. if Q = cy Vh-y. We say that the graph of a differentiable function y = f (x) is concave down in the interval (a. which. C are situated on a single straight line. 2] The Direction of Concavity. Points of Inftectlon. The Direction of Concavity. strikes C of mass m. Determine the diameter y of a circular opening in the body of a dam for which the discharge of water per second Q will be greatest. 890. Points of Inflection to. The centres of three elastic spheres A. . .•. A point (x o.b) or ot the interval (a1. For what value of n will the battery produce the greatest current? 889. 29). . having acquired a certain velocity. If Xl' XI' . where <fj is the electromotive force of one cell.b) [concave up in the interval (al. B·. re~pec- y I f I}€' b .Extrema and the Geonletrtc Appllcatlons of a Derivative '92 For the abscissa of the point of inflection Xo [Ch. 30).Y2 and x t = Y2· These points divide the number scale . The critical point of the second kind Xo is the abscissa of the point of inflection if {" (x) retains constant signs in the intervals x o-6 < x < X o and X o < x < xo +6. -.2 1 < x < Y2' 1 concave down when.00 < x < + 00 into three intervals: 1 (-00.xz . It will be noted that due to the symmetry of the Gaussian curve about the y-axis. The signs of y" will be. Xl)' II (Xl' XI)' and III (x z' +(0). Determine the intervals of concavity and convexity and also the points of inflection of the Gaussian curve y=e.x2 • Solution. The poInts 2) the ( Y2' ±\ Ye 1) are points of inflection (Fig. 31 Fig. where 6 is some posi- y tive number.2) e.Y2 and < x < + 00. x Fig.2 . We have o x a y' = -2xe. provided these signs are opposite. we take one ~oint in each of the intervals and substitute the corresponding values of x into y) Therefore: II the curve is concave up when curve IS 00< x < - . And it is not a point of inflection if the signs of fIt (x) are the same in the above-indicated intervals. 30 1ively. 3 of the graph of a function y = {(x) there is no second derivat ive {" (x o) -= 0 or t" (x o). Equating the second derivative y" to zero. it would be sufficient to investigate the sign of the concavity of this curve on the semiaxis 0 < x < 00 alone.x and y" = (4x 2 . we find the critical points of tHe second kind 1 X 1 = . + (this is obvious if. + . Example t. for example. Points at wh ich I" (x) == 0 or f" (x) does not exist are called critical points of the second kind. +. The tangent at this point is parallel to the axis of ordinates. If there are limits . Xl 894. Definition. Y = X2+12. 2°.. Sec. 3° Inclined asymptotes. If there are limits lim f(x) =k x I X~+CX) and rf (x)-ktx] = bt . when k 1 = 0. If a point (x. Y=X~3' 898. s if = -2 2 (x --t. y=x' -6x2 +12x+4. 899. y=(x+l)~. 896. Asymptotes to.Y) is in continuolIs Inotion along a curve {I::-= f (x) in such a \\'uy that at least one of its coordinates approaches infinity (aud at the salllC tinlc the distance of the point fronl sonle straight line tends to zero). We have: V -x+2. Vertical asymptotes. y = (1 +x 2 ) eX. Since y">O for x<-2 and y"<Ofor x> -2. y=x1lnx. x~a then the straight line x=a is an aSYlllptote (vertical asymptote). since the first derivative y' is infinite at X= -2. 893. we find that y" does not exist for x==-2. 900. y = arc tan x-x. Find the intervals of concavity and the points of Inflection of the graphs of the following functions: 891.2) . y=-x-sinx. a right horizontal asymptote). 892. y = V 4x'-12x. y=cosx. then this straight 1ine is called an asynzptote of the curve. If there is a nunlber a stich that Ihn f ( t) =-= ± 00 . Equating to zero the denominator of the fraction on the right of (I). 3.0) is the point of inflection (Fig. 3] 93 Example 2.Asymptotes Sec. lim X-++ 00 then the straight line y= ktx+ b1 will be an asymptote (a right inclined asymptote or. 31). 895. "9 - (1 ) It is obvious that y" does not vanish anywhere. it follows that (-2. Find the points of inflection of the graph of the function y= Solution. 897. .. the left asymptote is y= -x (Fig..CJ) x ba= lim (y+x)=O.... x-+-oo then the straight line y = k1x + b2 is an asymptote (a left inclined asymptote or.1 9_4 and Hrn [f (x) -krl = b2 . a -1 f x Fig.=--=lJm(y-x)==ltm X~+-:r. x-+. Example 2.. Equating the denominator to zero. when k 2 = 0. x -x V-I)Yx -1 ==0.. The graph of the function y = f (x) (we assume the function is single-valued) cannot have more than one ri~ht (inclined or horizontal) and more than one left (inclined or horizontal) asytnptote. Similarly. For x k 1 = lim ~ -~ + ~ y IX) . Find the asymptotes of the curve x2 Y= ¥xl-l · Solution. whenx---.-oo. t'~+oo x"-l 2 2 b... a left horizontal asymptote). x-+-CJ) Thus. Example t. the straight line y=x is the right asymptote.p_li_ca_t_to_n_s_o-=-'_a __D_e_'_lv_a_t_tv _e__ [ell. . we have k a= lim 1. x=-l We seek the InclIned asymptotes. . 1 Y x2 - .= lim x x-+-+ 00 ..p. we get two vertical aSYlnp- 1ote"· and x=1.E_x_t_'e_rn_a_Q_n_d_t_h_e_G_e_o_m_et_'_ic_A. -+ 00 we obtain 2 t x _ _ --1. Find the asymptotes of the curve y=x+lnx. 32 h~nce. Testing a curve for asymptotes is simplified if we take into consideration the symmetry of the curve.. 32).= -1. Y=-l -ex· x' 904. y= Y 3 x2 + 1 xl-l 911. Y = Xl 1 908. there is no inclined aSYlllptote. We have: k= Urn X-++(J) b= lim (y-x)= X--"'+ 1L=1. the curve has a vertical aSYlnptote x == c. y=e x • _sinx 912 · y.f (q» sin (p. 1 903. Find the asymptote of the hyperbol ic spiral r=-. x=t. the curve has a horizontal aSYlllptote y=c. . q> . Y=x 2 -4x-1-3. Y=Xi+9 · 905. When 1J'(t o)=oo and cp(to)=c. cos cp = f (cp) cos q>. Y=1J' (t). y sin cp ==.x). y=x-2+V~' -\1+ (X-2)2 · x 902. 914.x2 +2. ¥-++o the straight line x=Q is a vertical asymptote (lower).k({1 (t) 1== b. then we first test to find out whether there are any val ues of the parameter t for which one of the functions cp (t) or 1J' (t) becomes infinite. Y= Y xl-t x 907. Let us now test the curve onl y for the inclined right asymptote (since x> 0). 906. lim ['I' (t) . y = V xl--l. 00 Hence. If a curve is represented by the paratnetric equations x=cp (t). If cp (t 0) = 1J' (t 0) = 00 and lim '¢ «~») === k. Find the asynlptotes of the following curves: 901. Since lim y= -00. ==. Y = x 2 -4 · 910. === f (cp). If the curve is represented by a polar eq uation . When cp (to) = 00 and 1J' (to) = c. x2 909.x • 913. . y = 1n (1 -t. y=t+2arctant. I-+t o cP t -+ t n then the curve has an incltned aSYlllptote y==kx+b. then we can find its asynlptotes by the preceding rule after transfortlling the equation of the curve to the parailletric form by the fortnulas x =. while the other remai ns finite. a 915. x Urn In X= X -++ 00 00.95 Asymptute~ Sec. 3] Solution. y=e. The function 0(0. and so the critical points are only those at which y' and y" vanish. hence. +00). y") tn each of the indicated intervals. Graphing Functions by Characteristic Points In constructing the graph of a function.+oo thus. the straight lines x= ±I ~-H+O are vertical asymptotes of the graph. 3) Sec. To determine the signs of y' (or.1). tha tis. We have: . asymptotes. (-1. d) We find the critical pOints of the first and second kinds. Then find any points of discontinuity. ¥3) and (V~ +00). This b) The lim y= X~-l to a) The function exists everywhere except at the points x== ± 1. These elelnents hel p to determine the general nature of the graph of the function and to obtain a mathematically correct outline of it. and lim y = ± 00 and ± 00. Construct the gra ph of the function x Y= x2 _ 1 . is odd. (-3. Inonotonicity.1 and x = 1. bending points. constancy of sign. (1. From the symmetry of the curve it follows that there is no left-hand asymptote either. pOints at wh1ch the first (or. -1). points of inflection. only at pOints where the function y itself does not exist. respectively. (-1. V Solution. 00 X lim y= 00. etc. (0. first find its domain of definition and then determine the behaviour of the function on the boundary of this domain.3) and (3. 0). such as symnletry. it is sufficient to determine the sign of y' (or y") at some one point of each of these intervals. and find kl = lint X---++ bl = !I =0. (1. there is no right asymptote. and y"-in each of the intervals (-00. the second) derivative of the giveu function vanishes or does not exist. and therefore the graph is sytTlmetric about the point simplifies construction of the graph discontinuities are x = . From (1) and (2) it follows that y'=o y"=O when x= ± V3.0). x 2 -3 y' (I) 3 (x 2 -1)'" 2x (9-x 2 ) y" (2) 9V(x 2 -1)7 · V The derivatives y' and y" are nonexistent only at x= ± 1. x-. It is also useful to note any peculiarities of the function (if there are any). etc. 4.V3)t -1).00. (-¥3: . when x=O and x= ±3. that is. c) We seck inclined asymptotes. 1). . -3). Thus. respectively. periodicity. Example t. y' retains a constant sign in each of the intervals ( .Extrema and the Qeo/netrle Applications of a Derivative 96 [Ch. Sec 4] 91 Graphinl! Functions by Characteristic Points It is convenient to tabulate the results of such an investigation (Table I). the left-hand half of the graph is constructed by the principle of odd symmetry. Table I x 0 I II" (0.::ave han down I e) Using the results of the investIgation. +00) 1. calculating also the ordinates of the characteristic points of the graph of the function.37 -1- I I + V'2 - I + I Fund Ion Ol. it is enough to calculate only for x ~ 0.dcer ea~c'5..-r---r----1~~r--~-~--~I---__ J Fig. point FunctlOn Inrrca-. y . graph IS lOlll3ve up + 0 Point Function il1r:-rcn"cc. 33 4-1900 X . I - ±oo -- IeXIst non- - \lIon l~X I~t I 0 ----Conelu- I Point of mllcctlon I Function I t::a-.5 -t- 0 I + I I + + I + I I - Min.-"on. of graph illf1ec15 con.1ve do\\n o V3) I V3~1. ----. 73 1( V3.c~. grdph h del lOl1l. 1) 0 IJ y I - I I SlOlh 1(1.e'l. ttnui t I graph IS conCAve up 3 1(3. \ve construct the graph of the function (FIg 33). 3) I V3 ----=~1. It will be noted that due to the oddness of the function y. b) There are no discontinuities in the domain of definition. Putting y=O. lim X~+(SJ The right a~ymptote is the axis of abscissas: y =0. d) We find the critical points. Graph the function In x y=-. a) The domain of definition of the function is 0 < x < + 00./' = 2 In x-3 . c) We seek the right asymptote (there is no left asymptote. we find x= 1 (the point of intersection of the curve with the axis of abscissas). the curve does not . but ae. we construct the graph of the . the straight line x=O (ordinate axis) is a vertical asymptote. y x3 ' 'I' and yIP exist at all points of the domain of definition of the f unction and y'=O when Inx=l.00. I ·Imy= I'1m x-+-o x-+-o X ~Ience.ntersect the axis of ordinates e) Util izing the results of investigation. In addition to the characteristic points it is useful to Hnd the points of intersection of Fig.98 Extrema and the Geometric Applications of a Derivative Example 2. We form a tabl~.00): k= b= lim lL-o· X - X--++(SJ . 34 the curve with the coordinate axes. since x cannot tend to . .' . x Solution.unction (Fig. 34). when x =e*. when x=e. I-Inx Y =-x-2. we approach the boundary point (x = 0) of the domain of definition we have In = x . y=O. and have . that is. including the characteristic points (Table II). if =0 when In x = ~ • that is. graph l~ concave up .49 I 3 + I - I I i 2 JI--~O 33 3 e + - - I concave. 0' :un- e:::::: 2.- I I I I ~ (3-.In' t I 11" 1 RI . Funct decr . n I Funct ~ 1£"1(r • ~ IS. J:!f.oncave dov.lph lone-ave do\\n I I I J\\ax pOInt Of funl1.+00) 4.~ Table II • I I I u x i I I I I I 1 (1. 1) II I I I F . .72 -::::::0.f.n POlnt or inflec t lon Function decreases.37 e ! I I Conclu· Slons I I I i : I I I - II I y" I I I I I I I . 0 - I Funct II I I ..al asymptote (e. I I nonexist I I 0 2 I 1 j I -.lr It I" ~ conca\ c do\\.• graph l~ c. e+) I e- j I Boundary pOint of domain or dt. IIICrCtl"cl::. I f y I I -cc ! y' noneXlSt. I (0.: gTdph do\'"n I - I - I I I I - I 0 I I I I + I I I I . e) I I. I I I - -r -+- I I ctlOO 3 r I Vert1C. Y = 926.d .a l 916. y=(a+x. y=x -3x • 939. 958. Y= V-x' x2 x 952.x . y = (2·+ x 2 ) e-x~. Y= 4 x 945. y=V(X--2)2+V(x-4)~ 4 918. • 4x-12 4 21 x 1. Y= VI-x'. Y = xe.x In x 4~ 928. -. 949. 947. Y=-x-' 923. Y= x 2 -t. Y=-g-' 941. Y= 8111 X +COSX .1). intervals of increase and decrease. Y = (x-I)I (x+ 2). y = sin x + cos x. Y = sin a x + cos' x. 954. 934. ). 935. y=ln(1 933. y=(X-2):(X+4). Y= Vx 3 . x 2 -1 x +~x . 938. Y= V x+ I-V x-I.(X-4)1. Y = x 2 4-x 2 8 955. 1 963. 948.- V 919. VX2 +)-1 x 932. Y = 957. Y= (x -t 1) In:! (X-i. 942. 959. y=cosx-cos l x. 6x 2 _X 4 940.V~8-x. y == 2x -1. Y= x Y~2_4' 944. . 920. 16 930. Y=V~' 946. discontinuities. Y = (x • 2)2 V_x_. 961. and also the asymptotes. 956 · Y= In 2 - 1) +. 927. I 925.2-3 +e- X '1/' (x +1)1. -~.3' 8 950. (x 2 _ 5)a 125 . Y= Y . Y~ VS -i-x. Y=2"ln(j. +sin 2x .V4. Y= +3• x 924. Y=~4' x- = 2 951. 960 · Y= sin x -2-.:t2~ I . Y = xZ--4 . Y = In (x +1 x:J' V-x 4. 962. V. Y == x 2 (x-':4j • 931 · Y 3x • x x 929. Y= Vl-i2. Y= (X+4)2. 936.. a Graph the following functions and determine for each function Its domain of definition. 917. Y = fiiX . Y= 2 921 = x -2x+2 · Y x-I' x4-3 922. 937. Y= 4+x 2 943. y=e8X-x2-14. • 953.x. y=-xVx+ 3. the direction of concavity. y = In ( e + ~ ) .)eci-.Extrema and the Geo/netrlc Applications of a Derivative 100 [Ch. points of inflection of its graph. extremal points. then ds= b) x={. c) x=q> (I). here. 2x. y=¢ (t). • 976. y = cos x-In cos x. then ds= I~~I It II F'l + F'l dx= I~~I "dy. x=a (sinh/-I). then ds= V-(-~-)-2-+-(-~-)-2dt. y = x-2 arc cot x._C_u_. y = cos x· co:. 983.v_a_tu_'_e 964. and y=O when x=O. y=e arc tanx. 986. y = arc tan (In x). y=xarctanxo 982. vl+(~friy.(y).lnx. y-==asint (a>O). Differential of an arc. 985. 968.t • t 991. if the equation of the curve IS of the form a) y={(x). . Y= Inx-arc tanx. 967. 973.-2t.._-' I-x! y = In sin x. A good exercise is to graph the functions indicated in Examples ~~6-84Ho Construct the graphs of the following functions represented paralnetrica 11y. Y = x arc tan -x \vhen x =1=0 981. y =--= In sin x. Y= sin x sin ( 965. y===a(cosh/-l) (a>O). I 972. 981. !J = arc sin In (Xl + 1).S_e_c_._fi~l D--::ifJ:. 966. y = 2x-tan x.-e_'e_n_tt_Oa_1_0:-'_a_n_A_'_c_. y = earc sin Vx. Curv&ture 1°. x=l +e. y=x x • 975. y = 2+ arc tan x.21 • 992. x=-::al'o~3t. y)=O. Y = xx. y = sin x· sin 2x. V x::). Y == e. Differential of an Arc. 988. x=(1. 971. Y = arc cosh ( x x+ :) y = arc sin (1- 969. 989. The dIfferential of an arc s of a plane curve repre"ented by an equation in Cartesian coordinates x and y is expressed by the fornl ula ds === V(dX)2 + (dy)2. y=2t+-e. l01 978. Sec. 98~. Y= dIe ~lTl + ~ ). x y. x=te'. y=te. 977. y = In tan ( T-i)· 970. V F'2 + F'l d) F(x. 5. y==II_t-2t. vl+(tf then ds= dx . 987. 990. 979. y=x+ sinx. x I 974. ~ K = Ii rn ~\~ = da ~I\ ~ () /\') ds ' U IS the angle oetwe(:'!J the posItIve directIons of the tangent pOInt 1\-1 and the x-aXIS.. .=: As when ~V -~ M (FIt::. Thr curvature K of a curve at one of It. dIp Denot1n~ by ~ the angle between the radius vector of the point of the curve and the tangent to the curve at this point. . i. ds= y' (dr)t+(r drp)t = Y rt-l- (~)). \\hC're dt the o FIg. where a is the radius of the circle) and the straIght line (K=~ are lines of constant curvature. points AI is the 11n1lt of the ratIo of the angle between the POSltIVl~ directinns of the tangents at the points M and N of the curve (angle of COl1ftll6ence) to tlH. that I~. 1 R=iKl· The circle (K =~. e. 8 Denoting by a the angle fornled by the tangent (I n the direction of increasing arc of the curve s) with the positive x-direction. dy Sill u== ds.. Curvature of a curve.· length of the arc 1\1 N -. \ve get dx cos a::=. 35 The radius of curvature R is the reciprocal of the absolute value of the curvature.Extrema and the Geometric Applications of a Derzvatlve 102 [Ch. 2°. In polar coordinates. 35). \ve have dr cos A..==_ ds' t-' sin ~ =-= r~. ds ' . 2 \vhere . if the curvE' itself is represented bV equations in paralTIctric f orin) Example 1. \vhen the curve is given by the equation r=f (~). dr r =- dcp an d 3°.. Circle of curvature. The coordInates X and Y of the centre of curvature of th~ curve are conl puted fr0J11 the fornllJl 3'i . as P ---+ M and Q. y J 3) if the curve is represented by equation. x=<p (t). P and Q. and the centre of the circle of cur\'ature (the centre of curvature) ties on the nOrInClI to the curvr dra\\'n at the point M in the dIrectIon of concavity of the curve. in paralnetric form.. FInd the equation of th~ evolute of the parabola y =x~...x F: u F~ F~x F~y F~ F~ F11 0 K= (F't + F'2)3/ x . \vr have \vhcre . Y=f (x). Ftx. The radius of the circle of curvature is equal to the radius of curvature. then these fOrtl1Ulas yield pnralnetric equations of the evolute \\ ith parameter x or Y (or f.-.Il' + y'2)J/ .(1 .Y=y+--. then K- y" • . 1 + y'l q y The eLlo[ute of a curve is the locus of the centres of curvature of the curve.. If in the fonnulas for deternlining the coordinates of the centre of curvature we regard X and Y as the current coordinates of a point of the evolute. . y = ¢ (t). y) =-=0. x = dx rtt ' In polar coordInates. 1\-1. Curvature 103 We have the fo))owing formulas for computing the curvature in rectangular coordinates (accurate to within the sign): 1) if the curve is given by an equation explicitly.y'2)'/J ' 2) If the curve IS gIven by an equation implIcitly. then F..Sec 5] Differential of an Arc. X=x- y' ( 1 +" [1'2) . then K= x' x" I (x't y' \ . The clrcle of curvature (or osculati1t~ ctrcle) of a curve at the pOInt AI IS the lill1itln~ position of a circle dra\vn throu~h M ~nci t\VO other pOlllts of the curve. a). Fig. 995 y" = 2px (parabola). The normal Me of the involute r 2 is a tangent to the evolute r. Xl + y2 = a 2 (circle). a2 +b2 =1 (ellipse)..!.Extrema and the Geometr.ole and putting IOto wegetdd2~1 o=~cosh2!. 3' 6x Solution. We have -d =slnh2a x a R the derivative dd x to zero. a a a x u cosh 2 !. that is why the involute r 2 is also called the evolvent of the curve r 1 obtained by unwinding a taut thread wound onto r. x=O is the minimuln point of the radius of curvature (or of thfl nlaximum of curvature) of the catenary.4x'. a whence we critical point x=Q Computing the second derivative it the value x=O. 1 3 (XT Y=2+ )2/. the """--" length of the arc CC l of the evolute is equal to the corresponding increnlent in the radius of curvature CC..:::-::M. 40 • Vertices of a curve. with the positive x-direction~ by the tangent to each of the following curves: 993. x2 y2 . x x= a a x== a Therefore. I n place of the curvature K we can take the radius of curvature R= I ~I and seek its extremal points if the computations are simpler in this case.=J.. . The vertex of a curve is a point of the curve at which the curvature has a maximum or a minimurn. R=acosh 2 .(a > 0). we fornl the expression of the curvature K and find its extrelnal points. 36 a Solution. ~2X~ Equating find the <.cosh~. To dpternline the vertices of a curve. · The lnvolute of a curve is a curve for which the given curve i~ an evolute. Y = I +2 '. To each evolute there corresponds an infinitude of involutes.C1-MC. 36). we get sinh 2~=O. Find the vertex of the catenary x y=a cosh . (l thus. Example 2. Find the differential of the arc. which are related to dIfferent initial lengths of thread. X = . Since Y'=sinh!-andy. the point A (0. and also the cosine and sine of the angle forIned.-1 o=!>o. is. (Fig.. Eliminating the parameter x..c Applications of a Derivative 104 (Ch. a it follows that K= dR x and. we find the equation of the evolute in explicit form.. 994.. The vprtex of the catenary y = a cosh. hence. y = + 105 a 2 / S (astroid). 1016. 1015. y===a(l-cost) (cycloid). Xl + x!I + y2 = 3 at the poi 11 t (I. x=-= ~~_In:/ . 1009. 5] 996. Prove that the radi us of curvature of the catenary y = a cosh ~ is eq II aI to a segnlent of the nornla 1. r=!!. a y2{a = 998. 999. x=f\ 1I=t 3 at the point (1. al +. 1005. 1019. x=a(t-sint).. find the differential of the arc. At \vhat point of the parabola 1/2=8x is the curvature equal to O.. r'l. 1001.-4x ' -18x z at the coordinate origin. x2 ~~ • 1014. 1021.Sec.!(catenary). lOll. x=acos1t. 1018.12H? 1012. y=x4. y=a(sint-t~ost) involute of a circle). a cosh. a 2 + ·b 2 = I (ellIpse). r=aCP (Iogarithlllic spiral). a <:ofilpute the coordinates of the centre of curvature of the given curves at the indicated points: . Find the vertex of the curve y=-e"'. x=acosSt. (hyperbolic spiral). 1020. 1010. COIllpute the curvature of the given curves at the indicated points: 1006. . Find the radii of curvature (at any point) of the given hnes: 1013. x===a(cost-+ tsint). and also the cosine or sine of the angle formed by the radius vector and the tangent to each of the following curves: 1000.2 = 1 at the vertices A (a. x2 . (parabola). r=-a(l +COSfp) (cardioid). 1004. q> 1002. 1003. (cardioid). DiUerenttal of an Arc.=a 2 cos2q> (lemniscate). = aekq> (Iogarithlllic spi ral). y=x s (cubic parabola). 1). I). r 2 = 2a 2 cos 2ep at the vertices {p = 0 and <p == 1t. y=asinSt (astroid).l1008. b). r = acp (spiral of Archimedes). r = acos 2 . Curvature X 2fs 997. r=a sec 2 . 0) and 8(0. y=asin1t (astroid). 1017. Find the least value of the radius of curvature of the para bola y2 = 2px. 1007. x2 y2 1027. 1). a2 b2 = 1 (ellipse).]06 Extrema and the Geometric Applications of a DerIvative (Ch. y2 = 2px (parabola). 1028. 1029. 1023. 1030. Prove that the evolute of the logarithmic spiral r = aekcp is also a logarithmic spiral with the same pole. xy = 1 at the point (1. ayl = x' at the point (a. Write the equations of the circles of curvature of the given curves at the indicated points: 1024.tens t) is the Involute of the circle x = a co~ t~ y = a sin t. Y =. I). Prove that the evolute of the cycloid + x=a(t-sint). y=a(l-cost) is a displaced cycloid. . y = eX at the point (0.3 1022.a (sin t . Show that the curve (the involute of a circle) x = a (~os t + t sin t). y=x z -6x+ 10 at the point (3. Find the evolutes of the curves: 1026. 1). a). 1025. r dx arc~ln2 \ to.C 2°. Table of standard integrals. Basic rules of integration.. .!\nlx-al-t_C 2a x -1. tlrc tan - 1 t f-C==. . 1.C (x) d x -= F 2) ~ Af (x) dx ~. Sf (Ll. +c arc cot - \: a +C (a =F 0).x I.C 4) 1f ~ f In particular. (a#O).a Z 5 _(12 1 "t Ia-x dx .{"')ld\-'~fl(X)dx±~ f 2 (x)dx. ~ 0) . -f. Sf (x) d x --. 1) IfF' (x) =.A ~ f (x) dx.Chapter IV INDEFiNITE INTEGRALS Sec. . (a> 0).\"+1 l 1.-.. I I l. (a ~ 0).. (' y~_==lnJx+yx2-I:aJ+C (a#O).- 1 IZ-: II.(. JCd~x ~-= In I x I -. f (. VII.-.\ nd 3 (f' (t).2=-==2 In a+ ".\). y a 2 -x 2 5aXdx~: aX In a a +C (a > 0). 5 eXdx =e X a + c.\. V. . x +C =-arccos-+ x C VI. t h P 11 ~f where C is t ~onstant. j b) {(/.(. I =1= - C. 5---=:-- IV dx 1 x 2 -t-a 2 a 11 ! . the n (ll) du --:: F (u) b) d \: : II --: - -!r F (a.- a (l Jr X~:-_:. J x +£1 .] a a"-. when' A is a constant quantity. t ) \ "d \" . Direct Integration to.F ( \") -t. 3) ~ [f1{X)::l:f. nrbttrary all ~x) t. 1. XVI.. ('J(Vx-l-l)(x-VX-J--l)dx. 3 and the formulas of integration. ~ 5d 2 x 6dx. 1037.Indefinite Integrals 108 (Ch. X1.)4 . ~ (ax 2 +bx+c) dx= Sax dx+ Sb. - · 1044.V-. a "3 3 Applying the basic rules 1.~. 5 d~ X cos x =tan x-l-C. S x (x -+ a)(x -t b) £ix. ~ slnxdx=-cosx+C.ex -t. 2 - 2 . 1034.4 VIII. 2 1040. XV. 1045.dx . = - Sll1 Z x XIV.\'~ S(x:-=-\Il~ I JI"l lX. 5X2~ 5X2~ JV +- 1046.4-. 7' 10' dx 4 x2 • ~ V:~X2' __ JY2-t-x V2-x 1047. Scosh x dx= sinh x+ C. jl (V-a.l.V a~-. EX81npie 1. 2. r (X2+. + sec x 1+ C. ~ cosxdx=sinx+C. 5(nx)-n. find the following integrals: 1031.J . (aT -xTrdx. 5.5 1039. Xli 5s~:x=ln Itan ~ I+C=ln Iczosl'C x-cotxl+C.C.dx.:j~-2L dx.e 2 5 =- 2 dx x -f. lX. 1042. 1043. S(a + nx3)2 dx. l-n 1038. sin X2 x =-cothx+C.b -i x~ -1. 1032. SV2pxdx. . 1041 1033.dx+ Scdx= = a 5x dx + b 5x dx -f.dx. S sinhxdx=coshx+C. 5 .. S. S (6x +Rx+3)dx. 5c~:x = In Itan ( i + : ) 1+ C = In I tan .dh XVII. X II I. Jr cosdXh2 x =tanhx-{-C. --Y4_.d~ x cot + C. 1035. 1036. This type of transfornlatioll is called IntegratIon under the differential sign. 3. b) ) cothl X dx. a-x 1052** S2X-I-~d • 1053. III exatllples 2.du==-· 1 u 2 (5x-2) -+C=1 - 2 U 2 5 1 2 5 1 1 2 2-+C=. b) ) tanh l x dx. SX4~':~ I dx.r . IO Qr. ax+b 5x ++ a. ~ x dx 1 . Rule 4 consid(lrably expands the table of standard integrals: by virtue of this rule the table of integrals holds true irrespective of whether the variable of Integration is an independent variable or a differentiable function.Direct Integration Sec 1] 109 1049. x1e X" dx. Integration under the sign of the differential.J 1 + (x )2 J y- We iJllplied u =x2 . Sx-I (1 dx.. Y5x-2 S~~ = ~S (5x-2) -+ d (5x-2) = 5 1 1 = -5 S -.\grat VII.e of a tabular integral: ) f (q> (x) q" (x) dx = ) f (u) dll. 5aX::x' 1055. 2. . a) ) tan 2 xdx. 1 +. which were lIsed in EX31npi es 2 and 3. a) )cot'lxdx. d (x 2 ) 1 -Example 3. 5 5 Example 4. and 4 \\'(1 r(\duced the given integral to the following fornl before ITlaklng u.x 2 x. Exanlple 2. 1050. find the following in- tegrals: 1051**. V---=-2 I 2 =-2 In (xl + Yl+x~)+C. 1 dx. 3°.Y5x-2 f-C 5 ' 2" where we put u == 5x-2. 1058.:7sx2-t-fiX+7d • x +:~ )C. b) x dx = 2" d (x 2 ) and so on. 1054. \Ve took advantage of Rule 4 and tabular integral l.\ -1-1 5~~t £ix. ) 3 x ex dx..5 ~dx . are: a) dx == LJ~ing a1 d (ax -I b) (a 1 # 0).~ ~ eX I d (x") = ~ ex1 + C by virtue of Rille 4 and tabular intt. the basic rules and forIuulas of integration. 1056. SOlne COlnnlon transforrnattoll~ of d Iff~\rcntials. 1048*. and use \vas made of Rule 4 and tabular Integral V. where u = 'P (x). -x 1089.e-~ Ydx. ~ (et_e. 4 +X22 dx. x. 1068. S • +x 2 ) . ~ 42 -ax dx. S 1080..2xe dx. a -l 1092. J ax -1. Sx~. S 3x-t-1 dx.~. S(a-t b)-~:-b) x' 1065.7 dx. ". 1064. J JI r 2 tan !-.4 r . X2 x2 1095. S7X~~-8' 1067. r J5 yx dJl Yx' eX 1097. Sa~2dx. -t-l J Y 5x__ f' 2 x+3 1076.b a2x2 -t. 1096. a -x4 4 1081. S2:'~3' 1079. 1069. S dx.' 7 -5x 2 2x-5 1073. 1061. 1 +4x 2 fV(l+x~)ln(x+ V-I dx · 1087. SI . y I-y 1062. ~ ae. 4 107.r-.b2 dx. 1084.mx dx. S yx -4 1077.r-==== · Y x 6 -1 1083. Jr Y~' 7 +8~2 1072. S(e-~ +. 1063*. 1085 J rx- Yafltan2x d ·J 1086. r .dx. SYxt In x dx. 1059. . S bdy .bX)2 1091.t ) dt. ~ Va-bxdx. (aX -. 1088. SX'~:2dx. S(x ~ l)2 dx. r\ Y __ dx . S3x2 __ 2dx. eX . 1094. u>"b X 2X n 1093. 2 xdx -0 x. 1082. 1060*. . lCh.110 Indefinite Integrals S(a-l x b aYdX.l dx. 1070 S~X-t-6d 2 x -f-4 • 1071. 1090. 1075.r J y x x2 + 1 1078. (O<b<a). 3-2x 5x2 -t. J" eX dx. ~ e-ex'+J) X dx. / arci-x sin x dx.. llfC S3X~~-5' 1066. r . ~ x· 7 dx. dx X2 \ t. . J Ya" dx. dx. 1] III Direct I ntegrat ion 1098..3x tx..)'~.1r l'OS V--x Y-.d . t 132 · 5t an 1112. 2 1139.d II 29 · x. 1105. . j' sin xdxcos x . 1125. y cos X-S1I1 x 2 1131.Sec.rsin. 1138. ~coeaxdx. 1126. ~ 1110*. sin ax - xdx -2-2.1 dX . x Sttl- 1115 • 1116.: a cos 3x 1109*. 5 t.-2 l X Stn si t1 3x 2 l (cos ax dx ax)2 dx. 5sin (lg x) df · 1106.\ x .3 cos· x sin 2x dx. J cos 1134.I -==-==:t Sr Stan xdx. 1118. S~ 2x +3· 1123. e'dt . r~. l __ e-:. dx . ~sin·6xcos6xdx. 1100· · 1101 · 1122. • ) sIn (ax+b) · 5cos 2 V 1 -1. ~ sec· (ax I b) dx. a r dx 5 ytanx I 2 x lX. Va. 5 +adx I aX 2X • e-bx 5 1102.dx. 1114.b. ~ l 1113.-x.r) dx. + 1) dx. ~ sin (a + bx) dx.X. ~ x sin (1 . J 3 cos ( 51:. 1104. 5cota~bd"x."dx. III I. 1121. 1133. H20.x.t · 1108. 1099. S(e-~ + l)~ e~dx. 1103. sin 1130. a r SCinx\'2·. x sin .-C0S -. 1136. S (2 sinh 5x-3cosh 5x)dx. .be x dx. Sb-a cosec 3x d cot 3x x. Scot xdx. JCcos Vx 2 dx. 3 51cos4. ~ sinh! xdx. ~ eX 1119. j" . ~Sitl!xdx. 53 -t- 1128. ~ cos! xdx. X 2 X d "3 sec "3 x. 2 cot 3 x f x . 1137. cos ax d 5sin' ax:3x . ~ x cot (Xl · . 5cos - l-e' 11 07. J tans Stan Vx 1124. + sin ax)! dx. .sin 1135. X 1117. ~ (cos ax x 1127. ec2 x x (I-SilJ tan 3x. dX StanVx-l . 5X8~ 1147.b) + (a - 1160.\ 3 . cos 5sInh 1142. •J . d 1 1n( 1+x 2)-+_1 I -1"'\ 2 SUI 1157. 1155.. 5 y'-2 2 dx. 1156. 5 dx.~f X(I+x 2 )dx. dx 5 dx 1143. 1161. 1158.. l y 2X dx. y I-x" x sin 1170. 5SinX-C('SXdx... 1173. -3cot 3x dx. 5x + cos x dx. ) e.~'.\ V tan 2 x-2 . dx. 5-3x V4-3x 2 SeX~I' r dx J «(l -i.J'r x-I • 3. Find the indefinite integrals: 1145. 1146. dx. 1148.. SY4-tan s~_x • x I t 77. x cosh x . a dx. -=--h. v X' -1.JYex · 1152. ) a·1nxcosxdx. . 1174. ) cothxdx.r:.:-~ 1165. 1163.xdx 1144. S(2 +2XI~-I) 2X~~ I' SV + COS X 1169. J 1164. 11 59.Sill axdxcos ax • 2 b) Xl (0< b <a). 5! 1167. S. xl '" dx 1151. r I-sin x 5 t 154. . 5X. ~ x tl5-xl dx.ln' x sin 2x dx. dx.e x _2 dx. 1176.. sin x • X na x c.. 5 x dx • Slll X Z 1168. 5sin' . (GIL 4 1175.I V2Y dx. 5X. ) tanh x dx. \ t. ) tan ax dx. \ dx.x' 1171. t 166. eare t::ln x -1-X . 1149.. xdx ~r--=.Jndefinite Jntegrals 112 1140. 1141. J S 5:' /dx.3x 2 dx. X Sill 5-h-. Jr. 1162. 1 dx. ~ xe- t 150. dX ' x cos- SV Ix+ In x dx.Y2·t3x2 2-1. 1172. 1153. (1 -1. Integration by Substitution to. 2] 1178. S1 +cos dx 2 x· 2 1189. X+x dx.Integration by Substitution Sec. f (x) dx to the foraD . sin x cns x dx. Suppose we succeeded in transforming the integrand f (x) dx = g (u) du.\2 dx. 1179. . Sx (4 -ln dx x) · ) arc cos i 2 V 4-x 2 dx.sh 2 x · Sec. Find ) x Y-t-I d.tan x secl X dx. X • 1190. S f SI11 2 '- dx x CCJs 2 1185. ~ e. \ve will have: wIH'r~ ~ f (x)dx= ~ f [ljl (t» ljl' (I) dt. 1186. 1181. Solution. ) Vln (x+1+. S4+C05 cos 2x d 2x x. " Y2-su1" x 1183. EX31npie 1. where u = q> (x). \vhcnce x = t + 1 dnd 2 dx = 2t dt. t 182. V x-I. SOlnetilnes substitutions of the fornl u = q> (x) are llsed.t. 113 \ atanllA --Jx '" cr. (1) ThE' attelnpt is made to choose the function ({1 ill such a way that the right side of (I) beCOlnl'~ 1110rc convcl1i~nt for int~gratlon. 1180. Ssin C.t + fj>o) dt. 2. t is a new variable and (p is a continuously differentiable function. Putting x~qJ(t). Change of variable in an indefinite integral. SYsecseclx tanx+x I dx. It is natural to put t == Hence. 1184. 1187. ~ Xl cos (x 3 -+ 3) dx. l-x 2 afC SIll y 2 Vx + I) 1188. we put X~ll "-CC t... If an integral contains the radical V. It should be noted that trigononletric substitutions do not al ways turn out to be advantageous.3°. Sx eXJ 2 dx 2 =-21tn(u+V1+U2J+c=-21tn(x2+Vl+x4)+C.= ~ ~ + C = ~ 5 V "it 5 ~ Y-Sx---2 + C. see Sec... Trigonometric su bstitutions.. u =x 2 . du du = 3x2 dx.. i . we have already made use of this method in Sec. For more details about trigonometric and hyperbolic substitutions..x. J.2. SV 5x-2 = 1 dx = 5' duo du = Sdx.\2+ a2. 2°. Example 5. 1) If an integral contains the radical x= a sin t. 3. du = 2x dx. u = 5x. which are ~iInilar to trigonometric substitutions (see Exarnple 1209). =-21SVdU 1+ x 1+ S.tdt 4 u Exalnple 4. Find JY?+1d X 2 x. whence 2) If Va -x 2 an integral contains the radical whence 3) 2 . Actually.. x dX="2 . 1) nlay be sol ved as follows: Example 2. It is sometimes Illore convenient to make use of hyperbolic substit lltlOI1S. we put x=a tan t. that is. -=-d__. _du..(Ch. Exam pIes 2.. 4 Indefinite Integrals ll·t If ~ g (u) du is known. 9. whence V x2 + a2 = a sec t. S 2 du Example 3. u = x 3 .--. ~ g (u) du = F (u) +0. x2 dx = 3" . then ~ f (x) dx= F [q> (x)] +C. the usual thin~ IS to put Vx 2 -a 2 . . =~ SeU du = ~ en +c = -} eX' + C. 4 (Sec. \ Y cos \' e" 1200*. 1196. find the follo\vlng integrals: a ) I· I d.\ .~1I12~~ . j"' ~F..\ x 2 -1 . 1202. J\pplying the indicated substitutions. (1) x=-In(.\2 . Therefore. "' sin' x . dx • 1203. r dt J x Vl-t-.cos! t dt = ~ cos t dt = Sin! t cos t sin t . cos t cos t + sin 2 t 2 2 • = dt t cos 2 5 S S =--=\11 I tan t-{-sec t 1--.dx. find the f 0\ 1o\\'i ng integrals: 1192. r t -l- Jx •\ x /~_. Ytan t -1--1 ~=S sec t cos t -. fJ r __ 1==1/x ~ 1.I V ~. "' .. t 1194.r. . j . "Vl~. x dx . t :. 2) Solution. SIn2xdx.~ a! dx.-= ta n2 t cos 2 t SIn 2 t cos 2 t dt = sin! 2t i. I· y_. find the follo\ving integrals: .· X2 1201. y I-x· J Xl lr y dx 2-x2 · 1204*.5)10 dx. 2~+ 1 dx eX-l y.-!!:!..f ' cos x dx (. App) ying 5x 2 ---:3=I. Put x= tan t.r-dx._. \ I· l-ry_. 1W3. ~ x (2x -\. tan t x 2 1191.sin x.¥1-1-tan 2 t 1Sin t 2 .YTtTafiit +C=lnl x+ Y x + 11_ yx +1 +C. 1195. Applying trigonolnetric substitutions.115 Integration by Substitution Sec.t__ l\) . x -t. dx= 5yX2+T S f dx==- x2 . ' b) SeX~I' c) ~x(5x' --3)'dx. x V~ 2-~-2 x == 7 .I_-t-C=ln Itan t-I. Stl i ta hIe su bst itut ions. In 4x x 1199. ( ~ sinh 2t + t ) + C = 2 2 = a2 2 (sinh t cosh t + t) +C. Whence ~ Va2 +x2 dx= ) = a2 a cosh t·a cosh t dt = Scosh t dt = a 5cosh . Sec.116 '"defi/ute /ntet!ral. Integr~tion by Parts A formula fer integration by parts. 1205.. by applying the hyperbolic substitution x=asinh t. Since sinh t = - x a ' cosh t = Y·a-+x a 2 2 and x+ Va _+x 2 2 et = cosh t -1.a In a is a new arbitrary constant.a we finally get x a 5V q2+ X2 dx =2 ¥a 2+. If u = <p (x) and v = '" (x) are differentiable functions. 4 1206*. then ~ II dv = uv . Evaluate the integral ..\2+ 2 I n(x+ 2 -- Ya 2 +x 2 )+Cl' 2 where C. 1207. ~ C. Find putting x==acosht.. We have: Va 2 + x 2 = Va 2 + a 2 Sl11h 2 t=a cosh t and dx=a cosh t dt.~ v duo . 3. 2 1210.s rCll. 1209. Find ~ Vl1 +x 2 dx.t + 1 dt = a.. 1208. dx j YX(l-x) by means of the substitution x = sin 2 t.sinh t = . Solution. cus x t dx =--= e (sin x 2 + cos x) + c..IntegratIon by Parts Sec. Example 2. ~ eXcosxdx=eXslllx+excosx-~eXcosxdx.~ eX cos x dx. _ 2x+5) e. . ~ arctan x dx. 1215. 1222* ~ (x 3 +5x+6) I. one has to apply the furmula of Integration by parts several tirnes. S ~. whence 5 e·ot. filld the follo\ving integrals: 1211. Find ~ x In x dx.I Xl In x dx. x 5eX dx. ~ arc 1223. 1226. . ~ (x l 1220*. ~ 1228. 1217. integration by parts yields an equation froIll which the desired integral is determined. ~ sill xlix. to reduce a ~iven integral to tabular foro). ~ x arc Xl 1219*.. 1229.X dx. dv=xdx. ~ In 3 xdx. Applying the forrTIula of integration by parts. ~ xcos3xdx. 1225. 1218**. j xdx SUI1X. 1213. lienee.dx. ~ In xdx. 5111 x dx. 1221. dx Putting u=lnx. 1224. r 1230. we have dU=-X' 5 X2 xlnxdx=2 Inx - x2 v=2 Whence 52x=2 Int -4+ C. ~ x·2.~ e:' sin x dx= eX sin x-l- + ~ eX d (cos x) ~= eX 5111 x + e:c cos x. 31 111 Example 1. 3X e dx. \ x sin x co~ x d. 1227. 5xae-~ dx. ~ x arc tan x dx. ~ x sill xrlx.\ 121 2. t 2 dx x~ x2 Sometimes. In certain ca~es.x dx. 1214. ~ In (x + V 1 + x 2 )dx. Find ~ eX cos xdx.=. 1216.dX.:OS 2xdx. 5':. We have ~ eX cos x dx= ~ eXd (sin x) = eX sin x. Jr(X -1~xa arc tan 3x dx. 1251 *. reducing the quadratic trinomial to the form (1). Applying various methods.(2· r 9 . ~ Sarc :~11 x £lx. 1238. x cos x d Ssin! x [Ch. . 1253*. ~ e" sin xdx. dx. ~ (x 2-2x+3) In xdx. 5(X2~ (~n x) dx. r sin. (1) where k and 1 are constant~. ~ x 3 r x2 dx. 1250**. C x:!~ 1241. ~ 3" cos x dx. 1244. I 233. 4. find the following integrals: 1236. Integrals of the form 5 mx+n+ ax 2 -l. ~ (arc sin X)I dx. ~ eax sin bxdx.r _.bx d c x. it is best to take the perfect square out of the quadrattc trinomial. 2)2' . 1252*. ~ xtan 2 2xdx.. Sec. ~ cos 2 (In x) dx. we get the tabular integrals II I or IV {see Table). Xl 2 1)2 dx .e x dx. . x. 1247. J l--x 1237. then. 1246. ~Va2-x~dx. ~ sin (In x) dx. In2x r n I 242.x 2 dx. ~ x (arc tan xY dx. ~ eV~ dx. The principal calculation procedure is to reduce the quadratic trinomial to the for III ax! + bx+c=a (x+ k)2 + 1. 1248. The follo- wing substitution may also be used: 2ax+b=t. J I 249. I 243. 1254*.4 1234. ~ 1245.Jndefinite Jntegrals 118 1231. If m=O. Sx In : +. 1240. 1232. S7dx. I 239. Standard Integrals Containing a Quadratic Trinomial to. ~ V A -1. r"rt:~x dx. 1235. To perforrn the transformations in (1). ~ dx r y1 = 1 .+ c.dx-= \2.dl.n dx == ax 2 -t. 2 .In I x 2 _ x-l/- x _1 2 }V + n dx. Example 2.. J V~_ ('t..bx -1.S 2x+2 dx+2f dx j Yx x+3 +2x+2 2 Y x +2x+2 J Y (x+ 1)2+ 1 "» t 2 = Y x + 2x +2 +21n (x + 1 + Y x---·-+-2x-+-2) +O.=.arc tan 1r--+ C. (mb ) -2a ln / ax +b. r dx J Y2\-3x-2x2= I Y 2 £! r <~ o. The integral is finally reduced to tabuIII X 2°. The luethods of calculation J 1l. ' x--l ) .b) -12a 2a ax 2 bx + c + n1 2. 4x-3 arc Sill -5.!. If In ¥: 0.Sec 4] Standard Integrals Containlng a Quadrattc Trtnomtal 119· Example t.+c' dx and thus we arrive at the Integral discussed above.:.2a - 5axl+b.\:+c/+ n.x . 5 I 1 x. if a > 0.c ~ (n. Integrals of the form lar integral V. and VI.. if Example 3.\2+bx+c are sinlilar to thosE' analyzed above.l ii(2X--l)-..4 ----arc tan --+C= 2 V31 -4--4- y3T = 2 4x-5 y 31 y 31 lr.\2- I .nlb) dx =-= m (2ax -l. dX=. .. then from the nUlnerator we can take the derivative 2ax + b out of the quadrattc trtnolnial i mx -1. Example 4. 2- - 1'1 dt t 1 =- t. 5YI-2x-x 2dx= ~ Y2-(l+X)2d(l+x)= l-1-x .t + ~ I+ (x+l) Vx 2+1 = . the given integral is reduced to one of the following two basic integrals (sec examples 1252 and 1253): 1) a -x dx= . Sin (a > 0).4 . Y-I-2t+2t = rJ y ( d\ r . V x2+A dx= . Integrals of the form verse substitution S(mx+n) Vax +bx+c dx 2 (Ch. . Integral~ of the form ~ Y ax2 + bx + c dr.C. SX2+~:+5 · 1256. VA 2+A + ~ In I x+ Vx 2+A l+c.. r =-2- Pind the following inte6rals: 1255. By means of the in- 1 --=t mx+n these integrals are reduced to integrals of the form 2°.nite Jntegrals J20 3°. Example 6.2 1 +4 +c=. We put 1) ~ x +1• 2 1 x+l=T' whence dx:~ We have: 5 dx dt f2.:1 nit . --~ f 2 +V(+-lr+ 1 =-.Va -x +-..arc j-'1r-'2 2 2) 5 y 2 2 x 2 -2 a2 -=-ax -1. Sx ~2X · 2 . 4°..~ + -V t2.I ndefi. By taking the perfect square out of the quadratic trinOlninl.Cjlnll-X+x?t2+1)I+c. l+x y 1-2x-x2 +arcslO Y2 +C. Example 5. Find s+ (x Solution. Integration of Rational Functions to.' 2 1274. reduces to lIltehratlon of the proper ratlollal traetloll p (x) l Q (x)' (1) where P (x) and Q (x) are Integral polynollllals. and the degree of the numerator P (x) is lower than that of the denolninator Q (x). t 271. 1262. 1278. ~ ¥2-x-x 2 dy. 5. 1272. 1269.\2 -2· "\ (x-I) d-< (l J(x+l) Yx 1261. 1270. (x-l). 1275. IntegratIon of a rational function. 1263. If Q (x) == (x-a)'1 . (2) To calculate the undetertllined coefficients At.\ where a. S'" x 121 dx yxz--t-x-I . and a. 1260. .. 1279. 1265. . xdx Sx -4x cos+3x · 4 J 2 f} Sl11 2 1277. · · Q (x) =:=:i x-a + A a + (--)'X x-a +··· . A 2 . 1266. ••• .L1 Lz L). . ~2Xd. 5) 1259. 1276. A are natural nUlllhcrs (root lllultiplicities). X-6s1I1X+12 eXdx " ¥\+fx 1267. both sides of the identity (2) are reduced to an integral fornl. .\ Vc~ X + 4-~~~s~~.1 \ t' 1268. and then the coefficients of like powers of the variable x are equated (tlrst nlethod). ···+x-l+(x-l)2+···+(x l)1\. 1273. after takIng out thf whole part. then decomposition of (1) ·i~t·~ partial fractIons is justified: P (x) At A2 (-----)2 x-a +. •••• l are real distinct roots of the polyllotniiJl Q (x). ~ ¥x +2x+5dx. valent equation] x equal to suitably chosen nUlnbers (second Inethod). ~ Yx=?dx 2 1-2. 1_ SlIl \ dx · In x d\ Jx YI-lInx-li1 2 t' Sec. 1264.Integral ion of Rational Functions Sec. These coefficients may like\vise be detennined by putting {in equation (2) or In an equi. dA- V. The nlethod of undeiernlined coefficients. we get: -1=-8 2 -2.A. .8. puttIng x=O. and ~et· Hence. 8 2 ==1/ 2 • will have: O==A--B. We get 1= A. 8=-1. 1==2A.-B 2 . Le..=. . we put x=O in identity (4). Then. Further. we equate the coefficients of x 2 in identity (4).-1. we will have: J=A. (3) a) First "Iethod of deternunlnR the coefficients. applying the first nlethod. (4) When solving this example it is advisable to conlbine th e two methods of determining coefficients. O=A+8.Indefinite Integrals 122 [Ch. PuttIng x= 1 in identity (3). Putting x= -1.8 2 ) Eq uati n~ the coefficients of identical powers of x. 11=-1. Find xdx (x-I) (x+ 1)2=1. Find Solution. We rewrite identity (3) in the (onn x = (A 8 1 ) x 2 (2A + B 2) x + (A . putting x= I. we get: O-==A+B. we get I =C.-B 2 • Whence I A=-· 4 ' + + b) Second method of determtning the coefficlents. O==A-8.. Further. Example 2. and C=l. \ve Le.-8 2 .. S Solution. We have: Whence x==A (x-t-I)~-+-81 (x-I) (x+ l)+8 2 (x-I). Applying the second method. 8. 4 Example 1. We have: 1 x3 -2x 2+x 1 A 8 x (x-I)2=X-+ x-I C + (x-I)2 -and I=A(x-I)2+Bx(x-l)-f-Cx. \) J Q2 (x) (6) tJ where Q I (x) is the greatest dcri vativ e Q' (x).. x (x) and Y (x) are polynonriClls \vtth undetcrrnincd coefficients. Find Solution.\) (J 1 (.--.1 --drr tall z. COlll1110n Q! (x) diVisor of the polynoillial Q (x) and Its == Q (x): Ql (x).~~·t-1_rY(X) dx.J Id . •• . It is first advisable to represent the qua drahc trinonllal x' +( q_~2) + px + q in the forIII ( x + f r -r x+%=z.-life tan z= . A k .J . 5] Integration of Rational Functions 123' Consequently. PUltlfl~ .--~ lz~t") .\2+ px -f-q +. Find dx r J (x 3 -1)2· . x2+px+q=[x-(a-r-lb)] [x-(a--lb)J and AI' B I .Sec.:: 2 Ostrograd~ky 2 (Z2 . respectively. less by Ullity than those of Ql (x) and Q2 (x). If Q (.\'-1-2---:2.~ (2' 2 2°.. 02 az:=:. -2 (z:!+ I) --.alet a 1J (x -.-2 (2 2 + 1)z 1 z -1. whose degrees arc....--. and make the substitution Example 3. Q (.. 5 ~ dz (1:.-. Bk are undetennined coefJiclents \vhlch are detennlned by the nlcthods given above For k =-= 1. for k > 1) USl' is llHlde of the redLLctzoll Inethod.. Example 4.4 \' + 5) - ...J i~-n ·l. The undetermined coefficients of the polynonllals X (x) and Y (X) are cOlnputed by differentiating the identity (6). Sillce theil. 1= dx 5dx 5 dx I x-l+ (X_l)2=lnlxl-ln/x-11-x_lTC. The -1- 1) ~ aI ( t a 11 z --t. .\) has Illultiple roots.. here.t 1)£ - II (1 -t. 5-XI If the polynonllal Q (x) has cOlnplex roots a partial fractions of the fornl A1x+B. lIere.:3 2 (x 2 -t.z:!) - z'1.-1. then \P(X)d\.C :::. ~. the fraction (5) IS integrated din:ctIy.1) 2 \ -t. then Akx -t-R k (x2+PX-~_(/)k (5) will enter Into the expansion (2). + ± £b of mu It1phcity k.:-=-:.C)) _L C 2 I - 1 • fllcthod.!. Wl' gl\t C 2- 1 === J {2 2 t- 1 . J (Z2 +112 dz-== ell I 1 C dl I ~. 1=L{x 2 +x+I)+Mx{x-I)+N{x-I). D=O.a 3 3 F=-3" Sx dx-1 (7) 3 To compute the integral on the right of (7).!-I into partial fractions: x- lhat is. D+3C=0.=O. we decompose the fraction -.2-5x+6 dx 5ex + 1283. consequently.Indefinite Integrals 124 Solution. I dx 2 x S(x --1)2=-'3 x .I .1 5 (x . B= I 2 -"3. C=O. E-A=O. ~=AX2+Bx+C +5Dx2+Ex+F dx 3 Xl . x J x2 X 1 2 2x 3(x3 _1)+gln (x-I)! +3 Y3"arctan FInd the fol1owlng integrals: dx 1280. (8) L=}. x3 .I )2 Different i at i ng th is identi ty. E+2A=0. L-N=I. we get Equating the coefficients of identical degrees of x on the right and left lof (8). we get 'Or I (2Ax+B) (x 3 -1)-3x 2 (Ax 2 +Bx+C) {x -1)2 ' (x 3 _1)2 + Dx 2+ Ex+F x3 -1 1= (2Ax+ B) (x 3 -1)-3x l (Ax 2 + Bx+C) + (Dx 2 +Ex+F) (x3 _1).Y3 arc + + dx (x'-1)2 S SX 2 tan 2x 4.41 x . and. F-2B:. we find · L+M=O. we will have: D=O. (x+a) (x+ b)· 1281 2 -5t+9 • . whence A=O. 5 1282. B+F=-l. Therefore. Putting x= I.91 (x-I) {x+3)lx-4)dx• . £=0. -or I M--_o 3' 2 N=--a.I V3 +C +I Y-3 +C. S dx Xl - I= I 3 Sx -dx I - I 3 Sx +x+2+ I d x = 2 X I I =3 1n I x-ll-filn (x and 5 1 +x+ I) . Equating the coefficients of the respective degrees of x.I .\. dx I) ex 2X2 + 2) (x + ~) · -f. 4x 2 -t. X2_ \ -t.14 (x-tl):' (x-2) dx. cx+d 5 where 1312.! Certain I. 5 1296. 5(. \'3 1306. + 1)'"~X3 13 C 7 x + 8) 1310*·5 Xlx~Xt I)' dx.. 2x-3 1290. -x dx. 5 1308.6x+ 9 (X_t-1)2 dx • 5 x2-8x+7 1289. X Applying Ostrogradsky's method. 6. find the integrals: 1305.5 ( . 5 1309. find the following integrals: 1301 • 1302.1. dx -1.. 5lt~I\'. (-\2_ 1299. PI' ql' Pa' qa are whole l1ul1lbers. S 1295.1)2 5(x2-1-2~+-2){X2+2x+5)./ 1 dx. 5 1285. 5(x -3x+2)3 dx. (X-3)2 1298.2 J(. . 1)2' dx dx x3 .rational Functions 5X'+2 x3_5A2+4xdx. 5JI.See. 1313..x. 125 1297. 2 X3 1300. Int~grating 1314. -t-x+ I 1291.2 · 0 5xlS~ d'( . dx. 1292.4 1. 2 xl . 1288. 5 {x 1303.1 dx. 5 (x2 + I) dx. x 1294. 5r2 -t. 6] /ntegratinf. rm PI R [. R is a rattonal functIon and t ( P2 1 ax+b\q~ ex +d) .l2_2x+2 2dx .t d'(+ 1)4 . ~) 5 1304. n j Applying different procedures. 1)2' r-I 1286. 1293. J x I2 -2\. ( aX+b)Q.t4 {x3 Sec. 1311. 1287. 5 1307. x4-6x3+ 12~2+6 4-3X x3 -6-\2-t-12x-8 dx. 5 3x_IO)2 dt . · 5~ Xl AG • Certain Irrational Functions Integra Is of the f . 1 • -t. (1) . 5 (Xd~_ 1284.2 \ 2 ~.1)2 (x 2 + 1)2 • dx (t 4 _1)2' r (. r x-I " A -V j t.J _t·t+! +-=-_ dx.V2t-1 J 2. q2' Example 1.V 2x .. \) I·'VX" 1321..\2 -t..\) 1322.. 1)3' dx V-=-l '-1.YX-1-1 20 • I ntegraJ~ of the form Pn (x) SV·~a). x 2 r 2x ~.r----.:.. J 1316.\ \ 1324.dx.. x+2 t. t. ~__1=1 dx.V-· X . The C2z dz 3 d. S. £1. lb./3 _dx.([ 132:3.. an Integral of the fOrIn 2 J 2-1 == 1315. ~-.1. C J 'Ir "Y t. Find the integrals: r .\ -1 = 2 Z2_ Z =2S(Z+I+z r = zo1 leads to z dz = II)dZ=(Z+ll2+21nlz--ll+C= (1-1.. . JV·~·+l ~XV·(~ .--1 laJ9. 1318.bx -t.-2~+-b-x-+-c dx where Pn (x) is a polynol1llal of degree Put Pn(x) lrax2 ~\ f + bx + c dx~ Qn-l (x) (2) I tt lr r axl+bx+c+".b n --=2 cx+d I where n is the least cornmon lTIulti pie of the lllllnbcrs q.c (3) where Qn_l(X) is a polynolnial of degree (n-l) with undeternlined coefflcients and A is a num b~r. dx t 320.==. Fi nd J\----~--V 2x .\. \- 1 x -1. V\.=' dx r a. The coefficients of the polynomial Qn-l (x) and the nUlllber A are found by differentiating identity (3). J(x-l-l)2.. . ° \+ 1 'V\r i' .1 r ~ubstitution Solution.t--=-r . I ("2-.\ 1117.dx.\ J V2x-I.4 Integrals of form (1) are found by the substitution ax -l.3 -.V2x=f)~ -I-ln (V2x-1-1)2 +c.-.r-'==-. (' _d_t_ ~r-1----\ • v " J/.\• -- j '1 13~5.+l dx.Jnde{iniie Jniegra[s 12J (Ch. ... Integrals of the binonlial differentials ~ x ln (a + bt")P dx. \ve obtain Yx +4 2 Yx 3 2 2 +4+(Ax +Bx +Cx+D)x+ . A=-2 Hence. . J' )-"\" X6 1/--===== . J' .\-a Find the integrals: 1326. we Inake the substitution a+ bxn = n = zS. The integral (5) can be expressed in tenns of a finite COlnbll1ation of e1rl11entary functions only in the following three cases: I) If P i" a whole nUlnber. Here.4 A Y +4 . 1 C=-.Sec. 2) if 111-t-l is a whole llulllber. 1:J27. 6} Integrating Certaln Irrational Functions 127 Whence .\ r 1-1. nand p are rational Ilunzbers. 2 D=-=O. J r2dt ~ f 1329.:X 2 +bX+C' They are reduced to lntrgra\s of the form (2) by the substitution: _I_==t . I-Iere. Integrals of the form 5 (x-a)n . 1330. "If .\ 2 -f. 4°. I + P is a whole nUll1ber.'.. 3°. .r. Cheby~hev's conditions. 1328.\2_ ~-1-1 r~ 1r ~-==-"l dx.l2 and equating the coefficients of identical degrees of 1 A =4 ·' 8==0·. use is made of the substitution . (5) where tn. 3) if 111-1- n ax-n+b=zs.\4·~_4x2 =(3Ax2 +2Bx+C) Y x2 + 4 Multiplying by x. dx..x 2 1331. where s is the denonlinator of the fraction p. Integrating Trigonometric Functions 1°. 1) If m=2k+ 1 is an odd positive nUlllber. ~ 1333. 1 +~ . Exall1ple 1.- V + Vx-. 1334. it' 1336.:::::-11--13+ C.:~~:r dz= V V I-I- = 12 -x. n=T. ~ x" dx (2 133 7. 4 we have here Case 2 integrability. + x")-. --!"+l 2 m+ 1 1 n I P=3" .~ (I-cos" x)k cos n xd (cos x). 4 Find the integrals: 3 1332.· dx 1 J Sec.x dx " 'y 1+ x . \ -3 J Y. 11 (1 ) .1 10 x (I-sin" x) d (sin x) = sin x sin 13 x . Find 1 1 Solution.t 2 1 +. x ~. I = 5 where x- Z= ~(I +x+)+ dx= 12 5z~. " SV~. We do the sanle if n is an odd positive number. Here. ~4 S " ~ . dx= 12z (zl-I)3 dz Therefore. ~ s1n1o x cos'x dx = ~ 51. Hence. 1335. =2. nt= -2 . 7. then we pUf 1m• n = - ~ sin"k x cos n xd (cos x) = . i / . Xl (l + 2x") -2 dx. S(z'-ZS) dz =¥ z1-3z +C. 4 -------------~---=------------=---- Example 3. Integrals of the form ~ sinmxcosnxdx=I m• n' where nz and n are integers. dx .Indefinite Integrals 128 (Ch. The substitution 1 +x 4 =2 3 2 yields x= (z3-1)4.5. ] -I.+-+ 2 tan 2 _I-x +21n Itan ~ 1+ ta: 2 [- 2tan 22" . 7] 129 2) If m and n are even positive numbers.C.cosP. 5-.sins 6x ) + C 8 3) If m = parity. 2 2)[ sec 2 . .t .cos' . =ds ~ d: Sill 1~(1+tan2 .sin" x + 2 2 2 SI:: Example 3.!. an' x 2 d (tan x). Example 2. then 1 m. cos x sec 2 xd (tan x) = 5+ (I tan 2 x) d (tan x) = 1 =lan x +"3 tan 3 x+ C.-!. . the fotlo\ving integrals reduce to this case: 5 X=21'-~Jj~ -i) 54 = 5 5co~'~ x=s ((+ ~\ · dx ( x and d sintJ.2 - identical cosec'" x sec' -2 xd (tan x) = - 1) of 5 S1I11'. In particular. 1 cos! x="2 (I tran~ + cos 2x). seeS 1.cos 6x 1 -42 dX=8 5 Jr (sin 2 6x-sin 2 6x cos 6x) dx= ="815(I-COS12x. Example 4.!-dx=- 2 ~] d( tan i )= ~ 2 8 =i 5 tan -3 . 2 stn 2 6x cos 6x ) dx= = ~ (~_ sin 12x _ .)2 =8 +tan 5-1900 tan sX x dx sina !. ~ cos 2 3x sin' 3x dx = ~ (cos 3x sin 3X)2 sln 2 3x dx = = sin2 6x 1.Integrating Trigonometric Functions Sec..v are integral negative nunlbers 1 -t." = = 2 - 5 5( 24 ~t • and n = .::os' x \ 18 ~ =5 + V-2 2 (1 an x + tan 2 x) - (1 d (tan x) 2 ~+v --1 tan 2 x) t lJ.dx-= 2 tan-a 2x !-. 1 sin x cos x="2 sin 2x. then the integrand (1) is formed by tneans of the formulas 1 sln2 x= 2 (I-cos 2x). 5 dx 1363. ~ ~in' xdx. 1349. ~ sin~xcos'xdx. S 5 t an' x dx = 5 5 tan' x (sec' x-I) dx = ta~' x - tan' x dx = tanS x tanS x -=-3-. J350.cos a - ~ sins xdx. ) 2 1353. ~ sinS x Vcosxdx. Ssindx x · Scos'dx x · Ssin' cos' x d x x. sin x cos x 1354. ~ tan' 5x dx. ~ cote xdx. Find the integrals: 1338. sin l x cos 4 x · dx. 1356. S 1347. cos' x cos x 5sinsinx·--dx+ x 1 2 cos 2 X 1 2 e ---- 5cos cos x 5-cosdx-x= --dx+ 2 X 1 C : :2- 2 cos x +-2Inltanx+secxl+C. 1348. Example 5. 1344. respectively. 1340. 1362. 1346. ~ sec s 4x dx. x dx x • sin . 1345. integrals Jm n of the form (1) are evaluated by means of reduction formulas that are usually derived by integration by parts. 1360. 1341. are evaluated by the formula tan 2 x=sec·x-l (or. 1355. . where m is an integral positive number. 5 x. Ssin' dxcos' x · 1364. 5 Ssin' cos x d x x.Indefinite Integrals 130 [Ch. a x 4 cos' x d sin. 1343. cot l x = cosec 2 x-i). 2 ) sin (x+ ~ ) . x S dxx SYtan dx x· YSin cos' x · . ~ x sin' x'dx. cos' x = cos' x x= S == sin x S--=slnx dx. 1351. ~ cot' xdx. ~ cos' xdx. ~ sin' x cos' x dx. 1342. 1357. 1352. 1359. ~ cos' 3xdx. 1358. 4 1361. 1339.4 4) Integrals of the form ~ tan m x dx (or ~ cot m x dx). ~ sin! x cos' x dx. SsIn "x 2" cos sX 2 d x. 5) In the general case. 2 2 E 16 dx SSln x+cos x d xamp e .(sec2x-l)dx=-3--tanx+x+C. x dx sinS x · 5'l tan "3 + tan 4'X) dx. ~ sin 3x cos 5x dx. ~ sin mx sin nx dx and ~ cos mx cos nx dx. I . new Example 8. ~ ros(ax t. 32x dx. 1366. Integrals of the form ~ sin mx cos nx dx. Putting tan ~ =t. 1370. 2"x cos. Ssin 3" sln"3 30 • Integrals of the form ~ R (sin x. 1) By means of substitution x tan'j=t. cos x) dx.t2 2t slnx=l+t l ' COSt=l+tl' d 2dt X=I+t l ' integrals of form (2) are reduced to Integrals of rational functions by the variable t.2~ sin lOx + C. Ssin9xsin xdx= S~ [cos ax-cos lOx) dx=> =~ sin 8x. (n these cases the following formulas are usedl I) slnmxcosnx=~ [sin (m+n)x+sln(m-n) x]. 1369. 7] 131 2°.b)ros(ax-b)dx. whence . Find 5+ 1 dx sin x + cos x Solution. x. I 3) cos mxcos nX="2 [cos (m-n) x +cos (m + n) Example 7. where (2) R is a rational {unct ion. ~ sin x sin 2x strl 3x dx. xl. ~ sin rot sin (rot +cp) dt. we will have 2dt 1= S 1+t 2 2t I-t! 1+ l+ta+l+t l S· I • . ~ cos x cos· 3x dx. 1 2) sin mx sin nX="2 [cos (m-n) x-cos (m + n) xl.1ntegrating T rigonometrlo Functions Sec. Find the integrals: 1365. x 1371. x d Scos. ~ sin lOx sin 15 xdx. 1367 · 1368 · 1372. cos x). dt dx= ' 1+ t2 . we will have l (3) t2 sln x = 1 + tl dt = Y12 =/. 1378. Scos x + dx2 sin x + 3 1387. 5 . for example. 58-4sinx+7cosx' 1386. 5 1389*.. dx I 1379**.lr-2 tan x) + C. cos x I dx • 1385. 51+2t dt 1 S d (t y'2) t) =Y2 1+(tY-2)2 ( 1+ l+t J= ~ (l+t 2 2 ) 2 2 2 lr- arc tan (t . tanx dx I-tan x I 2 • S(I-cos sin x 5 X)I dx I sin 2x 1+ sin2 x dXI cos 2x Scos4 x + sin4 x d XI 1388. it is useful to apply artificial procedures (see. -cos x) e i R (sin x. 5 1374. In individual cases. 53s~nx+2cosx d 51+ 1381 *· + 3 cos x x. 1380. 1379). 2) + C = Y1 2 arc tan (.. . t 1 sinx= '1r ' cosX= '1r-2 2 f 1 t f I t + and + x = arc tan t.. Putting tan x= t. We note that the integral (3) is evaluated faster if the nurnerator and denominator of the fraction are first divided by cos· x. dx = I +dtt 2 • Example 9. 1390*. Find the integrals: 53+::0SX· 1373.( + 3 sin xcos x-cos x 1384*. Iben we can use the substitution tan x = t to reduce the integral (2) to a rational form. sin x. 1382* 5 1375. Sl+s1nx-cosx I-s~nx+cosx d X• I S1+:x . :X. · cos x d sin 2 x-6sillx+5 X. 5. 1377.sin x.sin x cos x 1383* dx sin x+ cos x ' cos x d 1+ cos x XI 1376. Find 51+ sin x dx Solut ion. dx 3 sin 2 x + 5 cos 2 X I SI-sin sin x d x XI 2 sin x 5 5sin!. dx (2-s1n x) (3-sin x) • I .[Chi 4 Indefinite Integrals 132 2) If we have the identity R (. Here. S tanh' xdx. (cosh 2x+ I) dx= ~ sinh 2x+ . 5 1400. 8] Integration of Hyperbolic Functions 133 Sec. Find ~ cosh' x dx.+c. 5smh xd. ==slnh sinha x x+. Ssinhl xcosh l xdx. 9. 3 Find the integrals: 1391. S sinh' x dx. 2 1399. 8. We have S cosh a x dx= Scosh l xd (sinh x) = S (I +sinh! x) d (sinh x) = . We have Scosh 2 x dx= 5. Example 2.Sec. sinh 2x. (1) . Find Solution. Solution. Scoth& x dx. The following basic formulas should be remembered: 1) cosh! x-sinh! x = 1. dx sinh! x + cosh! X • dx 5tan::-I · Ssinh x dx • Vcosh 2x Sec. Ssinh' x cosh xdx. 2) sinhl x = ~ 3) cosh 2 x = "2 (cosh (cosh 2x-I). S cosh 1398. Yaxl+bx+c)dx. 1402. 5sinh" :~OShl x • 1395. 1397. 1393.OSh x • 1396. 1392. 1394. x+C. & x dx. Integration of Hyperbolic Functions Integration of hyperbolic functions is completely analogous to the integration of trigonometric functions. 52 sinh x + 3 cosh x • 1401*. Using Trigonometric and Hyperbolio Substitutions for Finding Integrals of the Form SR(x. 1 4) sinh x cosh x = i 2x + 1). where R is a rational function. Example t. Example 2. 1 x+'~=-2. 2) ~ R (z. 3) z m sec t or z = m cosh t..+YXI+x+I)+ln 1 t) cosh t-l-- +C 2· X .} 3 (x + ~). Y3 sInh t..-sinh t 8 8 2 Since sinh t = . the integral (I) becomes reducible to one of the following types of integrals: I) ~ R (z. 3) ~ R (z.Indefinite Integrals 134 [Ch. and Y3 dX=-2. = . The latter integrals are.. we then have dx=sec 2 zdz and _rJ (x+ I)' Y(x+ dx S 1)'+ 1 sec! z dz = tan! z sec z J- Scos z dz = sin' z V x2 + 2x + 2 + X+I = __ .cosh t • -2. Find Solution. taken by means of substitutions: 1) Z= m sin t or z= m tanh t. 2) z=mtan tor z=msinh t..cosh tdt.-3. V ml+zl ) dz. V ml -Zl) dz. We have x! + X + I = (x + 2"1)2 + 4"3 · Putting Y3. = Example 1. We have x!+2x+2=(x+ I)!+ 1.slnht we get J= I) Y3 S( -Y3. 3 V Xl + and t=ln (x+ . Putting x+ l=tanz. cosh t = .4 + + Transforming the quadratic trinomial ax! bx c into a sum or difference of squares.cosh t dt= 222 3 }ras .1_ + C = SUI Z C. Find Solution. Y Zl ml) dz. =-8- sInh t cosh! t dt- 3 B 5 cosh! t dt = 3 Y3 cosh' t 3 (I . respectively.3' + 1 . 1428. elX -2 · dx Ye2~ +e~ + 1 • Xl 2) dx. 1421. 1426. 1418. 1414. ~ V3-2x-x l dx. In=5(xl~xal)n. r 1411. find I.V sing Reduction Formulas Sec. ~ V x l -2x +2 dx. ~ xe x cos x dx. l " J (xI-2x+5)1 1407. Sx arc cos (5x - 1420. 10. S 1417. dx 1413. ~ (Xl + 1)1 elx dx. SVx +x dx. Using Reduction Formulas Derive the reduction formulas for the following integrals: 1427. 1422. Ssin x sinh x dx. and II' C. J V 9+x dx. Sec. Integration of Various Transcendental Function s Find the integrals: 5 :. ~ x sin x cos 2xdx. 1424. ~ V Xl -4 dx. ~ V2 1410. In = ~ sinn xdx. -+. SIn' (x + VI + x') dx. S(x-I)Vxdx-3x+2 • 1412. 2 S(I + S dx VI-x' Xl) (l-x 2 ) dx Yl+x l • • Sec. 1425. dx.Xl dx. S(Xl +x+ 1)1 dx. " Xl 2 1406. 1409. 1408. ~ e2X sinl x dx. ~ Xl cosl 3x dx. I 1405.x 1415. 1423. find II and I •. 1416. 111 we finally have • 1 ( X + -1 ) J = -31 (X 2 + X + 1) -I -. ~ eX sin x sin 3x dx. SV x 1404. 11. 5 In ~ +. . 1419. 4 2 135 Y Xl + X + 1-~ In (x+ ~ + Yx l +x + 1) + Find the integrals: 1403. l -6x-7 dx. find 110" Sec. 1461. S x dx. ~ cos· x dx. ) lR=9 dx. J (x + 1) 2 1451 *. '"' dx j VI+X" Sy · 5x dx. ~ V x.2x 1448. Y dx1 -x' . 1+xt 1460. . 1 Sln 2 sinS ax 1464. 2 (1 ) I- XC • xdx 2 . cos x sInS x · 5 + yCOtX dx x • 1463. 1 . 1465·5sin2xd COSO 1& X. \ v x 1458. 5 dx. 1453.x dx. dx. 1459. Vcos 1462. Miscellaneous Examples on Integration S + x xdxV 1449. 2 dx. S Yl. 1456. 5 2 dx • (x 2 +4x) Y4-x 2 1452. /x xZ-1 4 .4 find I.4x 1454. In=S cosd~ x . ) xVx 2 +2x+2dx..Indefinite Integrals 136 1429. Sx 1457. and I". In = ) xne. [Ch.X4 • r x+ 1450. 1430. ) cosec s 5x dx.!. S Yx dx+x + 1 · 2 X 1455. 12. lX e . dx e!X _ 6e x (eX 1476. 1496. 1484... Sarc sin y"X dx. 1495.Miscellaneous Examples on Integration Sec. 1490.3x) sin 5x dx. ~ (x' -1) 1O-2X dx. 1500. ~ sIn x 2 1474. 1482. 1479. Ytan x+4 tan x+ 1 1470. Sx· arc sin ~ dx. 2X 5 1481. ~ sinh x cosh x dx. S dx sin x sin 2x • dx 1472. ~ (Xl . ~ l 1478. ~ xe dx. 1477. dx. S2sinx+3cosx-5 dx • S2 + 3dxcos! 1486. 137 l/ eX t 1 dx.!. ~ Xl In Vl-xdx. Yl-x S sinh x cosh x d sinh x+ cosh x x. S(tan x+dx1) sin x arc tan x dx. S seclx dx. \ y d x . cos 2 2 Scos! xdx 3x · S Yl-x dx. 1497. 1487. + 1) 4 S1_4xdx. 2 2 1475. 1480. X • Scos x+2 sin xcosx+2 dx sin x· 1471. 1467. 1499. ~ cos (In x) dx. 12] 1466. Ssin ( : -x )Sin(: + x) dx. YI +x 2 a x• + 13 dx. S (2 + cos x) (3 + cos x) • 1473. 1485. ~ x arctan (2x+3)dx. 1498. 1489. + ~ ) dx. 2 e2X -2e x • 1493. Sarc::" x dx. SI x I dx. S S I 1491. ~ x sin x dx. 1469. . ~ x'-e~' dx. sinh 2 1488. Stan' ( . 5(sin x +dxcos x)! • 1483. 1468. S'IX 3x sIn 2 cos 2dx. eX X 1494. SYu +slnax ax dx. 2 1492. 2. 37).. < xn=b is an arbitrary partition of this interval into n subintervals (Fig. n-l 11m b ~ f (M tui = ~ f (x) dx. Geometrical1y... 1. Let a function f (x) be defined on an interval a~x~b. is called the integral sum of the function !(x) on [a.. 37). The definite integral. &xi=xi+l-xi.. -. to zero. S" is the algebraic area of a step-like figure (see Fig. and the largest of them. dX'.. (n-l). 0 I =0 a (2) . max Ax. b].. i=O. The Definite Integral as the Limit of a Sum 1°.Xi+l.. 1. A sum of the form n-l Sn = ~ f (~i) AXi' (I) [=0 where Xi~6ie=. that is. 38 2°. Integral sum. The limit of the sum Sn' provided that the number of subdivisions n tends to infinity. y y o o1 Fig. . 37 to x Fig. and a=xo < XI < . is called the definite integral of the function f (x) within the limits from x=a to x=b.Chapter V DEFINITE INTEGRALS Sec. . n n (n + 1) (2n + 1) 6 X 0 Fig. 1] of a Sum 139 If the function f (x) Is continuous on la. n n n-l n-l ~ S'n = ~f (. = (: r y. and the ordinates x = 0. The definitions of integral sum and definite integral are naturally generalized to the case of an interval [a. those below the x-axis. we get the area of the step-like figure Sn=: (:Y l l+2 +3 +.. in which the areas of the parts located above the x-axis are plus. where a> b. minus (Fig. 2' n-+CJ) Example 2. 38). b) into subintervals and is independent of the choice of points ~i in these subintervals.i) dXi '\. xi+tl.. Form the integral sum Sn for the function f (x) = 1 +x on the interval [1. (~i)=l+l 9i 9i 9i +n. b). 39 . What is the lim Sn equal to? ai n -+ CJ) So I ' H ere. Solution.!. the limit of (2) exists and is independent of the mode of partition of the interval of integration [a. = [ 2 ( : YJ ·. Partition the base a into n equal y parts dX=~. the definite integral (2) is the algebraic sum of the areas of the figures that make up the curvilinear trapezoid aABb. and x = a (a > 0). Geometrically. 37). Y. +-=2+-.. I:! xr=-n-=--.. Yn=[<n-l):r· The areas of the rectangles are obtained by multiplying each Yk by the base Ax= ~ (Fig.)=58-. 39). n Summing. WIlence . Example 1.10] by dividing the interval into n equal parts and choo!ing points that coincide with the left end-points of the subintervals [xi... 1 1 Using the formula for the sunl of the squares of integers._81 n 2 2n ' lim Sn =-= 58 -. we will have Yl = 0.+<n-l)'). it is integrable on [a.. Hence (Fig. Find the area bounded by an arc of the parabola y =x2 • the x-axis. Choosing the value of the funcn tion at the beginning of each subinterval. 10-1 9 an d tfoi=xI=xo+t'A xi= 1 utlon. b). +n-l)= 1=0 1=0 =18+8In(n-l)=18+81 nZ 2 2 (1_-!. . n=n n + n 2 (0+ 1 + .!.~ ( 9i) 9 18 81 =~ 2+--.. Le. b).The Definite Integral as the Limit Sec. then tJ ~ f (x) dx=F (b)-F (a). then the function x F(x) = ~ f (t) dt a is the an\iderlvative of the function f (x). 1503. o o 5 V. A definite integral with variable upper limit. ~ x' dx. regarding them as the limits of appropriate integral sums: b 1 1501. 1 1506*. Find x f (x) = ~ sin t dt. 2°. ~ 2 x dx. Find the area of a curvilinear trapezoid bounded by the hyperbola by two ordinates: x = a and x = b (0 < a < b). ~ x'dx. we obtain S:'C lim Sn n -+ = lim a l (n-l) n (2n-l) _ al 6n I 3 • n -+ (J) a. ~ dx. b. F' (x) = f (x) for a ~ x <. +gt) dt. 1505*. If F' (xl = f (x). 2. and g are constant. Evaluating Definite Integrals by Means of Indefinite Integrals 1°. The Newton-Leibniz formula. and the x-axis. S(v. 1502. that is. a -! T 10 1504. passing to the limit.) Evaluate the following definite integrals. bl.[Ch.5 Definite Integrals 140 we find aSn (n-l) (2n-l) 6n l and. 1507*. o Sec. a t (t) Is . If a function continuous on an interval [a. ) 1520. dx 1516. n n n fl P lim IP+2 +.. S-to Applying the Newton-Leibniz formula..... Example 1. 1517.) 1519**. lim 'l-+a.4 dx. I = X x S cos (r) dt (x 1 x 1513. x 1 S dx X' • S et dl. F (x) = ~ 1n t dt (x> 0). ('. . Let b I= Find 5I:: (b > a> 1). find the limits of the sums: 1518**. F (x) vX' 0 1510. 1512. a dJ 1) da.+n n n ( +1 1+ +1 2+ .+n (p>O). 0 -I Using definite integrals. Find the integral s S y. F(x)= ~ Vltrdt. lim n-'a. Find the points of the extremum of the function x Y= sin t dt·III th e region . -I 1508. S:cos t dl. +n 11).. 1511. = Sr /2 dt.Evaluating Definite Integrals by Indefinite Integrals Sec. x> 0 . dJ 2) db· Find the derivatives of the following functions: x' X 1509. -x 0 - 1515. 2] 141 The antiderivative F (x) is computed by finding the indefinite integral Sf (x) dx = F (x) + C. 21 + . p n-+CIO nil > 0). find the integraisJ Sl+x· x 1 1514. +_+1 ). 1632. o 1523.1). I 1537. 5 :elX dx. sin 1 x!+3x+2' o 1528. SYx-2dx. o . o 8 1531. ~ (V 2x + Vi") dx. S sinh! x dx. dx J Y 5+4x-xl dx I VI-x·· 154. ~ sin 8 q> dq>. o 1 1543. 51+yrii dye 1 y2dy 1535. :t y 25+ 3x. n 1 8 5 X!+d:x +5' lMl.6 Evaluate the integrals: 2 I S(x l -2x+3) dx. ~ ~':~. o dx 5 e2 1538.Definite Integrals 142 [Ch. 1536. ~n x) dx. o n 4. S tanxdx. Scot' q> dq>. n trco::!x' In • VI' 2 11 P- 1533. Scos! ada. r I 8 1 1522. . n 5X!-~:+2 e 5Z8~ 1 • 1542. xdx I 4. o 1526. ~ sec l ada. y o o I 1525. 1544. :t 1540. 1 1529. dx x In x· e e 1dX X I · -2 1527. I o I dz. 5Y • +4 · n e 1524. Scosh xdx. 1 1534. -8 4. 1521. :t -1 1530. S 5 1 5 1539. x <. f (x). then by definition we put b c-e b a c+e r f (x) dx = elim~o Jr f (x) dx + e-+o lim r f (x) dx. b] and is continuous for a ~ x < c and c < x <. %-+rD converges. then 1) for m < 1 the integral (1) conver~es.. then the Infe· a 00. th. If f (x) ~ 0 and SF (x) dx 11m f (x) x m = A #r. J J a (1) If the limits" on the right side of (1) exist and are finite. Similarly. i. e. 3] 14~ Sec.. Improper Integrals 1°. b r J ~ b f lim (x) dx= a-+-rD -QO r f (x) dx J ri~ht of (3). 3.b and tegral (1) also converges (comparison test). F (x) when a <.Improper Integrals Sec. e. the improper integral is catted convergent.. i. otherwise it is divergent. 2) for In ~ 1 the integral (1) diverges. A #: o. If there is a continuous function F (x) on [a. If a function I (x) is not bounded in any neighbourhood of a point c of an interval [a. b ~ f (x) dx= a~~ Sf (x) dx.. F (x) and the integral gral (3) converges as well. 2°. If the function f (x) is continuous when a ~ x < 00. (2) a b If /I (x) I <. then we assunle b t:rJ r f (x) dx = J a lim b-+~ rJ f (x) dx (3) a and depending on whether there is a finite limit or not on the the respective integral is called convergent or divergent.I A '1= 0. b.. 2) for m ~ 1 the integral (3) diverges. A f (x)-Ii when x x -+ co. and OIl a b-+-+rD a -QO rD If II (x) 1<.: definition is correspondingly simplified. If f (x) ~0 r(x) and 11m x~c SF (x) dx converges. then 1) for m > 1 the integral (3) converges. then the in- a I c-x 1m ~-" A t= 00. then b ~ f (x) dx=F(b)-F (a). Integrals of unbounded functions. Integrals with infinite limits. A c-x liIJ when x -+ c. b] such that F' (x) = f (x) when x rI: c (generalized antiderivative). . When c = a or c == b. Test the following integral for convergence: (5) Solution. Example 2. Test the convergence of the probabllity integral (4) Solution.b +e.x2 dl:+ Se. 5ax-= 1 x2 - lim 2 -+ 0 J -52 2+ dx lim x 2 -+ 0 -1 5' dx . When' x-+-+ 00.Definite Integrals 144 (Ch. (1 )=00 2= 11m --1 ) + l'1m (1 --1 e x 2 -+ 0 8 e~0 8 and the integral diverges. while the second one converges. b GO 51+x dx o 2 = lim b-+GO 5I +x = lim dx 2 b-+(I) (arc tan b-arc tan O)=~ • 0 Example 3. we have 1 1 VxI+1 yxa(l+.J )=e. the integral (4) converges. 0 0 1 The first of the two integrals on the right is not an improper integral. 1 b-+~ J b-+GO 1 hence. since e. Example 4.a) Since the integral converges.~ ~ e. Test for convergence the eltiptic integral 1 dx S VI-x o 4 • (6) .1 . Example 5.x2 dx. We put 1 ~ ~ ~ e.xl dx= ~ e.x when x ~ 1 and (I) S b e-Xdx= Urn \ e-xdx= lim (-e.5 Example t. our integral (5) likewise converges. 5 5 I dx VI-xl • 1558. ' 2x: (I-x)" < 1. 5 dx tx· 5 5 1 (X) 1554.I mproper Integrals Sec. . d:.. for x -. Sx In· dx x • QI) 1559. /(I-x). -1 dx x!+4x+9 • -00 1 (X) xP • 1556. I 1561. . 5dx 1547. Hence. 1 we have Xl Since the integral converges. Applying the Lagrange formula we get I I VI-x. 1555. dx 1 (a> I). ~ cot X dx. 5 1557. 5 dx In 2 dx (x -1)2· x 0 x • 0 1 1 1550. 1552. 0 0 1 • 1549. 0 GO 1551. 0 (a> 1). Sx Inlx 1 a SdxXl • GO n GO 1553. . S7.4x a = V where x < 1 !. the given integral (6) converges as well. ~ sin xdx. Evaluate the improper integrals (or establish their divergence): 1546. S x dxInx 1 a GO dx 1560. = . -e» 0 I dx I +x l · (X) d : • 5 1548. 3) 145 Solution... The point of discontinuity of the integrand is x= 1. S l-x4 o 1571. Definite Integrals J46 CIO CIO 1562.x<. function continuous together with its derivative <p'(t) over a. S o xdx Yx + 1 • 5 1574*. dx S-'-----5xa· o 1566. 1575*. Change of Variable in a Definite Integral If a function f(x) is continuous over a<. o o CIO 1 Sa~t~nt dx. q) = ~ xp . 1 ex> Xl n I co 1570.b and x-q>(t) II. . and f [q> (t)) is defined and continuous on the Interval a<t<p.t". ~ e1563. 4. dx 5x'+ 1 • 1565.. Sec. kx (k dx > 0). 1572. X - o CIO 1564.[Cia.Pl where a==-cp(a) and b= cp (~). 5~ d x. Proye that the Euler integral of the fil st kind (betafunction) 1 B (p. 5(XI~ 1)1 • I integrals~ Test the convergence of the following 100 I dx 1567. I 5l:XX · 1573.. ) Vx+2 V x+x' · dx V=::e. Prove that the Euler integral of the second kind (gamma-function) QI) r (P) = ~ xP-1e- X dx o converges for p > o. .t (l-x)q-t dx o converges when p > 0 and q > O. For the integral b 2 l a x= sinh t. 4) 0/ Variable in a Definite Integral Change then 147 fJ b ~ f (x) dx = ~ f [q> (t)) q>' (t) dt. Then t=arcsin~ a P= and. Find (a> 0). t 580. S(I-cos 4!) dt = 2 0 0 a4 n ( 1 =8 t -4 sln4t ) 2 1 nat =16· o 1576. SYxdx+l -•. 1 Sf (x) dx (b > a) 's" 1579. 1581. x=2t-1. dx=a cos t dt. we can take a=arcsinO=O. Solution. ~ Vx+ Idx. a a Example 1. o I 1578. we shall have n a Sx 2 2 ~ x2 dx= Va 2 o a2 sin 2 t Ya 2 -a 2 sin 2 tacostdt= 0 =a& n n 2 2 Ssln 2 2 t cos t dt = ~& o n I Ssln 2t dt = ~. dx S YI-x· ' X= sin t. I I S f (x) dx. . consequently. arc sin 1= . Therefore. We put x=asint. x = arc tan t. t . Can the substitution x = cos t be made in the integral 2 ~ Vl-x1dx? o Transform the following definite integrals by means of the indicated substitutions: n a 1577.Sec. VI 15S8. dx. t Evaluate the following integrals by means of appropriate substitutions: 1 1587. 2/ 3 (x-2) I + S • dx.. 1 o 1 In S(l ~XX.5 ~Xcos X • o -I 1595. as a result of which the limits of integration would be 0 and 1t respectively. Sx Y x dx+5x+l • l a 1593. x-2=z'. 1nl 1584. Prove that if f (x) is an even function. ~ Vax-x· dx. then a a -a 0 ~ f(x) dx = 2 ~ f(x) dx. . x lr-- eX reX-l eX dx. S o I 2x+ dx Y3x+ 1 • Evaluate the integrals: s 1591. 5y~ S o 1589.Defil1ite Integrals 148 [Ch. 2/ (x-2) . dx S1+ V-' x o 18 1583. 1592. Applying the indicated substitutions. 1 +a sin1x S o l tan x = t.)I • 1594. SVex-l dx. Ins S ~I dx. +3 I 1590. o n t 585 · dt S3 + 2 cos t t o n I dx 1586. evaluate the following integrals: 4 1582. S. 5 indicate an integral linear substitution x=at + p. X'dx=2 5e-X'dx= se-. ~lnxdx. . 0 1605. ~ e. 5] But if Integration by Parts f (x) 149 is an odd function. 0 e 1600.Sec..Ssin -x xx d · arc cos x - S o 0 1598. 0 11. Show that n 1 I -dx. 6. co ~ eX sin xdx. ~ xe-xdx. I I ~ f(sinx)dx= ~ f(cosx)dx. Show that n 2!. Show that ~ ~ ~ 0 0 5e.ax cos bxdx I 0 1 co Sx'elXdx. evaluate the following integrals: n co I 1599. ~ e -axSIn bxdx 0 (a> 0). 1601. then a ~ f (x)dx = o. S x cos xdx. o 0 Sec. -0 1596. 1602. Integration by Parts If the functions u (x) and v (x) are continuously differentiable on the Interval [a. -~ V 1597. bl.dX. 1603.~ a a a v (x) u' (x) dx. (a> 0). (I) Applying the formula for integration by parts. then b b ~ U (x) v' (x) dx=u (x) v (x) b 1. 1604. ~ f (x) dx . If ... Applying repeated integration by parts. Sec.(x) and q> (x) are continuous for a ~ x m then b ~ b and.(x) b ~F(x) for a~x~b. (2) a where m is the smallest and M is the largest value of the function f (x) on the in terval [a.Definite Integrals 150 [Ch. From this derive that r (n + 1) = nl. then b ~ f (x) Ip (x) dx 0. evaluate I. besides. q)=~xP-J(l-x)q-Jdx. Evaluation of integrals.m = ~ sin m x cos n x dx. . if n is a natural number. and 110 • 1608. b ~ a b Ip (x) dx. (I) a If . lll. M ~ a Ip (x) dx. 6. ~ a F (x) dx..e.. evaluate the integral (see Example 1574) 1 B(p. o if m and n are nonnegative integers. Show that for the gamma-function (see Example 1575) the following reduction formula holds true: r (p + 1) = pr (p) (p > 0).e.. q> (x) ~ 0. Show that for the integral I ! I n= ~sinnxdx= ~ cosnxdx o 0 the reduction formula 1 n-l l n-I n=-n- holds true..5 1606*·. Mean-Value Theorem 1°. 1609*. Using the formula obtained. Find In' if n is a natural number. o where p and q are positive integers. Express the following integral in terms of B (betafunction): 2 I n... 1607.. we have that is.. x <. 1.. Evaluate the integral !!. o Solution. The mean value of a function. 2°. Example l.... ~ f (x) dx. respectively. /= ~ y 1+ ~ sin' x dx. sln l x E:. 1610*. I.. Determine the signs of the integrals without evaluating them: I a) ~ x' dx.91. o In C) SSi~X dx. (3) a The inequalities (2) and (3) may be replaced. if cp (x) == 1. = b 1 a Sf (x) dx a Is ca lied the mean value of the function f (x) on the interval a <. then b m (b-a) EO. b. by their equiva· lent equalities: b b ~ f (x) <p (x) dx = f (c) ~ a <p (x) dx a b ~ f (x) dx = f (~) (b-a).. Since O. -1 n b) ~ x cos x dx. 6) 151 In particu Jar.. o .57 < I < 1.c.Mean-Value Theorem Sec. The number b . M (b-a). a where c and ~ are certain numbers lying between a and b... prove that 5~d <_I_ 200n 0< lOon x x lOOn' . Determine (without eval uating) which of the following integrals is greater: 1 a) J SVI +x l ~ or dx 0 J b) 1 S Xl sin I x dx ~ x sin l xdx. -n~x~3t. SxVtanx. f(x)=a+b cos x. 1613.70. 1614. O~x~rc. Evaluate . 5sin x dx. 1620*. O~x~rc. I 1621. b 1611.Definite Integrals 152 [Ch. f(x)=x l o~x~ . Integrating by parts. )(. n -I & In 1619. or J I Find the mean values of the functions on the indicated intervals: 1612.67 and .the integrals: 11. or 0 0 I c) dx. 1 1616. 0 I Se%'dx Se"dx. Prove that 5V o ~ dx 2+x-~ · bt 1 ~ lies e ween -32 ~ 0. r- y2 0.. Find the exact value of this integral. o o n +1 1618. 58~XI. 1615. 1. 5lO+~cosx· o 1622. f(x)= sin! x. 1 1617. f (x) = sin 4 x. SV 4 t ~ Xl dx. and the x-axis (Fig. The sought-for area is expressed by the integral b X . Area in rectangu lar coordinates. the area of the curvilinear trapezoid bounded by this curve. 43 Solution. 7] 153 Sec.. (z) -2 Fig. by two vertical lines at the y 1( 0 a b 0 X Fig. (1) a 2 x Example 1. 40 J X Fig. 41 points x=a and x=b and by a segment of the x-axis is given by the formula b S= ae=..T he Areas of Plane Figures Sec. The Areas of Plane Figures 1°. 7. Sf (x) dx..x~b (Fig.. 40). 41). y y f x) I I I 0 g::f. 42 Fig. If a conti nuous curve is defined in rectangular coordinates by the equation Y=f (x) [I (x) ~O]. Compute the area bounded by the parabola y = -2 ' the straight I ines x = 1 and x -= 3. r y x o a Fig. where 11 (x) <af. Here. Evaluate the area S contained between the curves y=2-x2 and yl =x.=-1 and x 2 = 1. 44). (2-x -x2/ 8) dx= ( 2x-3 5 -I 15 S -1 If the curve Is defined by equations in parametric form x = q> (t). 42). By virtue of formula (2). Y = "I' (t) then the area of the curvilinear trapezoid bounded by this curve.3 xa) 1 =22 . the roles of the coordinate axes are changed and so the sought-for area is expressed by the integral s= • S(2-y_ y2)dy=4 ~. 43).(x») dx. Evaluate the area bounded by the curve x=2-y_ y l and 1he y-a xis (Fig. Solution. (3) (Fig. by tw ~ . Solution. (x) when a<x~b (Fig. Solving the set of equations (3) simultaneously. -I where the limits of Integration Yl = -2 and Y. we wilt then have: b S= S (2) [f2 (x)-f. we obtain 1 s= • I x' . if the area S is bounded by two continuo us curves y ='1 (x) and y = f 2(x) and by two vertical lines x =a and x = b. 44 x Fig. a Example 3.154 Definite Integrals [Ch. we find the Jinlits of integration: x. = 1 are found as the ordinates of the points of intersection of the curve with the y-axis. 5 Example 2. 45 In the more general case. are determined from the equations a=cp(tt) and b=cp(t 2 ) ['P(t)~O on the interval [tt.2 = a Z cos 2q> Find (Fig. and by t\\'o radiUS vectors OA and OB. Solution. 45) by using its parametric equations x=. Due to the synlnletry. Find the area of the ellipse (Fig. . hence. o~-._---X Fig. 46)... S=Mb. If a curve is defined in polar coordinates b~' the equation r =-= f (cp). n o ~ S 2 = ~ b sin a (- sin t) dt = ab n Ssin! t dt = ~b 0 2 and. a Example 5. The area in polar coordinates. 7] of Plane Figures vertical lines (x =a and x= b).. If in the equation x=a cos t we first put x:=: 0 and then x =a. t z ]]. then the area of the sector AOB (Fig. Example 4. is expressed by the integral f} 1 S = 2" Jr [f (cp)]2 dq>. we get the lilnits of integration i 1 = ~ and t 2 = O. 47). 46 which correspond to the val ues <PI = a and <PI =~. .. t.. 47 Fig..a cos t. 2°. and by a segment of the x-axis is expressed by the integral t3 S = ~ ¢ (t) cp' (t) dt. bounded by an arc of the curve. Therefore. where t 1 and t. the area contained inside Bernoulli's lemniscate . it is sufficient to compute the area of a quadrant and then multiply the result by four.The Areas Sec. { y == b sin t. 1636. the straight line Y= 1 and the vertical line x=8. 1635. the vertical lines x = a and x = 3a (a> 0) and the x-axis. 1632.s-~ . 1629.quadrant of the sought-for area: Whence S::: al. Evaluate the area bounded by the parabola Y= 2x-x· and the straight line y=-x. Y=i and the straight line y=2x. 2 1631. Compute the area bounded by the curve y = In x. 1627. the x-axis and the straight I ine x = e. and the straight line x == 1.. Compute the area of the figure bounded by the curve y=-=x 3 .. Compute the area contained between the witch of 1 Xl Agnesi y= 1+x2 and the parabola y= 2. 5 Solution. Compute the area bounded by a single half-wave of the sinusoidal curve y= sinx and the x-axis. 1638. 1626. Compute the area bounded by the curves ya:ae". Compute the area of a segment cut off by the straight line y=3-2x from the parabola y=x". 1625*. Conlpute the area contained between the curve y=tanx. 1628. By virtue of the symmetry of the curve we deter111ine first one .Definite Integrals 156 [Ch. . Find the area bounded by the curve y3 = x. II-a. Find the area contained between the hyperbola xy=m'l. Find the area bounded by the curve y=x (x-i) (x-2) and the x-axis. Compute the area contained between the parabolas yrc:r. 1633. 1634. Compute the area bounded by the parabola y=4x-x l and the x-axis. Compute the area contained between thQ parabolas x2 2 2 Y="3 and y=4-3"x · 1637. the x-axis and the straight line x=1. Find the area contained between the witch of Agnesi Y= 03 Xl • +a and the x-aXIS. t 630. 1623. Find the area bounded by the parabolas y"c:2px and x" = 2py.x·.. the straight line y=8 and the y-axis. 1624. Find the entire area bounded by the astroid x' +y' =a'.cos t). y= b sin' t. 1640*. Find the area bounded by the curve a2y2=x2(a2_xl). Compute the area contained between the circle Xl + y2 = 16 and the parabola x 2 = 12 (y-l). Find the area contained within the astroid X= a cos' t. the x-axis.a) and a tangent to it at its lower points.. 1648. 1642. x and the ordinate x=l (x>l).y 2=:8 is divided by the parabola y2=2x. Compute the area contained within the curve 1644.4).. COlllpute the area of the two parts into which the circle X 2 _t. 1650. 7) x2 al - T he Areas of Plane Figures 157 1639. y=a-b cos t (O<b:s. 1652. Find the area bounded by the x-axis and one arc of the cycloid X = a (I .Sec. 1645. Fi nd the area of the figure bounded by t"he hyperbola y2 b2 = 1 and the straight line x = 2a.. the x-ax is and the diameter passi ng through the point (5. Find the area bounded by the cissoid y2=2a ' -x and its asymptote x = 2a (a> 0). a :e the y-axis and the straight line Y= (e t + 1). 1643.. Find the area between the strophoid y2 = 2a-x and its asymptote (a> 0). 1641. x (x-a)2 1647*. Find the area bounded by one branch of the trochoid { x=at-bsint. { y = a (1 . Find the area between the equilateral hyperbola X 2 _ y 2 = = 9..sin t). . X 1646*. 1649. Find the area between the curve y=~. Find the area between the catenary Y= a cosh!. 1651. r = aq> (Fig. Solution. 8.2 r Fig. y. Differentiating the equation of the astroid. Y=l+t" 1655*. Find the entire area bounded by the curve = a l sin 4(p. 1654*. The arc length s of a curve Y=f (x) contained between two points with abscissas x=a and x=b is b s= ~ VI +U'2dx. 1658. The arc length in rectangular coordinates. = a sin 3(p. Find the area of one of the leaves of the curve r = a cos 2fp. Find the area bounded by the cardioid { X = a (2 cos t-cos 2/). Find the area bounded by the parabola r = a secl ~ and the two half-lines cp =: and cp =. The Arc Length of a Curve 1°. 1662. Find the area bounded by Pascal's limac. 1659*.=_L. x/ • . 1656*. Find the area of the loop of the foli urn of Descartes 3at 2 3ut x=l+t S . Find the area bounded by the curve x4 + y4 = Xl + yl. Find the area bounded by the curve . 6 1653. 1661. Find the area contained between the first and second turns of Archimedes' spiral. Find the area bounded by the curve ..on . we get II. = 2a cos 3(p and lying outside the circle r = a. 48). Find the length of the astroid x 2/1 y2 /1 :=a 2/a (Fig. 1664*. a + Example 1. 1663.sin 2/). Find the entire area of the cardioid f=a (l+cosq». Find the area of the ellipse f = 1+ ePcos q> (8< 1).Definite Integrals 158 {Ch. 48 = 2 + cos cp. y = a (2 sin t . Sec. 49). 1660. 1657.. 50) X = a (t . x = cp (t) and y = "I> (t). a where a and ~ are the values of the polar angle at the extreme points of the arc. 0 The limits of integration t 1 =0 and t l =2Jt correspond to the extreme poinfs of the arc of the cycloid. . We have dt =a (1. 50 Example 2. then the arc length s is P s= ~ V. 8] 159 For this reason. If a curve is defined by the equation r=f (cp) in polar coordinates.I +. y y x ---1~--------_----I"-'-. dx dy Solution.1 d<p.The Arc Length of a Curve Sec.. { y=a (I-cos t). The arc length of a curve represented parametrically. 2°. tI where t l and t 2 are values of the parameter that correspond to the extremities of the arc. we have for th~ arc length of a quarter of the astroi d: {s= SVI+ ~:. In S = S Val (I-cos 2n t)l+ a l o sln l t dt = 2a Ssin ~ dt =8a. Therefore. Find the length of one arc of the cycloid (Fig."-'-'- o 27Ca Fig 49 X Fig.sin t). then the arc length s of the curve is tJ S = ~ YX'I+y'l dt. If a curve Is represented by equations in parametric form.: o dx= 5:::: dX=~ ~ 0 Whence s = 6a.cos t) and dt = a sin t. 51). 51 Solution. . 1672. Compute the arc length of the parabola y = 2 from x =0 to X= I. dlp = a 3 3 o J. Find the arc length of the curve y = arc sin (e-'¥) from x=O to X= 1. =a sln l ~ (Fig. 1667. Find the length of the entire curve .!. 1670.a) to the point B (b.t = Va to x= VB... 6 Example 3. x Fig. The entire curve is described by a point as cp ranges from 0 to 3n.'= a sin t : cos : ' therefore the entire arc length of Ihe curve is an S = S'Vi a~ a2 sin' . sln . lying Vx between y=O and y= ~ .3 dlp = 3na2· 2 0 1665. . Compute the arc length of the semicubical parabola yl = x· from the coordinate origin to the point x = 4.!. Find the arc length of the curve y = eX lying between the points (0. 1671. Conlpute the arc length of the curve x=lnsecy. I) and (I .. Find the arc length of the curve x= ~ y2_ ~ Iny from Y= 1 to y=e. 1666*. We have . cos 2 . 1669.h)..!. 3 +a 2 sin' .Find the length of the catenary y = a cosh ~ from the a vertex A (O.e). Find the arc length of the curve y = In x from .Definite Integrals 160 (Ch. 1668.!. (r Sec. Find the length of the hyperbolic spiral r<p = I from the point (2"/2) to the point C/2. t 684.2). = 3.1900 . } y=a(2 sint-sin2t). Find the arc length of that part of the parabola . Find the length of the right branch of the tractrix z X= Va - yZ+aln la+ Y~I from y=a to y=b (0< b<a). The volumes of soB4s formed by the revolution of a curvilinear trapezoid (bounded by the curve !I ~f (x). Volumes of Solids 9) 161 1673. Find the length of the curve y= In (coth ~) from x=a to x==b (O<a<b). Find the arc length of the involute of the circle x=a(c?st+tsint). lying inside the circle r = a. 1680.cos 21). Find the length of the curve x ==. 1682. 1679.Sec. 1674. 1683. 9. = a sec!. 1675.and 'y-axes are 6 . Find the arc length of the curve cp = ~ + ~) from r = 1 to . The volume of a solid of revolution.} from t=O to t=T y=a(slnt-tc05t) · 1677. 1681. 1676*. which is cut off by a vertical line passing through the pole.a (2 cos t. Find the length of the evolute of the ellipse 1678. ·tbe x-axis and two vertical lines x = a and x = b J about the x. Find the length of the first turn of Archimedes' spiral r = atp. Volumes of Solids to. Find the entire length of the cardioid r=a(l+cos<p). Find the arc length of the logarithmic spiral r = aemcp . Find the length of the closed part of the· curve 9ay l = = x (x-3a)l. c obtained fronl formula (I). 52). by the formulas: b b ~ yldx. x=b. about the y-axis. Compute the volumes of solids formed by the revolution of a figure bounded by a single lobe of the sinusoidal curve y = sin x and by the segment 0 < x ~ n of the x-axis about: a) the x-axis and b) the y-axis. the y-axis and by two parallel1ines y=c and y=d. equal to b Vx=n ~ (y:-y:) dx a and b Vy=2n ~ x (YI-YI) dx. a Example I. In the Inore general case.). etc. may be determined from the formula d Vy=n ~ xldy. respectively. *) The solid is formed by the revolution. CI .and y-axes of a figure bounded by the curves Yl ='1 (x) and Y2 = f 2(x) [where fl(X)~f2(X)]. o :t b) Vy=2n ~ xsinxdx=2n(-xcosx+sinx)~=2nl. then in the foregoing formulas we must change the variable of Integration in appropriate fashion.[Ch. If the curve is defined in a different form (parametrically. o The volume of a solid formed by revolution about the y-axis of a figure bounded by the curve x=g(y). given above. of a curvilinear trapezoid bounded by the curve y=f (x) and the straight lines x=a. n a) VX=-n Ssin 2 n2 xdx=T. in polar coordinates. and the straight lines X=Q and x=b are.6 Definite Integrals 162 expressed. respectively. For a volume element we take the volume of that part of the solid formed by revolving about the y-axis a rectangle with sides y and dx at a distance x from the y-axis. the volumes of solids formed by the revolution about the x. Then the volume element dV y =2nxydx. by interchanging the coordinates x and y. I) Vx=n 2) V y =2n a ~ xydx*). a Example 2. Find the volume of a torus formed by the rotation of the circle x 2 + (y-b)1 =a2 (b ~ a) about the x-axis (Fig. whence b Vy=2n ~ xydx. and y=O. Solution. Example 3. 9] Solution.Va 2-xl )2J dx = -a a ~ Val-xl dx = = 4nb 211. about the polar axis.Val-xl and Y2=b+ Therefore.na'. 2 . n 5.Volumes of Solids Sec. Solution.~ n n 5 I . Vx=n ~ [(b+ Val-xl)I-(b. y -Q o X x a Fig 52 Fig. cp == ~ may be computed froln the formula p Vp = . We have Yl ==b. o Thls same formula is conveniently used when see{<ing the volume obtained by the rotation.1 alb -a (the latter integral is taken by the substitution x =a sin t). == a sin 2cp about the polar axis. Determine the volume formed by the rotation of the curve . of a sector formed by an arc of the curve r == F (rp) and by two radius vectors ~ =-~ a. of some closed curve defined in polar coordinates. n Vp= 2. 105 21p sin Ip dip = . about the polar axis.1 sin Ip dip = ~ o I na'5 siu l 0 n 5 I = ~ na S 3 sin 4 ep cos' (p dq> o == ~ . 53 The volunle of a solid obtained by the rotation. u Va -x2.1 sin Ipd Ip. . 1691. about the x-ax is. Find the volume of a solid formed by the rotation. Find the volume of a solid formed by the rotation.t. 1690.Definite lntegrals 164 [Ch. The radius of the base is R (Fig. of an area bounded by the catenary y = a cosh!. Solution. If S = S (x) Is the cross·~€ctional area cut ofT by a plane pprpendicular to some strai~ht line (which we take to be the x-axis) at a point with abscissa x. Be = 2"' yy tan a = Y2 tan Q. Find the volume of a solid formed by rotation. The equation of the circumference of the base is Xl y2 = R2. of an area bounded by the semicubical para bola y2 = x=s. about the x-axis. Computing the volumes of solids from known cross-sections. about the y-axis. 1687. 1686. y=O about: a) the x-axis and b) the y-axis. Find·. then the volume of the solid is XI V= ~ S (x)dx. and for the y-axis \ve take the dialneter of the base perpendicular to it. Therefore.1-=1 about the x-axis. Find the volume of a solid formed by the rotation. . of that part of the parabola yl = 4ax which is cut otT by the straight line x = a. 53). and the straight lines x == ±a. 1689. about the x-axis. the a x-axis. 1692. x=O. D~termine the volume of a wedge cut oft a circular cylinder by a plane passing through the diampter of the base and inclinpd to the base at an angle u. Find the volume of a solid formed by the rotation.. Find the volume of an ellipsoid formed by the rotation Xl y2 of the ellipse aa-+/. about the x-axis. Example 4. Find the vol ume of a sol id formed by the rotation of the same area (as in Problem 1689) about the y-axis. The area of the spction ABC at a distance x from the origin 0 is I 1 2 S (x) = area ~ ABC = 2" AB. of the curve y = sin 2 x in the interval between x = 0 and X=1t.he volumes of the solids formed by the rotation of an area bounded by the lines y=e x . of an area bounded by the x-axis and the parabola y= ax-r (a >0). and the straight line x == ]. Xl K 1 and x! are the abscissas of the extreme cross-sections of th~ solid. For the x-axis w? take th~ diam~ter of the base along which the cutting plane intersects the base. 5 2°.. the x-axis. 1688. the soughtfor vol urne of the wedge is where + V=2· ~ R R SyZtanadx=tana S(RZ-xZ)dx= ~ tanaR z• o 0 1685. k whose p~rallel bases are rectangles with sides A. 1696. 1700. 1701. Find the volume of a paraboloid of revolution whose base has radius R and whose altitude is H. and c) the axis of symmetry of the figure. Find the volume of a solid formed by the rotation of the astroid x=acos't. Find the volume of a solid formed by the rotation. 1699. h. about the x-axis. b) the y-axis. of a loop of the curve (x-4a) y2 = ax (x. and the altitude is h. 1695. Show that the volume of a part cut by the plane x = 2a off a solid formed by the rotation of the equilateral hyperbola x z _ y2 = a" about the x-axis is equal to the volulne of a sphere of radius a.3. Find the volume of a solid formed by the rotation of a figure bounded by one arc of the cycloid x=-a (t. Find the volume of a solid obtained by rotating the cardioid r=a(l +coscp) about the polar axis. A right parabolic ~egment \vhose base is 2a and altitude h is in rotation about the base. 1707. a -x 1698. which ar. of a figure bounded by the parabola yl = 2px and the straight line x = ~ . Find the volume of a solid generated by the rotation of the c)'ssoid y2=-2 AI about its asymptote x=2a. 1697.sin t).e parallel to the x-axis. Find the volume of a solid formed by the rotation. 1702. On the chords of the astroid x2/a+y"J/3. Find the volume of a solid formed by the rotation. of that part of the parabola yJ. Find the volume of a right elliptic cone whose base is an ellipse with semi-axes a and b. about the straight line y = . De~ermine the volume of the resulting solid of revolution (C:lvalieri's "lemon").Sec. y=a(l--cost) and the x-axis. 1694. = 4ax which is cut 011 by this line. are constructed squares whose sides are equal to the lengths of the chords and whose planes are perpendicular to the xy-plane. . Find the volume of a solid formed by the rotation. 1703. Find tHe volume of the solid formed by these squares. about the straight line x = a. B and a. 9] Volumes of Solids 165 1693. about~ a) the x-axis. of the area contained between the parabolas y = x2 and y= V"t. 1704. Find the volume of an obeli<.p. Find the volume of a solid formed by rotation of the curve r=acos 2 cp about the polar axis.2) (a >0). about the x-axis. 1706. 1705. and altitude h. y=b sin' t about the y-axis.=a2/J. . whi Ie its vertex slides along a straight line parallel to the stationary diameter at a distance h from the plane of the circle. The Area of a Surface of Revolution The area of a surface formed by the rotation. The plane of a moving triangle remains perpendicular to the stationary diameter of a circle of radius a.&.eg~ra_l_s _16_6 [Ch.. 1710. Find the volunle of the segment cut off froIll the elli py2 22 tic paraboloid 2p-l-2q=x by the plane X=Q. The base of the triangle is a chord of the circle. the centre describes an ellipse . 1711. C 2 22 x 2+ b! +2"= 1. about the x-axis.fi_n_ite_In_t-:. Find the volume of the ellipsoid a y2 Sec. Find the volume of the solid bounded by the hyperbox2 y" Z2 loid of one sheet a2+b2 -2"=1 and the planes z=O and z=h. A circle undergoing deformation is moving so that one of the points of its circumference lies on the y-axis. 55 I f the eq ua lion of the curve is represented differen t1 y. 1712.Zl = at and y2 + Zl == al..---__ --L o no 21Ca ___ X Fig. Find the volume of the solid (called a conoid) formed by the motion of this triangle from one end of the diameter to the other. c 1713. y . Find the volume of the solid generated by the circle. and the plane of the circle is perpendicular to the xy-plane. 54 -.. Find the volume of the solid bounded by the cylinders x" -t. .: +~= I. is expresse-d by the form uta b b Sx=2n Syd.D_e.. 5 1708. of an arc of the curve y == f (x) between thE' poi nts x = a and x = b. 1709. 10.x dx =2:t SY VI+y'ldx a (I) u (ds is the differen tial 01 the arc of the curve)..:. the area of the surface Sx is cbtained from formula (1) by an appropriate change of variables.y 2a o x Fig. 55). Find the area of the surface formed by the rotation of a part of the tangential curve y = tan x from x = 0 to x about the x-axis.Sec. from x = 0 to x == =: ' +00. t t t =4na 2 [ -2ncos2+2tcos2-4sin2+3s1n'2 0 3 1714. 56. From for- yx. we have y= . Find the area of a surface formed by rotation. Solution.~ dx. we y will have 2a S = 2n ds (Tea-x) dy • dYe 5 o Passing to the variable t we obtain J n S =2n j" (no-ot+a sin t) y (~~r + (~~r dt o =- o n 5 5(n = 2n (no-at + 0 sin t) 2a sin ~ dt = o n = 4no l o sin ~- t sin ~ + sin t sin ~) B FIg.. Solution. about the x-axis.. of an arc of the curve y::::z e-~. 1716. For the upper part of the curve. The dimensions of a parabolic mirror AOB are indicated in Fig. 56 tJ'" =8Jt (n. Find the area of the surface formed by rotation. It is required to find the area of its surface. o Example 2. dt = 4. 3. The desired surface is formed by rotation of the arc OA about the straight line AB.r. about the x-axis. of a loop of the curve 9y 2 =x (3-X)1 (Fig. the equation of which is x=na.4) at. mula (I) the area of the surface Jr31 a S = 2n . when 0 e=.x) Whence the differential of the arc ds= . . (3. 54). y==a (I-cos t) about its axis of symmetry (Fig. 10] 167 The Area of a Surface of Revolution Example 1.. 1717. Taking y as the independent variable and notIng that the axis of rotation AB is displaced relative to the y-axIs a distance na.x+ 1 (3 -x) r x 2 Yx dx = 33t. Find the area of the surface of a spindle obtained by rotation of a lobe of the sinusoidal curve y= sinx about the x-axis. x <. Find the area of a surface formed by the rotation of one arc of the cycloid x==a (t-sln t)... 1715. about the x-axis from a X = 0 to x === a. 11. Find the area of the surface forlned by rotation of the ellipse ~:+~2=1 about: I) the x-axis. thost' on th~ other side have the minus sign. Moments. Sec. F·ind the surface of a torus formed by rotation of the circle x 2 +(y_ b)2-==a 2 about the x-axis (b>a). . Guldin's Theorems to.=md. 1723. Find the area of the surface of rotation of the curve x={ y2_ ~ loy about the x-axis from y= 1 to y=e. m2 . 1725. b) the y-axis. Determine the area of the surface formed by the rotation of the lemniscate r 2 == a l cos 2cp about the polar axis. For the cases of geometric figures..cos t) about: a) the x-axis. = 2a (1 + co~ cp) about the polar axis. 1720. then the Rtatic moments Mx and My about 1he x.sin 2/). The static moment relative to the i-axis of a material point A having mass m and at a distance d fronl thz I-axis is the quantity M. 1726. If the massps continuously fill the line or figure of the xy-plane.6 1718. Find the area of the surface formed by rotation of one arc of the cycloid x = a (t . Fincl the area of the surface forlned by rotation. Static moment. d z . 1721 *.and y-axes are exoressed . Find the area of the surface (called a catenoid) formed by the rotation of a catenary y = a cosh!. about the x-axis. In a similar manner we define the statIc moment of a system of potnts rt'lative to a rlane. Determine the area of the surface formed by the rotation of the cardioid.sin t) and !J = a (1. 1722. dn is the sum n M l = ~midi' (1) I=J where the distanct's of points lying on one side of the l-axis have the plus sign. Centres of Grav:ty.. the density is considered equal to unity.respectively) a~ integrals and not as the sums (1). c) the tangent to the cycloid at its highest point. } y = a (2 sin t. The static moment relative to the I-axis of a system of n material ('oints with Inasses ml . mn 1ying in the plane of the axis and at distances d1 . ••• . 1724. ••• . Fi nd the area of the surface of rotation of the astroid a2/3 about the y-axis.168 Defin ite Integrals lCh. 2) the y-axis (a>b). of the cardioid %2/8 _~ y2/3 == x=a(2cost-cos2t). 1719. . Find the statIc th~ x-axis and two b My = a Example 1. 57) bounded by the straight lines: ~ Solution. (3) a monl~nts about the x.and y-axe3 of a triangle +f = 1.b 0 Fig.: ) . My=~X(S)dS o Y (dx) 2 + (dy)2 (ds = 169 (2) 0 is the differential of the arc). Here. 58 Fig.=-=-~ nllj~. mil IS the surn Il J. The nl0mellt of Inertia. about an l-a'Ci5.• • __ . vertIcal ltnes x= a and y.. l . Moment of inertia. I') the llulnber II-=. or a rn1fcrial point of rna~s m at a dIstance d fronl th~ l-aXl'i.g formula (3). about an I-axis. X ) J xli-a- dx= a2 b ti- o 2°.==. we obtain and a My=b {.fnd 2• Th~ moment of cnertia. o[ a systel1l of n nlalenal pOints with masses mJ . y = y (5). m:. 11] In particul ar: 1) for the curve x = x (s).b.Moments. CentrtS of Gravity. !---~ T---~ v b o J x a t--. Guldin's Theorems Sec. Is the arc length. x I --I· Sx r y I dx. we obtain b MX = ~ 5Y I y I dx. Applyir.. i . 57 2) for a plane fi~ure bounded by th~ curve y = y (x). y=b ( 1. y=O (FIg. x::=O. we have L wh~re the parameter s L Mx=~y(s)dS. b o 3°. Find the moment of inertia of a triangle with base band altitude h about its base. For the base of the triangle we take the x-axis. b Sy2 dx a y = --S- b where S = ~ Y dx is the area of the figure. we obtain h-y dm=b --dy h and b dl x =y"dm=7i y2 (II-g) dy. d l ••• . we get an appropriate integral in place of a sum. we have B ~ - static moments of the mass. f (a)l and B [b. Example 2. (y ~ 0) (Fig. the centre of g. Find the centre of gravity of an arc of the semicircle x 2+ y2 = a 2.5 where d1 . dn are the distances of the points from the I-axis. the y-axis (Fig 58).-5. f (b)]. 59). In the case of geolnetnumerically equal to the corresponding arc or area. Example 3. Utilizing the similarity of triangles. b X d!> A x= -. for its altitude. In the case of a continuous mass. The coordinates of the centre of gravity of a plane figure (arc or area) of mass M are computed from the formulas - My - Mx x=M' y=¥. Solution.= ~ B X b ~ Y 1 + (y')Zdx a - yds A y=--s ~b------ ~ Vl+(y')2dx a ~ y y 1 + (y')2 dx II b ~ V I + (y')' dx a The coordinates of the centre of gravity (~ Y) of the curvilinear trapezoid a ~ x ~ b. which play the role of elementary masses dm. where M x and My are the ric figures.ravity (~ Y) of an arc of the plane connecting the points A la. the mass M is For the coordinates of curve y=f (x) (a ~x ~b). Whence h r 1 I X--"= Ii J y2 (h-y) dy = 12 biz'. Divide the triangle into intlnitely narrow horizontal strips of width dy.. 0 <.Definite Integrals 170 [Ch.. a There are similar formulas for the coordinates of the centre of gravity of a volume. Centre of gravity. . y ~f (x) Inay be computed from the formulas b ~ Sxy dx - a x=-S. Centres of Gravity Guldin's Theorems 171 Solution. The area of a surface obtaIned by the rotation of an arc of a plane curve about SOIne axis lying in the same plane as the curve and not intersecting it is equal to the product of the length of the curve by the circunlference of the circle described by the centre of gravity of the arc of the curve.\' '-a 2 '" -a Z _udx - Va -x Z 2 _: 2a 2 . 59 1727. 2 Ya z x! and ds ~r y --=. Guldin's theorems. Theorem t. Theorem 2. -a l-1ence. 4°. a -a x Fig. x~-O.= Whence a u ' f ax J ds =-= Ya -x d M y::-:= X 2 Z t ---= 0. Yal-x! + (y')! dx. 1 _adx . a M- C adx J -V-a--x-' Z 2 na. . Find the static moments about the coordinate axes of a segment of the straight line ~+1[=1 a b » lying between the axes. We have Y -== Y-a-_-x- y' == _-_x_ 2. The volulne of a solid obtained by rotation of a plane figure about sonle axis lying in the plane of the figure and not intersecting it is equal to the product of the area of this fi~ure by the circumference of the circle described by the centre of gravity of the figure.:. OJ -a a -a (l Ya -x Mx== I'"' yds--. Y=-= 2 Fa.Sec J11 Moments. Find the centre of gravity of a hOlno~eneous right circular cone with base radius r and altitude h. 1740**. 1729. 1739**. 1732. Find the static moments of a rectangle. Find the centre of gravity of an arc of a circle of radius tJ subtending an angle 2a. about its sides. Find the coordinates of the centre of gravity of an area bounded by the first arch of the cycloid x=a(t-sinl). about the x. 1738**. 1737.Definite Ifltegrals 172 (Ch. I 1728. 1730. and the coordinates of the centre of gravity of a triangle bounded by the straight lines x+y=a. . with sides a and b. 1734. Find the static moment of the circle r = 2asin q> about the polar axis. about the x. Find the centre of gravity of a honl0~eneous hemisphere of radius a lying above the xy-plane with centre at the origin. x=O.b2 = 1 and the coord i na te axes (x ~ 0. y=a(l-cost) and the x-axis. and the coordinates of the centre of gravity of an arc of the astroid I I I x'-ty'=a'. 1731. y = a (I-cos t).and y-axes. lying in the first quadrant. Find the coordinates of the centre of gravity of an area bounded by the curves 2 r- y=x. 1733. Find the coord inates of the centre of gravity of the arc of on~ arch of the cyeloid X= a (t. Find the coordinates of the centre of gravity of an area x2 LI I bounded by the ell ipse Q2 -1.sin t). Fi nd the static momen~s. Y=~ x. I 1736. Find the coordinates of the centre of gravity of an arc of the catenary x y=acosh -a from x= -a to x=a. and y=O.and y-axes. Find the centre of gravity of a hemisphere of radius a lying ahove the xy-plane with centre at the origin. 1735. Find the static moments. y~O). Find the moments of inertia of the area of the ellipse ~: + ~: = 1 about its principal axes. 12.lloll~ some curve and the Clhsolute value of th~ velocity u =-1 (t) is :J known fUI1(uon of the time t. Find the path s covpred by the point In the interval of time T = 10 ~ec followIng the conlOlc. 1749. What is the Iuean velucity ~f mollon durlng lhis interval? . Find the mOlllcnt of inertia of a right parabolic segment with base 2b nnd altitude h about its axis of sYlnrlletry. tIl is Example 1. Find the moment of inertia of a circle of radius a about its d iarne~er. Applying Definite Integrals to the Solution of Physical Problems 1°. The velocity of a point is (J=O. 1745**.Sec. b) Find the Cl'n~rc of gravity of an area bounded by the curves 2 y2 = '2px and x == 2py.<R 2 ») that is. 1746**. 1748. 17fiO**. b) Prove by Guldin's theorem that the centre of gravity of a triangle is distant from its base by one thlrd of its altitude Sec. Find the moment of inertia of a homogeneous right circular cone \\lith base radlus R and altitude H about its axis.l. 121 ApplginR Definite lnieRrals to Solution of Phl/sical Probltms 173 1741. Find the mOITlents of inertia of a rectangle with sides a and b about its sides. 1747**. 1744. 1743. (R. a) Dcterrnine the po~ition of the centre of gravity of 2 2 l an arc of the astroid x 3 -~ {/3 =a7 lying in the first quadrant.· m/~ec. Flnd the Inonlent of inertia of a hotnogeneous sphere of radius a and of rnass M about its diaillcter.celnent 01 motion. Flnd the polar monlent of inertia of a circular rinq with radii R 1 and R. The path lraverfied by a point. 1742. then the path traversed by the pOint in an Interval of time (t l . the nlonlent of inertia about the axis passing through the centre of the ring and perpendicular to its plane. If a point IS In motion . a) Find the centre of gravity of a sen~icircle usin~~ Guldin's theorerTI. Find the surlace and volullie of a torus obtained by rotating a cIrcle of radIus a about 8n aXIs lying in its plane and at a distance b (l> >a) frOll1 its centre. Kinetic energy.lt'dt=O. v n ' is equal to n K= I L m~Vi.OI m and X=l kgf. in the limit.[Ch. mn having respective velocities VI' VI' ••• . The kinetic energy of a material point of mass nz and velocity v is defined as The kInetIC energy of a system of n material points with masses ml . the elemenfary . Example 3. Putting x=O. According to Hook's la\v the force X kgf stretching the spring by X m is equal to X == kx.'ro 2 h6 dr. We have: s= 10 t4110 O.S Definite Integrals 174 Solution. an integral in place of the sum (I). We have: dm = 211:' . =1 To compute the kinetic energy of a solid.= n. ••• . Find the kinetic energy of a homogeneous circular cylinder of density tJ with base radius R and altitude h rotating about its axis with angular velocity roo Solution. then by summing the kinetic energies of these particles we get. 60). 2°.-2. (I) I. then the work of this force over an interval [XI' x 2 ] is X2 A= Sf (x) dx. V = fro. Xl Example 2. ma. we get k=lOO and. The work of a force. where k is a proportionali ty constant.06 A= ~ 0. Whence the sought-for work is 0. Since the linear velocity of the mass dm Is equal to ki net Ic energy is v 2dm dK =. What work has to be performed to strC'tch a spring 6 Cln.h6 dr. if a force of 1 kgf stretches it by 1 cm? Solution. If a variable force X = f (x) acts in the direction of the x-axis. X==IOOt. the latter is appropriately partitioned into elementary parttcles (which play the part of material points). For the elementary mass dm we take the mass of a hollow cylinder of altitude h with inner radi us r and wall thickness dr (Fig.lT 0 =250 metres 5 o and V mean S = y= 25 mjsec. hence.06 =0.18 kgm 100xdx=50x21 o 0 3°. 60 FIt! 61 Example 4. 4°. The velocity of a body thrown vertically upwards with initial velocity Vo (air resistance neglected). The area of one such element (ignoring higher-order infinitesimals) located at a distance h frotn the surface is ds =-= 2xdh = 2 V.2 - h2 dlz. where y is the speciflc weight of the water equal to unity. where V is the specific weight of the liquid. To compute the force of liquid pressure we use Pascal's law. ~ S o 3 noo2fJR4h 4 . 12] Applying Definite Integrals to Solution of Physical Problems 175 Whence R K = 3tCl)2hfJ r d. which states that the force of pressure of a liquid on an area S at a depth of immersion h is p=yhS. Fi nd the force of pressure experienced by a selniclrcle of radiu5 r submerged vertically in water so that it5 diameter is flush with the \vater surface (Fig 61). is given by the . Solution. Fig. Whence the entire pressure is 1751. Pressure of a liquid. The pressure experienced by this elernent is dP =yh ds = 2yh V . We partition the area of the semicircle into elements-strips parallel to the surface of the water.2-h" dh.Sec. 1755. o't m/sec.":. What work has to be done to raise a body of massm from the earth's surface (radius R) to an altitude h? What is the work if the body is removed to infinity? . where t is the time and [Jo. l"he velocity of a body thrown vertically upwards with initial velocity Vo (air resistance allowed for) is given by the fornlula v=c. g is the acceleration of gravity. 1758. Calculate the w. 1753. Calculate the work that has to be done in order to pump the water out of a conical vessel with ver1ex do\vnwards. and c is a constant. CJ) are constants. What is the nlean value of the absolute magnitude of the velocity of the point during one cycle? 1754. Considering that when the rocket thrust is constant. find the velocity at any instant of time t. Find the altitude reached by the body.. Calculate the work to be done in order to pump wateJ out of a semispherical boiler of radius R =]0 m.17i Ie. 6 Definite Integrals formula v= vo-gt. Find the altitude reached at tim~ t = t 1 • 1756*. At what distance from the initial position will the body be in t seconds from the time it is thrown? 1752. 1757. Find the path covered by the point {roIn the commencenlcnt of motion to full stop. where t is the time that elapses and g is the acceleration of gravity. \vhere t is the time. A point on the x-axis performs harmonic oscillations about the coordinate origin. the length of the tank H and the radius of the base R. 1760**.~ t+arctan ~O). . 1759. Calculate the work that has to be done to pump the \vater out of a vertical cylindrical barrel with base radi us Rand altitude H. the acceleration due to decreasing weight of the rocket increases by the law J = a A bt (a.Jrk ne~dcd to pum~ oil out of a tank through an UP4Jcr opening (lhe tank has the shape of a cylinder with horizontal axis) if the specific weight of the oil is y.o. Find the law of oscillation of a point if when t = 0 it had an abscissa x=O.tan( . its velocity is given by the fornlula v = V o cos (J}t. the radius of the base of which is R and the altitude H. if the initial velocity is zero.. The velocity of motion of a point is v = ie. A rocket rises vertically upwards.bt > 0). 1769. The frictional force between a small part a of the base of the shaft and the surface of the su pport in contact with it is F = Jlpa. ° . Find the work done by the frictional force during one revolution of the shaft. y = 7. and the height h=~O ffi.-. respectively. Calculate the water pressure on the dam if w~ kn)w that the upper ba~e a=70 In. What work has to be don~ to stop an iron sphere of radius R = 2 mc~r~s rotating \vith angular velocity 0) = 1.. Find the preSSUie of the \vater.Sec. De~ermine the work perforrned in the adiabatic expansion of air (having ini1 ial volume Vo = 1 rna and pressure P.8 g. 1766.P . Calculate the kinetic energy of a Fig. \\'here P = canst is the pressure of the shaft on the surface of the su pport referred to unit area of the support. What work will be done if the second charge is moved to point XI = 10 em? 1762**. 1765**. on a vertical ellipse (wl1h axes 2a and 2b) \vhose centre is subnlerged in the liquid to a distance h. 62). 1770.::: b. Calculate tile kinetic energy of a ri~ht circular cone of mass M rotating \\lith angular velocity (1) about its axis. ) 1768. if the rad j us of the base of the cone is R and the aItitude is H. -= J em.. 62 disk of Inass M and radius R rotating with angular velocity (1) about an axis that passes through its centre perpendicu lar to its plane.lcal Problems 171 1761**. -= I I<gfjcm l ) to volume VI = 10 In'? 1764**. the lower bU5e b=:5Q m.000 rplD about its dianleter? (Specific \veight of iron. T\\'o electric charges eo = 100 CaSE·and e1 =200 CGSB lie on the x-axis at points X o = and x. A vertical trian~lc wIth base b and altitude h is submerged vertex dJwn\\'ards in wa1er sa that its base is on the surface of the water. Find the pressure of a liquid. \vhile the lTIaJur axis 2a of the ellipse is parallel to the level of the liquid (h. A cylinder with a movable piston of diameter D=20 em and length l = 80 em is filled with steam at a pressure l p = 10 kgf em • What work must be done to halve the volume of the s1eam with temperature kept constant (isotherlnic process)? 1763**. A vertical shaft of weight P and i radius a rests on a bearing AB (Fig. A vertical dam has the sh lP~ of a trapezoid. 1767*. 121 Appl"int! Definite lntellrals to Solution of Ph. 1771. Fl nd the water pressure on a vrrllcal circular cone with radius of base R and altitude H subln~rged in \va~el verlex downwards so that its base is on the surface of the walcr. whose sp~cific weight is V.tcnl J . \vhile Jl is the coefficient of friction. In the case of steady-state laminar now of a liquid through a pipe of circular cross-section of radius a. What is the force of attraction of a material rod of length 1 and mass M on a material point of lTIaSS m lying on a straight line with the rod at a distance a from one of its ends? 1776**. According to empirical data the specific thermal capacity of water at a temperature to C (O:s. 5 Miscellaneous Problems 1772.912 X 10. Find the mass of a rod of length 1 = 100 cm if the linear density of the rod at a distance x cm from one of its ends is 6=2+0. and the base a is great compared with the altitude 2b.178 Definite 1ntegrals [Ch. . 1773. on the y-axis. In studies of the dynamic qualities of an automobile.y)2J. 1775. where p is the pressure difference at the ends of the pipe. but the pipe has a rectangular cross-section.r . Show that the area S bounded by an arc of this graph. the velocity of flow v at a point distant r from the axis of the pipe is given by the formula P (2 2) t1 = 4fll a . and the reciprocals of corresponding accelerations a. Find the moment of the pressure of the wind striving to turn the door on its hinges. t :s. use is frequently made of special types of diagrams: the velocities v are laid off on the x-axis. The wind exerts a uniform pressure p gjcm l on a door of width b cm and height h cm.7 t 2 • What quantity of heat has to be expended to heat I g of water from 0° C to 100° C? 1774.Y) is defined by the formula v = 2~l [b 2-(b. Determine the discharge of Liquid Q (that is.001 Xl gjcm. 1777*.1 is the coefficient of viscosity.ing through a cross-section of the pipe in unit tinle). and 1 is the length of the pipe.. 100°) is c= 0. 184 x 10. The conditions are the same as in Problem 1776. 1778**. the quantity of liquid flow.5 t + 6. by two ordinates V=OI and V=OI' and by the x-axis is numerically equal to the time needed to increase the velocity of motion of the automobile from vJ to VI (acceleration time).9983-5. Here the rate of flow t1 at a point M (x. Determine the discharge of liquid Q. f.. Note. A horizontal beam of length 1 is in· equilibrium due to a downward vertical load uniformly distributed over the length of the beam. t 780. and of support reactions A and B (A =B = ~ ) t directed vertically upwards. where x is the distance from the left support and k is a constant factor. Find the bending moment M x in cross-section x. 1781*. The intensi ty of load distribution is the load (force) referred to unit length. the moment about the point P with abscissa x of all forces acting on the portion of the beam AP. Find the bending moment Mx in a cross-section x. 12) Applying Definite Integrals to Solution of Physical Problems 179 1779. that is.Sec. Find the quantity of heat released by an alternating si nusoidal current 1=loSin(~t-q» during a cycle T in a conductor with resistance R. A horizontal beam of length I is in equilibrium due to support reactions A and B and a load distributed along the length of the beam with intensity q=kx. . on. Express the volume of a cone V as a function of its gcneratrix x ann of its base radIus y . y). z Y This is the desired functional relation. if to ec)ch set of their values (x. Z is a surface (Fig. Similarly. 63).)2 f (X. Solution. The concept of a function of stveral variables. Frorn geometry we l<now that the volume of a cone is where h is the altitude of the cone.- Fig. The value oi the function z=t(x.-.b). we find :l2 Il2. Hence. x2 + . that is.(-3) 13 = -12· ~) if . Example 2. But h= Vx 2- y2. Find I (2. -3) and t ( I. y. y) in a givpn ron~t' there corre-. we define functions of three or more arguments.y) in a r£'ctangular coordi· nate system X.2. is drnoted by I (alb) or f (P) Generall y speakin~. Fa1ic Notions 1°. Substitutinq r= 2 and g= -3. 63 ~xy Solut.ponds a unique vatul' of z The variables x and yare called arguments or independent variubles. The functional relation is denoted by 2=f(x.Ch'lpter VI FUNCTIONS OF SEVERAL VARIABLES Sec. 1. + (_3)2 -J)= :l. Y.y) = -. the ~~eornetnc rerres€ntation of a runct ion like z -=-! (x. when x = a and y =: b. fxample 1.y) at a pOI nf P (a. Functional notation. A variable quantity 1 is called a single-valued function of two variables x. . The first tenn of the funct ion -2. !L. f ( 1. Slnlll(lrly. of i ~ 1 or Le.. In two C35es: definition of the entire function is shown in fig. y FIg 65 FIg.ich It takes on defInite real value~) I n the slnlplest cases. . x = thaI is. when { ~~ defined for . Example 3. x 181 we will have y )1 + 1+ ( X x 2 y2 ( 1I)' = '2xy . 65 and includes the boundaries of the domain. the rlotnain of definition of a function i~ a finite or infinite part of the xy-plane bounded by one or several Curves (the boundan.\yz-space. of the domain). Domain of definition of a function.y Solution. y Solution. poi nts I yi ng inside a The domain of defi64). in w:.x~2.s=.I E:. . or when { ~:z g. y.Sec. z) the domain of definit ion of the functlon IS a vol ulne in . The run~tion haC\ real values if 4_X 2 _ The I at ter I nrq ua I it Y is sat i ~fi('rl hy thr coord inates of circle of radil1\ 2 with centre ~t the coordinate orll!ln. The dom a In xy~O. Basic Notions I} Putting x= 1 and replacing y by Y) f ( I. y) WP undcr~tand a set of point~ (x. By the domain of definition of a function 'I == f (x. \ 1Lx)\ = f (x. Find the domain of definition of the function x z = arc sin 2' + V--). Find the domain of definition o[ the function 1 z= Y4_. for a funrtion of three variables u=f (x. 2°. 64 Example 4.. nition 01 the [unction is the InterIor of the circle (FIg y2 > 0 or x 2 + y2 < 4. 2·1 x y). IS The second term has rt'al values if g. y) in an xy-plane in which the given function is d~fin~d (that is to say.\2 __ 2 . 3). x 2 ). y. Y)=-2-. if 2 /(x. 6 30 • Level lines and level surfaces of a function.[Ch. 1783. Solution. 2 1787. Find f (x. ± I. at the points of which the function takes on a constant value u=C. z) is a surlace f (x. x Putting C=O. 66 1784. Example 5. ~ ) 1 I f (x. y) . xy 1786. if f(x. ± 2. . The equation of the level lines c has the form x 2 y == C or y =. x 1782. Find the value of the function xC + 2X 2y2+ y4 Z==--~2 l-x + at points of the circle x 2 y2 1788*. y) if f(x+y.. Find the values assumed by the function f (x. The level surface of a function of three arguments u = f (x. -1). -y). z) = C. y. + x-y y) = 1 at points of the parabola y = x . 1789*.. y) -== C (in an xy-plane) at the points of wh ich the function takes on one and the same value z=C (usually labelled in drawings). x--y)=xy+yl. y 1785 Find X2 _ y f (Y. if = R I _yl • f(~)= ~(y>O). . and construct the graph of the function F (x) = f (x. Determine !(x). Express the volume V of a regular tetragonal pyramid as a function of its altitude x and lateral edge y. y)===xy+-=-. . The level line of a function z = f (x. Fig. we get a fanlily of level lines (Fig. f( f. y) is a line f (x. 1(1.: I . f (-x.x). Find 1(1/2.F_u_n_c_ti_on_s_o~f_S_e_ve_r_al_Va_r_ia_b_le_s 1_8_2 . 66). Construct the level Ii nes of the function z=x2 y. Express the lateral surface S of a regular hexagonal truncated pyramid as a function of the sides x and y of the bases and the altitude z. x 11) c) Z=X 2 _ y 2. Find and sketch the domains of definition of the following functions: Z== a) z=Yi-X 2 _ y2. 1] Basic Notions 183 1790*. e) z=Yi-x 2 -!. xy I) 1 Z=-_· x +!J Y d) z=x-f-arccosy. Let z = VY + f (Vi-i). d) z=Yxy.Yi_ y 2. 1796. j) z = In (x 2 + y). m) I Z=V y-y . 1 I 0) zY sin (x 2 -+_ y2). Determine the functions f and z if 1791". b) z=arcsinxu. h) z==Y(xl+y2-a2)(2a2_x2_yl) (a> 0). x z=f(JL). c) u = arc sin x + arc sin y + arc sin z. c) z=ln(x+y). b) u = In (xyz): d) II = Y l-x 2 _yl_ZI. b) z = i-t_(X_y)2. x n) Z=x=r+y .. Construct the level lines of the given functions and determine the character of the surfaces depicted by these functions: a) z==x+y. Let Yi +yl when x= i. 1794. 1793. z= i) z= h. 1795.Sec. f) z=arcsin lL . + c) u=xl-t gl_ZI. d) z=f(y-ax). e) c) z = f <YXl + y2). b) u = Xl yl + Zl. 2X x 2 + y I. Determine ihe functions z if z==x when y= i. i) z=Yysinx. 1792. . f) z= I-lxI-luI. x g) z == y7=4 + Y4=iI. x-y 2 • 2 . Find the domains of the following functions of three arguments: a) u = ~rx + yy + V"z. Find the level surfaces of the functions of three independent variables: a) u=x+y-t-z. k) z = arc tang -+1 2 2 . f and z = xf ( ~ ). g) z= ~ . Find the level lines of the following functions: a) z=ln(x2 +y). .x l/ 00 ~ 1/-+0 d) Ii. y) as the point P' (x. x-+o "-~2 IJ -+ J ·. 0 XJ X-+OO X 2 _!J2 f) lIn ~+2- y-. x-.. d) 1 Z = l.n (1 y _X_ . xy .6 Sec. A function continuous at a pOint P (a.. A number A is called the limit of a function 2=1 (x. Find the following limits of functions: . x+y.-y . y) is called linl f (x.\. y) may cease to be continuous either at separate points (isolated pOlnt of discontinuity) or at points that form one or several lines (lines of dlscontinuity) or (at times) more complex geometric objects. x-~ox+y X X-+O + 1!.. Tile function will be m~aningless if the denomin3tor becomes zero. we have the inequality I!(x. oo. b)..)2 I l . Example 1. Find points of discontinuity of the functions: a) z = tn Vx~ b) + yl. f(x. sit] xy e) Ii n a) Ii itl (x~ T yl) sin! · C) I (Ln _ _ .x.. b) if 2 = f (x. Continuity and points of discontinuity.• >.. y)= { Vl-x"l.k 1798. But )(2_ y -==O or y=x 2 i~ the equation of a parabola.x 2 • 1797*.x2 _y2 z =cos -. h). 2.Q y-+b ran~e A function that is continllous at all points of a given is called continllO:. b) ILn --z-+2 t X .u ·t Y . Jn th is case we write lim f (x. Continuity 1°. x-+a y-+b 2°. find the discontinuities of the function xy+ 1 z=A 2 _y· Solution. the given function has [or its discontinuity the parabola y==. IJ . ) x. if for any B > 0 there is a tJ > 0 such that when 0 < Q < tJ..!.-y2 when x o 2 1799. wherE:' Q -== Ylx-a)1 (y-b)' + is the distance between P and p'. y) approaches the point P (a.. y)-AI<e. c) Z -- . y) = A..lS over this range A function f (x. +y"l. Hence.Funct ions of Several Variables 184 (Ch. • X ~o. Test the fotlowi ng III nction for conti nuity: + y2 ~ It when x"l.. y) = f (a. The limit of a function. > 1. - uX which is called the part ial derivat ive of the function z with respect to thevarIable x. Sec. Regarding y as constant. au -=x y2+1. cos! Y yl sin y u (11 = .Sec. Find the partial derivatives of the function z=lntan~. y) . 2x. we get the derivative OZ _ 10 ~ uX - 1m f (x + A". y) . Solution. Show that the function 2xy h I I 0 z= { Z--+ x y 2 W en x + y =F .x . YSlng Similarly. Partial Derivatives 1°.x Example 2. Example 1. X I y 2 z +2 . Find the partial derivatives of the follOWing fUJlction of three arguments: 1 U =X y 2 z + 2x-3y + z + 5.3.. A function f (x. y) Is called a homoleneous function of degree n if for every real factor k we have the equality f lkx. o when x=y=O is continuous with respect to each of the variables x and y separately. Y). one can use the ordinary formulas of differentiation. 3. holding x constant.. we get iJz 1 ax = t---X an . then assunling. y constant. I n similar fac. we will have iJz Y 1 tan y 1 ( x) 2x -2 =. ax iJu =. iJu iJy=2x 'y z-3. for example. y).· _ .f'% (X. 0) with respect to these variables together. ~t'~O y) A f (x. ky) ==: kill (x. Euler's theorem.cos 1 1 -2X g g • 2 Ii = ---:2X. DeOnltion of a partial derivative. but is not continuous at the point (0. If z =f (x. 31 Partial DeritJatives 185 1800·. iJz ' 2°.hion we define and denote the partial derivative of the function z w!th respect to the variable y It is obvious that to find partial derivativ~~. y Solution. (1. z=xy • x-y 1802. / Z= 1804. · . z=x'+y'-3axy. 1806. 1813.J (1. u=(xy)z. 1812. u=zxy. ox+ay+az:Z: 1825. Yx 2x+y2 • arc SI n JI z = In sin x+a yy I X _ x2 + y 2 y2 • .Y • ox Y uZ J826. 2. y)= 181 5. 1820. f (x. 1) and f y (2. it ~ = !+x I . z==ln{x+Vr+y'). S how t h at x ax dz · Y ay=2. 2. uy X y y-z .j · x!+y 1817. Find f1&(2. If z= In (xl+xy+yl). Find z = z (x.!L % 1 8 1 0 . 'x g.of degree n (Euler's theorem): xf~ (x.ax /ar + 1821. Fin d f~ ( 1. y).[. 2=arctan.. 1805. 0). '. x where r=VxZ+yZ+zl. 1818. X Z= x2+ y 2. 0). The following relationship holds for a homogeneous differentiable function .Functions of Several Variables 186 (Ch. y. 2. x " 1814. Show that ~+aau+~u=lt if u=x+x. 1819. 1808. Verify Euler's theorem on homogeneous functions in Examples 1816 to 1819: x+y 1816. f: x Vr---xY+ y . y) = V"j""'f'"'. 1) if f(x. Find the partial derivatives of the following functions: 1801. 1807. Z=x+y· 1803 • Z 1809.. If· u=(x-y)(y-z)(z-x). z)=ln(xy+z).if x=rcos<p and y=rsin<p. y) = Ax 2 + 2Bxy-Cy 2. f (x. Find ~(~~. sin 1811. acp az 1822. 0) if f(x. OZ OZ 1823. y). h 1824. z =: VX 2 _ y l.y)=lnJ!. Sh ow tat • If JL z=xy+xe%· au au au 0. y) = nf (x. Calculate :: . z=e = 1LX • Z= . y) + Yf~ (x. 6 A rational integral function will be homogeneous if all its terms are of one and the same degree. f(x. Sh OVl t h at x(jX+U ay=xy+z. y. The differentials of independent variables coincide with their incrernents. although it is discontinuous at this point. y) is the principal part of the total increlnent ~z. 1828. using the drawing. 4] Total DiUerential of a Function 187 1827. y). 1830·. are drawn planes parallel to the coordinate surfaces XOZ and YOlo Determine the angles formed with the coordinate axes by the tangent Ii nes (to the resulting cross-sections) drawn at their common point M. z) is computed fronl the formula du= ou dx au ou ax + ay dy+ iJz dz.oy" . y) is the difference ~z-= ~f (x. y) and f~ (x. 0). Find z==z(x. \vhich is linear with respect to the increments in the argulnents f1x and Ay. the total differential of a function of three arguluents u =/ t x. f 2 . Construct the geometric image of this function near the point (0. Similarly.d'l.y)=siny when x=l. 2°. 0 .0).Y . Find :~. that is. y) --=--' x y . If a function has a total differential. Total Differential of a Function to. y) knowing that + x 2 y2 ~=-x 02 uX and z(x. y+ ~y)-f (x.2. The total (or exact) differential of a function z==f (x. has partial derivatives f~ (x. determine their geonletrical meaning. f (x. 1829. 2 =1= 0 J ~+2 ' 1 X -1. if X=IJ==O ~. Through the point M(1.6) of a surface z=2x l +y. The total dift'erential of a function. . The area of a trapezoid with bases a and b and altitude h is equal to S=I/ 2 (a-l-b)h. y) is computed by the fonnula oz dz dz = 0)( . y) at the point (0. dx= ~x and dy= Ay.Ax. The difference between the total increnlent and the total differential of the function is an infinitesimal of higher order compared with Q= VAx 2 + li y 2. For the function f (x. Sec.dx -t. The total differential of the function z = f (x. :~. The total incretnent of a function z = f (x. A function definitely has a total differential if its partial derivatives are continuous. Show that the function I 2xy . :~ and.Sec. Y)·-=f (x-t. y)=x l ·t-xy_ y 2 find the total increlnent and the total differential. 4. then it is called differentiable. Total increment of a function. Example 1. 2 + yl xdr+ydy . Ax + ~x.: {( ~ y + ~y) = (x + AX)I + (x + Ax) (y + ~y) .\2 +x· Ay + y. r .06. axaz Ax+ay Example 3. 1.4 em'. two) variables the ordinary rules of differentiation holdJ a) d (u + u) = du + dv. =. compare them if a) ~x=l.y2) = ==.02'·01::::::: 1 + 0. y=3. y) = x 2 y find the total increment and the total diflerential at the point (I. if we increase H by 3 mm and ditninish R by 1 mm? Solutlol\.r l x + Solution. we have for a differentiable function z=flx. y a-= . The change in volume we replace approximately by the differential i 1 AV::. The initial value of the function z = 1'= 1.I.. f!z::::::: dz= yxY .u dv • + v du.t2 + Ay2. hence.~~l'). Example 4.r . Applying the total differential of a function to approximate calculations.3) = -IOn =:::-31. Solution.02.1 Ax+xY In x Ay= 3. y2 2 +y2.01.06= 1..1.(y + Ay)2.(x 2 + xy. y) f (J( + f!x.. the expression df=(2x+y)Ax+(x-2y)Ay is the total differential of the function. 1831.2. + l\X)2 + (x + Ax) (y + Ay) . 1--'::' Y r. Ay-fl y 3 = = [(2x+ y) Ax + (x-2y) Ay) + (Ax! + Ax· Ay.. 1832.1 + 100·0.\ + y2 3°.Ay·-2y.· r l! -t. the radius of thE' base R = 10 cm.Ax + A.(y + Ay)21. ~y=O. while (Ax 2 + tu·Ay-Ay 2) is an Infinitesimal of higher order compared with V ~. .yl Y X . Example 2.(Ch. How will the volume of the cone change.. Show that for the functions u and v of several (for example. Hence.02'·01 approximately. Compute 1. for ~ufficiently small Q= Y A\I+ Ayl. Yx . Ax=0.0. The altitude of a' cone is H = 30 em.:dV=3' n(2RH dR+R 2 dH)= = 3"1 n (-2·10·30·0. y) the approximate equality Az:::::: dz or For Az ~ az Ay. . ~f {x.dx+ r AI + y! 11"-'-dy 2 r .\.06. 2). Find the total differential of the function z= az 2 dz az x ax . dy=O. The desired number may be considered the increased value of this function when x= 1. ~y=-=2. suffirifn~ly small I Ax I and I Ay I and. Functions of Several Variables J88 Sol ution. For the function f (x. b) ~x=O. c) d ( ~ ) b) d (uu) = u dv = v dU.02+ 1·1n 1·0.01 =0. The volume of the cone is V = nR 2H.2x. We consider the function z =-= xY . Here. . 1850*. Find df(3.97)2~ b) V(4.\ 2 df (1. 1840. Z= 1+"1· . side b = 200m±3 m.05)2 + (2. x 1842. Y -t. How will a diagonal 1 of the rectangle change if the side a is increased by 4 rum and b is shortened by 1 mIn? Approximate the change and compare it with the exact value.y.Sec. A closed box with outer dimensions 10 em. r_ _2. c) sin 32°·cos 59° (when converting degrees into radius and calculating sin 60 take three significant figures. z==x'+y'-3xy. u = arc tan x~ • z 1847. 1837 . if theoriginal leng:h of t he radius is 20 cm? 1851. and 6 em is made of 2-mm-thiek plywood. Approximate: 0 . u=(xy+i-r 1846. z=sin x+cos y. 8 em. r AI 1848. 1849. . Find x 2 _l/! 1835. Y 1843.z)= . To what degree of accuracy can we conlpute the side c? 1854. Show that the relative error of a product is approximately equal to the sum of the relative errors of the factors.. 1839. 1834. 1= arc tan yx +aretan~. The central angle of a circular se~tor is 80 it is desired to reduce it by 1°. (O. 1853.y)=ln(l+i). 1841. a) (1. _ 1844. y 1838. z=lntan1!.yl). z == In (x 2 -~. Measurements of a triangle ABC yielded the following da1a: side a = lOOm±2m. Approximate the volume of nlaterial used in making the box. U = Vx 2 -t. z=x 2y'. 4.5) if 2 j(x. +y the other b = 24 cm.y2 + zj. The oscillation period T of a pendulum is computed from the fOflnula T= 2nyr _ I g . 2 1836.02)1. round otT the last digit)..y)=z. 4] 189 Total Differential of a FunctIon Find the total differentials of the following functions: 1833. if x Y f(x. angle C :== 60° ± 10. 1).. One side of a rectangle is a = 10 em. By how much should the radius of the sector be increased so that the area will remain unchanged. u = xyz . z=yx. 1845..93)1. f(x. 0 1852. Y = 'I> (i).ax ay dx · (2) Example I. then the partial derivatives z with respect to u and v are expressed as iJz = ~ ~ + ~ iJy iJu ax au ay iJu (3) . By how much will the angle a change if the point P(P o is fixed) moves to PI (x+-dx. If z is a composite function of several independent vari~bles. when determining T. y) is a differentiab Ie function of the arguments x and y. if t coincides with one of the arguments.v). obtained as a result of small errors ~l = a and ~g = p in measuring land g. C0I11- (I) In particular. if z=eIX + 2Y. x = cp (t). z = f (x. tI) (u and v are Independent variables). If z = f (x. Differentiation of Composite Functions to. y) is equal to Q. + Example 2. 6 where l is the length of the pendulum and g is the acceleration of gravity. 1855. where g=cp(x). y=t 2 • Solution. then the "total" derivative of the function z with respect to x will be: dz _~ + QZ eJJ! dx. The case of several Independent variables.x+2Y·3 (-sin t) eIX + 2Y ·2·2t =e'X+2Y (4t-3 sin t) =e (4t -:3')int).Funct ions of Several Variables 190 [Ch. ~=yeXY. where x=cos t. From formula (2) we obtain dz dx = yexy + xexy cp' (x). for insta~ce. The distance between the points Po (x ot Yo' and P (x. while the angle formed by the vector PoP with the x-axis is a. The case of one independent variable. lJ' (t)] may be puted from the form ula dz dt = ~ dx + az dy ax dt ay d t . Find the error. Find the partial derivative iJiJZ and the total derivative dz Ii x dx' z=e"Y. Find ::. 2°. y+dy)? Sec. which in turn are differentiable function~ of an independent variable t. Y=1I' (u.y). From formula (I) we have: dz ~ IcOSt+2t 3 dt =e. then the derivative of the composite function z=f (cp (i). Solution. where x= cp (u. 5. for instance x. Example 4. . we get: az. Find : : and z = t (x. (4) In all the cases consi dered the following formul a holds: az az dz= axdx+aydY (tile invariance property of a total dzgerential). av -fx + v F' u y) u-.. + ax + + that is. y) au and oz _ ' (x. Solution. therefore. (x I + y 2) 2x and ~= ~ aat =q>' (xl y2) 2y. we get OZ az Y -x ay = ycp'(xl + y2) 2x-xcp' (Xl y!) 2y = 2xyq>' (Xl + y2)-2xy q>'(x2 y2)ii5iO. Show that the function z == cp (Xl az +yZ) satisfies the equation az yax -x ay=O. u fy (x. y (x. Y= vt'+l. uy dt y Substituting the partial derivatives into the left-hand sid{) of th{) equation. 1856. az dz at ax=dl ax=CP . where x = 3t l . :~ if u = In sin . y=--V. y=lnt. where y). Find ~: if u=xyz. y) fj2 .(y. Applying fonnulas (3) and (4). 5] DiOerentiation of Composite Functions 191 and az az ax az ay au = ax au + au au . Example 3. Find x=ef . . y=lnt. where y 1857. 1858.Sec. the function z satisfies the given equation. 1 == f x (x. Solution. The function cp depends on x and y via the intennediate argutHent x2 + y2 = t. where x=tl+l. Find :: if z=~. if u x=uv. :~. z==tant. y). Find :. z = 1J' (x. v). z = R sin cp. 1863. and :. if Z = ltv.!. and y=x". if z=f(u. Find ~ and ~ if z = arc tan. where u == sin x... Find : . y where 1865. v=e xy • 1864. and :. %=H. y = R cos cp sin '1'. Show that if then u = cD (Xl + y' + 2 1 ). Find U= of Several Variables Z X 2+yl 1860. x 1862. ~-o oq>- au and 0lJ'=0. 1861. z=f (x+ay). Find :. Find :. y=R sin t. Find ~ if U = f (x. where x = R cos q> cos '1'. U). 1868. where f is a differentiable function.Functions 192 Y (Ch.. and ~ if z =xY . Y= u cos v. Show that if where y = cp (x).6 :~ if 1859. y. v = cos x. where u = xy + yx • 1866.where x=R cost. Find :: and Z X= u sin 0. if z=arc tan!. where y = cp (x). 1867. z). ~ if = f (u). then oz iJy =a az ax· . where u=xl_y'. v).+-y oz I iJy=!j2 az z · sa t IS 1871. y) ift a given direction I=PP. is az = iJl 7_ 1900 lhn f (PI )-f (P) P1P ~ 0 PIP • . The equations of Illotion of a material point are x = t. Their velocities are respectively 20 kmfhr and 40 knljhr. The side of of 5 mjsec. one moves northwards. rectangle x -:~ 20 m increases at the rate side y == 30 m decreases at 4 m/sec. Show that the function z = xy + xcp ( f· th e equa t·Ion x-a az -I-Y~=xy az sa t ·ISles x uy 1872. 61 Derivative in a Given Direction 193 1869. Show that the function w=f(u. y ~= t 2 . Show tl-)at the function z = y cp (X 2_ y 2) · f·les th e equa t·Ion x1 a.Sec. T\vo boats start out from A at one time. (ye::' ) ~+xy~=xyz. Derivative in a Given Direction and the Gradient of a Functioll 10 • The derivative of a function in a given direction. the other is the rate of change (X 1_ y 2) ~) + z. Show that the function z = eY cr satisfies the equation 1873. where u=x+at. What of the perilneter and the area of the rect- a angle? 1874. At what rate does the distance between them increase? Sec. the other in a northeasterly direction. The derivative of a function z=f(x. z:= t l • What is the rate of recession of this point from the coordinate origin? 1875. 6. v=y+bt satisfy the equation aw _ aW+ b ow -a ox oy · iJt 1870. 194 Funct ions of Several Variables [eh.' a(~) iJy y p =0. Here. 6 where f (P) and f (PI) are values of the function at the points P and PI If the function z is differentiable. y.2 • sin a = sin 1200 = -2-¥3 Applying formula (1). y o x Fig. 67).+0. The gradient of a function z == f (x. m=axcosa +aySlna. Find the partial derivatives of the given function and their values at th~ point P: axaz = 4x. (az) Ox p = 4. 67 In similar fashion we define the derivative in a given direction I for a function of three argunlents u = f (x. we get 1) OZ ( . oz = -6y.· The minus sign indicates that the function diminishes at the given point and in the given direction.. Find the derivative of the function z=2x 2 -3 y 2 at the point P (1. (2) where a. 2°. 0) in a direction that makes a 1200 angle with the x-axis. In this case all at = ax cos u du +au au cos ~ + au az cos y. The directional derivative characterises the rate of change of the function in the given direction. The gradient of a function. Example 1. lIr3" -=4 f 2 a1 2 -2-= . ~. y) IS . (1) where a is the angle formed by the vector I with the x-axis (Fig.a vector whose projections on the coordinate axes are the corresponding par- . yare the angles between the direction I and the corresponding coordinate axes. then the following formula holds: iJz iJz iJz . z). 1 cosu=cos 1200 = . Solution. Sec. 6] Derit.'ative in a Given Direction 195 tial derivatives of the given function: 02 02 grad z == ox I + oyJ· (3) The derivative of the given function in the direction I is connected with the gradient of the function by the following formula: 02 = pro)., gra d z. at That is, the derivative in a given direction is equal to the projection of the gradient of the function on the direction of differentiation. The gradient of a function at each point is directed along the normal to the corresponding level line of the function. The direction of the gradient of the function at a given point is the direction of the maximum rate of increase of the function at this point, tlUtt is, when l=grad z the derivative on its greatest value, equal to ~~ takes In sitnilar fashion we define the gradient of a function of three variables, u -= f (x, y, z): au iJu iJu grad u=iJ-x 1-+- uy ~ J + a- k. Z (4) The gradient of a function of three variables at each point is directed along the nonnal to the level ~urface passing through this point. Example 2. Find and construct the gradient of the function z.=;x 2 y at the point P (I, I). y 2 1 0 -~ - - --~ PI I I 2 3 X Fig. 68 Solution. Compute the partial derivatives and their values at the point P. ~=2xY; (~)p=2; (~)p=l. Hence, grad 7* z=21+J (Fig. 68). Funct ions of Several Variables 196 [Ch. 6 1876. Find the derivative of the function z = x!-xy-2y J 2) in the direction that produces an angle '()f 60° with the x-ax is. 1877. Find the derivative of the function z=x 8 -2x 1 y-+ xy2+ 1 .at the point M (1, 2) in the direction from this point to the point N (4, 6). 1878. Find the derivative of the function z == InVx! -~ y! at the point P (1, 1) in the direction of the bisector of the first at the point P (1, quadrantal angle. 1879. Find the derivative of the function u=x!-3yz+5 at the point i'A (1, 2, -1) in the direction that iorms identical angles with all the coordinate axes. • 880. Find the derivative of the function u = xy + yz ,- zx at the point M (2, 1, 3) in the direction from this point to the point N (5, 5, 15). t881. Find the derivative of the function u = In (eX + eY -t-~ e1) at the origin in the direction which forms with the coordinate axes x, y, z the angles a, ~, V, respectively. 1882. The point at which the derivative of a function in any direction is zero is called the stationary point of this function. Find the stationary points of the following functions: a) z==x!+xy+y2-4x-2y; b) z === Xl yl -3xy; c) u == 2y 2 -t- z!-xy-yz 2x. + + 1883. Show that the derivative of the function z=L taken x at any point of the ellipse 2x2 -1- y2 = C! along the normal to the ellipse is equal to zero. 1884. Find grad z at the point (2, 1) if z = Xl + yl- 3xy. 1885. Find grad z at the point (5, 3) if Z== Vx 2 _y!. 1886. Find grad u at the point (1, 2, 3), if u =xyz. 1887. Find the tnagnitude and direction of grad u at the point (2, -2, 1) if 1888. Find the angle between the gradients of the function Z= In ~ at the points A (1J2, 1/4) and B (1, 1). Sec. 7] Higher-Order Derivatives and DiUerentialb 197 1889. Find the steepest slope of the surface z=x! + 4y· at the poi llt (2, 1, 8). 1890. Construct a vector field of the gradient of the following functions: a) z=x+y: b) z=xy; Sec. 7. Higher-Order Derivatives and Differentials 1°. Higher-order partial derivatives. The second partIal derivatives of a function Z=== f (x, y) are the partial derivatives of its first partial derivatives. For second derivatives we use the notations :x (a:) t: ::~ = = x (x, y); a z ::::: f"Xli (x, ayo (DZ) ax = axay 2 y) and ~o forth. Derivatives of order higher than second are similarly defined and denoted. If the partial derivatives to be evaluated are continuous, then the result of repeated dtUerentlailon is independent of the order in which the differentzatlon IS perfornled. Example 1. FInd the second partIal derIvatives of the functIon Z ==- Flr~t Solution. arc tan !... y . find the first partial derIvatives: az 1 I y ax=~· Y=.\2+ y 2 ' l-t-y'l az 1 ay=~ ( + y2 x ) - !/t = - x x2+y". Now differentiate a second time: iJ'lz _ ~ ax 2 az - 2 ay2 = ax (_y_ )__ + y! - 0 ( ay - 2xy (x 2 x! x2 X) + y2 cJ2 Z 0 ( Y ) ox ay = oy x! + y2 = + y2)1 • 2xy + y2)1 • l 1.(x + y2)_2y.y = (xt (xl + y2)1 xl_gl + y2)2 • (x 2 We note that the so-called "mixed" partial derivative Inay be found in a <titTerent way, namely: 0 22 cJ2Z a( X) 1. (x 2+ y2)-2x.x x 2 _ y:! 2 2 ox ay ay ax = x + yl = (Xl + y2)1 = (x + y2)2 • = ax - Functions of Several Variables 198 2°. 2 =/ (x, [ell. 6 Higher-order differentials. The second diUerential of a function y) is the differential of the differential (first-order) of this function: d 2 z = d (dz) We similarly define the differentials of a function z of order higher than two, for instance: and, generally, dnz = d (dn-1z). If z = t (x, y), where x and yare independent variables, then the second differential of the function z is computed from the fonnula d 2 a~ a~ z= ax2 dx 2+ 2 ax ay dx dy a~ + ay 2 dyl. (1) Gener all y, the followi ng symbolic fornl ula holds true: dnz=( dX~ +dy a~rZi it is fornlally expanded by the binomial law. If z = f (x, y), where the arguments x and yare functions of one or sev- eral independent variables, then 02Z d 2 d! z~ax2 x Z . y 02 + 2 oxoydXd az 02 Z 2 az 2 2 +ay 2 d y +axdx+ayd Y• (2) If x and yare independent variables, then d 2 x=O, d 2 y=O, and formula (2) becomes identical with formula (1). Example 2. Find the total differentials of the first and second orders or the function z = 2x 2 - 3xy _ y2. Solution. First method. We have OZ 02 (jX=4x-3y, oy= -3x-2y. Therefore, dz iJz oz= ax dx+ ay dy= (4x-3y) dx-(3x+2y) dYe Further \ve have whence it follows that 2 a2 z dZ==ox2dx 2 2 a z dx dY+a02 + 2 axoy y Z 2 2 2 2 dY =4dx -6dxdy-2dy. Second method. Differentiating we find dz=4x dx-3 (ydx+xdy)-2y dy= (4x-3y) dx-(3x+2y) dYe Differentiating again and remelnbering that dx and dy are not dependent on x and y, we get d 2z= (4dx-3dy) dX-(3dx+2dy) dy=4dx 2 -6dxdy-2dy 2. Seco 7] Higher-Order Derivatives and DiOerentlals 1893. Find O~2~y if 1894. Find iJ~2;y if z=arctan 1895. Find t+y • l-xy ~~ if V x 2+y2 +zlo r = 1896. Find all second partial derivatives of the function u = xy + yz -+ zxo 97 F In 18. o d OIU ox dy 2 1898. a r-'1 n d iJxayZ o 2 az JOf °If z = sin (xy) ° 1899. Find f: x f: (0, 0), r/ (0, 0), f~'l (0, 0) if f(x, y)=(l-+ x)m(l +y)n. 1900. Show that 02 Z iJxiJy z if = iJLjiJx iJ2 z=arc sin 1901. Show that 0 2 2 iJxay = 02 Z ayox Y- x- y x if z=xY • 1902*. Show that for the function f (x, y) x 2 _y. = xy x--a-+y 2 • 199 Functions of Several Variables 200 [provided that f (0, 0) = 0] we have f ~y (0, 0) = - 1, f~x (0, 0) = 1903. Find Z [Ch. 6 + 1. if OIZ 02 ox axo!zoy , oy· 2 , z = f (u, v), where u = Xl + y., v = xy. 1904. Find ::~ if u=f(x, y, z). where z = cp (x, y). . o!z 02Z O!Z. 1905. Find oxay , ay2 If ax.' z = f (u, v), where u = cp (x, Y), v = 'I' (x, y). 1906. Show that the function u -= arc tan JLx satisfies the Laplace equation alll a2u -+--0 ax! ay 2 - • 1907. Show that the function 1 u=ln, , where r=V(x-a)2+-(y-b)2, satisfies the Laplace equation 02 U au 2 ox. -1- oy2 = o. 1908. Show that the function u (x, t) = A sin (a'At + cp) sin 'Ax satisfies the equation of oscillations of a string 02 U iJt2 = 2 a 0 !U ax2 • 1909. Show that the function u(x, y, z, t)= 1 y_ e (2a 1T t)3 (X-XO)2+(y- YO)2 +(% -Zo)l 4a 2 t (where x o' Yo, zo' a are constants) satisfies the equation of heat conduction Sec. 7] Higher.Order Derivatives and DiOerentials ~Ol 1910. Show that the function + 'i' (x + at), u = cp (x-at) \vhere cp and '¢ are arbitrary twice differentiable functions, satisfies the eq uation of asci llations of a string 2 a2 u 2 au . at 2 =a ox l 1911. Sho\v that the function z = xcp ( ~ )f- '!' ( ~ ) satisfies the eq uation Xl a2z + 2x iJ2 z Y iJxay ax 2 + Yay! iJlz = 0 . 2 J 912. Sho\v t hat the function u = cp (xy) + l/ xy¢ ( ~ ) satisfies the equation 1913. Sho\v that the function z = f [x -i- cp (y)] satisfies the equation az a2z az a2z ax ax ay = 1914. Fi 11 d u --=:; II (x. y) ay ax 2 • if a a-a x y =0. 211 1915. Determine the fornl of the function u=u(x, y), \\'hich satisfies the equation 1916. Find d!z if 1917. Find d 2 u if u = xyz. 1918. Find d1z if z=cp(t), where t=X 2_'t-yl. 1919. Find dz and d!z if Z = uf) x where u = -y , v = xy. Functions of Several Variables 202 [Ch.6 1920. Find d 2 z if z=f(u, v), where U=QX, v=by. 1921. Find d1z if Z = f (u, 1922. Find d'z if v), where u = xeY , v = ye". z = eX cosy. 1923. Find the third differential of the function z=xcosy-~y sinx. Determine all third partial derivatives. 1924. Find df(l, 2) and d2f(1, 2) if f(x, y)=x l +xy+y2-4Inx-IOlny. 1925. Find d 2f (0, 0,0) if f (x, y, z) = Xl + 2y 2 -1- 3z 2 -2xy -t- 4xz + 2yz. Sec. 8. Integration of Total Differentials to. The condition for a total differential. For an expression P (x, y) dx + y) dy, where the functions P (x, y) and Q (x, y) are continuous in a simply connected region D together with their first partial derivatives, to be (in D) the total differential of some function u (x, y), it is necessary and sufficient that + Q(x, iJQ ox iJP ==== oy . Example 1. Make sure that the expression (2x -t- y) dx -f- (x + 2y) dy is a total differentIal of some function, and find that function. Solution. In the given case, P=2x+y, Q=x+2y. Therefore, = I, and, hence, (2x+ y) dx ~=:: = au + (x+ 2y) dy = du = ou ax dx+ iJy dy, where u is the desired function. ou It is given that ax == 2x + y; therefore, u= ~ (2x+y) dx=xl+xy+q> (y). Buf on the other hand and Finally we have iJu oy=x+Q>' (y)=x+2y, whence <p' (y)=2y, cp(y)=y2+C u == x 2+xy + y2 + c. (2x+ y) dx+(x+ 2y) dy=d (xl+xy+ y2+C). Sec. 8J 'ntegration of Total DIfferentials 203 2°. The case of three variables. Silnilarly, the expression P (x, y, z) dx+ Q (x, y, z) dy+ R (x, y, z) dz, where P (x, y, z), Q(x, y, z), R (x, y, z) are, together with their first p~rtial derivatives, continuous functions of the variables x, y and z, is the total differential of some function u (x, y, z) if and only if the following conditions are fu Ifi lied: Example 2. Be sure that the ex pression (3x 2 + 3y-l) dx + (Z2 -t- 3x) dy + (2yz + 1) d2 is the total differential of some function, and find that function. Solution. Here, P=3x!+3y-1, Q =z2+3x, R=2I1z+ 1. We establish the fact that and, hence, (3x 2 -f-3y-l)dx+(z2+3x) dy+{2yz+ I) dz=du = au ~ uX au du dX+ - dY+- dz, ay az \vhere it is the sought-for function. We have au ax =3x2 + 3/j-l, hence, u= ~ (3x 2 +3y-l) dx=x'+3xy-x+q> (y. z). On the other hand, au a(j) 2 ~=3x+a-=z +3x, uy y au aq> -=2yz--1-1, az iJz - = \vhence °dq> =Z2 and aacp =2yz+ 1. The problem reduces fo finding fhe function y z of hvo variables cp (y, z) whose partial derivatives are known and the condition for total differential is fulfilled. We Hnd CI': q> (Y. z) = ~ zldy = yz2 + Ijl (z). :: =2yz + Ijl' (z)=2yz+ I, '1"(z)=I, 'I'(z)=z+C, that is, cp (y, 2)= yz 2+ Z+C. And finally, u=x' +3xy-x+ yz2+ z+C. 1933. y) (dx + dy) to be a total differential? 1940. 1932. -II dx-dye y2 II 1931.----11929. 1935. (3x 2y 2 3z) dx (4xy 2y -z) dy + (3x . + + + (~'-. What condition must the function f (x.3xy 2 -l.6xyz 8x2 Y -t. ¥X 2 +y2+ Z2 1938*. where A.(x 2+ 2xy + by 2) dy (Xl + yl)2 (ax 2 should be a total differential of some function z.1) dy -1. and find that function. ¥ x x-t.y)2 . 1928 • (x+2y)dx+ydy (x -1. y dx+x dYe 1927.z) dy + (x -r y -l. y) satisfy for the expression f (x.2z) dz. x dx + y dy + z dz . x+2y 2x-y dx -.3) liz.y I dy.204 Functions of Several Variables I-laving convinced yourself that the expressions given below are total differentials of certain functions. What must the coefficient A be for the force to have a potential? 1939. is a constant. (2x -t.2) dz. 1937. 1926. .y . (2xyz-3y·z 8xy 2 2) dx + (x 2 z. + + -r + 1934. 2 + 1936. Z--+ x y2 x . (cosx+3x1y)dx+(X 3 _y2)dy. Given the projections of a force on the coordinate axes X=-y(x+ y)2 .y2) dx . 1 x 1930.x dy). Find the function u if du = f (xy) (y dx -J. y=~ (x+ y)2 .y + z) dx + (x + 2y -r. Determine the constants a and b in such a manner that the expression + 2x Y -1.y 2 2 dx+ ¥ X Y+ y 2 2 dy. Convince yourself that the expressions given below are total differentials of some functions and find these functions. find these functions.(xly.2)dx f-(~-:2)dY+(~-:')dz. y) -= 3 (x 2 + y2)2.3·2x-= 6x [(x 2 + y2)2 -1]. y. y. 2x. defines y as a function of x. y) f~ (x./ (x. y. 2y-3· 2y == 6y [(x 2 + y2)2_1]. be found fro III the fonnulas oz at = F~ (x. if the equation F (x. y). oy = F ~ (x. 9. y) = . Sirndarly.y_x y_x{_'yt) d:!11 d ( x ) dx \ = _Y~ + x2 dx2=dx --y = - y2 y2 y3 2°. differentiate \vith respect to x the first derivatIve WhICh \ve have found. z) • y. may be found fronl the formula I (x. we find partial denvatlves == 3 (x 2 + y2)2. where F (x. The case of several independent variables. Flnd d. z) =0. generally speaking. !It z) F ~ (x. Differentiation of Implicit Functions 1°. y) i= 0. z) :p 0. y) • (1) found by successive differentiation of formula (I) 2 dy d y. where x and y. y. and. y)=O.and -d2 If x x (x 2 th~ + y~)' - 3 (x 2 + y2) + 1 == O. Solution.· To find the second derivative. z) is a differentIable function of the variables x. then the derivative of this irnplicitly defined function. \\lhenCl\ applyIng forll1ula (1). iJz ax QZ and ay • . defines z as a function of the independent variables x and y and F~ (x. !I. z) =-=0. then the partial derivatives of this irnplicitly represented function can. Denoting the left-hand side of this equation by f (x. y) 1. z) (2) Here is another way of finding the derivatives of the function z: differentiating the equation F (x. Example 1. ox cJy iJz Whence it is possible to detennine dz. y and z. z)' az F. therefore. \ve get dy f~ (x. y) 6x [(x 2 dx == --f~ (x.Sec./ (x. y. y) is a differentiable function of the variables dy dx = - Higher-order derivatives are f~ (x. provided that f~ (x. The case of one independent variable. I~ (x. If the equation f(x. \ve find aF of of -dx+-dy+-dz=O. 9] DiUerentiation of / nlplicit Functions 205 Sec.6y [(x 2 + !/)2_1] x + y2)2_1] = -y. taking into consideration the fact that y is a functlun of x' dy l. u. v)=O. we see that az 2x az 1-4y-z ax = y-6z ' ay = y-6z 3°. z)==6z-y. z)=2x. y.Funct ions of Several Variable') 206 Example 2. First method. y. ( x. x y dy +-au -a a- Example 3. z) = . z) p'z (x.(x. we get iJz ox = F~ (x. y. F. Denoting the left side of this equation by F we find the partial derivatives (x. The equations u+v=x+y. iJu au do av as functtons of x and y. u.(x. 2 Solution. (3) . v as functions of the variables x and y and the Jacobian aF of D (F. the total differential of the inlplicit function: dz =2xdx+(1-4Y-2)dy . y. 6 if x 2 _2 y 2 + 3z -yz + y== o. y. Dlfferentiating the given equation. y. A system of Implicit functions. z). find -iJ 'iJ--' ~ and x y ux ay . v) I au av :1= 0 ao ao ' au iJu then the differentials of these functions (and hence their partial derivatives as well) may be found from the following set of equations dF aF iJF aF ax dx + ay dy + au du + av dv = 0. y-6z Comparing with the formula dl=:. z)=-4y-z+l. Whence we detennine dz. Find :~ and :~ [Ch. v)=o defines u and. y.dX+~dY. that is. \ve obtain 2x dx-4y dy+6z dz-y dz-z dy+dy=O. { o (x.6z-y. F. ao ao ao { iJO dx +-a du·-+.v dv =0. 0) D (u. !I. y. z) 1-4y-z 6z-y Second method. y.(x. F. If a systelTI of two equations F(x. z) 2x F z. dz ay = - F~ (x. Applying formulas (2). xu+yv=l define u and t1 . iJ-v dv. If a function z of the variables x and y is represented paralnetricatly by the equations x=x(u. x du +u dx+y dv Solvin~ this system for the dIfferentials du and dv. y) D (u. v) and D (x. Differentiating both equations with respect to x. v) then the differential of this function nlay be found froln the following system of equations ax ax dX-a-dll-f--a dv. 91 207 Solution. (4) az az dz -== a-u du -1. By differentiation we find two equations that connect the differentials of all four variables: du+dv==dx+dy.DiOerentlation of Implicit Functions Sec. First method. we find the partial derivatives .(v -t. v). y=-=-y(u. az Knowi ng the differential dz ~ p az ox=p and dy-=.q· d~ +q dy. + v dy~. z=z(u. dv ~ (u x-y + x) dx + (v +x) dy • x-y Whence 4°. Parametric representation of a function.y) dy . we obtain \\'hence Sim ilarl y \ve find au v+y av v+x dy=-x-y' ay=x-y Second nlethod. we obtain du= _ (u + y) dx-t. II ay v ay dy = au da + av dv. v). { dy= 2u2du + 2v 2dv. From the first two equations we determine du and dv: du=. into formula (5). First method.6v). ax= Second method.6 Example 4. ax ax ox aZ=3u 2 au +3v2aV . az Find ox and ay· Solution. By differentiation we find three equations that connect the differentials of all five variables: dx=du+dv. 'Substituting the expressions :: and :.U 2(U-v)+ v '2(v-u) 3 ( 2 u-i-v).v)' ay 2 (v-u) . we obtain OZ 3 2 V 2 U iJx= U v_u+ 3v u-=v=-3uv. OZ _ 3 2 1 3 2 1 iJy. . we bave: <lz = 3u 2 2v dx-dy + 3v 2 dy. OZ .Functions of Several Variables 208 [Ch. dv==. The function z of the arguments x and y is defined by the (lquations x=u+v. dz -::::: 3u du + 3v dv.: -3uv dx -t-~ (u + v) dYe 2(v-u) 2 Whence oz az 3 -3uv. ay ay ay (5) Differentiate the first two equations first with respect to x and then \vith res pee t to y: l=au + av ax ax' au av { O=2Ll+2v ox ax' From the first system we find au v av u ax=v-u' ax=u-v Fronl the second system we find au 1 av oy 2 (u. y=U 2+V 2 .2vdx-dy 2 (v-u) . From the third given equation we can find az =3u 2 au +3v 2 av. z=ul+V S (u. ay="2 (u+V).dy-2udx. 2 (v-u) Substituting into the third equation the val ues of du and dv just found.2u dx = 2 (v-u) 2 (v-u) = 6uv (u-v) dx+3 (V 2_U 2 ) dy:=. 2y + 3 = O. 1948. Find ~ and ax az ay if x cos y -1. 9] 1941.y 2+2axy==O (a>l). (lfJ!) dx 1945.(a=i=O). Y is a function defined by the equation x!-t.2y' -f-. Find dx and dx 2 If y = x -1. l!)SO. · Find iJz ax and az au for the systenl of values x = .1.y cos z + z cos x = 1.Z3 .yX.209 Dtfferentiation of Impltcit Functions Sec. 1943. Find ~ if y = 1 -\. The function y is defined by the equation In 'Vx 2 -1.y2 =-= a arc tan. (ddx2y ) if x 2-2xy -l.e -xY) = o. Show that ~~ = 0 and explain the result obtained. Let y be a function of x defined by the equation x2 y2 ~+1J2=1. The function z is defined by the equation Xl + y2_ zl-xy = o.xy-ln (e xy -t.In y. · FInd axiJz and azay· 1949. sho\v approximately the portions of the given curve in the neighbourhood of the point x == 1. z = 1.11. . d2 y . The function z of the variables x and y is defined by the equation xa -1. y = 0. 1946. dy 1944.2 === O. x · dy Find dX and d 2y dx 2 • 2 · d dy d y I·f an d -1947 • F 111 dx dx 2 1 -+.y2 + x + y. Find and t=l 2 X=l Taking advantage of the results obtained.3xyz-. · FInd dy d 2y dx' dx2 day and dxS· 1942. where y is a function of x defined by the equation 'I\J(x. Find ::.Z2 = cp (ax + by +. Find dz and d2 z.8xz .bz) ax 1958. if In z=x+U+z-1. y-bz)=O. where F is an arbitrary differentiable function of t\VO argU111entS t satisfies the equation az az ay= 1.Z2 -. y ==. Show that a-a-a-=-I. dx' dxl' dx l or x = 1. f(x. dy dz d 2y d2z f Find ax. z)=O. care constants. y = 0. What are the first. z.0. Find dz and d2 z.z -t.and second-order derivatives of the funct ion z? 1937. Let the function z be defined by the equation x 2+ y2 -1. 1952. Z = 1. az az Show that x-ax+Yay=z. z is a function of the variables x and y defined by the equation 2x 2 + 2y2 -+. where (p is an arbitrary differentiable function and a.cz). 6 ax ayaz y. (cy. y)=O.210 Functions of Several Variables [Ch. y z x 1953. 1954.= 0 • Find dz and d2 z for the values x = 2. Show that the function z defined by the equation F(x-az. b. Xl + 2y l + 3z1 =4.:. F ( z' ==0. a ax + b X zy ) 1959. 1961. Show that az + (az-cx) dY az == bx-ay. 1960. 1956. if x 2 + y 2 + Z2 == al. t 1955.8 .:= I. The functions y and z of the independent variable x are defined by a system of equations Xl +yl-zl =~O. . . Show that the function z defined by the equation y = xcp (z) + 'P (z) satisfies the equation a2z(aZ)I_2~az JX2 ay a z +a Z(aZ)I_ O 2 axayaxay 2 ay2 ax - . z = cp (x y). ~z and adz. vX y z=cv. 10] 211 1962.!" y=b sin cp cos¢. x + y + z == b.if X=llCOSV. 1964.Change of Variables Sec. cos cp. The functions y and z of the independent variable x are defined by the following system of equations: xyz = a. if x=acoscpcos. y=usinv. y = .'P (u. find ~ and ~. 1967. dz. 1963. The functions u and v of the independent variables x and yare defined implicitly by the system of equations u-f-v=x. Sec. 1965. if x=eu+t'. y=eu-t'. 10. The functions u and v of the independent variables x and y are defined implicitly by the system of equations 2 2 u=x+y. Find du. y =. dv. v). d y. the derivatives in them should be expressed in terms of other derivatives by the rules of differentiation of a composite function. z = F (r. y= 1. if x=u+v. iJu a2u a2a a2 u iJu au 02U a2 u a2u ~'aY'~2'~~'~2'~'~'a~2'axaY'~2 au for x==O. The functions u and tJ of the variables x and yare defined implicitly by the system of equations x = rp (a. Calculate uv=y. d2 u. Z=llV. d 2 v. . a) Find b) Find :~ and ~. · az d az Find ax an ay· 1968. sin cp. d z. Find dy. q» where rand q> are functions of the variables x and y defined by the systenl of equations x = . z=csin'1'. v). Change of Variables When changing variables in differential expressions. y=u-v. u-yv=O. V· rln d au au au av a-' x -aY ' a-' x aY · 1966. z=uv. Regarding z as a function of x and y. c) Find dz. ddt ) +a 2ft~y~ 0 or Example 2. dx = d 2x dy 2 . r· Substituting these expressions of the derivatives into the given equation. ax taking y for the argument and x for the function. Transform the equation 2 2 d y dy a2 _ x -d 2+2xd-+2Y-O x x x x=+.(:. r·:. Transfonn the equation 2 d y (d Y )' dy x dx 2 + .) dx . Change of variables in expressions containing ordinary derivatives.(:. Example 1. Express the derivatives of Y with respect to x in terrns of the derivatives of x with respect to yl dy I {lX= dx . Express the derivatives of y with respect to x in tenTIS of the denvatives of y with respect to t.6 1°.dx dx dt d(dY) en dxdX = = -t 2 dy dt ' 1 -72 2 2y _ (2t dy + t 2 d 2 ) (_t 2 ) = 2t a dy + (-I d y dt dt dt dt 2 • dt +. We have dy dy en dy = dt = dx 2 d y dt 2 =!!.~=O. Substituting the expressions of the derivatives iust found into the given eiua tion and replacing x by we get 2y Y td ) 2 1 ( Y 1 t s (2ddt+ 12· dt 2 + "T _ tf)..(t!J!. rL-(dX)3 dy 1 1 dy dy .1 d 2x dy 2 = . we will have x ::~] +(dX)3. Solution. dy 2 dy d ( 2 dx = dx 1) ~ 1) dy d ( = dy :. putting Solu tion.dx =0.Functions of Several Variables 212 [Ch. _-. d:!x d y 2 .r sin cp d(p = dr .r sin cp + Putting into the ~iven equatIon the expressIOns lor x. we will have .y dx= x-y' by passing to the polar coorciinates rcos q>.1 + dy Example 3. dr dq> =-: r. \vhence dy ax . sin (P .r sin ep dep. + r cos q> sIn cos <p or. ~. r sin <p r cos q> ~. Considering r as a function of ep. q> iiq> d. y. and ~.-I r cos ep sin <p dr r cos ep dq> d(p = cos q> dr .Sec. X (dX)2 == o. X== y== r sin cpo (1) Solution. Exaillple 4. Solution. where a=x-at. 10] Change of Variables 213 or. fina 11 y. dy == sin ep dr -t. frorn formula (1) we have dx == cos <p dr .r sin q> r COS cp . Transfonll the equation dy x·+.r cos (V d(j). Take the equation of oscillations of a string (a i= 0) and change it to the new independent variables a and ~=x-t-at. Let us express the partial derivatives of II with respect to x and t in tenns of the partial derivatives of Ll with respect to a and ~. dr . 2° Change of variables in expressions containing partial derivatives. Applying the fonnulas for differentiating a conlpos1te function au all aa all at = act at + a~ \Vl~ get a~ at' au au aa all a~ ax = oa ax + a~ ax' .cos q> dq> . 51 nlpliHca Hons. after dr dcp - .r SI n q> . t' == for the new independent variables._J. Lc t us express tl Ie par t la ax an d ay fI partial deri.In t eftllS 0 f th e · 1 denva . takin~ II = X. and w=l. for the ne\v x Z x funetton. dv = dx2 _ dy• .dx+.!..Xi iJv I ( au . au Therefore. consequently.au . dw -_ ~2 _ dz .vatives ~~ and :~. To do this. On the other hand. Gz.Functions of Several Variables 214 lCh.6 a2 u aaa~ =0. differentiate the given relation- ships between the old and new variables: dII = d x. t·Ives az So Iution.-dy 2 dz=z and. Transfonn the equation x 2 aaZ X == ~_-.- + y2 aaYZ = Z2. x y~ x Z2 ow Ow dw=-du+-dv. Example 5. av ow iJw dx dz -du+-dv=--au av x2 Zl or Whence 2( -1 -dw I i)W) z aw --. Xl au x 2 av y2 cJz 1 ow 1 OW) ax = z Xi . . x Fl~ 69 Transform the following equations.x dx -=0 ' putti ng x == cos t. y y-X- tan ~==--­ 1 lL y' +x .2y a) -dx 2 3 (dY)1 =0 dx ' )! = O. Transform the equation 2 x 2 d y2 dx + 2x ~dx -f.Change of Variables Sec.!_ owall _. The tangent of the angle J! formed by the tangent line MT and the radius vector OM of the point of tangency (Fig. we get X2Z2 (. taking y as the are 1971 gUlllent: 2 d y -t. 10] and 21a oz Zl ow ay= y2 OV • Substituting these expressions into the given equation.y = 0 ' putti ng x === et • 1970. Transform the equation 2 d2 y dy (l-x )d>.2 . dx b) '!Jt dy3 _ 3 (d 2y2 dx dx 1972.. 69) is expressed as follows: .!x Ow) ou +Z2 2 Xl or aw=o OU oW = au Zl · 1969. . Transform the equat ion az 2 OX 2 - az 2 2ox ay az 2 + iJy 2 = 0. if u=x. Transform the equat ion az az Y ax -Xay = (y-x) z. the formula of the curvature of the curve y" K= [1 + (y')2]S/2 • J974. putting u~xy and v=~.216 Funct ions of Several Variables [Ch. in the polar coordi nates x = r cos cp. y = r sin cp. y and the new function w=lnz-(x+y)o 1979. y = r si n cp. v = JL for the new independent variables and z x w= -x for the new function.Y ay 2 =0. Transform the equat ion 2 2 az 2 02Z X dx z . 1975. y 1978. v=xo 1976.6 Transform this expression by passing to polar coordinates: x=rcoscp. Or t y tl=x. v=x!+y~. Transform the following equation to new independent variables u and v: az az x-+y--z=O ax ay . taking u = x + y. Transform the Laplace equation a u 02 a ox -to ay = 2 0 2 2 to the polar coordinates x= r cos cp. y=rsincp. 1973. V =~x +J. Express.. by introduci ng new independent variables u =xz + y2. 1977. Transform the following equation to new independent variables u and v: az az y--x-=O ax ay . Yo. When the equation of a surface is represented iIn pI icit Iy. OZ) At --2 . where' (x.1. Yo) Y -Yo == . . applying fornlulas (I) and (2). we will have z-1=2(x-2)+2(Y--1-I) or 2x-j-2y-z-I=O which is the equation of the tangent plane and x 2 2= = y ~ 1= z II. The equations of a tangent plane and a normal for the case of exp 1fcIt representation of a surface. Sec.The Tangent Plane and the Normal to a Surface Sec. I). z)=o. 11. Yo' zo) (Z-lo)=O (3) . (1) Here. Yo. y). Yo) (~X . Transform the equation 02 Z ax 2 + 2 ax02ay Z _L r 02Z oy 2 0 - .. Yo) and X. where w=w(u./ (x o.\1 (2. Z== f (x. The equa bons of the nOrInal are of the fOrIn X-x o . in a rectangular coordinate system. y) is a differentiable function. The normal to the surface is the perpendicular to the tangent plane at the point of tangency If the equation of a surface. Y.. v). Write the equations of the tangent plane and the nortnal to x 2" 2 the sur face Z= Y2 at t he po i 11 t .. which is the equation of the normal. 1 0 == f (x o.\ (xu. Equations of the tangent plane and the normal for the case of implicit representation of a surface.(x o. Z are the current coordinates of the point of the tangent plane. The Tangent Plane and the Normal to a Surface 1n. Let us lind the partial derivatives of the gIven function and their values at the point M oz ax=x. Yo) (Y .= '11 (xo. v=x-y.. Y. W=xy-z. y. f. AI =2 • Whence.~ (x o . The tangent plane to a surface at a point 1\1 (point of tangency) is a plane in wh ich lie all the tangents at the point M to various curves drawn on the surface through this point. Example t. the corresponding equations will have the fOrIn F~(xo' Yo' zo) (X-xo)+F. and F (x o. IS gi\ten in explicit form. ( ax dz (ayi!!) -=-'>'/ cy -~ . Yo.xo) +. 20 ) (Y-Yo)+F~ (X o. 2 0) 0 f the surf ace is Z- 2 0 =-=- f. Yo) Z-zo --I' (2) - where 4Y. then th e eq ua t ion of the ta ngent pIa ne a t the poi nt M (x o.Yo). F (x. . 2°. 11] 217 1980. putting u=x+U. Z are the current coordinates of the point of the nornlal. zo) =0. Solution. a. Example 2. x-O y-a z+a -3a 2 =-O-= -3a 2 or y=y 0 a=zt a.which are the equations of the normal. At what point of the ellipsoid c) x2 u2 t Z2 QI+/jZ+Ci=1 does the normal to it form equal angles with the coordinate axes? 1983. we get -3a1 (x-O) +0 (y-a)-3a 2 (z +a) =0 or x 1-z+a=0. 1981. y=a into the equation of the surface: -z'=al . 1982. Thus. Yo. whence z= -a. Find the z-coordinate of the point of tangency. Planes perpendicular to the x. F. R). we find the partial derivatives and their values at the point M: F~=3yz. Denoting by F lX. zo) F ~ (x o. Yo' [Ch. -2. which is the equation of thp tangent plane. 1984. Solution. z) the Jeft-hand side of the equation.6 (4) 20) which are the equations of the normal.=3xz. to the sphere X 2+y2+ Z 2=2Rz at the point (Rcosu R sin a. -a). Applying formulas (3) and (4). 4. 12) of the sphere r +!I + Zl = 169.3. (F~)M=O. and X-xo V-Yo Z-zo F~ (xo.5). (F~)M= -3aI • F~=3xy-3zl. Write the equations of the tangent plane and the normal to the surface 3xyz-z'=al at a point for which x=O.and y-axes are drawn through the point M (3. Write the equation of the tangent plane and the equations of the normal to the following surfaces at the indicated points: a) to ·the paraboloid of revolution Z=X 2+y2 at the point (1. (F~)M= -3al . Write the equation of the plane passing through the tan~ents to the obtained sections at their common point M. Show that the equation of the tangent plane to the central surface (of order two) ax" + by" +czl=k . x2 y2 Z2 b) to the cone 16+9-8=0 at the point (4. putting x=O. y. Yo' 2 0) F ~ (x o. the point of tangency is M (0.Functions 218 of Several Variables which is the equation of the tangent plane. y=a.4). 1989. R r:r. x2 y2 Z2 1990.0. zo) has the form axox + byoY -~. 1994*. zo)' where X o 7=. .y2 -I.Sec. Show that all the planes tangent to the conical surface z =~ xt ( ~) at the point M (xo' Yo. Sho\v that the tangent planes to the surface Vx + ~/y + -1 V Z-== Va cut off.c- are tangent at the points (0. (b I -1. x2 y2 Z2 1986. on the coordinate axes. 1985. 1988. Surfaces are called orthogonal if they intersect at right angles at each point of the I ine of their intersection. 1987. Show that the cone 2-1--b 2 =-"i and the sphere a c Xl -~ yl -t. On the surface X 2+y2'_ Z2 -2x=-=0 find points at \vhich the tangent planes are parallel to the coordinate planes. 1995. o)? 1992. Sho\v that the surfaces x 2 + !l 2 + Zl = rl (sphere). pass through the coordinate origin. 1\t \vhat angle does the cyl inder Xl -t. segJnents \vhose Slun is constant. Prove that the nornlal at any point of the surface of revolution z = f <V x 2 -l.yl) (/' =F 0) intersect the axis of rotation. Find the projections of the ellipsoid x 2 -1-. 1993. Draw to the surface x2+2!1-~-3z2=21 tangent planes parallel to the plane x -t.cZoz = k.( z - b2 + C2)2 = C2b'l.c) .10 surfaces at the point of their intersection. {fl. 1991. are I11utually orthogonal. 11] The Tangent Plane and the Normal to a Surface 219 at the point M (x o' Yo.y2 = R2 and the sphere (X-R)2 Hl + z· ~ R 2 intersect at the point M (~. c). y = x tan (IJ (plane).zl-xy-l = 0 on the coordinate planes. + b.4y -to 6z = o. The angle bet\veen the tangent planes dra\vn to given surfaces at a point under consideration is called the angle bet'lt'een 1u. '1'. Prove that the tangent planes to the surface xyz = m3 form a tetrahedron of constant volume \vith the planes of the coord inates. Dra\v to the ell ipsoid a2 -~ lJ2 + CZ = 1 a tangent plane \vhich cuts off equal segments on the coordinate axes. and Z2 == (Xl +!I) tan 2 '1' (cone). \vhich are the coordinate surfaces of the spherical coordinates r. 4-!-3. b)+Tj [fx(a. y) = 0... f~ (x. The particular case of formula (1). Taylor's Formula for a Function of Several Variables Let a function f (x. 2) = 0. (2) . (x. y)= (n [ (x . f~~~ (x. y) = -12y. Example: Find the increment obtained by the function f (x.x (n +. 2) = 6· 1 == 6. y)-f- . +_1__). y) =. Solution. f.2=-24.2)=3. y)= :. b) + R b) iJ y (y. y)]-t 2! [h fX. or M(x. (l ) b+ 0 (y-b)] < I). 12. !/). b) (x-a)+fy(a.. Y-1-0k). Similar formulas hold for functions of three and a lar~er nUlllber of variables. f~. y (x.1+3.a) [(x-a) :x + ~ I)! axa+ (y - aJn f (a.!=-21.~ (x. Y)+TI . 1 < 0 n (x. j(x.. Y) -1+ 2hktXY h ax + k ay a +k!ln+l f(x-I-ell. b) (y-b)]-t " 2" " 2 +2!1 lfxx(a. (0 I n other nota tion. b). y) + (n ~ 1) I d n Hf (x+ 011.Ok) (:~) uy Y)+~d2f(X.. \ 3 _ -2y'+3xy when passing from the values x= I. YJ=2+k.\ (1.6 F unct ions of Several Var iables Sec. f: t: x (x. 2): f~ (x.. is called Maclaurin's formula. y)= -6 y 2+3x. y) = . f~~y (x. I.l (x. [h 1 a.220 [Ch.2)=-6. y) == 0. f~~1J (x. y=2 to the values x1--=-I-l-h. f:~x( 1. 2) = 3. !I -+. f~:~{l.. y) = 6. 1. when a=b=O.. f~~y(l.. b) (x-a) +2fx l/(a. df(x. y) = .2=9. Then Taylor's formula will hold in the neighbourhood under consideration: 1 . y) have continuous partial derivatives of all orders up to the (n + 1) th inclusive in the neighbourhood of a point (a. y+k)=f(x.. 2) = 6. !(x+h. fuy (x t:. 2)=-12. First calculate the successive partial derivattves and their values at the given point (1. 2 " [hfx<x. ···+ niI where R n (x. f:y(l. y) = 3x 2 + 3y. 1:(1.12.6x. b) (x-a) (y-b)+fY/J(a. y)] + . f~y(l.b) ~r+J f (a + e (x-a). y)+kfl/(x. . . 2) = -12. The desired increment may be found by applying fOrn1l11a (2). b) (y-b) ]-t. Y)=f(a.. 2) = 0. y) + k2tyy " (x. + ~! dnf (x. + ti1t [a 0JIl f (x.J~( 1. y) c= 3. y)+ ./{I. y+k) in a series of positive integral 0 f han d kif po\ve rs f (x. 2)=TI [h. if f(X. Y)=f(l+h.2. Expand x I =--~ 1 -t Iz. 2000. Expand f(x i h. the function f(x.Z)~--=X2 + !/2-t-z 2 -2xy-2xz-2yz. Expand the follo\ving function in a Taylor's series in the ne i gh bou rhood 0 f the poi nt (1. -1) up to ternls of order three inclusive: .9+k{-21)]+ +~ W·6 + 2hk·3 -I. y) ~ e"t. y)=-x 2+2xy+3y 2_6x. Fi nd the increnlent received by the function f (x. Expand the function f (x. 2004.2bxy -t. t 999. y 1 =-= 1 -t . 1998. 2+k)-f(1. y=l to 1997. 2002. 12] 221 Taylor's Formula tor a Function of Several VarIables All subsequent derivatives are identically zero. Expand the follo\\"ing function in a Maclaurin's series up tot e rIllS 0 for de r f0 uri 11 c Iusive : f (x.k.'(.2y-4 by Taylor's fornlula in the neighbourhood of the point (--.1.yl + z! + 2xy-yz--4x-3y-z-j-4 by Taylor's formula in the neighbourhood of the point (1. k. 1). Expand f(x-+-h. we obtain: 1 Af(x. 2063. sin y. 2001. z) == Xl --t. y) = y.cy2. Expand the follo\ving function in a i\\aclaurin's series up to terms of the thIrd order inclusive: f (x. !I) == cos x cos y.Sec. Expand the follo\ving function in a Taylor's series in the neighbourhood of the point (1. y~ Il. y) = ax2 -1. y) = =-x 2 y \vhen passing frorTI the values x=l. y.k" (-24)] + ~! (h"·G + 3h"k·O+ 3hk"·O+ k l (-12)] = =9h-21k -~ 31z 2 + 3hk-12k 2 +h l -'2k l • 1996. and I. z-l I) in a series of positive integral po\\'ers of h.1). Putt}ng these results into formula (2).!I. 1) 11 P toternlS of ordcrtwo i I1C IllSi \'e: f (x. then the question of an extremum of the function at P (a. b). The case of a function of many variables. y) different from P in a suffici{'ntly sITlall neighbourhood of P the inequality f (a. b) is the nzaximunl of the function f (x. y) [or. then f (a. then there is no extremum at P (a. if A < (or C < 0). b) (0. Sec. f (a. 2°. c) if d 2f (a. b) = 0. Deftnition of an extremum of a function. the necessary conditions for the existence of an extrenlurn ° . b) > f (x. We fonn the dtscrimlnallt t: Then: 1) if ~ > 0. 6 2005. and a mininlunl. 2007. b). y) =0. b) is the Inintmum of the function t (x./ (x. then f (a. Sufficient conditions for an extremum. b) if d 2 f (a. y). y). b) a t the point P (a. 2006*.222 Functions of Several Variables (Ch. at the point of the extrel11tll11 P (a. In the general case. b) changes sign. 1. /"'(l+a)m+(l+~Y' JI 2 ' a ) arc t anl_~' if lal and I~I are small compared with unity. 13. accordingl y.01. b) > 0 for dx 2 + dy 2 > 0. b) == 0 and A = f: x (a. or df (a. y» is fulfilled. b) :-. C ~~ f~y (a. b) does not exist. y). b). b) _-. b). b). If A > 0 (or C > 0). Necessary conditions for an extremum. In similar fashion we define the extremum of a function of three or t110re vanables. Z is an implicit function of x and y defined by the equation z'-2xz+y=O. b) < 0 for dx 2 + dy 2 > 0. b).: = f~ (a. that is. y). which takes on the value z= 1 for X= 1 and Y== I.. namely a maximuln. The points at which a differentiable function f (x. Let P (a. B =--= y (a. b) rema ins open (which is to say. b) < f (x. it requires further investigation). y). df (a. 2) if Ii < 0. The foregoing conditions are equivalent to the folloWing: let f~ (a. or df (a. df (x.: o. Using Taylor's formulas up to second-order terms. Systenl (I) is equivalent to a single equation. then the function has an extremunl at the point P (a. The Extremum of a Function of Several Variables to.98. For a function of three or nlore variables. 4°.95)2. approximate a) vr03.onary points) are found by solving the following system of equations: t (x. y) may attain an extremum (so-called stat. !f for a 11 points P' (x. b). b) be a stationary point of the function f (x. We say that a function has a maximum (minImum) f (a. VO. b) . the function f (x. Derive approximate formulas (accurate to second-order terms in a and ~) for the expressions 1+(1.0 (1) (necessary cClnditions for an extremutn). y) -. y) t~ (x. Then: a) if d 2t (a.. b) IS not an ex tremum of f (\". 3) if l\ == 0. The generic term for maximum and minilnum of a function is ext retnu f11. 3°. y) -==0. Write several terms of the expansion of the function z in increasing powers of the differences x--l and y-l. then t (a. 13] The Extremum of a Function of Several Variables 223 are similar to conditions (I)..\ . y) -t. B = -6. y. C=(002~) = ) x Y PI Y PI =6. A = 144-36 > 0. To find the conditlonaJ extrE:'murn of a function (x.-I).Sec. 2) For the point P 2 : A -= 12. b). the conditional extrenllJ/ll of a function f (. otp iv -a=O' d y ===a-+ y y (2) q> (x. y) = 0 \\'C fonn the so-called Lagrarlge 1"6 110 y) fUllction F (x. L\. 1) For the pOInt PI: A=(Oa2~) \ X p. Let us find t~le second derivatives az az 02 Z =6x a-a x x y =6y. generall y speaking. Solution. ~ -= 144-36> 0. A < n. 8==-12. { xy-2=O. y) -= f (x. Pt(-2. and c) 3°.-2).\.A.~ 28 5<t. A > O. fronl which it is. y =-= l' zmin -~8+G-30-12=-28.8 2 for each stationary point.=AC-B 2 =36-144 < O. At PI thp function has a mininlUnl. y) -:: 0 with th ree unknowns x. 3) For the point P. C= 12. C = -12.ax + iv ocp _0 ox . y). y). P2 (2. Conditional extrenlum.30-t -t. In the shnplest case. I).2. ay a-2=6x.: A=--=-6. P. . The neces~ary conditions for the extrel11UI11 red uce to a systelll of tluC'e equations: aF _~ of aF of ax -. and we seek the ordinary extremUl11 of this auxiliary function. At the point P t the function has a nlaxhnuln equal to zmdx = -8-6 -t. where A is an undeternllned t1lllltipticr.2). given the relationship cr (x. A. A=36-144 <0. There extrelllUl1l. B ==6. ThIs tniniJTIUnl IS equal to the valu£' of the fune t ion for . Example 1.-12. possible to dete nnine these unknowns. Test the following function for an ex tremum: z=x·+ 3xy l-15x-12y. 4) For the point P4: A--=. =6.12 . 2 2 2 and fortn the discrirninant t:. there is no extremUI11 at the pOint PJ. C=-::-6.<p (x. = AC .. 2 oz =6xy-12 =0 dY or x!+y2-5=O. while the sufficient conditions are analogous to the conditions a). Z B=(oO!a =12. Thus. find the partial derivatives and fonn a system of equations (1): axoz ==3x + 3y2-15== 0. =() (coupling equation). y) IS 8 111axinlUl11 or nlininlunl of this function which IS attained on the condition that its 3r~unlents are related by the equatioll (P (x.(-I. Sol ving the systenl we get four sta tionary points: P J (I. 224 Functions of Several Variables [Ch.4x . however. The necessary conditions yield the following systenl of equations: -4+2AX==0. Solving this system we find 5 1v 1 =2 ' and 5 'A . y) has a conditional maxitnum.2' 2Since It follows tha~ .Y. if A > 0 (or C > 0) In similar fashion we find the conditional extrelnum of a function of three or Inore variables provided there is one or several coupling equations (the number of which. y) at a stationary point is positive.. As a particular case. We have :: =-4+2J. y). the problem reduces to finding the greatest and least values of the z-coordinate of the plane z=6-4x-3y for points of its intersection with the cylinder ~2+y2= 1 We fonn the Lagrange function F (x. if d!P > o. Example 2. Namely.y2::= 1 Solution. A obtained from (2) or the condition that dx and dy are related by the equation ~ dx+ :: dy=O (dx 2 +dy 2 =F 0). Find the extrelnum of the function z==6-4x-3y provided the variables x and y satisfy the equation x2 -1..A (x 2+ y! -1). then at this point there is a conditional maximurn of the function f (x. Geolnetrically. the funchon f (x.x. and a conditional minimum. if A < 0 (or C < 0). we have to introduce into the Lagrange function as many undetenll1ned multipliers factors as there are coupling equations. if d' F < 0. { x 2 + y2= 1. and a conditiona I minimunl. y) ~ 6 .=-3+2'J.6 The question of the existence and character of a conditional extremUln is solved on the basis of a study of the sign of the second differential of the Lagrange function: o2F a2 F olF 2 d F(x.3y -1. y)= ox! dx2+2axoydxdY+oy2 d y 2 for the given systeln of values of x. 3+2Ay-=O. nlust be less than the number of the variables) Here. y. if the discriminant A of the function F (x. C::. 5 = 1. x-t-y~-3 Solution. 0). 11--==-1. (Zsm)y=o=- ~ Shnilarly. 70 ~ 2lJ-x+ 1. When y=:. and \\'e get thc point M (-I.Q \ve get _1/ 2) Z==X2~-X. we find that (Zgr)v=o=6 at the at the point (_1/ 2. Example 3. and.'eral Variables Sec. y~O. If A=.. Determ inc the grea test and sll1allest values of the function z-=x 2+ y2_ xy +x+ y in the region x~o. the function has a conditional minimum at this point. and. 70). Greatest and smallest values of a function.=oo On the straight line x+ y=. 1) Let us find the stationary pOints: I '\\ z't == 2x-y l'y -to 1 =0. Thc indicated re~ion is a triangle (Fig. 6". 8-1900 . 3) Correlating all the values obtained of the function z. z5m = -1 at the stationary point M.v. point (-3.. When x-o \ve have Z=y2+ y .=. (32" .: and y=.The Extremum of a Function of Ser. Fig. Similarly . consequenHy. the function at this point has a conditional Inaxilnunl. consequently.-:[ at the pOint \ - nlctres coincides with (Zgr)x=o and (lgr).=o=--{. -3) and (-3. 0) Whl'1l x-ry=-3 or y-==-3-x \VC 3 \VC will have z=3x 2 +9x+6. and the problenl reduces to seeking the greatest and sll1allec..at the POl\1t (0.3 we could test the function for a conait ional extremUlll without reducing to a function of one argunlcnt. we conclude that zJlr= 6 at the points (0. -1) 1\t "1 the valuc of the function zl'1=-1 It is not absolutely necessary to test for an extrenlunl 2) Let us investt~ate the function on the boundaries of the region.~ • then d2 F < 0. (z~r)x+v=-a=6 find that (zsm)x+v=-a=. \\'e find that (Zgr)x=o= 6 at the point (0. Thus 1 16 zmax=6+ 5 Zmin =6- + 59 =11. 16 9 5.0). x=.{. Investigatin~. 13] 5 If A-==2' 4 x"=5 225 3 and Y=r.0.2:~ ) . -3). . A function thaf is differentiablc in a limited closed region attains its greatest (smallest) value either at a stationary point or at a point of the boundary of the region. (Zsm)".t valucs of thiS function of one argulIlent on thE' interval -3 ~ y ~ O.' then d2 F > 0. \\'hence x-=-l. .Z2 for -a2 +-b2 + -c2 = I (a >b > c > 0). z=xy 2022. for Xl + y2 + Z2 = 9. / x2 V y2 1 -7-1)2· 2 2014. z~O. 2028. . z = (x-I)2 + 2y2. y>O). Determine the conditional extrema of the following functions: for x+y= 1. Prove the inequality 2026. 2009. Find the extrema of the following implicitly represented functions: 2019*. U= xy 2 z 3 for x+y+z=12(x>O. 2012. x~o.=x+2y 2023. 2011. 2 y-x = ~ .2y for + 22 x2 2 Z2 x -t. 2010. y~O. z = x4+ y4-2x· + 4xy-2y· .z>O).Functions of Several Variables 226 [Ch. 2=X 8 y 2(6-x-y)(x>O. 2020. u = x. Hint: Seek the maximum of the function u=xyz provided x+y+z=S. 2029. z. u = xyz provided x -f-y+z=5. z=xy 2016. xy+yz+zx=8. U= x+y+z 3 if ~ y2 V-xyz. Z = I +x-y y 1 +X 2+y2 . 2018. z=(x 2+y2)e-ex +y2 ). 2027.y>0. 2024. z = cos l x+ cos l y 2025. x'-y2--3x+ 4y + Z2 + z-8=0. Z = (X-I)2 -2y'. Z = x 2 +y f or "2 x l/ -I."3 = I. 2021. 2013.y2 -J. z>O). 6 Test for maximum and minimum the follo\\ring functions of two variables: 2008. y2 2 02 2 u=x+y>O. U=X 2+y2 +z2. 2 =x2 +xy+ y2 -2x-y. Z= 1_(X +y2)2/ a• 2 2015. x 2+y2+ zl-2x+4y-6z-ll =0. y z 4x +-+-(x>O.xy + x -2z.. Find the extrema of the following functions of three variables: 2017. for X 2 +yl=5. Determine the greatest value of the function z = 1 +x+ 2y in the regions: a) x~O. We seek the maximunl of the function f (x. 71). Let us test the sufficiency conditions. Determine the greatest and smallest values of the function z=x'+y'-3xy in the region O~x~2. y. Finding the Greatest and Smallest Values of Functions Example 1. x+y~a (Fig. It is required to break up a positive nunlber a into three nonnegative nUIIlbers so that their product should be the greatest possible. -1~y~2. y)==x(a-x-2y)=O. the function f (x. According to the problell1. y) is considered inside a closed triangle x~o. Determine the greatest and smallest values of the functions a) z=x1y and b) Z=X I _ y 2 in the region x'+y2~1. x+y~ 1. 141 227 2030. y)=ay(a-2x-y)=O. 2032. y~O. y<:O. Let the desired ntllnbers be x.i) we will have the unique stationary point (. Determine the greatest and smallest values of the function z=sinx-l-siny-t-sin(x+y) in the region O~x~ ~. . 14. y x Fig. 71 Solving the systenl of equations f~(x. x-y~ 1. a-x-y. 31 O~Y~2· 2033. for the tntel'lor af the triangle. b) X~O. y~O. y) ==-:xy (a-x-y).Finding the Greatest and Smallest Values of Functions Sec. Sec. We have 8* . { fy (x. Solution. 2031. 2035. C=fuu (~. 2039. Inscribe in an ellipsoid a rectangular parallelepiped of greatest volume. Fil1d a point M (x. Of all triangles of a given perimeter 2p. Draw a plane through the point M (a. b. the sunl of the squares of the distances of which from three straight lines (x=O. From among all rectangular parallelepipeds with a given volume V. which And so at is to say that the product will be greatest. 2040. 2044. y). this maximunl will be the greatest vdlue. the SUITI m l P t P2 i-m2P2P2 +m aP ap2 ) be the least? 2042. 6 "(a a) 2 A= 'xx 3' "'3 =-3" a. A < o. 2034. rna· For \vhat position " of the point P (x. Represent a positive number a in the form of a product of four positive factors which have the least possible stun. ~) the function reaches a maximum. lipan being revolved about one of its sides. 2038. c) to [orIn a tetrahedron of least volume with the planes of the coordinates. find the one that has the greatest area. Since f (or. Find a rectangular parallelepiped of a given surface S with greatest volume. y) will the quadratic moment (the moment ()f inertia) of the given system of points relative to the point P (Le. y) = 0 on the contour of the triangle. Determine the outer dimensions of an open box \vith a given wall thickness ~ and capacity (internal) V so that the smallest quantity of material is used to make it.Functions of Several Variables 228 Consequent\ y. . by seeking the maximum of the function u ==xyz on the condition that x+y+z==a. find the one whose total surface is the least. all an xy-plane. [Ch.and the 3 aB greatest value is equal to 2f' Note The ploblenl ran also be solved by the l11ethods of a conditional extremum. if X= Y =a-x-y = a . " (a3' 3"a) = . y==O.3I a. For what dimensions does an open rectangular bathtub of a given capacity V have the smallest surface? 2036. generates a solid of greatest volunle. \vhich. 2041.. Given in a plane are three nlaterial points PI (xl' !Jt)' PI (XI' Y2)' P a (x a. x-y+ 1 =0) is the least possible. B=f xy " (a3' 2 a and 3a ) =-3 L\=AC-B 2 > 0. Find a triangle of a given perimeter 2p. 2037. 2043. Ya) with masses m m 2 . Find the axes of the ellipse 5xl + 8xy + 5y 2 = 9. then the quantity of heat released in unit tinle is proportional to J2 R. If a current J Hows in an electric circuit containing a resistance R. At what point of the elli pse x2 y2 £i2+V=l does the tangent line to it form with the coordinate axes a triangle of sIl1allest area? 2046*. Using the Ferlnat principle. 3) to the straight line x y z T= -3=2· 2050*. derive the la\v of refraction of light rays.Sec. Z2). The poi nts A and B are situated in different optical 111e(lla separated by a straight line (Fig. i3 light ill the first 111CdiuI11 is vI' in the second. 141 Finding the Greatest and Smallest Values of Functions 229 2045. Deterilline how to divide the current I into . 2. Find the shortest distance fronl the point 1\. according to which a light ray is propagated along a line A. Through \vhat points will it pass? 2049. 2052*. v 2 • Applying the Ferlllat principle. The beds of two rivers (in a certain region) approxinlately represent a parabola y=x 2 and a straight line x-y-2=O. 73). I t is required to connect these ri vers by a straight canal of least length. The Yclocity of A B I 13 C I Fig. Inscribe in a given sphere a cylinder having the greatest total surface. 2047.\1B \vhich requires the least tillle to cover. 2051.1 (1. 2048. derive the la\v of reflection of a light ray frolll a plane in a homogeneous medium (Fig. Ra. b) if Ii < 0. 2°. 74). y)=2ax+3x2 =0. Here. whose resistances are R1 .2y=O. 76) or of the second kind (Fig. a 11 isolated point if a < 0. then M is an isolated point (Fig. f~ (x o.Functions of Several Variables 236 [Chi 6 currents I l' 12 . so that the generation of heat would be the least possible? Sec. Example 1. Yo) = o. Singular Points of Plane Curves f (x. or an isolated point. f (x. then M is either a cusp of the first kind (Fig. Fig. y) ax 2+x·-y2. 74 Fig. c) if Ii = 0. Definition of a singular point. 77). A t a singular point M (xo. a cusp of the first kind if a = O. 15. Yo). then M is a node (double point) (FIg. . y)=. f~ (x. Yo) = 0. I a by means of three wires. Yo) be nof all equal to zero and then: a) if Ii> O. 75). f~ (xo' Yo) = O. let the second derivatives A == B= ':x f: y (xo' Yo)' (xo' Yo)' C = f~y (xo. Basic types of singular points. to. A point M (x(\' Yo) of a plane curve y)=O is called a singular point if its coordinates satisfy three equations at once: f (xo. R a . Solution. or a tacnode (Fig. Show tha t the curve y2 = ax! + x· has a node if a > 0. 75 When solving the problems of this section it is always necessary to draw the curves. 78). Let us find the partial derivatives and equate them to zero: == f~(x. :% (x. if a > 0./(X. 80).. but the coordinates of the point N do not satisfy th~ equation of the given curve. 81 Hence. 77 Fig. y) = 2a + 6x. then A < 0 and the point 0 is a node (Fig. 80 Fig. 81). A = 2a. 0). C=-2. . 76 Let us find the second deriva tives and their val ues at the pOint 0: . the point M is a cusp of the first kind (Fig. Hence. 79).: a. y=exists only when x~O. there is a unique singular point 0 (0. 0) and N ( . 8=0. y)=O. Hence. ':y(X.Singular Points of Plane Curves 231 0): This system has two solutions: 0 (0. then ~-=O. y)=-2. f. if a::--::O. y v y o o Fig. y x Fig. which is a tangent. the curve is symmetric about the x-ax is. 78 Fig. 79 X Fig. then 1\ > 0 and 0 is an isolated point (Fig. if a < 0. ~=AC-B2=-4a. The equation of the curve in this cas~ will be y2=X' or Y= ± YX'. Whfn solving the problems of this section it is advisable to Illake drawings. y2 (a-x) = Xl (cissoid). 2059. 2°. Sec. 2054. y. . b > 0) (conchoid). and at each point is tangent to sonle line of the given faillily. x +yl-3axy=O (folium of Descartes). (2) It should be pointed out that the formally obtained curve (2) (the ~o­ called "discriminant curve") may contain. (1 ) Eliminating the parameter a fronl the system (1). 2060. (Xl y2) (x-a)1 = b2x 2 (a> 0. which locus IS not part of the envelope of this f am j} y. in addition to an envelope (if there is one). 2061. y" = . The envelope of a family of plane Cllrvefi is a curve (or a set of several curves) which is tangent to all tinps of the given family. then till' Ilarametric equations of the latter are found fronl the system of equations { '. a)-D. CL):::::O dependent on a single variable parameter a has an envelope. y 2 _x 2 _ y" = o. y) =0. 16. Find the envelope of the family of curves x cos a+ y sin a-p =0 (p = const.Functions of Several Variables 232 [Ch. Exanlple. 3 2057. y. 2055. a) =~' fa (x. Determine the change in character of the singular point of the curve y2=(x-a)(x-b) (x-c) depending on the values of a. (x 2+y!)I=a 2 (x"_y2) (lemniscate). Definition of an envelope.x" +x". If a family of curves f (x. we get an equation of the f onn D (x. X 2 2058. p > 0). (Y_X I)2 = Xl. (a+x)yl=(a-x)x! (strophoid).(X. c(a~b~c are real). a 4 y! = a 2 x' - Xl. a locus of singular points of the given farnily. Equations of an envelope. + Consider three cases: 1) a>b. y. 2062. 3) a<b. 2) a=b. Envelope 10. b.6 Determine the character of the singular points of the following curves: 2053. 2056. Find the envelope of the family of straight lines Y --kx+.Y cos a -= O. we eliminate the parameter a: x 2 -1. 2063.P. the envelore of this family of strai~ht lines is a circle of radius p with centre at the origin. 82). { . 2064. This particular family of straight lines is a family of tangent lines to thIs circle (Fig.. y==psina.Sec. Squaring both equations and adding. Find the envelope of ellipses of constant area S whose axes of synlnletry coi Heide. Find the envelope of the family of circles (x-a)t + yt = a. Q. 2068. Form Solving the system for x and y. Find a curve which forms an envelope of a section of length 1 when its end-points slide along the coordinate axes. . Find the envelope of a family of circles of the same radius R whose centres lie on the x-axis. 2065. Find the envelope of a family of straight lines that form with the coordinate axes a triangle of constant area S. 2067. 2k (k is a variable parameter). we obtain parametric equations of the envelope x=pcosu. 16] 233 Envelope Solution.1/ 2 == p2 • x Thus.x sin a 1. 2066. . Th~ given family of curves depends on the parameter the system of equations (1): xcosa+ysina-p=O. y. z=z(t) are parametric equations of the space curve. Sec. b) semicubical parabolas 11 = (x-C)'. \vhere x. d) strophoids (a+x) (y_C)2 =r (a-x).VI" ( lIt + 2 t1 ( dy ) 2 df + (dZ) Cit 2 dt. 6 2069. y=y(t). z are the current coordinates of a roint of the curve. then the arc length of a section of it from t=t l to t=t 2 is S= dx ) Jtr. y o x Fig. 83 2070. 83). Arc Length of a Space Curve The diOerenttal of an arc of a space curve in rectangular Cartesian coordinates is equal to ds = V dx 2 + d!l2 + dz 2 . 17. Investigate the character of the "discriminant curves" of families of the following lines (C is a constant parameter): a) cubic parabolas y=(x-C)'. c) Neile parabolas yl = (X-C)I. The equation of the trajectory of a shell fired from a point 0 \vith initial velocity V o at an angle a to the horizon (air resistance disregarded) is Taking the angle a as the parameter.Funct ions of Several Varl abies 234 [Ch. find the envelope of all trajectories of the shell located in one and the sanlC vertical plane ("safety parabola") (Fig. . If x=x(t). \\'hich are its projections on the coordinate axes: a=a x (t) I +ay (t)J+a z <t) k.Llt) -a (t) dt . zo). 2077. 2076.v) + ~ dt 2 dt ( dt dadtz)1• The end-point of the variable of the radius vector r=r(t) describes in space the curve r=x (t) 1+ y (t) J+ z (t) 11. x2 Xl 2074. y=2sint. from x=O to x=6. Y=T' z=6 2075. 235 The Vector Function of a Scalar Argument 18] In ProbletTIs 2071-2076 find the arc length of the curve: 2071. x=2cost.0) to M (3.3.V. tangent to the hodograph at the corre· sponding point. x=t.S~c. z=t l • Find the mean velocity of motion between times t = 1 and t = 10.0) a 4 a-x to the point M (x o. The vector tUtlct ion a == a (t) may be defined by specifying three scalar functions ax (t). which is called the hodograp1J of the vector r. The Vector Function of a Scalar Argument 10 • The derivative of the vector function of a scalar argument. Xl = 3y. 18. Itit I=dt. ay (t) and az (t).At ~ 0 llt da x (t) i -1. Sec. Yo. from the point 0 (0. dr ds where s is the arc length of the hodograph reckoned from some Initial point. 2xy = 9z t=O to t=2. y=aarcsin~. The position of a point for any time t (t >0) is defined by the equations x=2i. Z=2~' from 3 2072. The derivative :~ is a vector. y=t'. z= a In ~ from the point 0(0. here. . l(da )2 + (da.0.2).day dt (t) J + dar (t) k dt dt· The modulus of the derivative of the vector function is x I 1= . z=-t 3t 2073. x=etcost) y=etsint. The derivative of the vector function a::::: a (t) with respect to the scalar argument t is a new vector function defined by the equality ~ _ lim a (I -1. For example. z=et from t=Otoarbitraryt. y=lnt. from t==O to t='Jt.0. I~~ 1= 1. (I) Detennine the trajectory of motion. Example t. 2=3i 2 • Eliminating the time i. we find the velocity dr ([[=. where m is a constant scalar.8ij+6tk and the acceleration d2r di 2 =-Bj+6k. 2 . 1) 2) d da db de dt (a+b-c)=dT+(jf-df .6 is the velocity vector of the extremity of the vector r. 2°. the velocity and accelera tion. The nlagnitude of the velocity is I~~ 1= Y(- 81)2+ (61)2= 10 I t I· We note that the acceleration is constant and l~ I::~I= Y(-8)2+ 6 =10. d da db 5) (if (aXb)=dtXb+aXdt. ~ (ma)=m d dq> ~~ 3) (if (cpa)=(jf d 4) (j[ . and :. then ~ =(/ =tv [Ch. The radius vector of a moving point is at any instant of Hnle defined by the equation r= i-4t 2j+3t 2k. Solution. y==-4i 2 . differentiatin~.~= ~~ is the acceleration vector of the extremity of the vector r. Basic rules for differentiating the vector function of a scalar argument. From (1) we have: x== 1.Functions of Several Variables 236 parametl~r t If the is the time. db (ab)=Fb+adt. d 6) F 7) a da a [ep (i)] =dfP ~~ dq>. we find that the trajectory or nlotion is a straight line: x-I y z -0-=-4=3"· Frolll equation (1). da a+ cp dt' da where cp (i) is a scalar function of t. · F' =0. if lal=const. The equation of nl0tion is r = I cos a cos rot -f. c) r = a cos t + b sin t. where r 1 and r l are radius vectors of two given points. . where t is the tinlC. with respect to the paranleter t.j sin a cos rot + k sin rot.Sec. 2081.. d) r=acosht-t-bsinht. the vectors a and b are perpend icular to each other. \vhile aO (t) is a unit vector. Find the derivative vector-function of the function a (/)::= a (I) a O(I). \Vhat are the nlagnitudes of velocity and acceleration and \vhat directions have they for time t = 0 and t = ~ ? 2085._~---------~----.::::.tSj+ k. t '-0 -T and t = ~ · 2084. 18] The Vector Function of a Scalar Argument 237 . and c. The equation of nlotion is r = 21 cos t + 2j sin t -J. l b) r = at -t. 2083.. 3) in length and in direction (general case). derive a formula for differentiating a fnixed product of three vector functions a. where a and (t) are constants and t is the tilne.-r t = (rl-r 1) t. Show that the vector equation r.. Construct the trajectory of nlotion and the vectors of velocity and acceleration for times. b. b=2ti-j+ t 3 k.3kl. is the equation of a straight line. and c are constant vectors. 2080. the velocity and the acceleration. for cases \vhen the vector a (t) varies: 1) in length only. Determine the trajectory of motion and the lTIagnitudes and directions of the velocity and the acceleration.. 2082. Using the rules of differentiating a vector funct i::>n with respect to a scalar argument.. c = . Deterlnine the trajectory of ITIotion. 2) in direction only.-------- 2078. Dctcrll1ine the trajectory of nlotion.4j sin t. Determine which lines are hodographs of the following vector funct ions: a) r=at+c. The eq uat ion of nlot ion is r =---= 3i cos t --~.tli -f. Find the derivative. b. t = 0. of the volunlc of a parallelepiped constructed on three vectors: a = i + tj+ t 2 k. the velocity and the acceleration. where a (t) is a scalar function.:. where a. 2079. Interpret geolnetrically the results obtained.bl. The Natural Trihedron of a Space Curve At any nonsingular point M (x. y=asin6. N \'=iNi . where 6 is the turning angle of the screw. 84): 1) osculating plane MM t M 2 • containin~ the vectors c:. 3) the buzofltlul MM" all of which are defined by the appropria te vectors: 1) T= ~~ (the vector of the tangent line). The equation of motion of a shell (neglecting air resistance) is where V o {vox. y. 2) B=dfX dt2 (the vector of the blllormal). = canst). 2088. Prove that if a point is in motion along the parabola y =~. 2) the principal normal MM z . 19. d 2r . and 3) rectifying plane MM l """ which is perpendicular to the first two J1lanes. B ~=iBi. then the acceleration remains constant as well. Deternline the velocity of the point. Find the velocity and the accelerat ion at any instant of time. 2087. v oy ' v oz } is the initial velocity. A point lying on the thread of a screw being screwed into a beam describes the spiral x=acos6. dr The corresponding unit vectors T -r=iTi. a is the radius of the screw. At the intersection we obtain three straight lines.. z == 0 in such a manner that the projection of velocity a on the x-axis remains constant (:. z) of a space curve r==r(t) it is possible to construct a natural trihedron consisting of three ITIutually perpendicular pI anes (Fig. z=h6. and ~:~. 3) N = Bx T (the vector of the principal normal).Functions of Several Variables 238 • (Ch. Find the velocity of a point on the circumference of a wheel of radius a rotating with constant angular velocity ro so that its centre moves in a straight line \vith constant velocity v O• Sec. and h is the height of rise in a rotation of one radian. 1) the tangent MM l . 2) normal plane MM:M •• which is perpendicular to the vector '::. 2089.6 2086. Z are the current coordinates of the point of the tangent. then the equations of the tangent have the form X-x TJI: Z-z Y-y (1) =r. v= I:: I. T y= :~-. d1r dtl == 2J +6tll. Ny.=r-. Example 1. the principal nornlal and the binor· n1al at this point. U=t 2 . . 19] The Natural rrihedron of a Space Curve 239 • may be computed from the formulas d'f tiS dr ~=dS . We have r=tl+tIJ+tl/l and :~ =1+2tj+3t 1k. (2) If in C'quations (1) and (2).-' Rectlfying plane N dr T=at Fig. Solution. we get the equations of the binornlal and the princij'al normal and. from the condition of perpendicularity of the hne and the plane \ve get an equation of the normal plane: Tx(X-X)1-Ty(Y-y)+Tz(Z-z)=O. T z= :: . \' and Ii of th(l curve x=t.'(. the osculating plane and the rectifying plane. ~4 wh('rc T x -= :. \\re replace T x ' Ty . N z . 1J=~Xv. T z by B. respectively. Bz and N".Sec. Y. If X. z=t' at the point t=l. . B". Find the basic unit vectors T. Write the equations of the tangent. l' p __ 31-3j+k 1+2j+3k Vl4 -- V19 . d2z = 2 "3 dx 2• . . B= dd~ 2 d Xdt~= I' I 1 j2 k3 =6i-6j+2ki N=BXT=I~ JO ~1~_221-16j+18k. 2 d y == - 2 3' dx 2 . Putting x= 1.Functions of Several Variables 240 [Ch. dy. a (x. dz} and d2r {d 2x. z) ==0. 123 Conseq uentl y. v= -111-81+911 V266 • Since for t= 1 we have x= 1. Example 2. If a space curve is represented as an intersection of two surfaces F (x. 2 nlay be considered independent and we can put its second differential equal to zero. we get dr T=d[ =1+2J+3k. we will have xdx+ydy+zdz=O. it follows that x-I y-l z-1 -1. 2==-2.= -3are the equations of the tangent. -2). X+Y+2~-=0 (3) at Its pOInt M (1. Solution. y== I. Differentiating the systenl (3) and considering x an independent variable. d 2 y+d 2 z=O. we get dy==-dx. z= 1. when t = 1. d2z}. y= 1. d 2r dr then In place of the vectors (j[ and dt 2 we can take the vectors dr {dx. Write the equation of the osculatin~ plane of the circle X2+y2+ Z2=::6. z) =0. dx+dy+dz=O and dx· + dy· + Y d 2y + dz 2 + Z d 2z -= 0. and one of the variables x. d2y.= -2.6 Whence. .1 are the equations of the binornlal and x-I -11 = 11-1 2-1 -8 =--9- are the equations of the principal normal. dz=O. y. y. x-I y-l z-1 -3-=-3 = . 1. y. For the screw line x = a cos t. Y == a sin t. -dx. 2090.y2 -t. the osculating plane is defined by the vectors and {D. ~ of the curve z=t at the point t = ~ . Deternline the direction cosines of the tangent line and the principal normal. therefore. that is. 19] The Natural Trihedron of a Space Curve 241 Hence. Write the equations of the planes that form the natural trihedron of the curve x" -1. x 2. O} {a. 1. since our curve is located in this plane. p of the curve at the point x == 2. - {dx. o} ~ dx·. Whence the normal vector of the osculating plane is B=I~a -{ ~I=-I-j-k -1 I and. its equation is . v. y=t'. -I. 2092. as it should be.k) at an arbitrary point.} or {I. Find the unit vectors of the tangent and the princi pal norillal of the conic spiral r = et (I cos t +j sin t i.y2 -}. Determine the angles that these lines Illake \vith the z-axis.I (x-l)-{y-I)-(2+2)==O. 8). and -1. 2095. y=sint. v. idX. 2).Z" = 6. . x+y+z=o. I}. Find the basic unit vectors T. Fornl the equations ot the tangent line.Z2 == 4 at one of its points 1\1 (1. z= 2x 2093. 2094. Find the basic unit vectors T. Z = bi write the equations of the straight lines that forln a natural trihedron at an arbitrary point of the line. 4. 2091. the normal plane and the osculating plane of the curve x=t. z=t' at the point M (2. x==:l-cost.Sec. y=x 2 . 3). Find the unit vectors tangent. K=-= R ~s-+o bas .1. Curvature and Torsion of a Space Curve 1°. the principal normal and the binormal of the curve t2 x=t. z=tV2 at the point t=O.2). y=e. Form the equations of the osculating plane. z=R sint for t= ~ . and binormal at an arbitrary point of the curve tC t' t2 X=4' Y=3' Z=2· Find the points at which the tangent to this curve is parallel to the plane x+3y+ 2z-10=0. X2_ y 2=3 at the point (2. Write the equations of the tangent and the normal plane to the following curves: a) x=Rcos1t. 2101. the principal normal. FOrlTI equations of the tangent. 2097. y2 t Z2 = b 2 at any point of the curve (x o' Yo. Find the equations of the osculating plane to the curves: a) x2_l-Y~+Z2=9. x' = 24z at the point (6. Form the equations of the tangent. z=bt at the origin. b) x 2 = 4y. of the origin. 1). 2103. y=R sintcost. By the curvature of a curve at a point M we mean the number 1 1·I mcp. 1. b) Z=X 2 +y2. y=x at the coordinate origin.f . x 2 =z at the point (I. and the binormal at the princit cos t. x+z=5 at the point (2. 2102. 2V3.2). 9). 9. z=2 at the point t = 2. the osculating plane. x=y at the point (I. the principal normal and the binormal to the curve y2=X. Find the equation of the osculating plane to the curve x=et . Compute the direction cosines of the binormal at this point.6 2096. 2098. zo). 2099 Find the equation of the normal plane to the curve Z=X 2 _ y 2. Sec. principal normal. c) X2 +y2+ Z2=25. c) x 2 + Z2 = a 2 . 2100.. Form the equations of the osculating plane. 20. y=-t. the pal normal and the binormal to the conical scre\v-line x = y=tsint. I.242 Functions of Several Variables [Ch. Curvature. \vhere t is an arbitrary parameter.j a cos t. then ~=I~~I· For the case of a general parametric representation of the curve we have dr Icn 1 d2rl ([t2 (1) I~~r R= 2°. If a curve is defined by the equation r=r (s). then drd 2rd ar ~= ± I ~1=dS(fS2(fSi 1 Q ds (ddsr)2' 2 where the minus sign is taken when the vectors ~~ and direction. Torsion. = . (2) dt X dt2 Example 1.I a dar cos t . R is called the radius of curvature. when not the same. By mean the number X of a curve at a point M we torsion (second curvature) 1 1.V.I a sin t .Sec. then 'Y have the same dr d 2r dar 1 Q= di (fl2 lfii (dr d2 r)2. and the pi us sign. If r= r (t). Solution. If r= r (5).j a sin I. where s is the arc length. d 2r dt 2 = . 0 T=-= 1m ~S-t>O Q ~s ' where 0 is the angle of turn of the binormal (angle of contingence of the second kind) on the segment of the curve AT. dt a Whenc~ 2 dr d r dt X dt 2 = I -a I sin t -a cos t aJcos t kb \ = I ab sin t -Jab cos t +a2i -a sin t 0 .I a sin t +J a cos t +kb. The quantity Q is called the radIus of torsion or the radius of second cart'at ure. f!s is the arc length of this segment of the curve. We have dr di = . Find the curvature and the torsion or the screw-line r= I a cos t + j a sin t + k bt (a > 0). 20) 243 Curvature and Torsion of a Space Curve where cp is the angle of turn of the tangent line (angle of contingence) on a segment of the curve MN. z=l-{1 is a plane curve. then the curve is a plane curve. a sIn t - Hence. 2109. 2107. 2110. Prove that the tangential and normal components of acceleration ware expressed by the formulas dv UJT = dtt'. 2108. Compute the curvature and torsion at any point of the curves: a) x = et cos t. Y = e' sin t. Prove that the curve x=1+3t-t-2t 2 . y. z=at (hyperbolic screw-line). y=a sinh Z t. 2106. where v is the velocity.2x -~.z = 0 at the poi nt (1. on the basis of formulas (1) and (2).Z2 == 1. R is the radius of curvature of the trajectory. 2105. the curvature and torsion are constants.Functions of Several Variables 244 and I-a [Ch. 'W y v2 = R \'.6 acos I a cos t 0 d d 2 dB sin t t b 2 . y=2-2t-t-5t 2 . == et . 1. Find the radii of curvature and torsion at an arbitrary point (x. find the plane in which it lies. y=sint. y2 . ~ and v are unit vectors of the tangent and principal normal to the curve.b2 • Thus. X = Xl = 6a 2 z. then the line is a straight line. 30 Frenet formulas: dv _ T+~ ds--7[ Q' dT _" ([S-7[' dP v ds=-Q"· 2104. b) x=a cosh t. 2=cosh t at the point t==O. 2xz = p2. Prove that if the curvature at all points of a line is zero. for a screw-line. Conlpute the curvature of the following curves: a) x=cosf. 1). b) x 2.y2 -t. Prove that if the torsion at all points of a curve is zero. 3p2y . we get yaz-tbi 1 a R (a2+b2)3/2 = a 2 +b2 and a2b 1 Q= a a2 (a 2 +b = a 2) b 2 -t. . z) of the curves: a) x 2 b) S = 2ay. dt~ dt~= -a c~s t -a sin t 0 = a b. = = ia cos t -1. Compute its accelera- tion w. at times t~O and t=l: 1) the curvature of thetrajectory and 2) the tangential and normal components of the. A point is in uniform motion along a screw-line . .Sec. 2 DetermiRe. 20] Curvature and Torsion of a Space Curve 245 2111. 2112.t j+ tSk. acceleration. The equation of mot ion is r= tl-t.ja sin t + btk with velocity v. y) over a bounded closed region S is the limit of the corresponding two-dimensional integral sum ~inuous ~ f (x. (x) (AB) and y = (J)I (x) (CD) [<PI (x) ~ q>1 (x»). Yk) !J.x. Yk) belong to S. y) dx (S) dy= ~ ~f (xi. Setting up the limits of integration in a double integral. each of wh ich intersects the vertical x = X (x• . The double integral of a confunction f (x. Direct computation of double integrals. y y D I o A A x X. 85 o c x Fig. In the region S. 2°.-xi' ~Yk=Yk+l-Yk and the sum is extended over tho~e values of i and k for which the points (xi. We distinguish two basic types of region of integration.Chapter VII MULTIPLE AND LINE INTEGRALS Sec. The integral (1) ma~ . 86 1) The region of integration S (Fig. 1.l!Yk. X < XI) at only one point (seE Fig. and x=x (XI> x. from below and from abovE by the continuous curves Y= (J).). lim max dXI ~ max dYl' -? 0 i (I) k 0 where ~xi=xi+. 85). 85) is bounded on the left and right by the straight lines x=x.-. the variable x varies from Xl to XI' while the va· riable y (for X constant) varies from Yl = q>l (x) to Ya = q>1 (x). The Double Integral in Rectangular Coordinates 1°. Fig. E. y) dx. y) dx dy = (5) her~.. Deternlinc the lilnits of integration of the integral H f(x. 87 Solution. (Y) ~ f (x. y) dx we consider y constant. y -2 Fig. y)dy. y)dxdy (S) . CPI (x) 2) The region of integration S is bounded from below and from above by the straight linE's Y=YI and Y=Y2(Y.. 1] T he Double Integral in Rectangular Coordinates 247 be computed by reducing to an iterated integral by the formula H X2 (5) Xl CPI (x) ~ dx ~ f(x. where x is held constant when calculating y) dy. y)dxdy= f(x. 1J'2 (Y) 'It 1J'1 (II) ~ dy ~ f (x. '1'1 (y) If the region of integration doe~ not belong to any of the above-discussed types. then an attempt is luade to break it up into parts. each of which intersects the parallel Y = Y (y.. we have Hf (x. As before. Y ~ Y2) at only one point (Fig.Sec. 86). Evaluate the integral 1 1= 1 ~ dx ~ o (x+y)dy. each of which does belong to one of these two types. > YI)' and fronl the left and the right by the cont inuous curves x = '1'1 (y) (AB) and x = '1'2 (y) (CD) ['1'2 (y) ~"" (y) l. Example 1. CPI (X) q>2 (x) ~ f (x. Example 2. in the integral '1'3 y. We have: H YI+X 2 2 f (x. The region of integration ABeD (Fig. 2125. y) dy. Y -I X2 Set up the limits of integration in one order and then in the other in the double integral H{(x. y)dy.2 ~ dx _ f(x. 87) is bounded by the straight lines x·-= -2 and x=2 and by two branches of the hyperbola y=¥1+x2 y-==-Yl+x2 and . it belongs to the first type. ~ dx ~ {(x. ~ dcp + Y/& • 1 51 x dy 2115. 2114 + 2y) dx. y)dy. y)dxdy= (S) ~ . o 2X 2124. y) dxdy (S) for the indicated regions S. a sin q> n 2 2116. y)dy. 2119. o X Write the equations of curves bounding regions over which the following dduble integrals are extended. ~ dy ~ {(x. Solution.7 if the region of integration S (Fig. y) dx. ~ 2120. ~dY ~ (x-l-2y)dx. ~ dx ~ {(x. 0 4. 0 11' I-X:' 1 7. ~ dy ~ (Xl o 1 -I 2 55 dx a (x dy o n -- 0 ! X x2dy 55 1 a cos q> 2 ~ dcp ~ r l sin l (p dr. and draw these regions: 2 2-y 3 2121. X+2 2126. -0 !!~_ I I X 4 1 1 X+8 1 2122.2 (we have in view the region containing the coo rdinate origin). ) dx I Vi-Xl_yl dy. y) dx. a 2118. 0 IO-Y 2 2123. ~ dy ~ {(x. that is. ~ dx x2 I 4 V 2S-X" ~ {(x. 87) is bounded by the hyperbola y2_ X 2= 1 and by two straight lines X= 2 and x=. ~ dx ~ {(x. .Multiple and Line Integrals '248 [Ch. dx ~ r dr. YI+X 2 Evaluate the follo\ving iterated integrals: 2 I 2113. 5dx 1 + y2 • ! y2-4 In I o I 2117. y) dy. 1) and 8 (--·1. B (1. 2135. d) y.0). 2129.1). y) dy. C(O. S is a circular ring bounded by circles \vith radii r == 1 and R == 2 and \\'ith cotTIITlOn centre 0 (0. I] 2127.. B (2. c) x 2 -t. 0). S is a circular sector DAB with centre at the point \vhose arc end-points are A (1. 2128. S is bounded by the hyperbola y2_x. A . y~l. 0). 5). y) dx dy (S) if the region S is defined by the inequalities a) x. 7). D(I. y)dy. 2) and A (1.f) A(~f) o x o x Fig 89 21 a2. Set lip the linlits of integration in the double integral ~ ~ f (x. S is a trapezoid with vertices 0 (0. ~ dx ~ I f (x. 2). e) y~x~y-+ 2a. Change the order of integration in the follo\ving double integrals: " J2. S is a triangle with vertices 0(0. A (2. 0 ~ y ~ a. 1) (Fig.= 1 and the circle x 2 I 1/ 2 == 9 (the region containing the origin is meant). 89). Q 2X . 88). ~ dx ~ f(x. B(-I. A(l. B (2. 1). 2131. y~O.8(1. 0). 0).~x. C(2. 2133. C(O. 1). S is a rIght para boIic segnlent AOB bounded by the parabola BOA and a segment of the straight line BA connecting the poi nts B (--.X o 3x2 2136.1.The Double Integral in Rectangular Coordinates Sec. S is a parallelogram with vertices A (1.(2. 2134. 1). 2) (F~ig. x 2137.-.0). 4). x~-l. o (0.!l 2 ~ x. 2130. 0). x-{-y~l. 1). b) x2-t-y2~a2.~~O. 249' S is a rectangle with vertices 0 (Ot 0). Y a _y2 ~ f(x. 0 2 n sin x 2144. ~ dx [Ch.2) B(O'1)a.. o a ~ f (x.l) e(G. ~dy f (x. y) dy. '1 ~dx ~ f(x.~A(f. y y £(0. 0). where the region of integration S is bounded (5) by the straight line passing through the points A (2. 1) A(2..O)X Fig.. 0). ~ I Y a2 -x2 dx S f(x. o YR2_X 2 R X ! (x. ~ dy S 1 2142... where S is a triangle with vertices 0 (0. ~ dx ~ f (x... l-y -Yl-y2 2 2a 2139. 2) and by the arc of a circle with centre at the point C (0. xdx dy. 91 Hxdx dy... y) dx. y)dy. 90 2146. 90). .. 1). I). and radius 1 (Fig. H (5) A(I. y) dy. o x Fig..Multiple and Line Integrals 250 a 2138. y)dy. 1).. y) dy+ ~ ~ dx R V2 0 f (x. y2 :& 2 Ym 2a 2140. o 0 Evaluate the following double integrals: 2145. Y 2ax-X 4 ~ d. y) dy.. o a -x a Y 2ax-x2 2 2141. and 8(0. B (0.. y)dx.x Sf 2143. A(I. SS:Id~~. 0) lying in the first quadrant. HVxy-yldxdy. where S is a curvilinear triangle OAB bound- (5) ed by the parabola y"=x and the straight lines x=O. " 1 S dx S y o S dy S 11 11 2 a cos y 2 C) 4 dy. where S is a parabolic segment bounded by (5) the parabola y= ~ and the straight line y=x. which is bounded by the x-axis and an upper sem icircle (x .The Double Integral in Rectangular Coordinates Sec. Evaluate the double integral ~ Sxy· dxdy. and 8 (1. + . 2149. -1). where S is a triangle with vertices (5) 0(0. 1). (S) extended over the region S. SSV :xd~ I' a -x-y 251 where S is a part of a circle of radius (5) a with centre at 0 (0. HeYdxdy. 1). Compute the integrals and draw the regions over which they extend: 11: I +cos x a) ~ dx 11: S if sin xdy. y= 1 (Fig. 2148. 1). 2153. 1] 2147. and 8(1. (S) if S is a region bounded by the parabola yl = 2px and the straight line x=p. where S is a triangle with vertices o (0. 2154*. 2151.0). cosx When solving Problems 2153 to 2157 it is abvisable to make the drawings first. x 2150. 2152. 91). o 0 b) Xl sin l ydx. HV Xl - yl dx dy. A (10. Evaluate the double integral ~ ~ xydxdy.2)2 y" = 1. (5) 0). 2158. y) in the 7 = ~ SSf (x. r sin Ip) r dr dip.252 Multiple and Line Integ rals [ChI 7 2155.). (S) -where the region S is bounded by the ax is of a bsc issas and an "arc of the cycloid x=R (t -sin t). Sec. (~) 2159. sin tp. Evaluate the double integral ~ ~ xydxdy. Hint.nlean value of a function I (x. y) of the circle (x-a)2+yl~R2 from the coordinate origin. cp). Find the mean value of the square of the distance of a point M (x. y) to polar coordInates (r. Find the mean value of the function region S{O~x~ 1. 2. Evaluate the double integral Hydxdy. when passing fronI rectangular coordinates (x. (S) -in which the region of integration S is bounded by the coordinate axes and an arc of the astroid x=R cos s t. rC~IOl1 S is the nUIuber y) dx dy. y = . y) == xyl in the I}. which circle is tangent to the coordinate axes and lies in the first quadrant. In a double integral. y) dx dy= ~~ ~) (r cos '1'. Double integral in polar coordinates. (I) . y=R(l-cost). y=R sinS t (O~ t~ . which are connected with rectangular coordinates by the relations x = r cos cp. 2157. 2156*. The . Change of Variables in a Double Integral to. O~y~ f (x. we have the formula Hf ~) (x. Evaluate the double integral dxdy SS Y2a-x' (S) where S is the area of a circle of radius a. and if the Jacoblan ax I-==D(x. Puttin~ x-=r~oscp. each of which is a region of a given type. (q» and '2 (q» [r 1 ({p) ~ r 2 (cp)] are single-valued functions on the interval a ~ rp ~ p. then the double integral nlay be evalualed by the forrnula f3 rJ (CM ~ ~ F (cp. rlrdr where F(cp. where '. which are connected with x. v) I III d Il d v (~') holds true The lilnits of the new integral are detef111incd fronl general rules on the basis of the typc of region S' Exarnple 1..(r sin tp)2 = Vi - (I .Change of Variables in a Double Integral Sec. y::. Y-=¢(ll. v) iJx ay au ay avav retains a constant sign in the region S. y) dx dy (~) it is required to pass froln the variables x. In pas~ll1g to polar coordinates. {f' (a. \v~ obtain: l'1" l-x 2 . if in the double integral ~ ~ f (x.. r) r dr dcp = ~ dcp ~ F (cp. == r) (q» and r = r2 (q». in both directions. (I. continuou~) correspondence behveen the points of the region S of the -. (S) where the region S is a circle of radIUS R == 1 \vith centre at the coordinate origin (Fig 92). If the region of integration does not belong to one of the kinds that has been exalnined. '" (u. Double integral in curvilinear coordinates.. y)== JU D (ct. r)=f(rcoscp. evaluate ) ~ Vl-x 2 -1J 2 dx dy. r sin cp). In evaluating the integral '\ (cP) we hold the quantity cp constant. y by the contInuous and differentiable relationshlOs x:. Solution."y-plane and the points of SOllIe region S' of the UV-plane. r) r dr. it is broken up into parts. 2] 253 If the region of integration (S) is bounded by the half-tines r ==Q and r =. '1 (q» a (S) '2 (cp) ~ F(cp. then the fonnula ~ ~ f (x. y to the variables ll. In the J110re gener al case.=.::rsincp. tI).y 2 -= VI - (r cos cp)2 . u) that establ1sh a one-to-one (and. v).= (~) ~ ~ f [rr (u.~ (a <~) and the curves .t dy. y) d. 2°. 93 2165. Passing to polar coordinates. ~ dx ~ f eVxz+yZ)dy. i- t . ~ dx ~ f (x. o 2162. (5) where S is a semicircle of diameter a with centre at the poin C ( J 0) (Fig. calculate the double inte gral ~ ~ ydxdy. and cp varies from 0 to 231. 1 2163. dq> (5) 0 0 Pass to polar coordinates rand cp and set up the limits of integration with respect to the new variables in the followin~ integrals: 1 1 2 2160.254 Multiple and Line Integrals [eh. x2 -I 2164. Hf (x. (5) where S is a triangle bounded by the straight lines Y=X. x y) dx dy. 93). where S is bounded by (s) + y2)2 = a2 (Xl _ {x 2 the lemniscatE y2). y) dy. y) dx dy. 92 Fig. o 0 Hf (x. y y x Fig. 7 Since the coordinate r in the region S varies from 0 to 1 for any cp. it follows that 21t 1 55 Yl-xZ-yZdxdy= 5 Sr Yl-rZdr=i n. y=-x y=l. 1 Sdx Sf ( ~) dy. 2161. = 2ax. evaluate the double inte· gral ~ ~ (Xl + yl) dxdy. Passing to polar coordinates. Evaluate the double integral of a function f(r. y) dy o (O<a<p uv=y. 2168. (5) where the region of integration S is a semicircle of radius a with centre at the coordinate origin and lying above the x-axis. evaluate the double integral ~ ~ Val-xl_yl dxdy. . Passing to polar coordinates. 2171*. qJ) = r over a region bounded by the cardioid r = a (1 + cos cp) and the circle r = a. 'CS) Xl y2 extended over the region S bounded by the ellipse aZ+/ii= 1 by passing to generalized polar coordinates: ax = y. Evaluate the double integral f f Vl-~-~dXdY. Transform c f3x ~ dx ~ f (x. (xt. (S) where the region S is a loop of the lemniscate (x 2 + yl)1 = at. 2167. evaluate ~ ~ Val-xl_yldxdy. ax and c>O) by introducing new variables u=x+Y. 2172**. (This is a region that does not contain a pole. Passing to polar coordinates. (5) extended over a region bounded by the circle Xl + y.Sec. r cos CPt b = r sin qJ. . evaluate 11 a 2 -x2 a ~ dx ~ o VXI+y1dy. Passing to polar coordinates.yl) (x ~ 0). 2] 255 Change of Variables in a Double Integral 2166.) 2169. 0 2170. 4 Compute these areas. The area of a plane region S is S=~~dXdY. Evaluate the double integral ~ ~ dxdy. Computing Areas 1°. cP (x) then b "I' (x). v=x-y in the integral t t ~ dx ~ f (x. dr. Area in rectangular coordinates. -] Va 2 _y2 a X+2 2 b) ~ dy x2 ~ dx. If a region S in polar coordinates rand q> ~ cp ~~. f (cp) ~ r ~ F (cp). Construct regions whose areas are expressed dy the integrals n arc tan a) ~ n 2 8 dcp sec cp ~ 2 . o 0 2174**. Change the variables u=x+y. b) ~ drp n a a (J+coc. a-lJ Evaluate these areas and change the order of integration. x == ar cos CPt Sec. then is defined by the inequalities a ~ S =~~ ~ r dcp dr = a (S) F (cp) ~ dcp f r dr. Make the substitution Y = br sin (p.256 Multiple and Line Integrals [Ch. 7 2173*. . (S) If the region S is defined by the inequalities a ~ x ~ b. (q» 2175. '\I> (x) ~ dx ~ S= ~y~ a dy. Construct regions whose areas are expressed by the in- tegrals a) ~ dx ~ dy. ~ a cp) r dr. (S) where S is a region bounded by the curve X2 a2 ( £_ y2 Y2)2 _ 2 - +b h2 k2 • Hint. y) dy. q> (x) 2°. 3. Area in polar coordinates. 2176. 2181. 2180. Find the area bounded by the parabolas y2 = lOx -J.25 and y2 = -6x + 9. Find the arpa bounded by the line y2 ) -+-4 9 X2 ( 2 _ x2 ----a· y2 -. Compute the area bounded by the straight lines x = YI x=2y. O<a<~). y2=ax. the parabola y2=4ax. Find the area bounded by the ell ipse (x. and tI.4 2185*. and put . x+3y=a(a>O). y2=~X(O< <a<b. Introduce lhe nc\v variables u and XY=U. l 2184. Introduce the new variables u. 2178. find the area bounded by t he lines x2 _t y 2==2x. x2 -i. and put x 2= uy. y2 === bx. xy = ~ (0 < a < b.3)2 + (3x -1. Find the area of a curvilinear quadrangle bounded by the arcs of the parabolas x 2=alj. Find the area bounded by the stra ight 1ine r cos rp = 1 a nd the circle r == 2. Hint. 2186. y2=VX. y2 =-= vx. y=x. O<a<~). Find the area of a curvilinear quadrangle bounded by the arcs of the curves !J2 = ax. .y 2=4x. y=O. x+v=a.257 Conzpu tintZ Areas Sec 8] 2177. r:ind the area bounded by the curves r = a (1 -~ cos (r) and r =-= a cos fl (a> 0). 2182. and the straight line x-!-y=3a. xy === a. Passing to polar coordinates. Hint. x 2=by. 9 -1900 tI.1)2 =-== 100. 2179*.2y t.4y. Compute the area bounded by the ellipse (Y_X)2 -~r = 1.) 2183. 2187. (The area is 110t to conta ina pole. Conlpute the area lying above the x-axis and bounded by this axis. Find the volume of a solid bounded by the elliptical paraboloid z=2x 2+y2+·t. z=O.O. 4. 0 2 (4 . reason geometrically to find the 0 value of this integral. Z Z y y B (~I. Compute its volume. 94 Fig. 95).2 2191.x . y) dx dy.0. 1) (Fig.x. 7 Sec. the plane x+y=l. 95 In Problems 2189 to 2192 sketch the solid whose volume is expressed by the given double integral: 1 I-X 2 2189. 2-X 2193. 8(1.O. ~ dx ~ o V 1.0) and C(O. Ccmputing Volumes Thp volume V of a cylindroid bounded above by a continuous surface z = f (x. A solid is bounded by a hyperbolic paraboloid z=xl_Jt and the planes y=O. and the coordinate planes. n 2192.y) dy. b£ 10\\' by the pie' ne 2 =.x .O) x Fig. and on the sidrs by a righ1 cylindrical surface.0 0). A(I.:Ki ~ Va! -r-y 2 dy. Sketch the solid whose volume is expressed by the ina tegral ~ dx o l/(l'2. 1. y). which cuts out of the xl/-plane a region S (Fig. 94). ~dX~(l-x-Y)dY. is equal to V= ~ ~ t(x. o 0 2-X 2 2190.O). x= 1.Multiple and Line Integrals 258 [Ch. 21i:S.~dX ~ (l-x)dy. 2 ~ dx ~ 0 (4 . Use a double integral to express the volume of a pyramid wilhvertices 0(0.y) dy. tS) 2188. . Set up the limits ot integration. 2194. Cornpute the volurne of a solid bounded by the xy-plane. a b a a 2211. Find the entire volume enclosed between the cyl inder 2 2 Z 2 X + y2 = a and the hyperboloid x + y3 . x2 2201. Z = ax. t Xl -t.. • • . In what ratio does the hyperboloid x 2 +y2_ z2=al divide the volUIlle of the sphere x 2 -+ yl + Z2 ~ 3a2 ? 2212*. x+z=6. and the cyl i. y>O). Z = O. y). 5. z==O. 2207. {x+-y=a.tin. xy= 1.z! = _ a .IJ2 -t. 2206. Find the entire volume contained between the cone 2(x 2 tif)-Z2=O and the hyperboloid x 2 +_y2_ ZJ. Find the volunle bounded by the suriaces 2az = x 2 + y'l. y=x. Compute the volunlc of a solid bounded by the xy-plane. y=O.!. Find the volunle of a sol id bounded by the surfaces z=x+y. z=O{x>O.C2 = 1. Computing the Areas of Surfaces The area 0 of a srnooth single-valued surface z = f (x.. =_a 2 • 2205.Z2 = 3a 2 • (The volume lying inside the parclboloid is nleanL) 2208. y==2 J. z=O.y2 == . 2204. z = r -r. C0l11pute thp VOlUI11e of a solid bounded by the xy-plane. C)mpute its volume. z=O.(x:l-f l/~). 2209.Com'Ju.y2. 2az Sec. y=2x. Deterlnine the volume of the ellipsoid x2 y'. az = y2. In Problems 2203 to 2211 use polar and generalized polar coordinates. Y = x2. xy=2.yl -t. the surface z =. +. x+y-j-z=a. z =0.\ 2 !l 2 x2 y2 x the paraboloid z=-2-1-2' and the cyltnder 2+-b2=2-. y = 1. the cylinder x 2 -t y2==2a~t and the cone Xl +y2=Z2.! th~ Afea~ Sec. 2198. 2200. x 2 +. Find the voluln~s bounded by the followiilJ surfaces: 2197. is equal to on the xy-plane is 9* .f faces 259 2196. 2203. Find the volume of a solid bounded by the paraboloid == x 2 -+ y'l and the sphere x 2 . 2199. 3x+y=a.tlx.nder x 2 + y2 = R J • 2210. y = -a x. A solid is bounded by the cylinder x 2 +-z~ =a3 and the planes y=O. Y=Vx.y2 = 2ax.. x 2 + c2 Z2 b = 1. y=x. 5) of S'1.2. 2a 2202.y 2 _ z2 = a2 • z=o.ae. Z2 £l2 -1.. whose projection th~ region S.. Z = o. y = 0. Z = px (a > Pl. 2220*. Find the area of that part of the surface of the cylinder x 2 + y2 = R 2 (z ~ 0) \\'h ich lies bet\veen the planes z = mx and z = nx (Tn> n > 0). Compute the area of that part of the surface of the cylinder x 2+ y2 = 2ax which lies between the xy-plane and the cone x 2-~ y2 = Z2. If S IS a region in an xy. Find the volurne and the area of the surface of the re111a in i ng part of the sphere. The mass and static moments ot a lamina.y2 + Z 2 =_ a 2 • 2217.. of the and is of the sphere of the 2218. Y)..y2 == R2 are of equivalent size. then the mass M of the lamina and its static .Multiple and Line Integrals 260 2213. The ax is of the open iug coincides with the diameter of the sphere. Find the area of that part of the plane ~ [Ch. 2223* An opening with square base whose side is equal to a IS cut out of a sphere of radi us a. 2216. C0111pute the area of that part of the hel icoid z=carctan JL \vhich lies in the first octant between the cylinx 2 ders x .y2 = Z2 \vhich is situated in the first octant bounded by the plane y -1. Compute the area of that part of the surface cylinder x 2 + if = ax which is cut out of it by the x 2 -t. 2222*. Compute the area of that part of the surface of the paraboloid y2+ z2=2ax which lies between the cylinder y2=ax and the plane x = a. 2219. Applications of the Double Integral in Mechanics to.plane occupied by a lanlina.y2=a 2 and X 2+y2=b 2. Sec. 6.z = a. 2215*. Compute the area of that part of the surface ot the cone X 2_!J2=Z2 which lies inside the cylinder x 2 +y2::::::2ax. y) IS the surface density of the laIn ina at the point (x. and Q (x. 2221*. A sphere of radius a is cut by two circular cylinders whose base diarlleters are equal to the radius of the sphere and which are tangent to each other along one of the dialneters of the sphere. Find the area of the surface of the sphere cut out by the opening" 2224*. COOlpute the area of that part of the surface cone x 2. Prove that the areas of the parts of the surfaces of the paraboloids x 2 +y2='2az and x 2_ y 2=2az cut out by the cylinder x2 -1. Compute the area of that part of the surface x2 11 2 sphere x 2 -+ y2 + Z2 = a 2 cut out by the surface £l2 + fj2 = 1.. 7 + f+ f = 1 which lies between the coordinate planes. 2214. 6] of the Double Integral in Mechanics Applications 261 moments Mx and My relative to the x. M X = ~ ~ YQ(x. Find the nlass of a circular lanlina of radius R if the density is proportional to the distance of a point fronl the centre and is equal to () at the edge of the lalnina. 2228. y.=2 relative to the x-axi~~. If the lamina is hornogeneous. 97).4 and y2 = . The coordinates of the centre of gravity of a lamina. equal to 1x= ~~ y2 Q(x. Find the coordinates of the centre of gravity of an area bounded by the cardIoid r=a(l-t-cos(p). (5) / y= ~ ~ 2 x Q (x. 2231. COlnpute the coordillat{)s of the centre of gravity of the area GlnAnO (Fig. (~) (S) My= ~ ~ xQ (x. y)dxdy. 96).and y-axes are "rxpressed by the double integrals M= ~ ~ Q(x. and its density at any point is equal to the distance of the point froln the leg 0 A. 2230. y) = const.. (2) (5) The l110ment of inertia of a lanlina rl'lative to the ongin is 10 = ~ ~ (X 2 +y2)Q(X. 2226. (3) (. A laIllina has the shape of a right triangle \vith legs 08 === a and OA == b. Find the coordinates of the centre of gravity of a circular sector of radius a with angle at the vertex 2a (Fig.and y-axes arc. 3°. The ITIOments of inertia of a laluina relative to the x. . (1) (5) If the lamina is homogeneous. which is bounded by the curve !J= 5inx and the straight 1ine OA that passes through the coord inate origin and the vertex A ( ~. respectively. then My - <x: Y> is the Mx x=M' Y=M' where M is the tnass of the lamina and At x' My are its static moments relative to the coordinate axes (see to). Compute the coordinates of the centre of gravity of an area bounded by the paraholas 1}2 = 4x 1. Find the sta t ic IIl0I11ents of the lamina relative to the legs OA and DB. 2°. The moments of inertia of a lamina.2x +4."» Putting Q (x. If C centre of gravIty of a lamina. y)dxdy. 2227. then in fonnulas (1) we can put Q = 1. x=2. 2225. y) dx dy. \ve get the geolnetric moments of lncrtJa of a plane figure. y) dx dy. y)dxdY=/x+1y.Sec. C01l1pute the rnoment of inert ia of a triangle bounLied by the straight linesx+y=2. !/) -== 1 in fornlulas (2) and (3). then Q (x. 1) of a sine curve. 2229. y) dx dy. JX 'Xi -+ max h") max ~zk 0 -+ 0 -+ 0 ~~~f(Xtl YJI i i k z")AXtAYJAz. 2233. passing through its vertex perpendicularly to the plCine of the square. and b) relative to its diameter. C )mpute the moment of inertia of a segment cut ofT the parabola y2 ==ax by the straight line x=a relative to the straight line y=-a. 7... 2237.(x. Triple Integrals to. 2238. 7 Multiple and Line lnfe/!ra!s 262 2232. relative to the x-axis. 2a 2 cos 2(J) rela tive to the axis perpendicular to its plane In the pole. Triple integrals in rectangular coordinates. 96 Fig. the density is proportional to the distance from one of its vertices. Compute the moment of inertia of an area bounded by the hyperbola xy=4 and the straight line x-t-y==5 relative to the straight line x=y. The triple inte~ral of the function !(x. C3mpute the moment of inertia of a homogeneous lamina bounded by one arc of the cycloid x=a(t-sint). . In a square la~in1 with side a. 2234i:. Sec. y=a(l-cost) and the x-aXIS. ~) extended over the region V is the limit of the corresponding threefold iterated sum: ~ ~ ~ .. Cumpute the moment of inertia of a square with side a relative to the axis. 2239*. C)mpute the nlOment of inertia of the lamina relative to the side that passes through tftis vertex.=.[Ch. Compute the moment of inertia of the area of the lernn iscate (2. Find the monlent of inertia of an annulus with diameters d and D (d <D): a) relative to its centre. y y o x Fig. 2236. 97 2235*. ~'. y. V z)dxdy dz=lim m.. Find the moment of in~rtia of the cardioid r=a(l +cos<p) relat ive to the pole.-:. als Evaluation of a triple Integral reduces to th~ succeSC.C2 = 1. -a 1/ x lJ2 + (2Z~ -= 1. O~Y~x. z =X (u. y. In both direLtiolls. Solution.!I.ts of the region of integratlon V 01 Xyz-spdce and the pOints of sOlne region V' of UVW -space. We have 1 1= x 1 XII Jt.nbc Sx l-~) dx=~na3bC. Evaluate ~~~ 2 x d x dy dz. whlr~ the functlon'\ q>. a dxdydz=. 0 0 0 0 0 r xI' =JdX.lve comput'. z bv the rl-'I at ion~ x = q> (u. z) d. 7] Triple 'nte.~.dY =J 25 0 0 \1..Jt1on of th~ three ordinary (on~fold iterated) integrals or to the COinputatlon of one double and one sin~le integral. and is equal lo l-~). w. a ~ ~ ~x 2 dx dy dz -= ~ a 2 x dx -a H') ~~ (5 \Ve therefore ti lla 11 y SSSx 2 -V l-~. contil1uJu. x 10 1 dX-=tj lod~=lll)' 0 0 Example 2.C -V l-~=nbc Syz=nb ~el 2 x2 S yz dx.!jec. 2) in one-to-one (lnd. Solution. v where the region V is defined by the inpqualities O~x~ 1. . Change of variables In a triple integral. w).. v. 2 ( -a (V) 2°. If in th~ triple il1h"\~rat ~ ) ) f (x. '1'. 2 to the variables u. y = 'I' (u. COlllpute J = ~ ~ ~ X 3y 2 z dx dy dz..) correc. Example I. which arl~ conncct~'d with X. v. w).t d y d z ( /) it is required to pass from th~ variables x.> ( const. X ar~: 1) continuous together with their pal tial fir~t derivatives. o r X'II" Xl 1/' xI 2. O~z~xy. v. II. v.. (V) t~xtel1drd x2 over the volulne of the ellipsoid Q2 -t- 2 11 Z2 b 2 +. w).pondence bet\veen the poi.Q2 'x= 2 \\ here S JlZ IS the area of the el1i pse ~ dy dz = u.. XII j dx 5dy Jx3y2z dz 5dx 5xV f Idy= = . ~. X(u. z) dx dy dz = (V) = ~ ~ ~ f Irr (u. x = r cos ¢ cos q>. (\I') z o~ __-+ h _ y x x Fig. 99). .264 Multlple and Line Integrals [ell. r (cp is the longitude. Passing to spherical coordinates. r the radius vector) (FIg. ¢(u. For a sphere. and r (radius vector) will be . w). 7 3) the functional deternlinant (Jacobian) of these functions ax au /= D (x. cp. we get I -= r. 2) for spherical coordinates cp. 98). v. compute 2 ~ ~ ~ yX 2 +y2+ Z2dxdydz. z ::::. 1) for cylindrical coordinates r.!l.= D (u. Z 'p the latitude. \vhere =. y sin (P. v. (V) \\there V is a sphere of radius R. 98 fIg. 'I' (latitude). then \\'e can lnake use of the formula ~ ~ ~ f (x. y. w)) III du dv dw. where x == r cos (P. Solution. y= r cos 'P sin q>. w) iJx ax iJu ath ay ay iJy au au dw au av iJw az az az retains a constnnt sign in the regIon V. Iz (FIg.II . = r sin ~" we have 1::-:: r cos 'i'. Example 3. w). the ranges of the spherical coordinates (p (longItude). y. 99 I n part Icular. v. v. y. A. Applications of triple integrals. then \ve can put y (x. z) z dx dy dz. The static l1Z011zenls of the body relative to the coordinate planes are M. y. z) ydxdydz. The volU1ne of a region of three-dimensiona I xyz-space is V=~~~dXdYdZ. (V) ly= ~ ~ ~ (Z2+ X2) Y(x. Evaluat ing triple integrals Set up the linlits of integration in the triple integral ~ ~ ~ f (x. z):. The l1Z011Zettts of inertia relative to the coordinate axes are I x= ~ ~ ~ (y2+ Z2) Y (x. (\/) The 1!laSS of a solid occupying the region V is Mc-~ ~ ~ ~ y(\'. Y. z) == 1 in the fornlulas for the coor(ltnates of the centre of grav1ty. 0 2 3°. z) dx dy dz.SPC 7J Triple Integrals 265 We therefore have n 2". Y. (\') (\') Putting 'V (x. z)dxdydz. y. ~~ ~ y x2-+!l -I.. z) dxdydz. (V) for the indicated rl'giolls V.. z)dxdy dz. z=~. z) dx dy d z . (V) The coordinates of tire celltre of gravlty are i\lyz A'zx - x="M-' A-1 XI' !I:::-:::~. Y. !/. z)= 1 in these fornlulas.22 dx dy dz = H') 2 R ~ drp ~ d1Jl ~ n 0 r r 2 cos 1Jl dr=1tR".(x. 0') Mz. lz= ~ ~ ~ (xt+yt)y(x. \\'e get the geoilletric Jnomen{s of inertia of the body. Y.dxdydz. y. y.!I. Y. z) is the density of the body at the pOint (x. (V) where y (x.\"= ~ ~ ~ y(x. z). Ml'z= ~~ ~ (V) .w ~= ~ ~ ~ . If the solid is homogeneous. (x. 2249. p 2248.x2. z=O. (V) \\. ~ 2 "V x a VQi":X2 o 2246. I-X dz j' dy 0 I-X-II ~ dx ~ dy ~ xyzdz. V is a cylinder bounded by the surfaces x 2 +y" =R 2 . 7 2240. . Evaluate where V (region ~ integration) IS the common part of the spheres x2+ y2 t.2 x2 -2 ( y=O. ~ dx ~ dy xdz. 2250. V is a cone bounded by the surfaces x=O. Sdx . V is a tetrahedron bounded by the planes x+y+z= 1.266 Multinle arrd Lire Integrals (Ch. 2241. 2242*.y2. z=O.~x2+y2 and the sphere x'+y2+z2~3a2.II + Z + 1)3 t (\I) \\lhere V is the region ot Integration bounded by the coordinate planes and the plane x+y+z=l.. Z2 + b = 2"' 2 C Z = c.\ 0 V-U2-_-X'l-_~y3 o I 2247.here V (the region or integration) is the common part of the palabolold 2a. Evaluate ~ ~ ~ (x + Y + Z)I dx dy dz. Evaluate 555 d"ldydz (x -t. Compute the following integrals: I 2244. I 1 r dx rdy r dz J0 0 J 0JVx+y+z+l . f • 4X-I/2 V-22245. V is a volume bounded by the surfaces z = 1. Z = o.. 2243.Z2 ~ R~ and x 2 + y2 + Z2 ~ 2Rz . z=H. Evaluate ~ ~ ~ z dx dy dz. 2~55.I-x2 first transforming it to spher ical coordinates.x' 2 ~ tlx a ~ dy ~ z V x 2 + y2 tiz. n') where V is a volume bounded by the plane z = 0 and the upper . first tra nsformi ng it to cy Ii ndrical coordi nates. x2 11 2 Z2 half of the ellIpsoId -2-t':-b = 1. 2256. Evaluate rs (£i2 + SJ x 2 Z2 ) -f. OliO first transfortning it to cylindrical coordinates. 7] 267 2251. . 2254. 2 +-2 II C 2252.C2 dx dy dz.yl) dz. Z2 = 2Rz. . 2257.l - ~ y2 (Xl -I.I. Evaluate 1":! \.x 2 ~ dy x. (\ ') \vhcrc V is a region bounded by the surfaces x:! -\.Triple Integrals Sec. evaluate ~~~ dxdydz.0. Evaluate ~H zdxdydz. y2 {j2 (V) "There l! is the interior of the ell ipsoid x: + ·b11: + 2:c = a 1. R). Evaluate lr 21x - 2r ~ ~ dx 1 A . -VR.I 4f~J_y~ ~ dy -V~ dz. n') \vhere V (the region of integration) is bounded by the cone 1z2 z! == R~ (x 2 t. Passing to cylindrical coordinates. 2253. Evaluate R ~ dx -R V R2 - V R.y2 ix 2 -I y2 -=-= Z2 and contaIning the point (0.y2) and the plane z == h. . Computing voluflles by means of triple integrals 2259. Applications of triple integrals to mechanics and physics x Fig. ~ Multiple and Line Integrals 2258. b ~ c) (Fig. O::=. O~y~b. evaluate the integral ~ ~ ~ V Xl +y2 + Zl dx dy dz. Compute the volume of that part of the cylinder x 2 + y2 = 2ax which is contained between the paraboloid r + y2 = 2az and the xy-plane. Compute the volume of a solid bounded by the sphere x 2 + y2 + Z2 = a2 and the cone Z2:. Compute the volume of a solid bounded by the paraboloid y2 z~ ~ b!+ CZ = 2 a and the plane x=a. if the denv sity at the point (x. the density varies in proportion to the + .Z2 ~ c2 .268 [Ch. Out of an octant of the sphere x 2 + y2 -I.. z~O cut a solid OABC bounded by the coordinate planes 1 (a~c. Find the mass and the plane : + : = of this body if the density at each point (x. 2267*. z) is (l(x. 2266. 100 2265.z~c.y2 -1. 2262"'. X ~~ 0. 100).-= x2 + y~ (external to the cone). Use a triple integral to compute the volume of a solid bounded by the surfaces y2=4a2 -3ax. Compute the volutne of a solid bounded by the xy-plane. B. the cyli nder x 2 -~. 2261*. z==±h.Z2 ~ x. y.y2 = ax a nd the sphere x 2 + y2 -t. z) is equal to the z·coordi nate of the poi nt. Passing to spherical coordinates. y. Compute the volume of a solid bounded by the sphere 2 2 X +y2+ Z 2=4 and the paraboloid X +y2=3z (internal to the paraboloid). Find the rnass M of a rectangular parallelepiped 0 ~ x ~ a. y~O. (V) where V is the interior of the sphere x2 -I.Z2 == a2 (i nterna 1 to the cylinder). In a solid which has the shape of a hemisphere 2 Xl + y2 Z2 ~ a t Z ~ 0. 2264. y" = ax. y. 2263. 2260**. z)=x-~y-~-z. Z~ C C. 2269*. Hence. \vhosc altitude is h and the radius of the base is a.. Find the nlonlent of inertia of a circular cylinder. Find the centre of gravit~' of the soli d.o'\:l . ~ > 0). Difl'erentiation with respect to a parameter. 2271**. f~ (x. we put a=~ in the latter equation. 2270*. ~):. 8] Improper Integrals Dependent on a Paranleler 269 distance of the point from the centre.= 1 1 1 -2 111 a+2"ln p=2"ln ~ a... We have 0= - ~ In ~ + C (~). Solution.Sec..eJ lJx1 o ! :::z _J. a) dx. evatuate 00 t e-l}x2 _e-.. F {a. 2272**. To find C(~). relative to the axis which serves as the diameter of the base of the cylinder. By diff<?rentiatinl! \vith respect to a paranletcr. a). Improper Multiple Integrals 1°. Sec. radius of basc. 8. et) dx == II Jr fa' (X. and density Q) relative to the diaIllcter of the base. Sho\v that the force of attraction exerted by a honl0geneous sphere on an external nlaterial point does not change if the entire I1laSS of the sphere is concentrated at its centre.. In the case of certain f('stnctlons iJllpOscd on thp functIons f (x. rind the forcc of attraction exerted by a homogeneous cone of altitude It and vertex angle t£ (in axial cross-section) on a l11aterlal point containing unit Inas~ and located at its vertex. Let 00 ~ . e-l'Jo~1 _e-~xl x ~). a) and on the correspondIn~ illlpro!Jcr Integrals \\TC have the Lelb111Z rule 00 r:r: d da JCf (x.. 1 Whence C (P) ="2 tn~.-. o Then ~ 00 aF (a. . Find the mOI1lcnt of inertia of a circular con~ (altitude. a. Improper Integrals Dependent on a Parameter.~ Ina+C(~). Find the centre of gravity of a solid bounded by the paraboloid y2 +2z 2 == 4x and the plane x= 2.. Iz. 2a 0 ~)==..dx==F(a. t\) = _ r xe-'lX I dx= 2a-!. 'I Example 1. aa Whence F(a.dx J (a x > 0. 2268. (2) ) E where Se is a region obtained from S by elinl1nating a snlall region of dla metfr E ihat contaIns thp roint P. y) i~ nonnegative l{(x. othcr\vise it is dIvergent. If a function region S. Passing to ~olar coordi nates for p :1= 1. then for the con· ver~ence of an t1l1rJop~r integral it IS nt'ce~sary and sufficient for the limit on the right of (I) to exist at least for one system of regions a that exhaust the region S. then the corresponding itnproper integral IS called cOllverR. If the inh'gratHI !(x. If (2) has a limit that dops not depend on the tYre of snlall r(3gions elimira t (3d from 5. we have 1(0)= 55 (~ 21t (1 5 ~d. y)~Ol. such reg:ons may be circles of radius f with ccntrp at P. y) dx dy= lim e -+ (S) Hf 0 (5 (x. Let a be a circle of radius Q with centre at the coordinate origi n. y)~O. y) dx dy.IO I-p 0 1(0)= 00 dcp=~ [(1+Q2)I-P_I]. y) dx dy= 11m (S) (J y) is continuous in an unbounded Hf (x. then we put Hf (x. For p = 1 we have . Example 2. b) A di\continuous function. If there is a limit on thp right and if it does not depend on the choice of the rpl!ion 0.270 [Ch. then Urn 1 (0)= lim o then -to S lim I (0) = ~l Q-+fD p- 0 2 P 1 2 (1+r )I. ~ e:t: and the integral converges. I-p and the integral diverges. a) An infinite region. UHn the limit on the ri~ht of (2) is not dependent on the type of regions elitninated from S. SotuHon. Test for convergence (3) where S is the entire xy-plane. (1) -'s(O) where (J is a finite region lying entirely within S. y) dx dy.. If a function f (x. If f (x. But if p> 1. 7 Multiple and Line Integrals 20 • Improper double and triple integrals. The concept of improper double integrals is readtly extended to the case of trirle integrals. the in1rroper integral under cons\deration is called (onverr. y) is every\\'here continuous in a bounded closed region S. oth( rWlse it is divergent.eHl.l)p =5 +~~ yElP = Sd<p 0 Q (l 0 21t o If p < 1. for instance. where 0 -+ S signifies that we expand the region 0 by an arbItrary law so that any roint of S should enter it and rel11ain in it. f (x. excppt the point P (a.enl. then we put Hf (x. b). u Find d) F (p). if Ql) f (x) = ~ e. I' (x). c) f (t) = sin ~t. I) 2277*. . if: f (I) = a) 1: b) t (t) = eat.Improper Integrals Dependent on a Parameter Sec.tf (t) dt. Find > that is. l"aking advantage of the rornlula 1 5xn-'dx = ~ (n > 0). The La place transfornlatlon F (p) for the function is defined by the fornlula f (t) c)CI F (p) = ~ e-t. 271 the integrat 1. Prove that the function +~ LJ C xf (z) = J x 2 + (y-zr' dz -00 satisfies the Laplace equation iJ2 u iJx2 iJ!ll + d!J2 = o. 2276. f (I) == cos ~t. · the integral (3) converges for p 2273. 8] 1 (0)= ISn dlpSQ I'd' 2=rtln(1+Q2). Thus. o compute the integral 1 ~ xn -'1nxdx. o 0 lim 1(0)=0). 2275. x 2274. +r O~CI) diverges.xy' dg (x > 0). Using the formu la Ql) 5eo eval uate the integral pt dt = i (P > 0). 00 00 o 0 Cdx S {x 2 +y2_I_a dr 2 )2 (a>O).x-. \vhere S is a region defined by the inequali(5) x~ ties Y~X2. 2286*.x sin mxdx (u > O.y2 . o Evaluate the following improper integrals: ex> cr. 1. Eval- o 0 uate / by multiplying these formulas and then passing to polar coordinates. o 2285.Y ' dy. 2288. 7 Applying differentiation \vith respect to a parameter) evaluate the following integrals: 00 2278. C arc tan ax d J +x x (1 x. Se-. 2) o 1 2281. X ~ dy ~ eu dx. dy. ~ > 0). ~ dx ~ o y2 1 2284. J 2287.272 M uLUpie and Line Integrals rCh.e-.x' dx may also be written in the form I = ~ e. ~e-ax Si:~X dx (u~ 0). 2283. o eX) 2280. The Euler-Poisson integral defined by the forl11ula 00 00 J= ~ e.. o 00 2282. Se-' X X . o eX) 2279. e-(HIJ) 0 0 Jr·rj »4dxdy _j. ~ > 0).i dx (u > O. Sln(I~)dX 2 x YI-x 2 (/u/< 1).e-.. Evaluate 00 00 00 Sdx ~ dy ~ (x2 + y2 ~ Z2 + 1)2 • o 0 0 . \ve have ~ f (x. The ltmit of this sunl. where S is a circle x 2 + y2 ~ 1. y. . IS c~lled a ILne I1ltegral of the fIrst type Il lim ~ {(xi.) (i == 0. .273 Line Integrals SefJ.. n) that break up the curve C into elenlentary arcs A1 .l. ~. 9] Test for convergence the improper double integrals: 2289**.' where S is a region defined by the ine· (5) quality x 2 -1 y2 C ~ 1 ("exterior" of the circle). dx dll where S is a square Ix!:=. (S) 2290.. Y --= 'l' (t) [a ~ t ~~]. Sec. 2. . y) ds = ~ {(x. (xi. ~ ~ In V Xl -+ y2 dxdy.':-. y... 1. y. then this integral represents the mass of the curve C. 555 (X2::'Y:~2)'1. \vhen n ~ 00 and l-I lliax f1s i ~ 0. Line integrals of the first type. where V is a region defined by the JV(X~y)21 2291*. y) ds C 1l"\X. C Ii y) ds = ~ f (cp (t). 55 (X~~d:2)'1. 9. These integrals are evaluated in li'<e fashion A line integral of the first type does not depend on the direction of the path of il1terzratton. Let us construct a system of points 1\1 .M i == ~si and let us fornl the Il integral SUln Sn -== ~ f (xi. er (x» Y 1 + (cp' (xW dx. if the integrand f is interpreted as a linear density of the curve of integration C. Iyl:=..i=1 the arc differential) and is evaluated fronl the fonnula b ~ f (x.) fls . a Also considered are line integrals of the first type of functions of three variables f (x. Yi) !'lsi = (ds IS ~ f (x. 5 2292.. (t» Y <p'l (t) + 1\J'1 (t) dt. 1... (S) 0') inequality x 2 -t. . z) taken along a space curve. a C In the case of paranletric representatIon of the curve C: x==<p (t). y) be a continuous function CJnd y = cp (x) [a ~ \ ~ b) be the equation of some snlooth curve C. Let f (x. Li ne Integrals 1°.y2 -1- Z2 ~ 1 ("exterior" of a sphere).. y) dy = ) (P (x. c where C is the contour of the triangle ABO wi th vertices A (1. Q (. y=lj)(t). q> (x»1 dx. C . a In the more genera) case when the curve C is represented parametrically: %=<p(t). B (0.x. the equation DB is x=O. This integral may he interpreted nlE. 0) (Fig 101). Evaluate the line integral ~ (x+ y) ds. and the equation OA IS Y = O. we have p ~ p (x. y) and Q (x y) are continuous functions and y = (f) (. Line Integrals of the second type. where t varies (roln a to p. 1). the equation AB is y= I-x. ¢ (t» ¢' (t» tIt. y) dy + ) (P (q> (t). y)} along the curve of integrdtton C Example 2. 1 Example 1. 0). 101 x varies. and 0 (0.'chanically 8S the work of an appropriate variable force {P {x. Evaluate the line integral ~ y 2 dx+x 2 dy. If) dx C b + Q (x.\) 1S a sInoa th curve C that runs from a to b as t y B A X Fig. Solution. Here. We therefore have ~ c (x+y)ds= ~ (x+y)ds+ ~ (x+y)ds+ ~ (x+y)ds= AB oA 80 I t I =) Y2dX+) ydY+) xdx= Y2 + I. ¢ (t)) q>' (t) + Q (q> (t). A line Integral of the second type clzanaes stgn when the dtreclton of the path of tl1teg ratton lS reversed. q> (x» + q>' (x) Q (x. If P (x. y) dx + Q (x. o 0 0 2°. C a Similar formulas hold for a line integral of the second type taken over a space curve. y). then the corresponding ltne tntegral of the second type is expressed as follows: • ) P (x.274 Multiple and Line Integrals (Ch. y)-u (.t 2) v.P I A1 (FIt! }(2). Then a long PoP I \ve have y =. y) dx+ Q (x. y) dx + Q (x. Solution. . %0 y) dg. If the inte~rand of a line integral of the second type is a total differential of some sIngle-valued function U ~ U (x. l/) and Q (x. y) dy = (X Ut Yo) x = Y ~ p (x. 1/) lronl its total diflcrcn1ial based on the lise of line integrals (\vhich is yet ~1t-:other method of inte~ratil1g a total differential). and along P 1"'1 \ve have dx:. For the contour of Integration C let us take a brok('11 11l1e P(. y). y). y) dx y) dy == U (x 2 . We have y=b sin t traversed o ~ y2 dx + x 2dy = c ~ {b 2 sin 2 t •( .Ko• Yo) = ~ Q (x o' y) dy + ~ p (x. Yo) dx+ Yo Xo SimIlarly. Uo ~ Q (x.275 Line III feR! als Sec. The case of a total dift'erentfal.' ir partial derlv~ttivcs af the first areier are continuous in S. integrating with respect to PtP"M.Q (x. if the contour of inte~ratioll C IS closed. y) U (x. then this line integral IS not dependent on the path of integration and we have the Newton-Leibn iz formula (xz.::." prcscllcc of conditton (3) (x. y) dy-=dU (x.l ~ Clnd 2) the functions P (r. y) dx. P (x. .'e give a 111ethod of Bnding a function U (x. y) dx -\. clockwise. and fOrlnulas ~ 1) and (2) may prove wrong (s('e Problenl 2. y) totfether \\'Ith th . y) dy =0 (2) C If ]) the contour of inteL!r:ltion C ic. where p(\ (X(I' Yo) is a fixed f oint and M (x. y) IS a variable pOI nt. tho? exi~tellce of the function U is the i. Yl) -U (XI' YI)' (1) (XI' YI) \vhere (XI' YI) IS the initial and (x 2 • y~) is the terminal point of the path In particular. that is.o 'unction U. Yo and dy -= 0.. we have y x U (x.: 0 We get: JiUed.lenticai fulJilrllent (In S) of the equality iJQ uP (3) ax =dy (see jnte~ration of total difrl"rentials) If conditions one and hvo are not ruldocs not guarantee the existence of a sll1l!le-vall1(. y).U (x o' Yo) =-= ~ p (x. then a neCCS"iarv and sufficiel1t co Hhtio 1 fa. Y2) ~ + Q (x. 9) where C is the upper half of the ellipse x=a cos t. n n 3°. P (x. the. cont:lined entirely within some simnlv-connected re£!iO. then ~p (x.a sin t) +a 2 cos 2 t • b cos t] dt = n 0 0 = -abo S sill' t dt + a'b Scos' t dt = ~ ab". 7 Example 3. 0) is an arbitrary constant.t 2+3xy+C. C 2) The work of a force. 0 where C ==. The fo lowing formula for area is more convenient for application: S= ~ § (xdy-ydx)= ~ C §x 2 d (~).276 Multiple and Line Integrals [CIl. Find U. then Green's fOrf1lula holds: 1P dx+Q dy= 55 (~~ -:) dxdy. Z). the work of a force field).U (0.Po (. Sol ution. Yo) I P. {Jo) llo o X X Xo Fig. 5°. Applications of line integrals. y. . y. z) (or. (4x+2y) dx+ (2x-6y) dy=dU. along a path C is z= Z (x. 102 4°.y) . J ) An area bounded by the closed contour C is S= -§ ydx= § xdy c c (the direction of circulation of the contour is chosen counterclockwise). y) are continuous together with their firstorder partial derivatives in the closed region S-t-C. Green's formula for a plane. If C is the boundary of a region Sand the functions P (x. y) = S4x dx+ S (2x-6y) dy+C=2 x2 +2xy-3y2+C o or 0 x y V (x. C (5) here tl'e circulation about the contour C is chosen so that the region S should remain to the left. (x. accordingly. 1/)= S -6ydy+ o S(4x+2y) dx+C=-3 y 2+2.xo. Then 1/ X V (x. having projec!ionsX=X(x. y M(x. Let Xo = 0. y. Yo = O. y) and Q (x. 2). Y=Y(x. y t- . 0).Sec. ~ V-X2-~-'_-y-2 lis. 2297. i\.).t connecting the points 0 (0. where C is an arc of the logarithmic spi· c ral r ~ ae l1lrp (Ill> 0) froln the point A (0. where C is the first arc of the cycloid x = a (t .00.. where C is a segment of the straight line cr x i. z = t ' (0 ~ t ~ 1]. where C is the right-hand loop of the lemc niscate r" = a" cos 2<p. 91 277 Line Integrals expressed by the integral A= ~ X dx c +Y dy +Z dz.:. 2296. c Y = . ) (x +y) ds.C()~ f). \ .' z. where C is the contour of the square I x I + I y I = a c (a > 0). ~ y'ds.r -2 ds . 2299.:) ) Xdx-j-Ydy-!-Zdz= zd (\"1. . Lille Integrals of the First Type Evaluate the follo\ving line integrals: 2293. ". . -=--=z. th. 111' Zl) is the Initial ano (x 2. 2).y) ds.e. a) to the point 0 (. 2 II r x 2295. J xy ds. . ax ij. zd Z2) is th~ tefl111nal point of the path. Vi' where (\"1' Yl' ~ (t 1 . i.sin t). If the force has a potential. \vhere C is an arc of the involute of the c circle x::-:-a(eost i-tsint). z) au iJU -=-=Y. c y = a (1 . ) (x 2 + y2)2 ds. 0) and A (1. is equal to (X J • II'!" A= (X. if there exists a function U ==U (x.-~-. \\'here C is a qua rter of the ellipse £i2 + ~ = 1 c 1ying in the first quadrant.. !J=a(slnl-tcost) \0~t~21t]. (a potential functIon or a force function) such that iJU -=)(. ) (x 1. where C is an arc of the curve x = t. 2300. 2294.d dU=U(X 2'Y2' Z2)-U(X"y. Y'2. 2298. ~ xy ds. Z. iJy dz then the wort" irrespective of the shape of the path C. y=a(l-cost) [O~t~jtl. if the linear density 01 it at each point M (~. 0. of the first 1urn of the screw-l ine x ==. ~ (x l -2xg)dx+ (2xy+y2)dy. 2302. ~ -V 2g -1. 2311. z=x. 2305. where AB is an arc of the AB parabola Y= x 2 from the point A (1.Zl ds. Find the arc length of the conic screw-line C x=ae t cost. y = ae l sin t. y=asint. b)? B.Multiple and Line Integrals ~78 2301. Deterlnine the Blass of the contour of the ellipse ~~ = I. 0) to the point A (a. z=bt) if the density at each point is equal to the radi us vector of this point. ~ 2xgdx-x!dy taken along different paths emanating VA from the coordinate origin 0 (0. find the mass of the first turn of the screw-line x = a cos I. 0) and terminating at the point A (2. 2307. act on a Blass nl located at the point A (0. 2308. z = ae t from the point 0 (0. Z == 0. Y =. Line Integrals of the Second Type Evaluate the follo\ving I ine integrals: 2310. where C is the circle Xl + gl +Zl = a" Z= 1 c x=y. Find the area of the lateral surface of the para bolic cylinder g=iX.bounded by the planes z=O. 2312. 4). Z = bt. where C is an arc of the first c arch of the cycloid x=a(t-sint). 2306.a cos t. 7 ~c xI +d~ + zI ' where C is the first turn of the screw-line y = a cos t. Determine the coordinates of the centre of gravity of a half-arc of the cycloid x=a(t-sint)..a sin t.y) dx \. . a). x dy. y=6. 2304. Find the moment of inertia. about the z-axis. y=a(l-cost) which arc runs in the direction of increasing parameter t. 1) (Fig . With what force will a mass M distributed with uniform density over the circle x 2 -+ y2 == a 2 . 1) to the point B (2. x=O. 0. bt . 2303*. ~ (2a. X [eh.•l 03): a) the straight line OmA. 0. y) is :: + ·equal to 'y . 2309. Y = a s j nt. sin xdy taken along the segment AB of the 2316. the axis of symnletry of which is the y-axis.O) Fig. 2i9 b) the parabola OnA. where C is the upper half of the ellipse r: x ~ a cos t. where C is the right-hand loop 01 the lenlnlscate . th (x+ u) dX-(X-L/) dy 2 :Y counterclocl{wise. 2318.. 3) a) xdy+ ydx. AH bi~ector of the second quadrantal angle. d) the broken line OBA. 2 (3.. ) b) ) (0.- (I. Y = b sin t traced clocl<wise. 2) x-ax is). ~ 2xydx +-x'dy as in Problem 2312. + y2 ==al y cl~ 1) t-----II-~ A (2. 2313. the axis of symmetry of which IS the x-axis. 0) (2.z =. Evaluate the line integrals \vith respect to expressions. 103 2315. c) the parabola OpA.Lin~ Sec.l~ 2) (I. I) xdx+ydy. 1) x o B(2.. 9] I nte~ralf.0 COS 2(p traced counterclocl<wise. y dx. 1) d) J.x dy y2 (along a path that does not intersect the . x + taken alonQ the circle x 2 y2 . ~ cosy dx. ~ y'dx+x'd!l. OA 2314*. e) the broken line OCA. point A is 2 and the ordinate of B #X!l(llt~~~: IS if the abscIssa at the 2. (0. which are total ditTerentials: 2317. dIll. c) ) (x+y)(dx+dy). I) ( .. oi) (2. 0) 2320. x+y+x+y + . (a. y) dx (l-x-y) dy]. (x 2 • Y2) f) ~ cp (x) dx -\.1) d) S (v x2x + y2 + Y) dx+ (-V x2Y+ y2 -I.7 (x. I) the straight line y=-x).I ) straight line y = x). Yl) 2319. (1. 2321. Find the antiderivative functions of the integrands and evaluate the integrals: (a. y) S e) dx-t-dy (along a path that does not intersect the x+y (~.'i' (y) dy. (-2. Find the antiderivative function U if: a) du=(2x+3y)dx+(3x--4y)dy. then 1c f (Xl + yl) (x dx+ y dy) = O. -1) (I. 0) b) S x ~~=~)~x (the integration path does not intersect the (0. 2322. Show that if f (u) is a continuous function and C is a 'Closed piecewise-smooth contour.x) dYe (0.)tYdY (the integration path does not intersect (I.Multiple and Line Integrals ~80 [CIL. c) du = eX . I) c) S (x+~. b) du = (3x l -2xy y2) dx _(Xl -2xy+ 3y2) dy.~. Co~pute Xl y2 taken clockwise along the quarter of the ellipse 02+ fi!= 1 that 1ies in the first quadrant. .+) straight line x-~y=O).Y [(1 + x -t. 0) ~ a) (x t + 4xy 8) dx-+ (6X 8y l_5y t) dy. + d) du= ~ !!JL. (x". Evaluate the line integrals of the total differentials: (6. transfOrlTI the line integral l=p1lx2 -1 y 2dx+y[xy+ln(x-+ VX 2+y2)] dy. (1. l z = R sin a.Line Integrals Sec 9) 281 Evaluate the line integrals taken along the following space curves: 2323.xy dz (the xyz integration pa th is si tun teo (1.\2 _1_ y2 -1.z dy +. . C. where C is a tum c of the screw-l ine = a C?S i. (1. . 2325. 1. situated on the side of the xz-plane \vhere !I> O. 1) (:s. 0) ( x.1. 8) ~ a) xdx-'r-ydy-zdz.Z2 (0. b.y2 --1- Z2 = 2Rx. Green's Fornlula 2327. where C is the ci rc1e p c x = J? cos a cos t . = const). 2. (a. Y dx +. ~ (y-z)dx+(z-x)dy+(x-y)dz. = corresponding to the variation of the parameter t froln 0 to 2n. d) 1/. 1. 2324.4 circle x 2 -1. 0.1) i 11 the fi rs t oc tan t). where OA is an arc of the 0. c) ~ b) yzdx+zxdy-i-xydz. Using Green's forlnula. ~ xydx+yzdy+zxdz.. C where the contour C bounds the region S. -I. Y . 2326.x dz. J Y = R cos a sin i. traced in the direction of increasing parameter. l z bt. Z = x. s) c) S x cl~ I_-_~~ . f xy=a slnt..0. -3) (cr.) 5 Xli liZ dx + Zt dr/-I. Verify the re~ult ob1ained by conlputin~ the inteeral directly. while the area. -1). Find (x + y) dx-(x-y) dy directly and by applying Green's ff AmBnA fornlula. 2335*. n)] ds.3). 2330. Consider t\VO cases: +b a) when the origin is outside the contour C. 2332*. 2331. -where s is the arc length and n is the outer norl11al. c taken along the contour of a square with vertices at the points A (1. provided the contour is tra<. \vhere ds is the differential of the arc and n is the outer normal to lhe contour C. Find ~ eXY[yZdx 1-(1 +xy)dy].y sin (X. evaluate the inLcgral ffe _xZy dx + xyz dy.:ea countercl()~kwi~e. . is drawn through the points A (1. Applying Green's formula. 2329. Applying Green's forlTIula. is equal to S.if) dx + (x +y)Z dy. evaluate J= ffc 2 (r -t. b) when the contour enrircl(\s the origin n times. x dl~-!l2d!. Show that If C is a closed curve. 2334. if the points A and B AmB lie on the x-axis. where C is the circle x 2 + y" = R 2 traced counterclockwise. Evaluate the integral ~dx-dy ::f x+y . then l ). bounded by the integration path AnlB and the segrr~ent AB. 0) and B (2. 0) and D(O. ffc cos (X. whose aXIs is the y-axIs and whose chord is AnB. 1). C(-l. n)ds=O. Applying Green's formula.0). find the value of the integral I = ffe [x cos (X. \vhere C is the contour of a triangle (traced in the positive direction) with vert ices at the points A (1. 2333**.7 Multiple and Line Infellfals 2328.282 [Ch. A parabola AmB. 3). n) +. B(O. I). 2) and C (1. B (2. Evaluate tI. ~-~--=. A circle of radius r is rolling without s1idin~ along a fixed circle of radius R and inside It. Analyze the particular case of r == R (cardioid).y)3 = axy. =-= V Xl + y" + z" (Newton attractive force) and the material point moves from position A (a. 2342*. 2346. b. Z=. 2344. 2338.).28~ Line integrals Sec. Applications of the Line Integral Eva1 uate the areas of figures bounded by the following curves: 2336. Z-=-I1Zg (force of gravity) and the material point is nl0ved fronl p0sitiol1 A (Xl' y" Zl) to position B (x~p Y2' zs). y=bsint. Assuming that ~ is an integer. 2340. l'he ellipse x=aLost.- . Find the work done by an elastic force dIrected towards the coordinate origin If the l11agnltudc of the force is propJrtional to the distance of the point flOIll the origlll and if the point of application of the force traces counterclock\vise a quarter of the ellipse ~+~~= 1 Iyin~ in the first quadrant. 2337. Find the \vork done by the forLe of rravity when a material point of nlass nl is Il10ved lrOll1 position . .2xy=O (a> 0)."\ \vork that the fIeld does when a Inaterial poi I1t traces clock\V1Se a quarter of the circle x 2 -t. = ~ (astroid). The cardioid x=a (2cost--cos2t). Analyze the particular case when . to position B (x 2 . y=a (2sintsin 2/). I yi ng i 11 the first quadra nt. Assunling that !i. 9] D.1 (xl' !it' z. Find t:le p)tential rUl1~tion of a for~c R {X. Y. A circle of radius r is rolling \vithout sliding along a fixed circle of radius R and outside it. 2339*. b) X ~ _~t:.y2 == RJ. find the area bounded by the curve (hypocycloid) described by some point of the nloving circle. c) to infinity. 2343. tt:_. The curve (x -t.. y-==asin 3 t. 2345. Y=-=O. r y = _Jl~. find the area bounded by the curve (epicycloid) described by some point of the nlc>villg circle. A field is generated by a force of constant nlagnitude F in the positive x-direction Find th(. is an integer. Y2' Z2) (the z-axis is directed 'vertically upwards). where J1 = canst and . Z} and deternline the work don~ by the force over a given path if: a) X=O. The astroid x=acos 3 t. A loop of the folium of Descartes x 3 +y:l-3. 2341*. 2°. y. z) be a continuolls function and z=-=q> (x. y) a smooth surface S. z) dS = lim . y.SSf [x. The value of this inte~ral is not dependent on the choice of side of the surface S over which the integration is performed.) Mi. z). z) dS =. q> (x. O~y~l. where S is the surface of the cube O~x~l. Let f (x. every strdight line parallel to the z-axis intersects the surface S at only one point.k 2 x. If P = P (x. (0) Example t. !I. S n-+oot=J Yi' Z. where k = canst (elastic force). cos y}. z) are continuous functions and S+ is a side of the snJooth surface S characterized by the direchon of the normal n {cos Cl. Y. If a projection a of the surface S on the xy-plane i~ single-valued.7 X = . Surface integral of the second type.k 2 z. Q = Q (x.Multiple and Line Integrals ~84 [Ch.! surface tntegral of the second type is expressed a~ follows: R = R (x. 10. lj. O~z~l. The surface integral of the fIrst type is the liruit of the integral surn n SS f (x. y) dx dy. lj. Z = . (hen 1he correspondinf. COlnpute the surface integral H s (x+y+z)dS. The desired surface integral is obviously three tilnes greater and equal to ~~ s (x + y + z) dS = 9. Sec. YP Zi) belongs to this elelnent. z). Y = _k 2 y.~ f (Xi. that is. . S y. Surface Integrals to.. the point (xi. y)] VI + cp: (x. then the appropriate surface integral of the first type Inay be calcula ted froln the formula SSf (x. while the terminal point is located on the sphere c) x2 + + Z2 = y2 r 2 (R > r). Surface integral of the first type. Let us comput(3 the sum of the surface integrals over the upper edge of the cube (z = 1) and over the lower edge uf the cuue (z = 0): J 1 J 1 11 H SS (x+ y) dxdy= H 00 00 00 (x+y+ 1) dx dy+ (2x+2y+ 1) dxdy=3. y) + CP~' (x. cos ~. where ~Si is the area of the Ith element of the surface S. and the initial point of the path is located on the sphere 2 Xl + y2 + Z2 = R . and the mctximum diameter of elements of partition tends to zero. S~ P dydz+ Q dz dx+ R dx dy= s+ H (P cosa s +Q cos ~ + R cos V) dS. this integral reverses sign. Z = O. then the direction cosines of the nonnal of th is surface are determ ~ned fron] the form ulas 1 1 iJF 1 cos a = 15 ax ' cos ~ = D ay . 3°.! __ cone 2-1-'2--l2=-=0 [O~z~b). R == R (x. of the surface.:= a. y. z). y. 2351. 10] When we pass to the other side. \ve then have the Stokes' fornzula \v ith 1 c P dx + Q dy + I? dz = =55s [(iJR .iJ iJy Q ) cosa iJz Q + (ap cos ~ + (a . + Zl = a 2 (z ~ 0). Xl H yl Z2 Qi-I. x -I. J .iJP) cos v1 dS az _ aR) 0'" ax iJy . where S is the sphere x· + y2 -1. Stokes' formula.b2+ CZ = 1. where S is the external side of the surface of the henlisphere x" + !It.285 Surface Integrals Sec. z). (4 ') H yzdydz+xzdzdx-l-x!Jdxd!J. H s Xl dy dz -+ yl dz dx + Zl dx dy. y. Evaluate the follo\ving surface integrals of the first type: 2347. and the dircctton of the nonna) is defined so that on the side of the nornlal the contour S IS traced counterclockwise (In a right-handed coordinate ~)steln). Fi nd the l1lass ot the surface of the cu be 0 ~ x ~ 1 o ~ y ~ 1. If the functions P = P (x. \vhere cos a. 2350. ~ ~ V x 2 -I.y -t. F {X. Q = Q (x. where S is the external side of the ellipsoid s y = 0. cos~. S-. 2352. dS. and the choice of sign before the radical should be brought into agreemenf the si de of the surf ace S. z) = 0. 0 ~ z ~ 1. Evaluate the follo\ving surface integrals of the second type: (I 2349. z) is equal to xyz. zdxdy. cos y = D oz ' of iJF where D=± Y(~~r+(~:r+(~~y. cos V are the direction cosines of the norrnal to the surface S. y.z =-. ~ ~ (x' -+ y'.y' dS where S is the lateral surface of the s ~\ 2 t/!' z'.Z2 = a 2 • ~ 2348. z) are continuously differentiable and C is a closed contour bounding a hvo-sided surface S. if the surface density at the point M (x. y. If the surface S is represented implicitl y. s where S is the external side of the surface of a tetrahedron bounded by the planes x = 0. Multiple and Line Integrals 286 [Ch. 2358.7 2353. U. 0). 0). find the given inte~rals and verify the results by direct calculations: 2356. where C is the circle c x Z+y2+ z2=a 2. 8 (0.z = 1. where ABCA is the contour of ARf.yz) dx (yl . Determi ne the coordi nates of the centre of gravity of a homogeneous parabolic envelope az=x' +y2(0~z~a).yZ = 1. c + (Zl - xy) dz. 2359. 2354.x dz. and P -= P (x. x -1. 2355. . Q = Q (x. a). 11. z=a(sint+cost)[O~t~2Jt]. Ij. 2360. 0. The Ostrogradsky-Gauss Formula If S is a clo'ied smooth surface bounding the volume V.4 ~ ABC with vertices A (a. (V) p. C (0.pxdx+(x+Y)dy+(x+y+z)dz. R = R (\'. y. then we have the OstrogrlJdsky-Gauss forlnula SS (P cos a+ Q cos ~ + R cos oS where ers a. :I 2357. X-i"-Y+z==O. cos yare the direction cosines of the outer normal to the Applying the Ostrogradsky-Gauss formula. (y + z) dx + (z + x) dy + (x + y) dz. z) dre functions that are continuous togpther with th-'ir first partial denvatlves in the closed region V. Find the mOinent of inertia of a part of the lateral surface of the cone z = V x 2 + y2 [0 ~ Z ~ h] about the z-axis. 0. y=acost.zx) dy c b) y dx + z dy -+. In what case is the line integral 1=:1 Pdx+Qdy+Rdz c over any closed contour C equal to zero? Sec. where C is the curve c x=asint. :ft Applying Stokes' formula. z). transform the following surface Integrals over the closed surfaces S bounding the \ .cP y 2dx-t z2dy-t-x 2dz. Applying Stokes' formula. y. :Ic (y-z)dx+ (z-x)dy+<x-y) dz. cos S surrac~ y) dS= 55S(:: + ~~ + ~:) dxdydz.. where C is the ellipse Xl -1. z). transform the integrals: + a) p (Xl . then is any fixed H cos (n. 2370. s 2362. H xy dx dy -I. cosp. z =.Zl dxdy.S x2 Z2 55s (~cosa + :~ cos ~ + ~cosy) dS.O. where S is the exters na I total surface of the cone [O~z~b). 236ft Prove that if S is a closed surface and direction.yl dz dx + Zl dx dy. cos)' are direction cosines of the outer lormal to the surface S).. where S is the external s side of the surface of the cube O~x~a. s where n i5 the outer nOftTIal to the surface S. rIll xc('sa+~~ ~~c()S Y dS. y=O. Usin~ the Ostrogradsky-Oauss formula.yz dy dz -+ zx dz dx.Zl cos y) dS.ec. 2367. ~ ~ Xl dy dz -1. cosy are the direction cosines of the outer normal to the surface S. ~~xdydz+ydzdx+zdxdy. O~y~c. I) dS = 0. s where (~oSC. where S is the external side s of a pyranlid bounded by the surfaces x-+-y-l-z=a. where S is the external ~ side of the sphere x!_t-y2--1-z2=a2. cos~. V + y2 + J. 2368 ~ ~ (Xl cos a -j yl cos ~ -I. ~ 2363. 2364. )) Xl dy dz + yl dz dx -I.yl dz dx = Zl dx dy. compute the following surface integra Is: 2~63. 2361. . 2366. x=O. ~ ~ Xl dy dz -I. I)rove that the volunle of the solid V bounded by the sL1rface S is equal to V = -} 55(x cos a + y cos ~ + z cos v) dS. 11] The Ostro[!radsky-Gauss Formula 287 /olume V (l'osa. O~z~a. The divergence of a vector field a (P) == a>. s If S is a closed surface bounding a volume V. Fundamentals of Field Theory 10. y.28R Multiple and Litle Integrals [Ch. then au au au cos~+dz au cosV ar=gradU. 7 Sec. cos ~.i t.=== \jU. y. y. k · tl I d· iJel x day iJa z ~"7 +ayJ+a IS lescaar Iva=-iJx+ay-+az==:va. where P is a point of space and r= xi + yJ+ zk is the radius vector of the pOint P. and az = at: (x. cos p. where \l=i ~ +j:+kO is the Hamiltonian orerator (del. y. Divergence and rotation. 2) are projec. The flux of a vector field a (P) through a surface S in a direction defined by the unit vector of the normal 11 {cos U. Flux of a vector. where ax == ax (x. z). Gradient. cos V} to the surface S is the integral ~~ s an dS = ~~ s an dS = SS(ax cos a -I. if it depends on the time. 3°. z). or nabla). and has length equal to au === -. 12. The vector lines (force lines. . 4°. an Y ax ay \dz If thr directIon is given by the unit vector I {cos a. The surfaces f (x. A vector field is defined by the vector function of the point a == a (P) == =: a (r). 6). it is called nOttstationary. and n is a unit vector of the outer normal to the surface S. lj. is vX' dz vy called the gradient of the field U = f (P) at the given point P (cf./"'(aU)2 + (aU)2 + {UU):!. ChI VI. z) is a point of space. au ay J + az k. In coordinate form. where C = const. 2°. z) = C. z). z The rotation (curl) of a vector field a(P)=axl+ayJ+azk is the vector rot a= y) l + (aa. a~axi+avJ+azk. are called level surfaces of the scalar field. Sec. cos V}.ay cos ~ -+ az cos y) dS.I=gradtU=ax cosa+ dy (the derivative of the function U in the direction I). ay = ay (x.tions of the vector a on the coordInate axes. The gradient is in the direction of the normal tl to the level surface at the roint P and in the direction of Increasing function U. y. wh~re P (x. A scalar field is defined by the scalar function of the point u = f (P) = f (x. Scalar and vector flelds. flow lines) of a vector field are found from the following system of differential equations dx dy dz -=-=ax ay az A scalar or vector field that does not depend on the time t is called stationary. The vector au grad U (P) = ox l-~ au. then the Ostrogradsky·Gaus:s formula holds.« _ aaz ) J + (oay _ aa x ) (iJadyz _ aa ~ ~ ax ax ay k == \!Xa. AB in particular. which in vector form has the form 1c a dr= ~s~ n rot adS. the direction of the vfctor should be chosen so that for an observer looking in the direction of n thf circulation of the contour C should be counterclock\vise in a right-handed coordinate systelll.a+ ay2+ az 2 is the Laplacian opera"tor 2371. in this case the flux of the vector through any closed surface is zero. If C is closed. If the field is at the same tinle potential and solenoidal. 6°. where tl = V2 = ch. that is.289 Fundamentals of Field Theory Sec. The vector tiel d a (r) is called potentlal if a=grad U. then the line integral (I) is called the clrculuilon of the vector field a around the contour C. Determine the level surfaces of the scalar field U = f (r)t -= y 2 -t-Z 2 • What will the level surfaces be of a field U = F (Q). rot a=O. where Q = y2? where 10-1900 r Vx· + Vx. given in a sirnpl y·connected domain. For the potentiality of a field a. If the closed curve C bounds a two·sided surface S. Circulation of a vector. (V) 5°. where n is the vector of the normal to the surface S. the work of a fteld. then Stokes' IOrtrlula holds. Potential and solenoidal fields. A vector field a (r) is called solenoidal if at each point of the field div a =0. then ~ a dr = V (B)-V (A).] wh ich in vector form is tan dS= s H~ diva dxdy dz. 12. the circulation of the vector a is equal to zero: 1c a dr=O. it satisfies the Laplace 8 2 U a2 u a2 u a2 a2 a2 equation O~I + ay 2 + az" =0. + . then div (grad U) ==-0 and the potential function U is harnlonic. In that case there exists a potential U defined by the equation dU = ax dx + a v dy + a z dz. If the potential V is a single-valued function. that is. where U =1 (r) is a scalar function (the potellttu/ of the field). or tlU =0. The line ultegrai of the vector a along the curve C is defined by thl: formula ~ adr= ~ Qsds= ~ axdx+Qvdy+azdz C C (I) C and represents the work done by the field a along the curve C (as is the projection of the vector a on the tangent to C). it is necessary and sufficient that it be non rotational. Find the vector lines of the field a = . z) in the direction of the radius vector r of this point.. Determine at what points the gradient of the field is perpendicular to the z-axis and at \vhat poi nts it is equal to zero. 1. 2374. to: a) r.. Find the magnitude and the direction of the gradient of the field U = Xl + y3 --1. where C1l is a constant. Find diva for the central vector field a (P) = f (r)!. In what case will this derivative be equal to zero? 2381. What will the level surfaces be of this field. 2375.y2 + Z2).Z3 . r 2378.. Find the gradient of the scalar field U = cr. c) div(Ua)=grad U·a+ Udiva. Show that straight lines parallel to a vector c are the vector lines of a vector field a (P) = C. • where r = VXl + y2 + Z2. d) gra d ( V . 2376. Find the derivative of the function U = a + Yb-. where c is a constant vector.V grad U -U V2 . d) f (r) (r = V x 2 -t. Evaluate div ( l:!-) _ x: f) . r ..Multiple and Line Integrals 290 (Ch. +:. where Cl and Cz are constants. where C1 and C1 are constants. I). if U is equal.. Find the derivative of the function U = ~ in the dir rection of I {cos a.royl-~ roxj. b) r'.3xyz at the poi nt A (2. where c is 8 constant vector. lIn what case will this derivative be equal to the magnitude of the gradient? 2380. 2377. and what will their position be relative to the vector c? 2379. y. 2383. Derive the formulas: a) grad(C I U+C 2 V)=C 1 gradU+C a gradV. e) grad cp (V) = (P' (U) grad U. r z U = arc sin · y x2 + y2 2373. cos~. 2382.V grad U. cos V}. respectively. Derive the formulas: a) div (C1a l + C2 a 2 ) = C l div a l + C2 div a 2 . where C is a constant vector. c) J. grad V . Evaluate grad U. b) grad (UV) = U grad V -t. b) div (Uc) = grad U· c. 7 2372. at a c given point P (x. c) grad (V 2 ) = 2U grad U. Determine the level surfaces ot the scalar field · . b) rot (Uc) = grad U· c. Ill: 10* . 2392. 2391. 12] 291 2384. Find the potential U of a gravitational field generated by a material point of mass nz located at the origin of coordinates: a = . \vhere c is a constant vector. . Find the flux of the vector a=x 3i+y3j-t-z sk through: x 2 + 11 2 Z2 a) the lateral surface of the cone ~ ~ H2' 0 ~ z ~ H.y2 for the surface. Y = R sin t. Evaluate the line integral of a vector r around one turn of the screw-l ine x = R cos t. prove that the flux of the vector a = r through a closed surface bounding an arbitrary volume v is equal to three times the volume. it is directed to\vards a fixed point 0 and depends only on the distance r from this point: F=f(r)r. located at the coordinate origin. Using Stokes' theorem. c) rot (Va) =grad U·a + U rot a. 2393*. O~z~H. 2395. b) the total surface of the cone. Find the potential U of the fiel d. 2390. where c is a constant vector. Find the divergence and rotation of the field of linear velocities of the points of a solid rotating counterclockwise with constant angular velocity (0 about the z-axis. 2388. 2389. Prove that div(rota)=O. respectively. z = hi from t = 0 to t = 231. Show that if a force F is central. that is. Using the Ostrogradsky-Gauss theorem. 2396. where f(r) is a single-valued continuous function. then the field is a potential field. rot aI' where C 1 and C2 are consta nts.~ r. taking the hemisphere z == VR 2 _x2 . 2387: Evaluate the rotation of a field of linear velocities ttJ = (t}·r of the points of a body rotating with constant angular velocity ro about some axis passing through the coordinate origin. Evaluate the divergence and rotation of the gradient of the scalar field U.. + C 2a 2) = C1 rot at + C.r of a point of mass ln. + i+zk along the circumference x2 + y2 = R2~ Z == 0. Find the flux of the vector r through the total surface of the cylinder X2+y2~R2. Evaluate the divergence and the tlux of an attractive force F = . 2397. equal to: a) r. evaluate the circulation of the vector a == X 2 y 3. Show that the potential U satisfies the Laplace equation ~U = o.Fundamentals of Field Theory Sec. 2386. Derive the formulas: a) rot (Cta. b) rc and c) f (r) c. 2385. through an arbitrary closed surface surrounding this point. Evaluate the divergence and the rotation of the vector a if a is. 2394. Find out whether the given . and find U if the potential exists: a) a= (5x·y-4xy) I + (3x l -2y)j. Prove that the central space field a = f (r) r will be solenoidal only when f (r) = r~ t where k is constant. b) a=yzi+zxj+xyk.vector field has a potential U. Will the vector field a = r (c x r) be solenoidal (where c 15 a constant vector)? . c) a = (y+ z) 1+ (x+ z)j+ (x + y) k.292 Multiple and Line Integrals [eh. 7 2398. 2400. 2399. is called the remaInder of the series. If the series (J) diverges.. . to take a geometric progression: {a ¢ 0). necessary and sufficient that for any positive nUI11ber e it be possible to choose an N such that for n > lV and for any posi tivE' p the following inequality is fulfilled: \a n +l +a n + 2 + . the series is then called divergent. for purposes of cOlnparing series. n-+<r> The converse is not true. For conv~rgencc of the series (I) it i-.Chapter VIII SERIES Sec.. The convergence or divergence of a serit's is not violated if we add or subtract a finite nunlber of its terms. · +an The quantity S = of the series. then (2) diverges as well. Number Series It). 1. and the series 00 bl +b 2 +··· +bn +··· = n=1 ~ bn (2) converges. ... A number series ~ al +a2 +··· +an +··· = ~ an (1) n=1 is called convergent if its partial sum Sn=al+ a2+ has a finite limit as n --+ 00. If a series converges. It is convenient. If 0 ~ all ~ bn after a certain n = no. then Iirn an = 0 (necessary condition for convergence). +an+pl <8 (Cauchy's test). If the limit lim Sn does not exist (or is n~ lX) infinite). \vhile the number lim Sn is then called the n -+ Rn == S-Sn ==a n + 1 + all + 2 Sllln 00 +. Fundamental concepts. a) Comparison test I. 20 • Test~ of convergence and divergence of positive series.. then the series (1) also converges. . Example t..2 2 + 3. I· e . Example 4.-) =1 2" . sInce ··t I s genera I t ertn In n. if an ".... since lim n -+ QO (_1_ .. . +2 n -n + . n' \vhich is a divergent series. +~+ 11 ••• n d iv~rges..n diverges.• 1) =-:1=0 12 ' 11m· (1 --:2n -1 n n -+ aJ whereas a series \vith general term .. I cOllverges.IS grea t er th an th e correspon d ing ternl . since he re 1 atZ == n·2 n 1 < 2'i' while the geometric progression whose ra t10' ·IS q = 21 ' converges. The series In2+~+ 2 3 . SI nce 1 1 + 73 + "5 + ·. The series 1 1 o Iverges. The series 1 1 1 2-1 +2 2 -2 +2 3 -3 converges.-!. The series 1 1 1 t:2 + 2. and the harrnontc sert es 00 1 L.294 Series (Ch.n · 2 n while a series with general terln ¥I 1 + .11 of the hannonic series (which diverges).. 1 + 11 • 2n +··· converges. then the series (I) and (2) converge or diverge at the sanle time.=:: 1.2 3 -1. .!..bn). If there exists a finite and nonzero linlit Hln an H ~ 'X btl (in particular.· + 2n-1 ~. b) Comparison test II.8 which converges for I q I < 1 and diverges for I q I . Example 2. Example 3....-!. Let a" ~ 0 (after a certa in n) and let there lim a =q. J1 295 Number Series c) D' Alembert's test. I + 2n 1 ..=--pr-' 0n+l= 2n + 1 and 1 lint an + t = t1 -. When q= 1.I (2n .4n 2 1-2n ' .1) 1 . rnollot0111Cally decreasin~ and continuous for x:. We have 1 an :. 2n+ 1 2n-l a.hnt . the series (I) and the integral (X) ~ f (x) dx a converge or diverge at the saine time. and diverJ. · + (-2'-1--1)-2-n +··· Solution. Solution. Example 5.oo 1-. c) Cauchy's integral test.!2· '2n Hence. Test the following series for convergence convcn~es 1 1 1 1 r:2 +3 ·4 + 5· 6 + . 00 n (2n -t... n/ rz-. and diverges if q > 1. If atl = f (n).::sJ V be a Ii mif tz Then (1) converges if q < 1. d) Cauchy's test.= 2 n-. then it is not known whether the series is convergent or not. the given series converges.Sec. the question of the convergence of the seri~s relnains open. By l11eans of the integral test it may be proved that the Dtriclllet sertes (3) if p > 1.1 .. and diverges if p ~ I. Test the convergence of the series 1 3 2n-1 5 2" + 22 + 23 + ···+ ~ + ·. where the function f (x) is posi ti ve. ~ atl lim t1 -..!:es if q > 1.:::a~l. Here.:: -(2-n---1)-2-/~ 1 1 1 4n 2 .1) 2 2n ~. The convergence of a large nllIuber of senes l11ay be tested by conlparing \vith the corresponding Dirichlet series (~3) Example 6. If q = 1. n -+ an (X) Then the series (1) converges if q < 1. Let an a limit >0 (after a certain n) and let there be lim an+t=q. 3°.(3)3 2 Solution. Example 8. The series l-~+~. This series converges condi tionally. Lei bniz test If for the aIterna t i ng series bt -b 2 + ba-b~+. (32)2 .. -+ ~ then (5) converges. (3"2)2 + (3)' (4)4 (n)n "5 + -::.. Bu t if (1) converges and (4) divE:'rges t then the series (1) is called conditionally (not absolutely) convergent.. In this case t for the remainder of the series Rn the evaluation IRn'~bn+1 holds.8 Since the Dirichlet series conver~es for p=2. For instance t (1) converges absolutely if I lilll t1 n + 1 n -+ (X) an I< 1 or . (4) composed of the absolute values of the terms of the series (I)t conver~es. converges. since the series I I 1 1+"2'-3+···+/1+··· diverges (harmonic series).. .. 2) n lim b'J = 0. Test for convergence the series 1- n )n "5 + (4)4 7" + . Tests for convergence of alternating series.. Example 7....1 11 -+ ~ 2 _ ~ 2' n -+ n the serieR converges absolutely./1 a I < lim n -+ ri 1. · + 2n -. the divergence of (1) does not follow from the diver· an .-1 + .. > It then not only does gence of (4). If a series I aJ '+' a2 1+ ..._= Jim _l_=~ ex> 2n .*+ .. + I an 1+ .. +(-1) ~ (211. (5) (btl ~O) the following conditions are fulfilled: 1) b 1 ~ b2 ~ b. then (1) also converges and is called absolutely converJIent.~ • •• .1 lim _11.1 + ··· 1+ Since lim n -+ (Xl V(_n_)n = 211.Series 296 (Ch. But if Jim an +1 I > 1 or lim n -+ (X) I an n -+ VI (X) (4) diverge but the series (1) does also. Let us form a series of the absolute values ot this series: th~ terms of .+ ... +<_l)n+I. (Xl In the general case. since the conditions of the Leibniz test are iulfilled.. . For investigating the absolute convergence of the series (l)t we can make use (for the series (4)] of the familiar convergence tests of positive series. . it follows that on the basis of comparison test I I we can say that the given series likewise converges. .. the series 1 1 1 1 1 1 1-5"+"2..Ib n (i2 == . Operations on series...52 +"3. and only if..4i + . Sk=. -t. that lc). 4°.+ a +. hence. Series with complex terms A series with the general term cn =an + 00 -1. the senes 1 1 1 I . if a. 1·1 1 I) + 3 +···+ k1. 00 On the other hand. For the convergence of an alternating series it is not sufficient that Its general term should tend to zero.Sec Number Series I) 297 Note.5k + . the monotonic variation of the absolute value of the general term has been violated). 2 then ka.. here. + ka n + . converges.+···+5 II Sk=1+ 2 and tiln S~ = k -.(15+S..ka 2 +. the Leibniz test is not necessary for the convergence of an alternating series: an alternating series may converge if the absolute value of its general ternl tends to zero in nonmonotonic fashion Thus. 00 k ' (S~ is a partial SUITI of the harrrtonic series). =S. Indeed. the series with real terms ~ an n=l :'¥) and ~ b'l converge at the same time. although the Leibniz test is not fulfilled: though the absolute value of the general term of the series tends to zero. lim k -+ S2k = 00. +an + . S2k=S~+S~. it does not do so monotonically.. where . diverges despite the fact that its general term tends to zero (here. if the series ~ I en I = ~ V a~ + b~l' I1=J n=J whose trrlllS are the moduli of the terms of the series (6).. of course. \vhereas the 7J limit litll S~ exists and is finite (S~ is a partial sum k -. 50. 1 + (2n - 1)3 converges (and it converges absolutely). in this case 11=1 00 ~ 00 C" n=l r:IJ = ~ an + i ~ bn .22 + 3i . . a) A convergent series may be multiplied terlnwise by any number k. = kS.. The Leibniz test only states that an alternating series converges if the absolute value of its general term tends to zero monotonically. "=J (6) n=1 The series (6) defini tely converges a nd is called absolutely convergent.I) converges if. +k. Of the convergent geo- Xl metric progression). for example.... Thus.. an =[3+(_1) /J J".. · (7) (8) a J +a 2 +··· +a n =8 1 .. =8 2 we mean a series +··· + + (at ± btl (a 2 ± b2) (an ± bn ) -1. In Problems 2411-2415 it is required to write the first 4 or 5 terIllS of the series on the basis of the known general term an. + Ii "5 +.. ). 1. 5" + 2" 5" . If the series (7) and (8) converge absolutely... 2" .. Write the simplest formula of the nth term of the series using the indicated terms: 1 I + 7I + .4. 2401. + cn + . +anbl(n=l. 1+2+4+8+ .. 1 I 2407. bl +b 2 +··· +b.~ +3+++5+~ +- ...t. an =n 2 +1 1 · 2414.. +(_I)n-J + .298 Series (Ch.1-+-3"-1. 1 1 1 1 2+4+6+ 8+··· 234 2403.••• = 8 1 c) The product of the series (7) and (8) is the serifs c1 ± S2· + c + .·· . .1 -t 1 ... .20I .7+1.4."6 +121 -1.1 -~- 2410. 2.. 3n-2 2411. n 2 1 (2)2 +"31 (2)3 I (2)n 5 + ..1 1..7. 1+ .. 1-1 + 1-1 + . This property is absent if the series COIlverges conditionally.1 .5 2402. (9) 2 \vhere cn=albn+afGbn_t+ . alJ= 2-t-(-I)n 2413 an = n2 .10·i - 2409.. d) If a series converges absolutely.. its SUITI remains unchanged \vhrn the terms of the series are rearranged. then the series (9) also converges absolutely and has a sunl equal to SIS2. 2412... 2+8111 nn) cos nrt 2 n! • Test the following series for convergence by applying the cOlnparison tests (or the necessary condition): 2416.3·5·7 2408.8 b) By the sum (diUerence) of two convergent series +. 1+r:4+1. (-l)nn -2n - ( 2415..J + . ·· ~417..301 + 42I + I 1·3 1·3·5 1. I] . + . 1 1 + -3' + . .-Series .2r1 .42 + 42._+'. + 3 )' 21 -}. + .)3·4 + ·. 1)2 I follo\vin.n + ··· 1-+..· · · -r.3 -1. .. ... f 11 3 1l f f 1 . 2 3 2422.---V· -J.·· -t.VTO + . y"2 . 3" -1.· · · (_1)" r 1 1 1 VTo.5 -:1-..1) r 11 3 \ Using d'Alenlbert's test. 8 1 I 2· 5· 8 r31l ..-.S c.( 311-1 Test for convergence the posit ive series: 1 2431.19 + ··· + 2n + 1 -t. r: 243 n.• l'est for convergence. + (:3n _ V2 ~3 2 4 f 1 t. · -r (V2)n • • • ~crles .· · · -j..1 ) It 2429. T+ :3 5" -f. +.1 2425....1) I 2 4_') 8. 2433..· · · 3 + '21 + 52 + 10 ···-r nn+1-r . + ·..-_+ .1 -t. 22 -r 52 f tt 1 1 + 82 -1.2 ••• · · . 22. 1 + 2'... + 2n t. 2 I I 2 3 2436. · • I I 2430...52 + ··· + (Il + 1)2 (11 +2)2 +··· for conver- ..~ -~.'.· · .( 52 )3 + (-8-1- + fl )2/1-1 + ·...... test the gence: -~ VIl V3 I' 1 2426.-...1·5 r:S:-9 -. -1.. -I-.. 1 2432. · · .10/1 +1 + + V~+..1·5·9 (411-3) . 1 1 1 2+4+6+ 2421..... 2 2434. h+~ +~I + 24 t 8...2 -~. r (Il + 1) 2 2 2 2 -J.(n -f. · -.-Numb:!r . r -1(Il 1.33 5 7 2n..3 2427.VlO + Vllj.· · · -}. · 2420..R + is -t.1 ..3 ) 2 ( 4 ) 3 ( 11 -1. ) - 1·2 y f 2 2423. j •• • 2 I 2·5 + 2· 5. -.1 +32...... 7" l I '1+ 1 -l-2n -+... 'T -J.. 1 II 1 1 3 -1. 2424...-299 2 3 4 n+l 2419. 1 +.1)2 - I 1 I" · · · 1~4+4\+/1O+ '"+(311-2)\311+1)+''' 1 4 9 11 3 + 9 -t. I + ··· 2·3 . -1. 2n -1 t-. llsing Ca uchy 's test: 2 ( ........~ +-2 Y-25 + . .. 3! 5' --(2n -1 ) r -•• 1-4 1·4..... +G:~:ri+ .2.. n ~2 n• 1 n' ~ 2459_ L. + 7 + .. 1-5 .+ 2-4-6 ..+( l~ ). (2n-l) 44 23·"4+4._.:.. (998+ 2n) 1·4-7 . I + n=l Y n (n 1) • . _4n (21)2 (U)2 2444.8 SerifS +( ~ ). (4n-4) (4n -2) + --· +1-1l-21 ... ~. L sin n=J (X) 1 ~n-lnn-lnlnn· n=J n=a (X) co 2 . 2 1 2440.11-21+ + .• 11 2! 3! nl 2+1 +2 +1 +2'+1 + ---+2 n + 1 + ... · 2456_ ~ L In ( 1 +{ ) . (10n-9) + 24 48 ~-I_~+1. n=1 ~rnn· n=2 L n:..3-5-7-9+··· +1-3-5-1-9 (4n-3) 2450_ L arc sin +.12+··-+ 4-8-12 . 2441_ 2 2 1 2442. - + 1. 1000·1002-1004 . 1000 + lOO~:~002 + lOOO·11~~. +(2n)1 + .8.. (311-2) (fin -7) (6n-4) (an-Il) (8n-7) + ••• +.. n 2455. (4n -3) 1. -+ (n-l)! + --. _.5 2 447 · 2 2-4-6 + · -.I! 24 49.. .5-9 + . 2437· : I 2438.8+4.300 (Ch. I • 2458... 2457_ ~ 2451.. .. n I (i)I+~+(l~)t+ .. (n!)! 2-5·8 2-5-8 T+ 1.. 2439. + .. _. 1 1·3 1·3-5 1-3-5.. 1004 + ••• 2 2446. + ···+ (3n~ I ) n +·. 1 8 27 n' e + e + e' + ··· + en + ··· 2 4 2n 1 +2 +3' + . 'IJ n=l (X) .5-9+ 1 -.2 00 2454_ ~ 1 1 ~nln2n­ n=2 ~ 2453... L n lin n I. (31)2 "21+41" +61+ . 2 4 2n 1 +rr +2f + . 2445.9 1·4·9 n2 1 +1-3-5+ 1. Lin n J co 00 2452.... Prove that the series ~ '\. and rorq>l. . .1 2476. For convergent series.. CJ) CJ) 2466. ~ 1 ·4. if P = 1. . n=1 fIn • eX) 2464..7 2 + 13 - 2474 5 2472..."2 . + ••• + Y + ...1) ••• 2479..4-··· 7 L( ~r - l)n-1 2n+1 n(n+1) 11 2 1234 2475.Sec.U.. 3nil"nl • '1=1 Vii V-· 1) (5 ~ n -1) ~ ~e"nl 2468*.5+3. n=1 CJ) 2462. 1 ) (_1)"-1 2470.. + ··· +(_I)n-l~~~':+I)+ 2· 5· 8. I: 1 • n=2nlnn+Ylnln 00 1 n1 I: nn • n=1 1: 2"nl nn • 2465.9+7. 1) Number Series 301 CJ) 2460.. I: (1- ~) cos n=1 • CJ) 2469.1. + . 1 + . (-1 ) · 2" -r 234 2 yf-I 3 3-I 4 Y4.11-··· +(-1) n_ J I· 4· 7 l.. test for absolute and conditional convergen~e. 7 ··· ··· + . (_1)1l-1 l-yr2 + Y3...4" + ( ""15 ) 2 - (7 ) a 10 2478. + Yn1 + 91 - 2473.4 3 + (-1)"-1 n"l. 1. and for q ~ 1.4 -to 1f + 16 - ••.. eX) L n V n 1eX) 2463. (In . + 3.... (-I)n-I n ••• + 6n-5 + .5. +(-1 ) n n+l (n+l) Vn+l-l ( 211 -t 1) n 311 -t. + +11 -z-n. .9. L n=1 Yn(n+l)(n+2)· 2461. 3 · [:2-2.3 +.~ fiPTiiQIi 1 •• n=2 1) converges for arbitrary q. if p=l: 2) diverges for arbitrary q. if p>l. 1-3"+5-····f. Test for convergence the following alternating series.7_ 2 2· 5 1 1· 4 2· 5 . YiI-· n=J L. ~_3..2n-1 1 2471.1 + .1I (~in - 2) (2n+5) + ... if P < 1. + +(_l)n 3 2477... L n =1 (2n - 2467. 1-1. It an . 2484*.. ( a2k . ).. n n=J 00 2488. 2481. 32k - J ' + . 2k~1' 1 1 l-Y4+1+··· Yk+l+l ' 1 1 I 1 + 2-33 + 2 -35" + . L n (3~i)n. ~ 2482 · ~ . 1 d) 3-1--1-7-5+TI-g+ . which converge conditionally and which converge absolutely: 1 a) 1 1 1 Y2~1-Y2+1+Y3-1-Y3+1+Y4( 1 b) 1. Convince yourself that the d' Alembert test for convergence does not decide the question of the convergence of the (X) series ~ an' where 1l=1 (k=l.4k~3 ) • Test the following series with complex terms for convergence: 1 L n (2+i~~ 2 f7J 2485. • L n (2/-1)n 2486: n=1 3'Z f7J • 2489. 2.. a 1 a 2k = - ~k) . n.. whereas by means of the Cauchy test it is possible to establish that this series converges.:l Ly~n + I · I1=J ~ 2487.. 8 Series 302 ~ 00 ~(_I)nlnnn.1r-.~( ..[Ch. Convince yourself that the Leibniz test cannot be applied to the alternating series a) to d)..l)n . L (n+i)1 Yn • n=l . Find out which of these series diverge. L~' n=J 00 f7J 2490.. 1). .3 a 1 - Yk+l-l' 2k-l- a _ 2k-- 2 ( 1 1 - 2k-l - 1 c) J -3 +~-32 -t- (a 1 1 zk - 1 _1_ 2k - 1 ' 1 1 5-33 1 = a _ _ 2k - 1).·.1 = 4k l ' a2k = . n=1 n n=1 J' It 2483. . Will the sum of the lengths of these segments be finite? 2494. + 2 + . Will the sum of the lengths of the segments mentioned is replaced by the in Problem 2493 be finite if the curve y = -. What can you say about the signs of these errors? 2502*.~-. 1 2 by the sum of its first n terms. Estimate . Between the curves y = da and y = II • ~ and to the ri~ht of their point of intersection are constructed segments parallel to the y-axis at an equal distance from each other. 1] 303 ~ ~ 4 1 ~ 1 29.. Choose two series such that their sunl converges their difference diverges. ( ~ ) -I. L Il=J ile a...2 . Form the series (1 + ~ +{ + . ... Estimate the error due to replacing the sum of the series ~ -I. + <_:>n + .n • ~ 2496. Form the product of the series \\yh . Given the series 1 -t. 2492 n=1 • ~ [ n (2 .~! ( ~ ) a + . II=J 2497.2 .and 11 Y 11 L LJ . ~[n+(2n-l)i]2. Does the series forlned ~ L by subtracting the series 00 'ill ~ I from the series n=l L ~ converge? ft=-J 2498. Form the difference of the divergent series ~ 1 ~ 21l-1 n=l (L and L 2~ and test it for convergence. + ~ ({ )n +..i) + 1 ] ~ n{3-2i)-3i n=1 2493. the sum of the first five terms. x I curve y==-? x 2495... Does this series converge? 1 2501. LJ )2..Number Series Sec.3\ + . the error committed when replacing the sum of this series with the sum of the first four ternls.. . Form the sum of the series Does this sum converge? rSJ r:n n=1 n=J L 1t nand L <-11:.. 2 n=1 Does this product converge? 2500. r:r: 2499. 11. Rn (x) = S (x) -Sn (x) is the remainder of the series..4 + .3+3. = 1. How many terms of the series (_l)n-l n does one have n=l to take to cOInpute its sum to t\VO decilnal places? to three decimals? .. The set of values of the argument x for which the functional series fl(x)+f2(x)+·.. Functional Series 1°. Estimate the error due to replacing the sunl of the 5eries 1+2 ( ~ + 3 ( + . ~:~ 2507. + fn (x). n-+~ where Sn (x) = f I (x) + f 2 (x) + .. estimate the accuracy of such an approximation for n= 10.r yn-I by the sum of its first n terms. 2505**..Series 304 [Ch. 8 2503.. and x belon~s to the region of convergence. In particular.. 2.. 00 L 2506.n2 + ·· · 2 by the sum of its first n terms...·+ftl(x)+.000. is call(lld the surn of the series. In particular. Find the sum of the series r:2~-2. Sec.. The function S (x) = lim Sn (x).. 2504**.. + 1 +n(n+l)ot . + . (1) converges is called the region of convergence of this series.Vi)+ . +n ({ + . · 2509. r i.Ii + ··· by the sum of its first n terms. Estimate the error due to replacing the sum of the 5eries 1 1 1 1 I I 1 + 2! + 3! + . Estimate the error due to replacing the sum of the series 1 + 2i + 3 + . .Vx)+ . estimate the accuracy of such an approximation for /1.. +(2k+ VX_ 2k .. How many terms of the series ~ (2n +n I) sn does one n=l have to take to compute its sum to two decimal places? to three? to four? I 1 1 2508*. Find the sum of the series Vi+(Vx-Vx) +<Vx. Region of convergenc~... -I.·. 305 Functional Series Sec. 21 I n the simplest cases, it is sufficient, when determiping the region of convergence of a series (1). to apply to this series certain convergence tests. holding x constant. Converges Diverges Diverges i««««,«(1f(~9 -J -1 0 Fl~. 1~4 x 1 Example 1. Determine the region of convergence of the series + + (x + 1)2 + (x + 1)1 + x I 1· 2 2 . 22 3 ·2' Solution. Denoting by u" the general ... + (xn+. 21n)n + .. • t~rm n (2) of the series, we will have n lim Itln+11_lim Ix+ll + 2 n _lx+ll n-+ etJ I un' - n-+ etJ 2n+ J (n + 1) I x In 2 · J Using d' Alembert's test, we can assert that the series converges (and converges I 11< 1, that is, if -3 < x < I; the series diverges, if X! absolutely), if IX!Il> I, that is, if -co <x<-3 or I <x< co (Fig. 104). When x=1 i ii - i- + + + ... , which diverges, and when X= - 3 -I + + ... , which (in accord with the Leibniz we get the harmonic series I we have the series test) converges (conditionally). Thus, the senes converges when - 3 ~ x < 1. 2°. Power series. For any power series Co -1- C1 (.\ - a) +C 2 (x-a)2 + ... + Cn (x-a)n + ... (3) (C n and a are real nunlbers) there exists an interval (the interval of convergence) I x-a I < R with centre at thr point x= u, with in which the series (3) converges absolutely; for I x-a I> R the series diverges. In special cases, the radius of convergence R may also be equal to 0 and 00. At the end-points of the interval of convergence x=a :1: R, the power series may either converge or diverge. The interval of convergence is ordinarily determined with the help of the d'Alembert or Cauchy tests, by applying them to a series, the terms of which are the absolute values of the terms of the given series (3). Applying to the series of absolute values I Co I + I c1 II x-a 1+ ... + I cn II x-a In + ... the convergence tests of d' Alembert and Cauchy, we get, respectivel y, for the radius of convergence of the power series (3), the formulas R= lim 1 jYlc" I and R=liml~1 n n-+oCD c +1 • n~(f) However, one must be very careful in using them because the limits on the right frequently do not exist. For exanlple, if an inflniture of coefficients clI Series 306 (Ch.8 vanishes [as a particular instance, this occurs if the series contains terms with only even or only odd powers of (x-a)], one cannot use these formulas. It is then advisable, when determining the interval of convergence, to apply the d' Alembert or Cauchy tests directly, as was done when \ve investigated the series (2), without resorting to general formulas for the radius of con· vergence. If z=x+ ty is a complex variable, then for the po\ver series CO+c 1 (z-zo)+c 2 (Z-ZO)2+ ... +c n (z-zo)n+... (4) (cn=an+ib n, zo=xo+iyo) there exists a certain circle (circle of convergence) I Z-Zo 1< R with centre at the point z=zo' inside which the series conver~e5 absolutely; for I z-zo I> R the series diverges. At points lying on the cir· cumfprence of the circle of convergence, the series (4) may both converge and diverge. It is custoInary to determine the circle of convergence by nlcans of the d' Alembert or Cauchy tests a pplied to the series I Co I + I ct '·1 z-zo I+ I c2 1·1 Z-Zo 12 +... + I en 1·1 z-zo ,n+ ... , whose terms are absolute values of the tenns of the given series. Thu~, for exaInple, by means of the d' Alelnbert test it is easy to see that the Circle of convergence of the series z+ 1 +(Z+ 1)2+(z+ 1)3+ 1.2 2.2 2 3.23 ••• +(Z+ l)n+ n·2 rl ••• is determined by the inequality I z+ 11 < 2 [It is sufficient to repeat the cal~ culations carried out on page 305 which served to deterIlline the interva I of convergence of the series (2), only here x is replaced by z]. The centre of the circle of convergence lies at the point Z === -1, while the radius I? of thl~ circle (the radius of convergence) is equal to 2. 3°. Uniform convergence. The functional series (1) converges unifonnly on some interval if, no Inatter \vhat e > 0, it is possible to find an N such that does not depend on x and that when n > N for all x of the given Intervdl we have the inequal ity I Rn (x) I < 8, where RIl (x) is the renlai nder of th~ given series. If Ifn(x)l~cn (n=l, 2, ... ) when a~x~b and the nunlber series 00 ' ~ en converges, then the functional series (I) converges on the Interval n=t lat h' absolutely and uniformly (Weierstrass' test). The power series (3) converges absolutely and uniformly on any interval lying within its interval of convergence. The power series (3) may be ternlwise differentiated and integrated within its interval of convergence (for , x-a I < R); that is, if Co +c (x-a) + 1 C2 (x-a)! + ... +en (x-a)n + ... = f (x), (5) then for any x of the interval of convergence of the series (3), we have c1+ 2c 2 (x-a) x x x ~ codx + ~ c1 (x-a) dx+ ~ c2 %0 Xo +... + nCn (x_a)n-l + ... =1' (x), x (x-a)2 dx+ ••. + ~ cn (x-a)n dx+ ... = (6) Xo Ko 00 = ~ en n=:o (x_a)n+t (x o-a)n+l n +1 x S = f (x) dx Xo (7) Functional Series Sec. 2] 307 (the nunlber X o also belongs to the interval of convergence of. the series (3»). I-Iere, the series (6) and (7) have the same interval of convergence as the series (3). Find the region of convergence of the series: rn 00 2510. t~x, L n=1 2518. "f) 2511. L(-l)"+I~. X n 11=1 L (2tl~ 2519. . . VOn YJ _1_ Inx· n (_1)"+1 11=1 L. sin(~-l~ (211-1)2 • 2520. L (x--2),i' .L. (11 2a+ 1 -t- 1)5 IZ .= 1 'Y) "f) 2521. 11=1 X 2" • II='J L. YJ .L. 2 sln:)ijx . n ' d ~ 2522. 11=1 'r. L c:;,.\~ . OCJ ~.,~ 2523 . ...... "rz· 11=1 X 11=0 Cfj 00 2516. L.(-l)l1tJe-Il~ln.~. 2524*. 1l::0 L (XII +2'!X II ) · 11=1 Cfj 2517. (_1)"-1 1l·:3 1l (x-5)"· n=o 2515**. I) X" • 11=1 YJ ~~ 2512. ...... 2514. n! x n · '1=1 00 2513. 1 L 00 L~ x" • 2525. L x n • 11=-1 11=1 Find the interval of convergence of the power series and test the convergence at the end-points of the interval of convergence: Yl 2526. 2531 . -n 2532. 00 x~n-I 2533. 211-1 • L. 2 00 2529. n 00 1 W x - 1 (4n-3)2 • - 2534. L(_I):-I L n!x n n==1 00 00 n=l x" 11=1 n=1 2530. L (_l)n (2n + 1)2 L n! • 11=0 'J:) 11=1 + 1)5 X2n 2n -t- 1 • 00 L n·2" ~" · L 11=1 2528. (tl 11 =0 '1=0 2527. L Y.) ~.., " .... X. XI i . 2535. L'~ nn • n=l • Xr&. 308 Series (Ch.8 co ( n 2536. ~ 2nl )2n-. + II=J 2537. L 3/1 x 2 n1 • r· 2553. ao nn • co 2554. ~ nl (x+ 3)11 ~ J lI L n! 2557. n n=l ~558. L 2559*. 3/ · n·g" • ~561. (_1)/1-1 (x_2)zn • 2n n 2 2562. • co n=1 1 (x - 1)". i: (- 1)" V~2 X n n=o +1 co 2549. ~~ (x+3)n L t:1 ( + ~ )"I X (x-2)n. CD 2550. nn 2560. n=l n=J (x+2)n' (X • n S GO ~ L CD • 00 2548. (n-r I) In (n+ 1) • fl.=J ~ (X_I)211 n=l X (_l)n+1 CIJ L. (_1)"-1 (x-S)" n·3 • ~ L (x-2)n L x::. co (x-3)Z" X n=l 2547 L (n+l)ln(n+I)· n=J crJ '1 ==J (x+ I)n (n+ I) In 2 (11+ I) • n=l x nl • L 2n~n:n" • n=J L • CX) 2556. crJ 2546. L n=1 x U1 • n=1 2545. • co n=l 2544*. nn n=1 2555. 2543*. (2n_I)2n (x_I)n (3n-2)2fJ 11=2 2542**. L (_I)n+1 X X x ~ n·3 n ·lnn· 2541. ~ • n=l n=J 2540. ~ ~n·4n (x- 2)11 ~ (2n-I)2 n • n 2539. ~ nlx X+5)211-1 2552. ~ • ~ n~l (~ ~ 2551. CX) co co x n n=l /1=0 2538. L( co ~~ ~ (3n-2) (X-3)1I ~ (n+l)!2/J + 1 ClO nn (x + 3)n. 2563. L(-I)" n=0 t2n (x-3~ +1) Y II -t 1. Functional Series Sec. 2] 309 Determine the circle of convergence: ex» 2564. L inzra. n=o ao ,.., 2569. ao 2570. ~ n=l ao (z-2i)" n·3 n • ~ Z211 + ni) zn. 2567. ~ 2n • n=o n=o (1 +2i) + (1 + 2i) (3 + 2i) z + ... + + (1 + 2i) (3 +2i) . .. (2n + 1 + 2i) zn -I- .•• Z Z2 1 + l - i + (I-i) (1-2i) -t- · · · 2565. ~ (1 2568. ao 2566. ,..., · · · + (I - z" i) (1 - 2i) ••• (I - ni) + ··· L Cn~2~ir z", n=o 2571. Proceedi ng from the definition of uniform convergence, prove that the series 1 +x+x2 + ... +xn + ... does not converge uniformly in the interval (-1, 1), but converges uniformly on any subinterval within this interval. Solution. Using the fortnula for the sum of a geometric progression, we get, for I x I < 1, xn + 1 Rn'(t) --X'I+I + ,xfJ + 2 + ···-I-x· --Within the interval (-1, 1) let us take a 5ublnterval {-I +a, I-a), \\'here a is an arbitrarily small positive number. In this subinterval I x I ~ I-a, Il-x'~a and, consequently, (l-a)n+1 I R,. (x) I c=; . u To prove the uniform convergence of the given series over the subinterval (-l+a, I-a], it (TIust be shown that for any £>0 it is possible to choose an N dependent only on e such that for any II > N we ",'ill have the ine· quality I R n (x) 1< e for all x of the subinterval under consideration. (I_a)n+l Taking any e>O, let us require that < e; whence (l-a)1J+ I<ea, a that is, n + I > h~~l(~~) [since In (I-a) < 0) and In (ea) . In (ea) . n>ln(l_a)-l. Thus, puttIng N In(l-a)-I, we are convinced that when n > N, I Rn (x) I is indeed less than e for all x of the subinterval [-I a, I-a] and the uniform convergence of the given series on any sub· interval within the Interval (-I, 1) is thus proved. As for the entire interval (-I, I), it contains points that are arbitrarily . . xn + J • closetox=l.andsince hm Rn (x)=lIm -1 -=ClO, no nlatteJ how large n IS.. %-+1 x .... J -x (n + 1) In (I-a) < In (ea), + Series 310 [Ch.8 points x will be found for which Rn (x) is grea ter than any arbi trarily large number Hence, it is impossible to choose an N such that for n > N we would have the inequality I R" (x) 1< 8 at all points of the interval (-1, 1), and this means that the convergence of the series in the interval (-1, 1) is not uniform. 2572. Using the definition of uniform convergence, prove that: a) the series X x2 1 +Ti -t- 21+ xn · .. -1- 11T+ .. · converges uniformly in any finite interval; b) the series x2 X4 ,,\6 T-2"+3-·· · -t·converges uniformly throughout (_l)"-J x ::n n the + ... interval of convergence (-1, 1); c) the series 1 1 1 +¥+3x+ 1 .. · -t;;X+ ... converges uniformly in the interval (1 -I-i), (0) \vhere B is any positive nUlnber; d) the series (X 2 _X 4 ) + (x<l-x )-t-(X6 _X8) + ... + (X21J_X211+2) +_ ... 8 converges not only within the interval (-1, I), but at the extremities of this interval, however the convergence of the series in (- 1, 1) is nonuniform. Prove the uniform convergence of the functional serIes in the indicated intervals: CD 2573. l: :: on the interval [-I, I]. n=l 00 2574. ~~ sin nx ~ 2n over the entire number scale. 1l=1 00 2575. ~ (_l)n-J ;nn on the interval lO, 11. Applying termwise differentiation and integration, find the .sums of the series: 2576. x2 x2 2577. x- 2 2578. x3 xn x+ 2 +3"+·· · +n T x3 ... n + 3- .. · + (-l)n-l nx + ... Xl XS x:. n - I x+3"-rs-l-··· +2n-l-t- ... Taylor's Series Sec. 8] x 2n - x5 Xl 311 J + .. · 2579. x- 3 + 5 - .. · +(_l)n-J 2n-l 2580. 1+2x-l-3r + ... (n-t-l)x n + ... 2581. 1-3x2 -t- 5x 4 - • • • +(_l)n-J (2n-l)x 2n - 2 + ... 2582. 1·2-t-2·3x+3·4x2 +n(n+ l)x"-I+ ... Find the sums of the series: + + ... 1 2 3 2583. -+ 2-13+ x x X x9 XS -1- 9 2584. x+ 5 11 .. · + xn+ .. · X 4n _ 3 + ·· · + 4n-3 + ··· 1 1 1 (_l)n-J 2585*. 1- 3 . 3 + 5.32-7.3:1+ ... +(2n-l)3 1l 1 ~ 2586. 2+22 5 J + ... 2n-l +2"3+ .. · +2'-z+ .. · Sec. 3. Taylor's Series to. Expanding a function in a power series. If a function f (x) can beexpanded, in son1e neighbourhood I x-a I < R of the point a, in a series of po\vers of x-a, then this series (called Taylor's series) is of the form ._, r (a) ,(nl (a). 2 1(.t)-f(a)+1 (a) (x-a)+2t(x-a) +···+-n,-(x-a) n +... (1) \Vhen a=-O the Taylor scri~s is also called a Maclaurin's series. Equation (1) holds If \vhen I x-a I < R the rellzainder ternz (or sin1ply remainder) of the Taylor series RrI (x) = f (x) - [ f (a) L 1 kl(a) (x-a)k1 tl (k) -+ 0 k=J as n --+ 00. To rva I uate the r C111 ain der, Rn (x) = (x a)"+J (~t-l )1 r 0 ne n + J) [a can n1 a k e USC 0 f + 0 (x-a»), the forin uIa \vhere 0 <0<1 (2) (Lagran£!e's lo"n). Exanlple t. Expand the function 1(x) = cosh x in a series of powers of x. Solution. We find the derivatives of the given function 1(x) = cosh x. I' (x) =-= SInh x, f" (x)Ul=:: ccsh x, {'" (x) = sinh x, ... ; generally. f(n) (x) = cosh X, if n is even, etnd , ) (x) == sinh x, if n IS odd. Putting a= 0, we get 1(0) = 1, 1'(0)=-=0,1"(0)=-1,1"'(0):.=0, ... ; generally, [<")(0)=1, if n is even, and I(n) (0) =- 0 if n is odd. Whence, from (1), we have: coshx=1 x2 x4 x2n +21+41+··· + (2n)! +... (3) To determine the interval of convergence of the series (3) we apply the: d' Alembert test. We have . I 11111 2n x +2 n ~ 00 (2n ,\ 2n +2)1 : (211 )t I· = ltm n -+ etJ (2n x 2 +1) (211 +2) 0 Serie~ 312 [Ch. 8 for any x. Hence, the series converges in the interval - 00 remainder term, in accord with formula (2), has the form: < x< 00. The xn + 1 Rn (x) = (n + 1)! cosh e x, if n is odd, and ..\n+1 Rn (x) = (n • sInh ex, if n is even. + 1)! e > 1, it follows that e~x+e-6x I cosh e x 1= 2 ~ e' x I, Since 0> I sinh e x 1= le~x_e-~xl 2 <; el x I, Ixl n + 1 ~,. and therefore I R n (x) J.,;;;; (n + 1)1 e' x I. A series with the gen~r81 term nl converges for any x (this is made immediately evident with the help of d' AleInbcrt's test); therefore, in accord with the necessary condition for convergence, . I xl n + 1 11m - - - = 0 lZ -+> 00 (n +1)! ' and consequently lim Rn (x) = 0 for any x. This signifies tha t the sum of the n-+oo sert~s {3) for any x is indeed equal to cosh x. 2~. Techniques employed for expanding in power series. Making use of the principal expansions I. X e =1 xn x2 X +fi+2f+'" + iiT+'" (-00 . x ~ I x5 n II. slnx=n-31+5f- ... +(-I) x 2 x" < x< (2n+l)I+'" x 2n 11 I. cos X == 1 - 21 + 4f - ... + (_I)n (2n)1 + ... IV. (1+>mt-l x - + mTI x + m(m-l) 21 x + ·.· (-00 <X< 00), (- 00 < x < 00), 2 ... +m(m-I).~.!{I1l-n+l)xn+... V 00), x2n + I ~2 Xl xn (-I <x<I)*). In(l+x)::,:::x-~+3- ... +(-l)n-ln+'" (-I<x~I), -and also the formula for the sunl of a geometric progregsion, it is possible, in many cases, simply to obtain the expansion of a given function in a power series, without having to investigate the remainder term. It is sOllletim('s advisable to make use of termwise differentiation or integration when expanding a function in a series. When expanding rational functions in power 'S~ries it is advisable to decompose these functions into partial fractions. *) On the boundaries of the interva1 of convergence (i. e., when x = - 1 and x = 1) the expansion 1\9 behaves as follows: for m;:;: 0 it converges abso· lutely on both boundaries; for 0 > m > -1 it diverges when x == - 1 and ~onditlonally converges when x = 1; for m E;;-1 it diverges on both boundaries. Tuy/or's Series Sec. 8] 313 Example 2. Expand in powers of x *) the function 3 t(X)=(I_x) (1 +2x)' Solution. Decomposing the function into partial fractions. we will have 1 2 !(X)=I-x+l+2x' Since r:IJ _1_= 1+x+x 2 -l- ... I-x ==,~ xn (4) ~ n=o and tT- I ~2X = 1-2x + (2X)2_ • •• =- L n n (-l) 2 x n , (5) + (_1)n 2 n + 1 ] x n . (6) n=o it follo\vs that we finally get f(x) =- ~ x n +2 ~ (_l)n 2 n x n = ~ [I n=o Il=O n=o The geometric progressions (4) and (5) converge, respectively, when I x I < 1 1 1 and I x I < "2 ; hence, formula (6) holds for I x I < 2' i. e., when 1 1 -"2<x<"2' 30 • Taylor's series for a function of two variables. Expanding a function of two variables f (x, y) into a Taylor's sertes in the neighbourhood of a point (a. b) has the fornl f (x, y) =f (a, +(y-b) ~ + -it [(x-a) ~ + (y-b) ~] f (a, b) + ~1 [(x-a) r b) f (a, b) +... + ~ [(x-a) ~+ (y-bl ~r f(a. b) +... If a = b = 0, the Taylor series notation is as follows: a [ (x-a) ax + (y - a] b) ify I (a, b) IS then called a ,Haclaurtn's sertes. at (x, y) ax (x-a) + at (;~ x=a iJJ (x~a)ax+(y-b)ay a [ f(a,b)= a 2 1(x, ax" y ) Her~ (7) the (y-lJ); x=a y=b 2 tx+ y=b y) (x-ar~ -I- x=a IJ=b (x-a)(y-b) x=a y=b + a2f(x, y) (y-b)2 and so forth. ay! x=a IJ=b *) Here and henceforward we mean "in positive integral powers". 2589. Applying various techniques. 2601. 2606.8 The expansion (7) occurs if the remainder term of the series iJ a]k {(a. X2 2595. sin! x.v here 0 < e < 1. cos (x -t. aX (a> 0).x cos 3x. 2605. In (2+x). sinh x. 4 -4 • 2611.{ {(a. xe.2X • 2600.a). 2594.Y) Rn(x. Expand the indicated functions in positive integral powers of x. ) 2590. 2612 x . sin2 xcos 2 x. ( 1(.3x + 1 2 x · x -5x+6· 2609. (I + x) e.• 2613. write the expansion. (I + eX)I. 3x-5 2593.~14 Series [Ch. find the intervals of convergence of the resulting series and iJlvestigate the behaviour of their remainders: 2587. 2610. e • 2596. (l+x)ln(l+x). x.ln(l+x-2x 2 ). -x Vs + .b)+ n k=1 8S n --+ 00. arcsinx. In 11 +x . Applying differentiation. x 9+x 2 • 1 ¥4-. expand the given functions in po\vers of x and indicate the intervals in which these expansions occur: 2 2608. 588 2 · Sin x +4· 2591*. Making use of the principal expansions I-V and a geometric progression. and indicate the intervals of convergence of the series: 2x-3 2598 • cos 2 x. b) } -+0 L kI1 [ (x-a)ax+(y-b)ay Rn(x. 1 2614. y). y)=f(x. 2592. arctanx 2607. -x 2597.ln(x+VI+x 2 ).\2 · 2602. cos2x. in powers of x. si n 3x -t. of the follo\ving functions. and indicate the intervals in which these expansions occur: 2604. x2-4x + 3 • 2599. (x-l)2. cosh' x. The remainder term may be represented in the form 1 [ a a] n+1 f(x. 2603. . expand the following functions in po\\'ers of x. Y)+(n+l)! (x-a)ax+(y-b)ay x=a+~ (x-a) y=b+'J (y-b) . 2·~. a 2628. x 2616. 2637.. Sho\v that for computing the length ot an ellipse it is possible to make use of the approxinlate formula s ~ 2na ( I . to the order of x\ it 111ay be taken that the string hangs in a parabola y=a 1. SSi:X dx . Expand 2634. Expand lnx in a series of powers of :+. under its o\vn \veight. dx o o Write the first three nonzero terms of the expansion of the following functions in powers of x: 2620. SVI-x". A heavy string hangs.. Expand Vi in a series of powers of x-4. x 2632. \vhilc q is the \veight of unit length. in a catenary line y == a cosh ~. 2622. tanhx. f(x}===5x 3 -4x 2 -3x+2. In (Xl 315 n (I ~X) dx • o x o 2619. sec x. 2623. x-i. r x 2615. Expand cosx in a series of powers of 2 . Expand -. 2629.l q tension of the string. eX sin x. ~e-x2dx. 2639*. Expand f(x+h) in a series of powers of h 2630. Sho\v that for small x.~) . where e is the eccentricity and 2a is the major axis of the ell ipse. Expand In x in a series of po\vers of x-I. series of powers of x series of powers of Expand eX in a series of pqwers of x + 2. 2636. Expand cos x in a series of po\vers of x. Expand J. 2627. Expand the function x 3 -2x 2 -5x-2 in a series of po\vers of x -I 4. Expand X2+~X+2 in a X2+~X+7 in a +4. \vhere a =. 2618. 2621. 2631.T. in a series of po\vers of x-I.O~X. tan x. 2635. x 2633.Taylor's Series Sec 3] + 3x + 2). x 2617. 2624. 2626*. 2638. x +2. 2625. lncosx.!i and H is the horizontal c. el. in a series of po\vers of x + 1. have to be taken to find the number e to four .. 2644. How many terms of the series x e 2 x =1+ 11x '-21+ . putting a = 5. in order to calculate cos 18° to three decimal places? 2645. Find out the origin of the approximate formula r Va' +x~a+. How many terms do we have to take of the series x2 cosx= 1.. . Calculate to two decimals by expanding the function V 8 -l. How many terms of the series x2 In (1 +x) =x-2"+ .Series 316 [ell... if we make use of the series arc tanx=x- Xl 3 XS +5-·. vr V . 1+% x 2641. and estimate the error..x in a series of powers of x. Expand . r" in a series of powers of 1 +" • . ...21 + .a (a>O). ' do we have to take to calculate In 2 to two decimals? to 3 decimals? 2648. 2649.·' by taking the sum of its first five terms when X= I? 2643*. How many terms do we have to take of the series • Xl slnx=x-3f+ . evaluate it by means of V 23 . To what degree of accuracy will we calculate the number ~. 8 2640. to calculate sin 15° to four decimal places? 2646. 2650. What is the magnitude of the error if we put approximately e~ 2 1 1 1 + 2f+3f+41? 2642. Calculate the number ~ to three decimals by expanding the function arc sin x in a series of powers of x (see Example 2606)... Calculate 19 to three decimals.decimal places? 2647. 2 266 t.x2 dx to four decimals. of the following functions and indicate the regions of convergence of the series: 2660. 3] 317 Taglor's Series 2651. l-x+ It l+x-y 2665. Find the increment Qf this function when passing from the values x= 1.OOOI? 2652. Evaluate ~ e.01? O. 2667.01? 0. o lID 2658. 2663*. y) = x' -2y' -1. Y -f k) in po- \vers of hand k.J sin yXx dx to three declma o 1/4 2657. Expand the function eX +Y in powers of x-2 and y + 2. y=2 to the values x == 1 + h. find the region of convergence of the resulting series and investigate the remainder.OOI? O. Evaluate ~ VI +x l dx to four decimals. 2656.001? 1/2 Sinx 2653. Expand f (x + h. !I = 2 + k. In(l-x-Y1-xy). Evaluate . o 1 2655. I s. 2668. Y-2· . Evaluate ~ Vxcosxdx to three decimals. WI ite the expansions.2bxy + cy". For what values of x does the approximate formula x2 cosx~l-­ 2 yield an error not exceeding 0.Sec. Evaluate S-x-dot to • I s. in powers of x and y. f (x. + y • 2662*. Expand the function sin (x y) in powers of x and + 1(. o r 1 . For what values of x does the approximate formula sin x~x yield an error that does not exceed 0. Expand the function cos (x . sin (x + gl).y) in a series of powers of x and y. o 2659.3xy. 2666. sinx·siny. Evaluate ~ Vi eX dx to three decimals. f (x. ~664"'. y) • -xy = ax" -1. four decllna o 1 2654. arc tan . bn=n f(x)slnnxdx If x is a poinf of discontinu ity. at each discontinuity . ) Clnd n an = ~ St (x) cos nx dx o (n = 0. Fourier Series to.. •. •.•. 1. We say that a function f (x) satisfies the Dirichlet conditions in an interval (a.+0)= lim f (. b) if. Sec.b2 sin 2x + . Incomplete Fourier series. ). then in formula f (x) is even li. n) satisfies the Dirichlet conditions at any point x of this interval at which f (x) is continuous. the function 1) is uniformly bounded. (1 +X)I+Y. ( 1) where the Fourier coefficlents an and bn are calculated frorn the formulas I an=n- n n -~ -n S I ...x) bn=O (n= 1. = .[Ch. f (. where M is constant. I. 2) has no more than a finite number of points of discontinuity and all of them are of the first kind [Le. Jl). Dirichlet's theorem. (n=l.8 Series 318 Write the first three or four terlTIS of a power-series expansion in x and y of the functions: 2669. If a function (l) f (x)]. ••.xt and x = xt. e. that is I f (x) I ~ M when a < x < b. eX cos y. of a function f (x). belonging to the interval (-1t. then the SUITI of the Fourier series S (x) is equal to the anthl11Ctical mean of the left and right limits of the function: I S (x) =2" [f (x-a) At the end-points of the interval S (-n) = =S + f (x + 0)).2. I (xt) =2" [f (-xt + 0) + f (1t-O)].•. 2. Dirichlet's theorem asserts that a function f (x).S . 2670. finite limit on the right 8 8~O ~o 3) has no more than a finite number of points of strict extrenlum.+e) (e > O)J. f(x)cosnxdx(n=O. . the function f (x) has a finite limit on the left f (~-O) = linl f (~-e) and a f (. 2. 2°. 4. x = . may be expanded in a trigonometric Fourter sertes: f (x) = ~ +a 1 cos x + b1 sin x + a 2 cos 2x -1. + an cos nx + --1. in this interval. ).. 2. which in the interval (-xt.bn si n nx -1. ).... xt) may. Consider the special cases: a) a = b = 1. f (x) = x 2 • 2676. l At the points of discontinuity of the function f (x) and at the end-points ±l of the interval.dx (n=. be continued in the interval (. 2671. b=O. ). 4) 319 If a function f(x) is odd [i. If a function {(x) satisfies the Dirichlet conditions in SOlTIe Interval (-I. c) a=O. d) a=l. f (x) = si nh ax.re.. 1. 2678. +ancos-l-+bnsln . t . .e..re. c2 = 1. {(-x)=-{(x)]. f (x) = eax. b) a = .• ) and n bn = . then an=O (n=O. cos 1- -I (2) I 1 bn=y 5f -I . Expand the function f (x) = rr 2 x in a Fourier series in the interval (0.Jt. nxtx 1 2 . it may be expanded in the interval (0. the SUlll of the Founer series is defined in a manner X= similar to that which \ve ha've in the expansion in th~ interval (-xt. ). where l 1 an = T 5{ (x) nnx dx (n = 0. construct the graph of the function itself and of the sum of the corresponding series [outside the interval (. 2672. 2. (x) SUl. at our discretion. = --1.. f(x)= { c2 \vhen O<x<1t.1 b = 1. ••. f (x) = cosh ax.2 •. hence. 2. . X == n). deterrnine the surTI of the series at the points of discontinuity and at the end-points of the interval (x = . 2673. xt) In an incomplete Fourier series of sines or of cosines of nlliltir-le arcs. nrtx •.. 2nx T + b1 sin T +a 2 cosl +b 2 sin -l-+' H nxtx . ). n). f (x) = cos ax. •. f (x) = { bx when 0 ~ x < n. 1) of length 21. 5f (x) sin nx dx (n = I. n).n. I. b=l.Fourier Series Sec.l +.. 2679. 0) either as an even or an odd function. 3°. then at the discontinui ties of the function belonging to this interval the following expansion holds: a0 {(x) ~-2+al cos xtx 2nx xtx .. the limits of integration in fornlulas (2) should be replaced respectively by a and a-1-2l Expand the follo\ving functions in a Fourier series in the interval (. 2677. a+2l) of length 2l. 2674. Consider the special case \vhen c. f (x) = sin ax. n) as well]: CI when -n<x~O.. Fourier series of a period 21. 2675. 2n). ax when -. o A function specified in an interval (0.n < x ~ 0. In the case of an expansion of the functton f (x) in a Fourier series in an arbitrary Interval (a. f (x) = eax .. 31). in the interval (0.32 - 2) 1- 1 42 -r ··.. 1) 1 + 22 2 2683. ~ interval (0. o \vhen h < x< 31.. 2687.. Sketch graphs of the functions and graphs of the sums of the corresponding sel ies in their domains of definition. in • n { o when 2 < x < n.. .. · n x when 0<x~2' 2685... in the when 0 < x 0=.. . f (x) = x.320 Series (Ch. f (x) = { 1 when 0 < x ~ h. 2688. .. 2684. Find the sum of the following series by means of the expansion obtained: 1 1 1 -t.. f (x) = Xl.. Expand the function f (x) in sines of multiple arcs in the interval (0. l 0 when i o=. 1 1 1 1 b) 1+5-7-n+13+17.. n)..32 + 52 + ··· 2682. Expand the following sines of multiple arcs: 2686. 1 1 1 +"7-n+ 13. in the interval (0. .3 1 c) 1-5 + 51 - 1 7 1 + .. x (n-x). sin i· Expand the following functions. Use the expansion obtained to sum the number series: 1 a) 1. n)..x<n. b) of cosines of multiple arcs. Tt). in cosines of multiple arcs: 2689. 2681. f (~) = f 1 1 when 0 < x < ~ 1 2Z'. into incomplete Fourier series: a) of sines of multiple arcs. f (x) = J l n-x when ~ <x<n. Take the functions indicated below and expand them. f (x) = f (x) = f (x) = X functions. Find the sums of the follo\ving number series by means of the expansion obtained: 1 ) + 3 + ·· ·. 8 =: 2680. . sin x.1 < x < I). prove the equality 2694**. Expand the following function in cosines of multiple arcs 10 the interval (~. f (x) = 2x (0 < x < 1). \ 3-x when 2<x<3.x ) I then In the interval (-n. 2-x when I <x< 2. Expand the rollo\ving functions in l:ourier series in the Indicated intervals: 2695.Fourier Series Sec. 2703.!.. 2698. 2701. Using the expansions of the functions x and Xl in the interval (0. f(x)=x 3 (0< x<2n). n) represents an expansion in CaStrIeS of odd lTIultiple arcs. 2696.x ). n) in cosines of nlultlple arcs (see Problems 2681 and 2682). COS x when 0 < x ~ n { i. f (x) = ~21 I _. Expand the rollo\vin:{ functIons. is even and we have then Its fourier series in the interval (-n. f(x)=lO--x (5<x<15). 2. n) it is expanded in sines of odd multiple arcs.+ x ) = -I ( ~ . 2693.' -cosx when 2"<x<n.. in the indicated intervals. f (x) = x 2692. and b) in cosines of multiple arcs: 2699. 2700. 02h when 2h<x<n. in incoInplete Fourier series: a) In sInes of ~llultlple ales. 11-1900 .. x when O<X~ 1. f(x)=e x (-l<x<l). Prove that if the function f (x) 1( . f(x)= { f (x) = I 3 when 2" < x :e. f (x) = 1 (0 < x < 1).when 0 < x ~ 2h. 3): 2702. f (x) = I x I (. x) and if the function f (x) is odd and f ( ~ -I= f ( ~ . 41 2690. f (x) = x (0 < x < I). 2697. f (x) = { 2691. . Find the differential equation of the family of parabolas y:=. . Initial Conditions 1°. An equation of the type F(x. dn<D . y)(n)==o.l. By ass\~n­ ing definite values to the constants Cit . y.-t. Solution.. get a differpntial equation of type (I) whose general integral in the corresponding region is the relation (2). is called the solution of the equation.. we get: y'=2C. If the solution is represented irnplicitly. Conversely. .:=0. y" -t y == . (D (. d<D dx=O. which contains n independrnt arbitrary constants Ct . Differentiating equation (3) twice. Cn and is equivalent (in the given region) to equation (I). we. which converts equation (1) into an identity.(x-C z) and y"=2C t • (4) Eliminating the parameters Ct and C2 from equations (3) and (4). y) . generally speaking. . (3) Solution. Forming Differential Equations of Families of Curves. dxn=O. while th~ ~raph of this function IS called an integral curve.sin x sin x === 0. CIl in (2). The Integral (1) (x. y' . ••• . Ct (X-C Z)2. C t . IS called the general tntegral of this equation (in the respective region). if we have a family of curVf\S (2) and elirninate the raranl· eters CIt ••• ' Cn from the system of equations (1)=0. 1. .. Check that the function y = sin x is a solutton of the equation y"+y=O. . .Chapter IX D1FFERENTIAL EQUATIONS Sec... . we get fJarLlcul£lr tntegrals. is called a diUerential equatton of order n. The function y=(p(x). y. \Ve have: y'=cosx. consequently. Verifying Solutions. Basic concepts. then it is usually called an lntegral Example t. Cn) === 0 (2) + of the differential equation (1). (I) where y=y (x) is the sought-for functi l1 n. y"=-sinx and. we obtain the desired differential equation 2yy" =y. . . Example 2. y' (0) =:: - Solution.(AI +. 2 t Y=x. If for the desired particular solution y ==Y (x) of a differential eq uatto n y(n) = f (x. Example 3. cos rot + C2 sin cot. y' = Cle x -2C 2e. (x-2y)y'=2x-y.I ) (x o) = y~n -I) are given and we know the general solution of equation (5) y=cp(x. ••• . Show that for the given differential equations the indicated relations are integra Is: 27\11.x -~ C 2e 'A x. ••. we obtain 1=C1 +C 2 . :. Y = 2x • 2707. .. y. Cn). Cn)' then the arbitrary constants CI . 27 to.. Initial conditions. . y"+y=O. We have: Putting x=O in (6) 2. b) lJ=x 2 e". J 11* . y~ == cp~ (x o. Cn are determined (if this is possible) Yo -= cp (x o. 2708. y=3sinx-4cosx. y=5x 2 • 2705 · Y '2 =X 2 +y.y) dx -t. for which y (0) = 1.e x + C2e. y'. X = C. (x l.x dy -= 0. CI Xi 2706. xy'=2y. y(n-J)) (5) the inltlal conditions y (x o) -= Yo.t'ertfyinR Solultons Sec 1) 323 It IS easy to verify that th~ function (3) converts this equation into an identi ty. y" .: + Ct)2 X = 0. Cn). (7) -2~CI-2C2' \\'hence and. 2709. a) y=xe". xl-xy ~ y2=::C". y"-2y'+y=O. CI .2X . y(n . !I = CleA. y' (x o) = y~. [roln the system of equations ••• . Determine \\J"hether the indicated functions are solutions of the given differential equations: 2704. Cit . .. 2°.. hence. Find the curve of the family Y= C.2X fOflnula~ (6) and (7). Cit .A2 ) y' + A1 A2 Y = 0. ••• . In":'= 1 +ay 2715. . y) = f (x. (a !!IS a parame ter). or . 9 2712. 2728.2 . y" (0) = . y). y=x-l-Ce'. A differential equation of the first order in an unknown function y. y = Gle. y) dy= 0. Y = C\ cos 2x+ C 1 sin 2x. 272 ( )2 2 2.!. y) dx+ Q (x. is of the form y'=f(x. y (0) = 0. y). II 2720. 2732.Ie 2 • 2719. -C 2X-t-C e-x . 2725. By solutions to (2) we mean functions of the form y=cp (x) or x='\I' (g) ahat satisfy this equation. y(n)=l. 2713.. y) are known functions. Y (0) = 5. y' (0) = 1.x + C2 ex + C aelx. y = ex!. (x-1/+1)y'=1. In certain cases it is convenient to consider the variable x as the sought-for function. y'(n)=O. 2730. : \ · 2727. 2721. = (x . The general integral of equations (1) and (1 ').=2+Ce. (1) where f (x. are arbitrary constants): 2714.DtUerential Equations 324 lCh. Form the differential equation of all parabolas with vertical axis in the xy-plan~. 2. X 2 _ y2 = C.. y=Cx. y-yo = px 2717. and to write (1) in the form x'=g(x. 27 18. g) and Q (x. y=(C I -l-C 2x)e 2X .. 2723. Sec. 2724. y) is the given function. xl+y'=C 1 • x (Yo' P are parameters). the differential equations (1) and (1') may be written in the synlmetric form p (x. JI. Taking into account that y' = :~ and x' =:. Y= In (xy). 2731. First-Order Differential Equations 1°. y(O)=O.. C I .y ' . C a . (xy-x)y" + xy'2 + yy' -2y' = 0. y= Ce. y'(O)=l. (1 ') 1 where g (x. y=(CI+C~x)ex+Ca· x 2726. For the given families of curves find the lines that sati5fy the given initial conditions: 2729. Form the~ differential equation of all straight lines in the I xy-plane. y=C l sin(x-C 2 ). 2 2C 2716. (2) where P (x.2. Form the differential equation of all circles in the xy-plane. Form differential equations of the given families of curves (C. y). Y = x. C.a C 2 2) Y . solved for the derivative y'. Types of first-order differential equations. y 2 -t. Using the method of isoclines. Curves f (x.~ . is of the form <D (x. make approximate constructions of fields of integral curves for the indicated differential equations: 2733. 2734. 105). The set of directions tana=f(x. 20 • Direction field. y) is called a direction field of the differential equation (I) and is ordinarily depicted by means of short lines or arrows inclined at an angle Q. By constructing the isoclines x = k (straight lines) and the direction field. y y' = 1 °f. where C is an arbitrary constant. y' = . it is possibl~. 2736. By constructin~ the isoclines and direction field. USing the method of isoclines. C) =0.x. x-y 2737. y) =-= k. construct the field of integral curves of the equation y' :=2X.Sec. to give a y rough sketch of the field of int~gral curves. Exalnple 1. y'=xl+y•• . y' =x+y . y' = .yl. The family of parabolas Xl y=-+C 2 is the ~eneral solution. at the points of which the inclination of the field has a constant value. 2735. 21 First-Order DiOerential Equations 325 equa Hon (2). Solution. we obtain approximately the field of integral curves (Fig. y. are called isoclines. in the simplest cases. regardin~ the latter as curves which at each point have the given direction of the field. equal to k. 326 Dlfferenttal 3°. Cauchy's theorem. If a function Equations f (x, y) [Ch. 9 is continuous in some region U {a < x < A, b < y < B} and in this region has a bounded derivative /~ (x, y), then through each point (xo, Yo) that belongs to U there passes one and only one integral curve y=ep (x) of the equation (1) (ep (xo)=Yo]' 4°. Euler's broken-line method. For an approximate construction of the integral curve of equation (1) passing through a given point Mo (xo, Yo)' we replace the curve by a broken I ine wi th vertices M i (xi, Yi), where Xi+l=Xi+L\ Xi, Yi+l=Yi+L\Yi' AXi=h (one step of the process), AYi = hI (xi, Yi) (i = 0, I, 2, ..• ). Example 2. Using Euler's method for the equation find y (I), if y (0) = I (h We construct the table: = 0.1). , xi I 0 0.1 0.2 0.3 0 I 2 3 5 .6 7 8 9 10 Yi 0.5 1 1 1.005 1 015 1.030 1.051 0.6 1.077 0.7 0.3 0.9 1.0 1.109 1.148 1.194 1.248 o4 4 I I L\ xiYi lJ;= 20 0 o 005 0.010 o 015 o 021 o 026 o 032 0.039 o 046 0.054 Thus, y (1) = 1.248. For the sake of comparison, the exact value is 1 Using Euler's method, find the particular solutions to the given differential equations for the indicated values of x: 2738. y'=y, y(O)=l; find y(l) (h=O.l). 2739. y'=x-t-y, y(l)=l; find y(2), (h=O.l). 2740. Y'=-l~X' y(O)=2; find y(1) (h=O.l). 2741. y'=y_2x, y(O)=l; find y(l) (h=O.2). II Sec. 3) DiOerential Equations wlth Variables Separable Sec. 3. First-Order Differential Equations Orthogonal Trajectories with Variables 321 Separable. I". First-order eql.lations with variables separab Ie. An eq ua tion wi th variables is a first-order equation of the type ~eparable y' = t (x) g (y) ( 1) or X (x) Y (y) dx+X 1 (x) Y. (y) dy=O Dividin~ (I') both sides of equation (I) by g (y) and multiplying by dx, we get dll) = f (x) dx Whence, by integrating, we get the general integral of equag(y hon (I) in the fonTI 5gdy (y) =Sf(X)dX+C (2) Similarly, divining both sides of (\quclhon (I') by XI (x) Y (y) and integrating, we get the general integral of (1') in thp form (r) dx + SY (Y) dy = C Y (y) S,X.X (x) I (2') 1 If for some value y=Yo \ve have R (Yo) =0, then the function Y=~/o is also (as is directly evident) a solution of equation (I) Similarly, the straight lincs x -=ll and y -== b will be the intet!ral curves of equation (1'), if a and b 3re, respectively, the roots of the equations XI (x) =0 and Y (y) =0, by the lcft Sldc'i of whlch we had to divide the inItial equation. Example 1. Solve the equation , Y == - 11 ~ . (3) In particular. find the solution that sathfies the initial conditions y(l) =2 Solution. Equation (3) may be written in the tonn dy y dx=-x Whence, separating variables, we have dy =_ dx y x and, consequently, lnIYI=-lnlxl+lnC" where the arbitrary constant In C. is taken in logarithmic form. After taking antilogaritluns we get the general solution (4) where C = ± CII When dlvldlng by y we could )Olie the solution y=O, but the latter is contalned in the formula (4) for C=O. [Ch. 9 DifJerential Equations 328 Uti1izln~ the ~fven Initial conditions, we get C = 2; and, hence, the de· sired particular solutIon is 2 y=-. x 2° Certain di tfertntial equationc; that reduce to equations with variables separable. Differential equation~ of the fonn y'=f(ax+by+c) (b#O) reduce to equations of the form (I) by means of the substitution u=ax+bu+ e, where U IS the new sough t-for functIon 3° Ortho~onal traJectoric:s ar~ curves that intersect the hnes of the given family <1>(x,y,a)=O ta IS a parametpr) at a right angle. If F(x,Y,y')=O is the diflerentIal equation of the family, then F (X. y, - ~,) =0 Is the differential equaHon of the orthogonal trajectories. Exalnple 2. Find the ortho.Jonat trajectories of the family of x2 + 2// 2 =- a2 • Solution Difff'rentiatlng the equation (5), we find the tion of the famll y x -f- 2yy' -= O. ellips~s dlder~ntJal (5) equa- y x Fig. 106 Whl\nce. repla("n~ y' by - y" we get the differential equation ef the orthogonal traJeclones x_ 2y =O or u,=2y. 1/ ., x Integrating, we have y = ex! (lalnlly 01 parabolas) (Fig. 106). Sec. 31 DiOerential Equations with Variables Separable 329 4°. Forming difl'erential equations. \\'hen forming differential equations In geometrical problems, we can fr€quently make use of the geometrical meaning of the deriva tl ve a s the tangent of ~n angle fornled by the tangent hne to the curve in the pos'tive x-direction. In olany cases this makes it I=ossible straightway to cstabltsh a rplationship hptween the ordinate y of the desired curve, its abscls~a x, and the tanllent of the angle of the tangent line y', that is to say, to obtain the diffe.ent\al equation. In other Instances ~see Problems 2783, 2890, 2895), u~e is made of the geometrical significance of the dpfinite integral as the area of a curvihnear trapezold or the length of an arc. In thi~ cac;e, by hypothesls we have a simple lntegral equation (since the desired function is under the '\ign of the Integral); however, we can readily pass to a differential equatlon by dlfterentiatIng both ~ides. Example 3. Find a curve passing through the pOint (3,2) for which the segnlent of any tangent line contained between the coordinatE:' axes is divid~d in half at the pOlnt of tangency. Solution. Let M (x,y) be the mid-pOint of the tangent line AB. which by hypothesis is the point of tangency (the pOlnts A and B are points of lnter')ection of the tangent line with the y- and x-axes). It is given that OA =2y and OB = 2x. The slope of the tangent to the curve at M (x, y) is dy OA !I (jX=- VB=--X This is the differential equation of the sought-for curve. Transforming, we gel d·\+~=o x y and, conscquentl y, lnx+lny -=lnCorxy==C. Utilizing the initial condition, \ve determlne C =3·2 ==6. Hence, the desired curve is the hyperbola >.y =.:. 6. Solve the differential equations: 2742. tan xsin 2 ydx-t-cos 2 xcotydy=O. 2743. xy' -- !I = yS • 2744. xyy' =:-: 1_x2 • 2745. y-xy' = a (I -t-x 2 y'). 2746.3 p x tan ydx-f-(1-e X )scc 2 ydy=O. 2747. y' tan x =!J. Find the particular solutions of equations that satisfy the indicated initial conditions: 2748. (I + eX) y y' = eX; !J =- I \vhen x = O. 2749. (xy2.t-x}dx-4-{x 2 y-y)d!J=O; y=l \vhen x=O. 2750. y'sin x=ylny; y:.=l when x=;. Solve the differential equations by changing the variables: 2751. 2752. 2753. 2754. y'=(x +y)2. [I = (8x -~- 2y + 1)2 • (2x + 3y- I) dx -t (4x + 6y- 5) dy = O. (2x-y) dx+ (4x-2y -1- 3) ely = O. Dzt!erential Equations 330 [Ch.9 In Examples 2755 and 2756, pass to polar coordinates: 2755. y' = ~-x y • 2756. (X 2+y2)dx-xydy=O. 2757*. Find a curve whose segment of the tangent is equal to the distance of the point of tangency from the origin. 2758. Find the curve whose segment of the normal at any point of a curve lying between the coordinate axes is divided in two at this point. 2759. Find a curve whose subtangent is of constant length a. 2760. Find a curve which has a subtangent twice the abscissa of the point of tangency. 2761 *. Find a curve whose abscissa of the centre of gravity of an area bounded by the coordinate axes, by this curve and the ordinate of any of its points is equal to 3/4 the abscissa of this point. 2762. Find the equation of a curve that pa5ses through the point (3,1), for which the segment of the tangent between the point of tangency and the x-axis is divided in half at the point of intersection with the y-axis. 2763. Find the equation of a curve which passes through the point (2,0), if the segment of the tangent to the curve between the point of tangency and the y-axis is of constant length 2. Find the orthogonal trajectories of the given families of curves (a is a parameter), construct the families and their orthogonal trajectories. 2766. xy=a. 2764. x 2 + y2 = a 2 • 2765. y2 = ax. 2767. (x-a)! r- y2 = a2 • Sec. 4. First-Order Homogeneous Differential Equations 1°. Homogeneous equations. A differential equation p (x, y) dx+ Q (x, y) dy=O (1) is called homogeneous, if P (x, y) and Q (x, y) are homogeneous functions of the same degree. Equation (I) may be reduced to the form y'=f (~): and by means of the substitution y=.-:XU, where u is a new unknown function, it is transformed to an equation WIth variables separable. We can also apply the substitution x == yll. Example 1. Find the general solution to the equation Sec. 4) First-Order Homogeneous Differential Equations 331 Solution. Put y=ux; then u+xu'=eu+u or dx e -ud u=-. x Integrating, we get u=-ln InE..., whence x C y=-xlnln-. x 2°. Equations that reduce to homogeneous equations. If (2) and 6=1::~:I¥o, y=v+~. where are found from the following system of equations, then,puttinginto equation (2) x=u+a, the conc;tants a and ~ ala+bl~+cl=O, a2a+b2~+c2~O, we get a homogeneous ditTerenttal equation in the variables u and u. If 6=0, then, puttIng in (2) a1x-t-b 1y=u, we get an equation with variables separable. Integrate the ditTerential equations: 2768. y' =- JL -1. 2770. (x- y) y dx-x1dy = O. x 2769. y':= _ x+x y • 2771. For the equation (Xl +y2) dx-2xydy=0 find the family of integral curves, and also indicate the curves that pass through the points (4,0) and (1,1), respectively. 2772. y dx i- (2 Vxy-x) dy = O. 2773. xdy-ydx=Vx 2 -f-y 2 dx. 2774. (4x 2 + 3xy + y2) dx + (4 y l + 3xy + x 2) dy = O. 2775. Find the particular solution of the equation (x 2 -3!1)dx+ -t- 2xyriy = 0, provided that y = 1 when x = 2. Solve the equations: 2776. (2x-y -+ 4)dy+(x-2y+ 5)dx=O. , 1-3x-3y 2778' x+2y+ 1 2777. Y == l+x+y · · Y =2x+4y+3 · 2779. Find the equation of a curve that passe~ through the point (1,0) and has the property that the segment ~ut off b) the tangent line on the y-axis is equal to the radius vector of the point of tangency. 2780**. What shape should the reflector of a search light have so that the rays from a point ~ource of light are rel1ected as a parallel beam? DitJerential Equations 332 (Ch. 9 2781. Find the equation of a curve whose 5ubtangent is equal to the arithmetic mean of the coordinates of the point of tangency. 2782. Find the equation of a curve for which the segment cut off on the y-axis by the normal at any point of the curve is equal to the distance of this point from the origin. 2783*. Find the equation of a curve for which the area conta ined between the x-axis, the curve and two ord inates, one of which is a constant and the other a variable, is equal to the ratio "of the cube of the variable ordinate to the appropriate abscissa. 2784. Find a curve for which the segment on the y-axis cut off by any tangent Iine is equal to the abscissa of the point of tangency. Sec. 5. First-Order Linear Differential Equations. Bernoulli's Equation 1°. Linear equations. A differential equation of the form y' + p (x). y Q (x) =.::; of ( I) de~ree one in y and y' is called linear. If a function Q (x) =:; 0, then equation (1) takes the form y' + p (x) · y = 0 (2) and is called a homogeneoLls linear differential equation. In this case, the variables may be separated, and we get the general solution of (2) in the form -J P(x) dx y=C·e. (3) To solve the inhomogeneous linear equation (1), we apply a method that is called variation of parameters, which consIsts in first finding the ~eneral solution of the respective hornogeneous linear equation, that is, retationship (3). Then, assumin~ here that C is a function of x, we seek the solution of the inhomo~eneous equation (1) in the form of (3). To do this, we put into (1) y and y' which are found from (3), and then from the differential equation thu5. obtainerl we determine the function C (x). We thus get the general solution of the inhomogeneous equation (1) in the form -f P (x) dx y=C (x).e • Example l. Solve the equation y' = tan x·y+cos x. Solution. The corresponding homogeneous equation is y' - tanx·y:=:O. Solving 1t we get: 1 . Y ::-c.-ccsx (4) 5J Sec. 333 Bernoulli's Equation' Considering C as a function of x, and differentiating, we nnd: y =_1_ • de cos x dx sin x + cos 2 X .c • Putting y and y' into (4). we get: I de sin x C de - - . -+--.C=tanx.--+cosx or -=cos 2 xt cos x dx cos 2 X cos X 'dx whence C (x) = 5 cos 2 xdx= } x+ { sin 2x+C\. Hence, the general soJution of equation (4) has the form y= ( hon ~ X+{ sin 2t+C\ ) • CO~X · In solving the linear fquation (1) \ve can also make use of the substitu .. y==uv, (3) where u and v are functions of x. Then equation (I) will have the form lu' -t- P (x) u 1v -f- v' u -= Q (x). (6) If we req U Ire fha t u' -1- P (x) u == 0, (7) then fronl (7) we find ll, and fronl (6) we find v; hence, from (5) we find y. 2). Bernoulli's equation. A first-order equation of the form y' + P (.()!J -= Q (x) y7., where a ~ 0 and a '# 1, is called Berflolllll'~ equation It is reduced to a linear equatioll by nleans of tlll substitution z == yJ-'1. I t is also lossib1e to apply dire('tly the substitution y-::=.uv, or the Iuethod of variation of paranleters. Example 2. Solve the equation l 4 y'==-y+x x lIrf .tl. Solution. This is Bernoulli's equation. Pulting \ve ~et U'I1+I1'U=~ UI1 +x ¥IW or v (U'- ~ /I) +v'u=x ¥iii1. To detennine the function u we require that the relation u,_i u=Q x he fulfilled, whence we have Putting this expression into {8), we get V'.,,4=xYvX". (8) 334 DiOerential Equations whence we find v: 11=( ; lnx+c y. lCh.9 and, consequently, the general solution is obtained in the fonn ~ lnx+C y=x4 ( r. Find the general integrals of the equations: 2785. dy _J!.. = x. dx 2786. ~dY x x + 2yx = x' . + 2787*. (1 -+ y2) dx = tV 1 y2 sin y-xy) dYe 2788. y 2 dx- (2xy 3) dy = O. + Find the particular solutions that satisfy the indicated conditions: 2789. xy' + y-e" = 0; y = b when x === a. 2790. y ' - -x -Ylz-l-x=O; y=O when x~o. 2791. y'-ytanx=_l- ; y=O when x=O. cos x Find the general solutions of the equations: +.JLx = xy2. 2792. dc!J!. 2793. 2xY~-!I +x=O. X - + (x- ; 2794. y dx x1y) dy=O. 2795. 3xdy=-=y(1 +xsinx-3y 3 sinx)dx. 2796. Given three particular solutions y, Yl' Y2 of a linear equation. Prove that the expression Y2-Y rernains unchanged for Y-Yl any x. What is the geometrical significance of this result? 2797. Find the curves for wh ich the area of a triangle formed by the x-axis, a tangent line and the radius vector of the point of tangency is constant. 2798. Find the equation of a curve, a segment of which, cui off on the x-axis by a tangent line, is equal to the square of the ordinate of the point of tangency. 2799. ,Find the equation of a curve, a segment of which, cut off on the y-axis by a tangent line, is equal to the subnormal. 2800. Find the equation of a curve, a segment of which, cut otT on the y-axis by a tangent line, is proportional to the square of the ordinate of the point of tangency. . Find the general integr al of the differential equation (3x 2 + 6xy 2) dx + (6x 2 y + 4y') dy = 0. Find the equation of the curve for which the segment of the tangent is equal to the distance of the point of intersection of this tangent with the x-axis from the point M (O.. satisfies the equation o a ay (JLP)=ax (flQ)· The integra ting factor . Integrating Factor 1°. Exact differential equations. then.Exact DitJerential Equations.ox = F I (Y). Integrating factor. y) = 0 and is then called an exact dtUerential equation. X' + 3x y2 + y4=C is the sought-for general inte~ral of the equation. whence = 12xy and...a)... y) is determined by the technique given in Ch. The function U (x.. I r the left si de of equation (1) is not a total (exact) differential and the conditions of the Cauchy theorem are fulfilled. then ~t = Jl (y).. The general integral of equation (1) is U (x... then there exists a function . y) dx+ Q (x..4y S) ox := Here. au .. from this we get q>'(U)=4tj' and q>(f)=Y"+C. VI. the equation is of the form dU = O. . Whence it Is found that the function . 6. 2°. Is readily found In two cases: 1 aQ ) 1) Q ay . 9).. y) (integrating factor) such that )1 (P dx+ Q dy) =dU. 8. 6] 335 2801.\1 3x! + 6xy l au . Example I. y) dy=O (1) the equality ~: = ~~ is fulfilled. Sec.= 6x"y + 4y'· ay' and +6xyl) dx+ cp (y) =x l Differentiating U with respect to y. since y a (6(2 y -t. y) = C.ax = F (x).= ax u= ~ (3. or froln the formula x U= II ~p (x. then equation (1) may be written in the form dU (x. (2) . Integrating Factor Sec. Th·· IS IS an exac t d'ff 1 eren t'la I equa t·lon. ~aU = 6x!y -t <p' (y) = 6x!y + 4y' (by Y hypothesis).. we find + 3X1y 2+ cp (y). If for the differential equation P (x. =. We finally get U (r. = JL (x).. y) =XS + 3X 2y 2 + y4+ CClt consequently.. (x... hence. y) dx + ~ Q (xo• y) dy Yo Xo ~sce Chi VI I t Sec. a(3x 2+0 6xy2) == So 1Ut •Ion. Sec. Exact Differential Equations. (OP 1 (OP OQ) 2) -p ay .. \vhich satisfies the initial condition y (0) = 2. Q iJy ax and ln~ . =x... (Xl + y2 2x) dx -{. • 2 y' y4 2807.ox . 2803.3xy 2 + 2) dx .t (y): 2808.. (x+ y2) dx-2xy dy = O. Yx dx + (y'-In x)dy = O. Find the particular integral of the equation (x+e~) dx+ef (1-~) dy=O. we get the general integral Find the general integrals of the equations: 2802 (x+ y) dx + (x -1.. + 2805. Here P = 2xy + x 2y + l/3 • Q = Xl + y2 and 1 (OP OQ) = 2x +x!x!+ + y2-2 y2 Q oy .2xy dy = o. 2xdx+ y2 -3x dy-=O. 9 (2XY+X2y+~)dx+tx l +y2)dll=O... (Xl -:.t oy = J.t = I.y sin y) dy + (x sin y -~ y cos y) dx = O.tP) iJ(~Q) ay=--ax- or t = 1.336 Dif1erential Equations (CIl. \ 2+ Y2 2806.! (ap _aQ)dX=dX .t= J.(3x 2y . oP iJQ J.t ax +Qd. dx t it follows that dJL=. 2809. = eX. JL=r.. (x cos y .. 2811... 2810. ... xdx-ydy= x dy-y dx . Integrating it...2y) dy = O. hence. Stnce o(J. we obtain eX ( \ 2xy + x 2y + ~ 3 ) dx-{-e" (Xl + yl) dy -===0 which is an exact differential equation. Solve the equation a Solution.t (x). Example 2. Multiplying the equation by IJ. J. Solve the equations that admit of an integrating factor of the form ~ = 11 (x) or I.. 2804.. y(l -~-xy)dx-xdy=O.y2) dy = O. a singular integral is the cnvelope of a fal1lily of curves (3) and Inay be obtained by elinlinating C frolll the s~'steln of equations <D~(X. (3) \vhere (D.(x. h'1s the fOrln <I> (x. y).Oerential Equation.r. DifTerentiatin~ th~ general integral \\'ith respecf to C and eliminating C.» /lot Solved for Derivative 337 Sec.=.y. 7.0. y') =0. Besides. there may be a slnRular lllte{!ral for equilt10n ~1). \ve find the singul ar in tegral y+x=O. (4) or by eliminating p=y' fronl the system of equatIons F~ (. Multiplying. p) = o. Solving for y' we have t\\·o hOlllogeneous cquations: y' =- 1 + 1/ 1+ ~. C) cD 2 (x. y' = - lit'fined in the region x(x+y) 1- Y 1t ~ I >0. First-Order Differential Equations not Solved for the Derivative to. 11. (x.y). C) === cD. generally spt'aking. (2) Thus. Exalnple 1.\ hich for example is of degree two in y'.l by solving (1) for II' we get two equations: y'=f.y. we get the general integral of the given equation (2x+ y-C)2-4 (x 2 +xy):::0 or (y-C)2=4Cx (a family of parabolas). D. in each casc.~ec. First-order dlft'erential equations of higher powers. 7) Fir~t-Orde. y. The general integral of equation (1) then. y. a check is necessary. F (x.C)=O. Yt 2 +xy=O.C)=O <D(x. and cI>2 are the general inte~rals of equations (2). (5) We note thflt the curves defined by the £1quations (4) or (5) are not alwavs solutions of equation (I). thert'fore. Solution. FJ nd the general and sln~ular integrals of the equation XII'2. y.r 2xll'-Y~0. If an equation F (x. !J. (1) . generally speaking. through each pOint M" (xo. (It ll1ay be verified that y +x = 0 is the solution of this equation. Yo) of some region of a plane there pass two integral curves. C) =0. Geometrically.) . the. y. y'= f2(X. p) =-. the general integrals of which are or (2x+y-C)-2 yx 2 (2x+ y-C) +2 +xy=0. 4y. we have the general solution x2 x2 y=(X+C)2. Solving a differenti al equation by introducing a parameter.= · 2813.) 2y. then x and yare deternlined from the system of equations a". we J 2 obtain the singular solution: g=X • (It may be verified that g=T is the 4 solut ion of the given equation. putting p into the given equation. then the variables y a'ld x may be determined from the system of equations l=aq>+aq>dp p ay a dy' x = q> (y. y'). which was cancelled out. If a firstorder differential equation is of the form x= cp (y. y'). Substituting into the original equation. we get p= ~ and. \\'here p=y' plays the part of a parameter.2 -x U . p). Differentiating the general solution with respect to C and eliminating C. if y='I' (x. Example 2. which is the same singu lar sol ution. 1 0 28 t 2· y . we have p=. Find the general and singular integrals of the equations: (In Problems 2812 and 2813 construct the field of integral curves. Integrating we get p=x+C.2pdp _p_x dp +x dx dx or ~~(2P-X)-(2P-X). Similarly.Z-9x=O.. Making the substitution y' = p. 2°. Find the general and singular integrals of the equation 2 y=y'2_ xy' +x2' Solution.9 It is also possible to find the singular integral by differentiating xp2+2xp-y-=O with respect to p and eliminating p. I . or~=l. p). we get g=~.x (x+C)+2 or Y="2+CX+C2.xp +x2 · Diffelentiatiog with respect to x and considering p a function of x.) If we equate to zero the factor 2p -x.338 Differential Equations (Ch. we rewrite the equation in the form 2 y=p2. a"'dp p= ax + ap dx' y='I' (x. 2y' • Sec.2 p )'. yy. then we get Cla. y y' 7-= P we have y = 2p\ + ing dy by p dx. Puttin~ --2y'x+. Besides. Its general solution IS of the fornl y.2_(xy+ l)y' ·tx=O.'I' (C) (a farnl1y of ~tralght lines).4y==x. Clairaut's equation.). If In equation (1) p =-.2. · ... we will have I p *. 2 + '2 2818 · Y=Y . An equation of the forrn y = xfP (p) -1.2 -2xy' + y = o. solve the equations: 2820.Sec 8] The Lagrange and Clairaut Equations 339 2814. there [nay be a Singular solution that is found in the l1~ual \vay. • 2821 eX-~ 2819..: ql (p). 2817. (1) \vhere p==y' is called Lagrange's equatIon Equation (1) is reduced to a linear equation in x by differentiation a nd taking into con'Sideration that dy -= p dx: p dx == cr (p) dx -1.raut's equ.'I' (p). x=siny'+-lny'. Solve the equation ) y Solution. where p is a parameter and f (p) ..'P (p). llting and replac- . x= 2 (In p+C). Introducing the paranleter y' =p. 8. P (3) different.atlon y::= xp -1. yy. Find the integral curves of the equation y'!-t-y"=1 that pass through the point M (0. g (p) are certain known functions. (2) If p ~ cp (p). There IS also a parttcular solutiO'} (envelope) that results by clilninating the parameter p fronl the systenl of equ~t1011~ x==.[xcp' (p) + '1" (p)] dp. y=y'" +2Iny'.+y. The Lagrange and Clairaut Equaticns to.'P' (p).-: Cx -t. . 2816.~ dx 2 1 dp=-px+ pl· ~quation. 2°. then from (1) and (2) we get the general solution in parametric form: x= Cf (p) + g (p). { y==PX+~·lP)· Example. Lagrange's equation. y --= [Ct (p) + g (p)] q) (p) + 'I-~ (p). 2815. we get p dx == 2p dx or Solving this lineal + 2x dp. 9 Hence. Firld the curve for which the area of a triangle formed by a tangent at any point and by the coordinate axe\) is constant. Y 1·/-1-'2- "" 2825 • Y + y') x + y". y=2px+ ~ · To find the singular integral. Miscellaneous Exercises on First-Order Differential Equation') 2833. y 2830. \ve form the system 1 1 y=. the general integral will be f . Y = xy' ~. 2 Y=p con~equently. Solve the Lagrange equations: 2822.2 ' 2826 . 2831. . b) (x.2px+-. g) Y == xeY '.y) y' = y2. Determine the types of diflerential equations and indicate methods for their solution: e) xy' -t.DiOerential Equations 340 (Ch. 2 Y (2x-}. Find the curve it the distance of a given point to any tangent to this curve is constant. . 0==2x. . therefore. y =. 9. . d) y' = 2xy -t. 2828. 2832.!. Find the general and singular integrals of the Clairaut equations and construct the field of integral curves: .y).y'. y= ± 2 V2x.) . h) (y' -2xy) Vy=x l . Y == (1 2 -' P. y=xy +y . 2823 · 2829.. y-y + J' y. 2827..J. Find the curve for which the segment of any of its tangents lying between the coordinate axes has constant length l~ Sec. a) (x + y) y' =-= x arc tan 1!. Putting y into (3) we are convinced that the function obtained Is not a solution and. equation (3) does not have a singular integral.2 p p In the usual way. c) y' = 2xy + Xl.y = sin y.2 X= l (In p+C).x (y' + 2824. Whence 1 x= 2p z' and. Y = xy' + l/rl 1---(y-')-2. x f) (y_xy')2 = y'd. y=xy' +y'. =- I. xdY-I ydx=-~y!dx... .1.y2) (e 2. yeY = (yS -t 2xeY ) y'. 2835. . n) (xy' + In x) dx = y2dy. dx-e Y dy) -(I -1. y'_y2x 2 1 =1..2 · Y. 2dx+ . xy . 2852. 2856.O.. y 2834.. 2836. 2857. i) y' = (x +_ y)2. 2846. eY dx + (xeY . 2841. V11-x dx = O. xdy-ydx=y dx. Solve the equations: (x-ycosf)dx+xcos~dy-=O. X . V!I 286 t. 2849.~t·Order DiOerential Equations 341 1) (x 2+ 2xy 8) dx -1+ (y2 + 3X2y2) dy = 0. x y'2_l-3xYY'-l- -r ¥X 2+y2 2860.y) dy =-= O. y'+y cos x == sin x cos x. a (xy' 2y) = xyy'. 2838.tan ~ . x 3dx-(x 4 t.sin y) 2864.ye dx = O. 9] Miscellaneous Exercises on Fir. 1 2859. yy' + y' = cos x. x VI x 2854. x 2 (y + I) dx+ (xS_I) (y-l) dy== O.2 2851 . m) (x ' -3xy) dx + (x 2 + 3)dy=O. y' = ( 1 +~. xy' . y' (x -J-. xy(xyl-t-l)dy-dx= =0..V -ay'.-. x dx+ y d~ + +2y 2=O.x+ I-x = O.p2 • dy 2858. 2 2842.. 2845. . y' (x cos y -i.~.\ I\.O. + 2867.Y -tl· - + 7='. 2848. Y ="A(l-+-Iny-Inx). k) (x 2-xy) y' = y4.:lr 2850. _yl) dy. 2855. (2e X + y4) dy- == 1. (l-x )y'+xy=a.\. 2840. xy3 dx (x 2y 2) dye ' 3. Y =-= xy2 In x. xdy--ydx 11 z 0 =. 2847.y=-=.+ (2 xy 2 . y' = Y -I.a sin 2y) = I.y=-= . y =2 \x+y-I · 2866. b) x In ~ dy-ydx == O.p -J. Y t}E = . j) X cos y' + y sin y' =-= 1. dy. 2 2868. :y -j ~y · 2863. y 2843. /!. 2837.y) dx x dy =.2y) dll = O. A . (x 2 y-x 2 -l-y--l)dx -t-(xy+2x-3y-6)dy=O. 2853. 2844. (y + 2 )2 2865. 2862 =2 ' _ .. y = xy' -l. 2839. (1 -t. xdx = . Y == xy' -1.y' In y'. a) (x.Sec.y3). . Find a curve knowing that the area contained between the coordinate axes. c) y==O for x=l. 2879. 2892. 2888. y= 1 for x= 1. 2883. 2885. eY (y' + 1) = 1.=y+l. b) y=O for x==O. cut orr by the tangent on the x-axis. is equal to the length of the tangent. 2873. 2887. y=O for x= 1. Find solutions to the equations for the indicated initial conditions: 2876. a) y=O for x=O. tan + (x + 3xy l/ Xl 3 1) dx = O. a) y= 1 for x= 1. ) 2881. 2878. Find a curve whose angle formed by a tangent and the radius vector of the point of tangency is constant. 2891.9 DiOerential Equations 2869. + +x 2871. 2yp :: = 3p·. b)y==l forx=O.Val 2872.3y 2) dy =-. xyy'8_(X 2 +y2)y'+xy=O. 2xyy'+X I _ y 2=O. Va 2 x 2 dy --I. y'-2y=-x 2 . (x'-l )3/2 dy 2870. eX-Yy' = 1. 2889*. x 2877. Find the equation of the curve if it is kno\vn that the curve passes through the coordinate origin. y. cot xy' + y = 2. !J=-l for x==o. the segment of which. xy' =y. The sum of the lengths of the normal and subnorlna\ is equal to unity. Find a curve. Find a curve if we know that the sum of the segnlent~ cut off on the coordinate axes by a tangent to it is constant and equal to 2a. Find the curve passing through the point (0. 2 ) dx= O. y == 2 for x == O. 2886. 2875. y = xy' y~ 2874.o. Y=4 for x==O.(x+ y. 2884. + + . I). 4y B. 2890. y == 0 for x == 0. b) y=O for x=O. y' +y=2x. 2882. for which the subtangent is equal to the sum of the cooldinates of the point 'of tangency. Find a curve knowing that the area of a sector bounded by the polar axis. (3x 2 2xy_ y 2) dx+ (x 2-2xy. dy x dx -y = a. y' +y=cosx~ Y=2 for x===O. by this curve and by the radius vector of any point of it is proportional to the cube of this radius vector.342 lCh. I 2880. this curve and the ordinate of any point on it is equal to the cube of the ordinate.. . 2xy'=y: a) y=l for x=--= 1. The area bounded by a curve. The quantity of solution in the tank at that instant \vill be 100 t litres. particularly in ph\sical problelns. If the concentration is uniforlTI. Therrfore. Find the curve if we know that the mid-point of the segnlent cut off on the x-axis by a tangent and a normal to the curve is a constant point (a. Water IS rntcnng the tank at the rate of 3 htres per minute. conscquent\ y. 0). The constant C is found fronl the fact that \\ hpn t = 0. and the mixture is f1owin~ out at 2 Idres per Jllinute. Find the equation of this curve if it is known that the latter passes through the point (0. and the ordinate of sOlne point of the curve is equal to the length of the corresponding arc of the curve. the tank will contain X= lOf~~~ ~ 3. of an aqueous solution containing 10 kg of salt.2x. 2896. the t'Oncentralion c = 100 + t kg per Ii Ire. Separating variables and integrating. 91 Miscellaneous ExercIses on First-Order Differential Equations 343 2893. 2894. which con~ists in the fact that approxirnate relationships between infinltesinlal i"crenlent~ of the desired quantities (these relationships are accurate to infinitesinlals of higher order) are replaced by the corresponding relationships between their dlfft·r~nttals. 1). of which the segment of the tangent contained between the coordinate axes is divided into half by the parabola y2 =. \ -= 10. 2895*.000. Find the curve for which the area of a triangle formed by the x-axis. not affect the result. Let the quantity of salt in the tank at the end of t minutes be x kg. Find the curve whose normal at any point of it is equal to the distance of this point from the origin.Sec. we obtain In x= . . Duril1~ tinle dt. it is frequently advisable to apply the so-called method of differentials. tha tis. the coordinate axes. Find the c~rve. This doe. a tangent. This is the sought-for differential equation. and the radius vector of the point of 2 tan~ency is constant and equal to a • 2897. C= 100. A tani< contains 100 litre. When forming first-order differential equations. The concentration is maintained unlfonn by stirring.. 2dt litres of the solution 110\\'S out of the tank (the solution contains 2c dt kg of salt).21n (100+ t) In C or C + x = ( 100 + t) I . The concentration c of a substance is the quantity of it in unit volume. Problenl. then the Quantity of substance in voltllTIe V is eV. At thr expiration of one hour.9 kilograms or salt. I-fo\v tTIuch salt \vill the tank contain at the end of one hOll r? Solution. + x and. a change of dx in the quantity of salt in the tank is given by the relationship 2x -dx=2c dt = 100 + t dt.. During what tirne will a body heated to 100 cool ofT to 30 0 if the temperature of the room is 20° and during the first 20 minutes the body cooled to 60°? 2904. 2903. During what period of time will the water filling a hemispherical boiler of diameter 2 metres Aow out of it through a circular opening of radIus 0. known that the disk began rotating at 100 rpm and after one minute was rotating at 60 rpm.92 kgf on 1 em! at an altitude of 500 metres. by which the rate of cooling of a body is proportional to the difference of temperatures of the body and the ambient 1l1edium. is brought into a room at constant temperature (a degrees). heated to To degrees. The quantity of light absorbed in passing through a thin layer of water is proportional to the quantity of incident light ann to the thickness of the layer. By how much will the band increase in length due to its weight W if the band is suspended at one end? (The initial lenglh of the band is I. 2899*.344 DiOerential Equations (Ch. The rate of disintegration of radium is proportional to the quantity of radium present. Find the percentage of radium that has disintegrated after 100 years. 9 2898*. If one half of the original quantity of light is absorbed in passing through a three-metrethick layer of water. Find the relationship between the temperature T and the time t if a body. 2902.) 2901. what part of this quantity will reach a depth of 30 metres? . The rate of outflow of water from an aperture at a vertical distance h from the free surface is defined by the formula 0 v=c¥2gh. When solving Problems 2902 and 2903. 2900*. 2905*.1 m in the bottom. Prove that for a heavy liquid rotating about a vertical axis the free surface has the form of a paraboloid of revolution. 2907*. where c ~ 0.6 and g is the acceleration of gravity. The retarding action of friction on a disk rotating in a liquid is proportional to the angular velocity of rotation_ Find the relationship between the angular velocity and tillle if it ic. make use of Newton's law. Solve the same problem for a weight P suspended fr0l11 the end of the band. 2906*. According to Hooke's law an elastic band of length 1 increases in length klF(k=const) due to a tensile force F. Radium disin1egrates by one half in 1600 years. Find the relationship between the air pressure and the altitude if it is known that the pressure is 1 kgf on 1 cm 2 at sea level and 0. 2910*. The case of direct integration. we get .) = f (. assuming y' = p. Solving the latter equation as a linear equation In the function p. Determi ne the current i at tilne t if e ==. Higher-Order Differential Equations to. \\'e get an equation ot an order one unit lower: F (x. The bottom of a tank with a capacity of 300 litres is covered with a nlixture of salt and some insoluble suhstance. i/.. p'):=O..es of reduction of order. Find the particular solut. 1) If a dJ1Yerential equation does not contaJn y cxpl1cltly. 2909*. then Y=~dx~ . y")=O. Cac. F(x.Order DttJerentlal Equations 345 2908*.Sec. y' =0 \vhen x=O. for instanc~. whence xp'+p+x=O. Solution. The air resistance to a body falling with a parachute is proportional to the square of the rate of fall. then.. we have !I' = p'. Example 1.E sin rot (E and (J) are constants) and i = 0 when t = O. 10. ~f(. Assunling that the rate at which the salt dissolves is proportional to lhe ditference between the concentration at the given time 8nd the concentration of a saturated solution (I kg of salt per 3 litres of water) and that the given quantity of pure water dissolves 1}3 kg of salt in 1 minute. resistance R and self-induction L is made up of the voltage drop Ri and the electromottve force of self-i nduction L :.. find the quantity of salt in solution at the explraiion of one hour..on of the equation xy" +y' +x=O. 101 IJigher.\). Find the limiting velocity of descent.--" 1 'me's 2°. Sec.. If y(/. that satisfies the conditions y=O. lhe electromotive force e in a circuit with current i. Putting y' =p. p.\)dX+CIXII-I+C1Xn-I+. +Cn' '----. p.Ct Putting y=-O when x= 0. Find the particulnr solution of the pquation yy"_y'2=y4 provided that y= 1. We have obtained an equation of the Bernoulli type in p (y is considered the argument). y'=O when x=O. we have 1 arc cos - t} ± x = C2_ I Putting y=l and x=O. we find C 2 -= O. w~ obtain x:! y=-4"t. we find p= ±y VCI +y2. for instance. Integrating. p~~)=O. y")=O then. Hence. putting y' =p. \ve get an equation of all order one unit y lower: F(y. Le. Solving it. y'. 2) particular If a differential equation does not contain x explicitly. p= ± y V y2-1 or :~= ± y Vy"-1.-O. Put y'=p. \ve obtain C2 =O. Solution. From the fact that y' =p=O when y= 1.DtUerenltal Equations 346 [Ch 9 From the fact that y'=p=O when x=O. x p=-2 or dy x dx=-2' whence. . lienee.. we have C1 = -1. Example 2. Hence. whence -=cosx or y y=s~cx. F(y. y" =p dd P . integrating once again. we have O=C. C1=O. the desired solu lion is 1 X2 Y = _-4 . then if=pddP and our equation becomes y I IJP : : -p"= ~/·. 2 _ y. 2914. yy" . y'" _ yy" = y2 y '. 2919. x 2920. Find the general integrals of the following equations: 2932. y' = 1 for x = 1. 3y'y"=y+y.y=2' y'=l tor x=l. _ O.+y. 2929. y" = = 1 _ y.r y =xy. y = I. y=O.·i_~1. y=2. y=-2. y" 2922. y"y' = 1.~ y. 10] 347 Solve the following equations: !-. y + 1) y" - (x -t. xy"=y' In y'. 2925 · y +-. 2935. y'=1 for x=O. 2916. y" = 2912. y = i ' y' = 1 for x = 1. 2931.Z = 2yy". y2+ y '2_2yy"=O. 2944. 1 + y. y' = 1 for x = -} . y =:= 1. !I = 1. y'=O for x=O.2)y' + x+ + 2 = O. y". 2930.y' yZ y.l. yy" = y. y=I. xy"= VI . 2926.Z + y"Z = 1. Find the particular solutions for the indicated initial conditions: 2928. y'= I for x=O. 2917.J+yy"=O. yy" + y. V + 2933.. 2915. 2934. y"= -2~S' 2921. x1y" -t-xy' = 1. . y'=3 for x=O. 2927. 2942.2.2+ I ==O. yy"=y. y=l. 2940. y'=2 for x=O.ol In y = + o. y'=O for x=O. 1 .y' (1 + y') = 2911. yy. y=O.:i. 2924. y" = ~ ( 1 + 1n ~'). y' = 1 for x = O.Z -t. 2913.Z.+y. yy' = V y2 + y.J = y. yy" y. . y' (1 + y'2) = ay". (1 +x 2)y"-2xy'=O. y=l for x=O and y=O for x=-l. 2943. (x O. yy"-t-y'2=I.2. 2918. x yy. Fi nd sol utions that satisfy the indicated conditions: 2936. xy'" + y" = 1 + x. 2941.Z y" _ y' y".. 2938.2 " ' (ll-x 2 )y"+-y. xy"·t y' == 2923.Higher-Order DiUeTential Equations Sec. y"_y'2+ y'(y_I)=O.z=O. y"(1 +lnx)-+-x· y'=2+lnx. y= I for x=e J • 1 1 2939. y=O for x= 1. 2937. yy" = yly' + y. x. xy"=y'.2. yg" -y'. Choose the integral curve passing through the poi nt (0. y' = xy.yy" -1. The frictional force is J1N.. 1 f 1 2949 • Y = Y .e' 2951.2. y' =xy" y" _y"z.y. y". 1. Y = . where N is the force of reaction of the plane. Pind the curves whose projection of the radius of curvature on the y-axis is a constant. + 2957. y" = 0. y'=O for x=O. 0) and tangent. at it. y'=4 forx=l. We may consider that the air resistance in free fall is proportional to the square of the velocity. y = 0. y' = 1 for x = O.a = 0.2=4y l. Find the equation of the cable of a suspension bridge on the assumption that the load is distributed uniformly along the projection of the cable on a horizontal straight line. y'=e for xO-=-.3 -t. 2962. the ends of which are attached at two points and which has a constant load q (including the weight of the thread) per unit length. Y"+. (x+l)y"+xy. 2956. 2946. y=l. y' = I for x == I. 2963.2) y'.2=O. Find the curves of constant radius of curvature.2eJ'2 Y'-2yy. Find the law of motion if the initial velocity is zero. 2955. Find a curve whose radius of curvature is double the normal. 2952. y= 1.Z = 0. 2958. The weight of the cable is neglected. 2960.1 = 4y". 2948.yl_y. 1 -t.2==y'.. Find the position of equilibrium of a flexible nontensite thread. 2965*. 2961. Find a curve whose radius of CUIvature is equal to the normal. 2959. Find a curve whose radius of curvature is proportional to the cube of the norma 1. yy'y":=: y. y' = 2 for x= O.. g'y. .yy') y" = (1 y.9 2945. 2yy" -t. + Solve the equations: 2954. (Hint.. y=l. y == 1. to the straight line y + x = O. 2y' + (y. 2yy"-3y. ". A heavy body with no initial velocity is sliding along an inclined plane. 2964*.y"l. + 2947. y' = 2 for x = 2. = 0.4"' y = 2 or x = . 2950. (1 -t. y' = 1 for x = o. y = 0.2 -6x).2 -t y. and the coefficient of friction is Jo'.) 2966*. Find the law of motion if the angle of inclination is a. y = 1.348 DiOerelltial Equations [Ch.y. y=-2. 2953. . (x) y~n-I) + C: (x) y~n-.. The functions Yl=q>l (x). such that C. otherwise. 111 349 2967*.) + + C. 2°.. Y = q>n (x) are called linearly dependent if there are constants C. CI . (x) and the right side f (x) has the fonn Y=Yo+Y.ith continuous coefficients P. (x) Y. YI' . The resistance of the water is proportional to the velocity and is 10 kgf at 1 metre/sec. +C n (x) Yn' where the functions C. . .Linear Differential Equations Sec. (x) Y2 C. where Y. +C~ (x) y~n-I) = 0.. . •. (x) Y. + C~Y2 + ··.v) y(II-I)+ ••• + Pn (x) Y =f (x) (2) + \\. where Yo is the general solution of the corresponding homogeneous ~qua lion (I) and Y is a particular solution of th~ given inhomogeneous equation (2). Solve the equation xy"+y'=xl • (4) . . y". Cn not alj equal to zero.. . 11.l (x) Y. Inhomogeneous equations. rz) IS (I) of the fornl y = C1Y. . Yn of the homogeneous equation (I) is kno\vn. Linear Differential Equations 1°. these functions arc called linearly t11dependent. Example. -1.•• .y.. + C1Y2 + .(x) (i=l. 2. then the general solution of the corresponding inhornogeneous equation (2) Inay be found from the formula y= C. -t.C... If the fundam~ntal system of soluHons 'il.) + C~ (x) y~n-.+. . + CnYn e= 0.. + CnYn.. The general solution of an inholnogeneolLs ltnear different ial eq ua tion y(n) PI (..) = f (x) J +C~ (\)y~==O. +C 1 (x) Y. + C~ (x) y~ c~ (x) y~t1-2) + + +C..•. .(x) (i=l.11) are determined from the follo\vjog system of equations: C~ (x) Y. Homogeneous equations.P n (x)y=O with continuous coefficients P. lin are linearl y independent sol utions of equa tion (I) (fundamental system of SO/Ii/tons).. = 0. Y2=q>2(X).·. How long will it be before the velocity becomes 8 m/sec? Sec. . 2. (3) (the method of uariation of parameters). . A motor-boat weighing 300 kgf is in rectilinear motion with initial velocity 66 mjsec. The general solution of a homogeneous linear differential equation y(1I)+Pl (x) y(fJ-1) + . ••• . 1 ~ + C~ (x) y~'-. . 2970. 1. h) sin 2 x. e2 " . . Y2 = cos x.9 (5) Hence. e) x. c) 0. Yt=x . we obtain x Whence X' x' x' CJ(x)=a+A and C2(x)=-31nx+g+B and.2x 2 . 1. we get y=C 1 In x+C 1 • [Ch. x d) Yl=e . Xl . Forming the system (3) and taking into account that the reduced fonn of the equation (4) is y" + y' =x. knowing its fundalnental system of equations: a) Y1 = sin x. b) xs. Ya=e"co5x. 2. Yt=exsinx. g) sin~. b) Y1= eX. y. Yt=x!.~=l=-I) Y"IX:l=2. cos x. consequentl y.~=l=O) y'I. where A and B are arbitrary constants. d) x.DiOerential Equations 350 Solution. 2968. el x . « 2969. x. f) e". x -+. It may be taken that Yl=lnx andY2=1 and the solution of equation (4) may be sought In the form y =C 1 (x) In x + C2 (x). Solving the homogeneous equation xy"+y'=O. find its particular solution y that satisfies the initial conditions YI.=x' . Test the following systems of functions for linear relationships: a) x. Yt = xe x . cos! X. x + 1. x + 1. 2 c) YI=X. x!. Knowing the fundamental system of solutions of a linear homogeneous differential equation Y1=x. x' Y=g+A Inx+B. 1. Form a linear homogeneous differential equation. Sec. 2) y -= e'l. 12] Linear DiOerential Equations with Constant Coefficients 351 2971*. =a-t-~t and k2-=-a-~1 ((3:f:. then Y=reuxQn(x). 0). The function Y may be found by the method of undetermined roefficl~llts in the following simple cases: l. Solve the equation x! (In x-I) y" -xy' -t. then we put Y =e'J.:l= k~.=secx. where Yo is the general solution of the corresponding equation (I) without rl~ht side and deternlined from fOrJnulas (1) to (:3). (x) sin bx].>"· + C~ek. Inhonlogeneous eljuations. X2Y"1-Xy'-y=x2. == k 2. The general solution of a linear neous differential equation y"+py' -+qy=f(x) lllhoI110~e­ (3) 01 ay be wntten in the form of a sunl: Y=Yo+ Y. Homogeneous equations. where Pn (x) is a polynomial of degree n. Linear Differential Equations of with Constant Coefficients S~cond Order 1°. + C 2x) if k.Y = 0. By the method of variation of parameters. cns PX+C 2 SlIl ~. and 11 2 are roots of the characteristic equation cr (k) === k' + pk +q ==.Cle". 2975. . knowing its particular solution YI = X. A second-order llncar equation \\'ith constant coefficients p and q WI thou t the right si de IS of th~ fOrIn y" + py' +qy =0 (I) If k. Solve the equation y" + !)' y' +y = 0. <p (a) ::1= O. y'" -t-y'::. x1y" -xy' = 3x'. 1f a IS a root of the characteristic equation l2). 12. = -x2972. x Qn (x) \\hcre Q. 2.\) if k. then the general solution of til ree \va ys: ~quation (2) (1) is \\ritten in one of the following 1) y -.0. <p (a) =0. f (x)=eGX (P n (x) cos bx+ Q". "here r is the multiplicity of the root a(r=l or r=2). .::. If a is not a root of the characteristic equation (2). 3) y -=e?'x eC . (x) is a polynomial of degree n with undetermined coefficients.X (C.. I ar so It· si n x . f (x) =e ux PII (x). solve the follo\ving inhomogeneous linear equations.ts par t leu u Ion U. 2974*. Sec. . and Y is a particular solution of the given equation (3).x if k . that is. that is. 2973. 2°. k nowlng I. and k 2 are real nnd k. we obtain: 2e 2X (4Ax+ 48 + 4A)-e 2X (2Ax+ 28 + A)_e 2X (Ax-j. The right side is of the form I {xJ =eIJx IP n (x) cos bt + Qm (x) sin bxJ. then we put Y = eax [SN (x) cos bx -1. r= 1). is the multiplicity of the roots a ± bl (for second-order equations.TN (x) sin bx). Hence.9 DifJerential Equation..~~) • and the general solution of the given equa tion IS I Y= CIe X +C e-7 (+ e~. and equating the coerticil.0. consequently. 1 he right side of the given equation is f {x) = =4xe2X =tOXp"(x).I and us~d k. a = I and n=-l. since a cOIncides \1Jith the doubl~ root k=-l and. The characteristic equation k 2 + l=-O has roots III = i and k 2 = . since n=1 and 1-=. The general solution ot the corresponding honlogeneous equation will Isee 3. Find the general solution of the equation y"-2y' +u=xe"". sin x. Ye f '>' ( : x. Thus. where.l =0 has roots III =. Solution. then y = xr eax [S N (x) cos bx + TN (x) sin bx].>nts..k . 2X Can.B) = 4xe~x. where a=O and p = 1J be· Yo=C.. B =0. is In the general case. where SN (X) and T f:I (x) are pol ynomials of degree N-max {n.352 (Ch. But if q> (a ± btl = 0.x ( S-x-2"5. Y=e 2X (Ax+8). l-Ience. . Example 3. 4 28) 2 Example 2. cos x+ C. Find the ~eneral solution of the equation 2y"-y'-y=4xe2". m}. The characteristic equation 2k 2 . DiOeren· tiating Y tWice and pu tting the derivatives into the given equa tion.= . the method of variation of parameters (see Sec. the general solution of the given equation will be \\rltten in the form y=(C I +Cax) eX +~ x'e". The particular solution is Y =x 2e" (Ax+ B). Solution. we helve 5A =4 and 4 28 7A+58=O. . substItuting into the equation.I . The charactenstlc equation k 2 -2k+] ==0 has a double root k = 1 The ri~hl side of th(:l equation l~ of the {orin f (x) =xex . 11) to solve equation (3). her~. Exall1ple t. Solution. The general solution of the corresponding homogeneous equation x (first type) is 1/ 0 = Cle" +C 2e z. Find the Jl:en~ral solution of thE:' equa tion y" + I/=X sin x. r=2.cplling out e and equating the coefficients of identical powers of x and the absolute ternlS on the left and right of the equality.s If q> (a ± bi) :F 0. whence A=S and 8=-25. Diflerentlating Y h\ice. we obtam A = i. b = 1. y"-9y=O. y'=O for x=O.-qY=--. b== 1. 12-1900 . Y n + is t he so1uti 0 n 0f equa t ion (3). y" + 3y' +. 30 • The princi pie of superposition of sol utions. x cos x. . y = 1..y' == O. y" + 4y = 0. y" + 2y' = 0. ••• . y = 0. y"-4y' +2y=0. N= I. Y =x [(Ax+ B) cos x+ (Cx+ D) sin xl (here. + In (x) and Yi (i = 1. 2984.. C=O. 2978.4y = 0. y" -5y' 2991. 7 =3.Y == O. Fi nd the particular sol utions that satisfy the indicated conditions: 2987. y" + ky = O.. y" . y=a.. 2983. y' = 8 for x = 0 2988. 2990. r= I). y" -I. y" -5y' -J.. 2989. Y = y" + y'. a 2992. If the righf si de of equation (3) is the sum of several funct:ons I (x) -=11 (x) +'2 (x) + . y"·t 4y' + 13y=O. -4A=1. D=4· Therefore. 2977. we equate the cOE'fficlcnts of bolh sides in cos X. y=O for x=O and y=O for x= 1. 2979. a=O. y' = 2 for x = O. Y = 1.fi(x) (i=I. Y= 5. b) y"+9y=cos2x. 2985. P n (x) =0. y'-y 2981. y"+3y'=O~ y=O for x=O and y=O for x=3. and x S1n x. y' = 0 for x = 0 2976. 2980.4C=O. n).. sin x. Indicate the type of particular solutions for the given + i Ilhomogeneous eq uat ions: a) y"_4y=x 2 e2X . QIn (x) =x. cos x+C 2 sin x-"4 ros x+T sin x. y" . 3. /2] Linear DiUerential Equations with Constant Coefficients 353 where a =0. y' = -1 for x = o. y" n 2 y=O. The general solution is Xl x y= C. We then get four equations2A+2D==O. 2982.2y = 0. -2B+2C=O. 2993. 1 1 mine A = -4' 8=0. Y= --:rcosx+-:rsinx. then the sum y=Yl+ Y 2+ . 2. 2994. 2986. [rom which we deterx2 x . Differentiating twice and substituting into the equation. Find the general solutions of the equations: + 6y = O.Sec.2y' + 2y = O. g" + 2y' + Y= o. n) are the corresponding solutions of the equatIons y"+py'. .. y"=Y2. 2. To this side there corresponds the particular solution Y. y" . y" + y' = sin 2 x. ~:~+<illx=ASinpt. 3013. Solve the equations: 3020. y" . 3018. 3021.2y' -+ 5y = eX cos 2x. 2999.y =-=: eX. 3019. y" ~t y' = 5x + 2e x • y"-y'===2x-1-3e x . 3015. y" + y' -2y = 8 sin 2x. Consider the cases: 1) P=/=<ilj + 2) P = 0>. y" + y' -6y = xe2X • 3003.I) e" + xelX~ f) y" . yIP = xe" -t. 3004. y" 2y' y= eX +e. 3025. 3008. y"-2y' IOy= sin 3x +e x • + + y" -4y' + 4y = + 2e IX + ~. y" -y' + y = xl·t 6.4y' + 4y = Xl . y"-2y'=e 2X +5. 3000. . y"-2y' -t-y= sinx+sinhx. 3010. y"-2y' =x 2 -1. y' = 1 for x = O. Find the solution of the equation y" + 4y= sin x that satisfies the conditions y = I. Solve the equations: 3007. Find the solution to the equation y"-2y'=e1x +xl . 3006. 3001. y" + 4y = 2 sin 2x-3 cos 2x+ 1. 2996.x . 2997. y"-y=2xsinx. y"-3y' =x+cosx.l that satisfies the conditions y=-lr.2y' + 5y = xe" cos 2x-x1 e" sin 2x. 3024. y" + 2y' + y = e2X • 2998. y"-2y' -8y= eX -8 cos 2x. y" .2y' + 2y = 4e x sin x. y" . d) y" + 2y' 2y= eX sin x. y" -By' + 7y= 14. e) y" -5y' + 6y = (xl-t. 3023. 3017. 3014. 3005. 3009. 3022. y" y = cos x. 3016.9 c) y"-4y' -t-4y= sin 2x+elx . 3002. y"-2y'+y=2e x . y"_4y=e2X sin 2x. y'=l for x=O. + Find the general solutions of the equations: 2995. 3011. y"-7y'+12y=-e4 ".y. y" + 9y= 2x sin x+ xelX • . 3012.354 DiUerential Equations [Ch. 3030·. y" + y = cot x. Let the increase in the length of the spring under the action of one load in a state of rest be a and the Jnass of the load. to the increase in its length and is equal to 1 l<gf when the length increases by 1 cm. x 3038. e-" + 2y' + y ="7. y"-2y' =3x+2xe". 3028. dt 2 tion is x=CtCOS V: t+Ctsln V~ ~ t. mg d~ where. A load weighing 2 l<gf is suspended fron) the spring. 3034. i12· . 3027. a) y"-y=tanhx. cosx 1 snx 3037. Denote by x the coordinate of the load reckoned vertically from the position of equilIbrium in the case of a single load.and. The ~enerat ~olu- The initial condItIOns Yield x=a d\': and dt =0 when t =0. solve the following equations: 3036. y" -2y = 2xe x (cos x-.nd the law of motion of the load if the upper end of the spring performs a vertical harmonic oscillation y=2sin30tcm and if at the initial instant the load was at rest (resistance of the medium is neglected). y" +y= ex. obviously. Fi. In. whence C. =-ag x. A load weighing P=4 kgf is suspended from a spring a. Solution. 3033. y" -2y' 3035. The force stretching a spring is proportional. yn + y = _1_.Sec. y"-4y' +4y=xet ". · 3041*. y" + y = 2x cos x cos 2x. y" + 2y' -3y = 2xe-·" + (x -t 1) eX. 3031. 3032. 3029. Two identical loads are suspended from the end of a spring. f"ind the period of oscillatory motion of the load if it is pulled downwards slightly and then released. 12) Linear DiDerential Equations with Constant Coefficient~ 355 3026. y"-2y'-3y=x(1 +e'X ). =a and C. 3040*. consequently.nd increases the length of the spring by 1 em. Then d2 x m dt 2 =mg-k (x+a). k=a. y" + y = _1_.sin x).=O. and so X=QCOS V: t. b) g" -2y = 4x 2 exJ • 3039. y" + y = tan x. Applying the method of variation of paranleters. Find the equation of nlotion that will be p~rfornled by one of these loads if the other falls. 1e'l.=xe'l. A long narrow tube is revolving with constant angular velocity ro about a vertical axis perpendicular to it.. Ya=xe'l...•. YI = xek" . how long will it take for the entire chain to slide down? 3044*.=e'l. . . the initial velocity of the baH was zero. then to the latter there correspond 2m linearly independent solutions of equation (I): 1I. The fundamental systelTI of solutions Yl' Y.Xcos~x. A particular solution of the inhomogeneou§ equation y(n) -t a1ycn -1) + .x sin ~x. Y m =xm-'ekx ...xcospx. Y2m=X m . then to this root there correspond m linearly independent solutions of equation (1): Yl =ekx .=e'l. b) at the initial instant the ball was located on the axis of rotation and had an initial velocity vO• Sec.. t) if k is a real root of the equation (2) of multiplicity m.. considering that a) at the initial instant the ball was at a distance a from the axis of rotation. . 2) if a ± ~l Is a pair of complex roots of equation (2) of multiplicity nl. Yf. . linear Differential Equations of Order Higher than Two with Constant Coefficients to. ...Xslnpx. 13. 2°.1y' + anY = f (x) (3) II sought on the basis of rules 20 and 3° of Sec. A ball inside the tube is slidin5 along it without friction. Homogeneous equations. Find the law of nlotion of the ball relative to the tube. .1e'l. Un of a homogeneous linear equation with constant coefficients y(n)+a1y(n-I)+ •. 3043.356 DiOerent fat E quat ions [Ch.• Yam_l=x m .. Find the law of motion of the point knowing that the distance between the centres is 2b. If the motion begins when 1 m of the chain is hanging from the support. . A chain of length 6 metres is sliding from a support without friction. at the initial instant the point was located on the line connecting the centres (at a distance c from its midpoint) and had a velocity of zero. 12.. 9 3042.x cos ~x. A material point of mass m is attracted by each of two centres with a force proportional to the distance (the constant of proportionality is k). y. + all .Xsin~x. +an_ty' +any =0 (1) is constructed on the basis of the character of the roots of the characteristic equation (2) Namely. Inbomogeneous equations. -6054. ':1 3064. . + An -1 (ax + b) Y + AnY -= f (x). 357 Euler's Equations the general solutions of the equations: y"'-13y"+12y'=O. 3061. 3057.n (n -1) en-I) 3049.b -=e t .Sec..yIP = x" + I + 3xe-. (1) where a. sin (In x) + 1. IS called EIller's equation.AI (ax -t. + yIV_2y"'+y"=e x • ylV -2y'" + yIP = Xl. 14. -J.. Consequently. y'" +.t r!:!!. . + )·2 . 3052.dt l -l.+y=l. 3058. 3055. the given equatio 1 takes on the fonn d!y dt. 3058. putting ax -1. AI' . y'V -2y" + y = O.. 3060. y"'-3y"+3y'-y=O. y" ==a!e-~t 1_ Y' (ddIe! d!/) dt' dt' :s -:st (d " ~ d 11 dl l -ae dl:J. ylv+2y"'+t/'=O.} yCn-I)-t_ y'" +y=O. 3056. Putting x==-e t . ylV + aly" = O. whence or y=C I cos t +C" sin t + 1 g= C1 cos (In x) + C. 3067. 3059. Eul er's Equations A linear equation of the form (ax + b)n y \ll) 1. y'" + y" y' + y = xex' y"'+y'=tanxsecx. 141 Find 3045. yIV+y'=O. ylV -6lJ" + 9.b)n-l y (n-1l + . 3066. ylV -2y" = O. 3046.. Then 91 = ae.. we get + dy= _tdl/ dx edt' dll/= -2t(d!Y_ d l/\ dx" e \ dt" dt ) ." = O. y'" -y = xl-I. 3063. Find the particular solution of the equation y'" +2y" +2y' +y=x + that satisfies the initial conditions y(O)=y'(O)=y"(O)==O.y'" = cos 4x.. ylV -t.. 3048. b. y(n) -+!!. y'" _y' = O.8y"+ 16y=O.2 dt 2 3 y II' _ l ) an d so for th and Euler's equation is transfornled into a linear equation with constant coefHclcll ts. ylV t. 3062. Y +~1 y' +y=O. 3050. 3047. 3065. fxarnple 1. ~olve the equation x"y" + xy' y = 1. Sec. ylV _ally = O. ylV -f-4y==O. An_I' An are constants. Solution. Let us Introduce a nc\v Independent vanable t.. ylv+2y""+y=O. 3051. 3070. 3072.. Solve the equations: 3068. x·y"-4xy' +6y=x. x Y' Y 3074. (1 +X)I y"-3(1 +x)y' +4y=(1 +x)·.. .. Ys =xt% In x cos (~ In x).n-u+ . y"=k(k-l)xk - 2 • Substituting into the given equation. 4y = O. . + + 3073. We put y'=kX k - y=xk. 3069. . 2 d2 y X dx l + 3x dy dX + y = O. the general solution will be Y = Ct x2 + Ct. Ya= x fJ ln x·sln (~ In x). y" = 2~. •••• Ym =xk (In x)m-l..xy' +. x 2 y" -f.. (3x --1. y'. we get a characteristic equation from which we can find the exponent k. Y2m-l = x fJ (1n x)m-l cos (~ In x). Example 2. then to it correspond m linearly independent solutions Yl =xk.ln x. Ys =xk (In X)I.2) y" 7y' = O. If k is a real root of the characteristic equation of multiplicity m. Y1 = /l.. . 1 . YI = xt% sin (~ In x). xSy'" -3x' y" 6xy' -6y = O. 3071. Y2m = x fJ (In x)m-l sin (~ In x). Solution. + An_1xy' + Any=O the solution may be sought in the form Y=~. Solve the equation rt x 2y -3xy' +4y=O. x x 3075. . after cancelling out xk • we get the characteristic equa tion Solving it we find Hence. y"+-+"!=O. 3076. y(n) found from (3). If a ± ~i is a pair of complex roots of nlultipllcity m. (2) (3) Putting into (2) y.DiOerential Equations 358 [Ch. 9 For the homogeneous Euler equation xny(n) + A1xn-1y. then to it there correspond 2m linearly independent solutions Yl =x~ cos (~ In x). 2 x y"-xy'-3y=O.x 2 In x. The set of fornlulas (3) and (4). y. Solve the system :~ +2y +4z= 1 +4x. that is. Substituting function (4) into formula (3). we get a second-order equation with one unknown function 1/. We differentiat-e the first equation with respect to x: dly +2~+4 dZ=4.6x 2 -4x+3. :~) (3) into equation (2). we arrive at a secord- order equation in one unknown y: 2 dy dx' + dx dy -6y=. CI' C2 ). 15. we determine the function z without new integrations.z). d2g of af dX 2 =ax+ayf+ af az g. we differentiate one of them with respect to x. Sec. To find the solution. z=q> ( x. Example.Systems of DiUerential Equat ions Sec. . Find the particular solution of the equation x 2 y" -xy' + y= 2x that satisfies the initial conditions y= 0. where y is replaced by 'i'. Putting z into the equation obtained after differentiation. y' = 1 when x = 1. lt51 359 3077.:~ -2Y ) dz 3 2 1 3 1 dy havedx=2x+x+4-2Y-4dt(. dx=g(x. for instance. We have. z). Solving it. (2) Determining z from the first equation of the system (1) and substituting the value found. yields the general solution 01 the system (1). Systems of Differential Equations Method of elimination. WIll and then . of a system of the fonn dy dz (1) dx=f(x. . we find Y = '" (x. y. dx 2 dx dx From the first equation we determine frOln th esecond we and :~ Z=} ( 1 + 4x. (4) where C I and C2 are arbitrary constants. solved for the derivatives of the desired functions. for example. of a normal systenl of two first-order differential equations.z ="2 X • Solution.y. { dz 3 2 dx+ y . z=O when x=O. x-y +z.Ex-y' z-x isolate the integral curve passing through the point (I. -2). { * dx dy dz. dt =Y dy y2 { :=T' dx=2 Y . dz dx=-Y. ~=z. dz d~ +2x--y +2e'=O. sin x. Y= 1 when t =0. d~ +2y+4z = eX. { dz 2 dx + x2 Y = In x. . 1. t dt = 3082. x=O. ~ + 2y + z = = x+y z' dy =. I dz dx=x+y+z. { dx -4y. ~=-3Y-z. =2x. 3080. y-. d!J+ = 1 d" Z . Z.3z = -x. 3089. dx 3086. :~=y+z.=x. 3079. dz { dx = dl/ 3087.9 Solving it we find: and then 1 ( 1 +4x---2y dY ) = -C e2X +-!e C _ 3X _ _ 1 X2 z=1 4 dx 2· 4 We can do likewise in the case of a system wi th a larger number of equations. b) ~= dy =dz. d2 3090. 2 { d z dx 2 - y .z. d dy l :. Sol ve the syste ms: 3078. 3084. ~. t dt=x+y. c) ~ y-z dt=x+z. y=O. ( ~7=Y' 3081.Y + 3z = o. { dz :~=Y+5z. { :~ +3y+4z 3085. dx-Y-z=x.DifJerential Equations 360 (Ch. 30 88 · a) X 1+-3 2z ' xy 2=21y = -y2 dz 3083. -4x-y + 36t =0. dz { dx -J.2z = cos x. y' (x o) = f (x o' Yo) and the subsequent dprivatives y(nl (x o) (n =. .. .. y). If y=Yo. on the left of this equation. ]-x (CO+c1x + ....2. Find the solution of the equation y"-xy=O.:O. y' =y. ] ==:0.. We put whence. +3. Find the equation of motion if we take the air resistance as proportional to the velocity. for x:=. 3092*. 16. differentiating. ) are successively found by difT(r~ntiating e'luation (2) and by putting Xo in place of x Example 1. Sec. Solution. (3) n=o where y (x o) = Yo. I Collecting together.Sec. y(n~~xo) (x-xo)n. we get y" = 2. Find the trajectory.x +. We can also seek the solution of the equation y'-:=f(x.lc.+ + (n + 2) (n + I) en + 2xn +. then in some cases Its solution may be sought in the form of a power series: co y = ~ Cn (x-xo)n. ) are found by putting the series (I) into the equation and equating the coefficients of identical powers of the binomial X-X o on the left-hand and right-hand sides of the resulting equation. 2 1 1 Substituting y and y" into the given equation. +cnxn + . A shell leaves a gun with initial velocity V o at an angle a to the horizon. The motion begins from point A at a distance a from the centre with initial velocity V o perpendicular to OA... A materia I point is attracted by a centre 0 with a force proportional to the distanc~. y(xo)=--=Yo (2) in the form of the Taylor's series «J y(x) = L. we Will .2c. 3..+ (n + I) nCn +lX n . + 3·2cax + .. Integration of Differential Equations by Means of Power Series If it is not possible to integrate a differential equation with the help of elementary functions. 2... we arrive at the identity [2·1c. the terms with identical powers of x and equating to zero ihe coefiiclents of these powers...+ (n + 1) ncn + Xn .1 + + (n +2) (n + I) cn + 2x n + .. + n (n-I) cnx n . + n (n-I) cn x n . (1) n=o The undetermined coefficients cn (n = 1. 16] Integration of DiOereniial Equations by Power Series 361 3091**.. . ' " Yo. ·3k (3k + lr (k= 1.-co=O..7+···-J.. . cak+1 = Consequently f X' xci x·k y=co ( 1+ 2.. (or x =O. For the examp'le at hand.4. Y" =y. -Yo 2 Yo a Y=Yo+Y ox+2f x + 31 x + . Yo = Y (0) = 1. y=yo for x=l. . generally speaking. test the solutions obtained for convergence.=O. = .6. + 2.3.etc.6 +. y'=X _ y 2. 3098. ..6.7 .3c4 -cJ =O.3. sJvely 2 y= 1 +x+2f x 2 2 a +31 x + .. With the help of power series. y' =+ yl I Xl. (3k-l) 3k' Cak+2=O ~ 3. this solution may be written in final form as y=1+x+2(eX -l-x) or y=2eX-l-x. c'=3. • 4.. Example 2. y'==y+x2 ..'14 CJ 5 . y=-2 for x=O. + We have yo=l.DiOerent ial E quat ions 362 [Ch. y=O for x=O.3. find the solutions of the equations for the indicated initial conditions.. 3093.00 < x < + 00. In Examples 3097. Testing the resulting series for convergence is.5.. (3k-l) 3k +.3 + 2. We put " ".9 have Co 3·2c.• 3k(3k+l)+··· .. Find the solution of the equation y' = x y. y~=O+l= 1.. The procedure is sim ilar for differential equations of higher orders. 3099.4c.2.3. it is readily seen that series (4) converges for .6 . cJ=O. y =. fiIn d y" = 1+y. ). we succes.4·6·7· . x7 X4 +C 1 ( x.. Consequent 1y... y'=2y+x-l. Generally. = 2..4... y=O for x=O. 3094. COIDplicated and is not obligatory when solving the problems of this section.2.3. C. 3101.4+3. (4) Y..3' and so forth. Differentiating equation y'=x+y.kq.. 1 ) + ) x+3.5. (l-x)y' = 1 +x-y.-c.. 3096. ~ cak 2.. ' Yo= " 1+ 1=2.. 3095..5·6 . c4 =4. where Co = Yo and C1 = Applying d'Alembert's test.. 3097. Solution.. t) =0. 17._---~--. The desired solution u is represented in the form of a series arranged according to these particular solutions: (I) n=l The coefficients condit ions. y" X 2 3102. x + 2. In the simplest case.y' + y = 0. 3101 *..+J:. y' = 0 for x = O. A transversal displacement u =U (x. y~l.- r'--' 1 x Fig. Solution.. en which reInain undetermined are found from the initial Problem. there is an infinite set of such solutions un (n= l. dx dx dt + x cos t = 0. . u (I. 107) and its points had zero velocity.. each of which represents the product of functions that are dependent on one argument only. xy" + y = 0. y=l. y. dt = 0 for t = O. Problems on Fouriefs Method To find the solutions of a linear homogeneous partial differential equation by Fourier's method.Y'-t-y=O. y"+xy=O.. =a :2 au ax 2 2 (2) ' = To (To is the tensile force and Q is the linear density of the Q string). Y = 0. 3100*. Find the form of the string at ti tne t if its ends x =0 and x= I are where at. It is required to find the solution u = u (x. ). 3099.. t) of the points of a string with abscissa x satisfiE:'s. y'=O for x=O. 2 Sec. u O~--_-. y' = 1 for x = o. first seek the particular solutions of this special-type eq uation. 107 fixed and af the initial instant. the eqnation au 2 att. which are linearly independent among themselves in any finite number and which satisfy the given boundary conditions. the string had the form of a parabola U = ~: x (i-x) (Fig. t) =0 (3) . y'=O for x=O. t) of equation (2) that satisfies the boundary conditions u (0.Problems on Fourier's Method Sec. y = 1. t = 0.. at time t. 2. x = a. 17) 363 3098*. C = 0 and sin Al = 0 (since D cannot be equal to zero at the same tinle as C is zero). O) = ~ A k sin -1. where k is an mteger.• k=J it follows that.. iJt CJ) 0) ~ kan ..- Ak . kant) .== j2 x (I-x) k=J and au (x. (4) We seek the nonzero solutions of eauation (2) of the special form U = X (x) T (t)..t + Bk sin -1. kant Sill .- • k=J whose sum obviously satisfies equ3 tion (2) and the boundary conditions (3).x + D sin ).Sill -.. where A. k:rrx 4h (x. T (t) = 0 and X" (x) + ). Solving these equations.). From condition (3) we have X (0) = 0 and X (I) = 0.IX (x) =0.B k sin -lk==J c:::iO. Ut (Ch. 3•. we get T (t) = A cos aAt + B sin a). knx U= ~ Ak cos -1-+Bk s1n -1. 0) = j2 x (I-x).AI. B. . we find two ordinary dIfferential equations: T" (t) + (aA)2.. equation (5) is possible only when the ~eneral quantity of relation (5) is constant Denoting this constant by . Since CJ) au ~ kart ( at = ~ -/. Putting this expression into equation (2) and separating the variables. hence.9 (x. 0) =0. . X (x) = C cos ). kan ) .t. We construct the series CJ) ~ ( kant . knx uk= ( Ak cos-/. we obtain CJ) U ~ . For this reason. We choose the constants Ali ann Bk so that the sum of the series should satisfy the initial conditions (4). D are arbitrary constants. It will readily be seen that we do not lose generality by taking for k only positive values (k = 1.. - k~nt) knx + Bk cos -1sin -l. Let us determine the constants. 2.Differential E quat ions 364 and the initial conditions 4h u (x. we get T" (t) X" (x) (5) aIr (t)= X (x) • Since the variables x and tare indeprndent. by putting t = 0.X.t sin -llhat satisfies the boundary conditions (3). C. Ak = ~n . To every value Ak there corresponds a particular solution kan .. krrx ~ -l. and All = 0 if k is even: SOSin-. is pulled out to a distance h=2 cm at point x == 50 cnl at the initial time. De. (2n+ l)nx (2n -1. at time t.=O. and the points of it had zero velocity. had the form of the sine curve u = A sin 'IT/x. 17] 365 Hence. and is then released without any impulse. and then released without Impulse. Find the lorrn of the string at time t. l . 4. 3106*. t) of a cross-section of the rod with abscissa x satisfies.Bk=T knx o The sought-for solution will be (2n 00 _32h~ u- n3 L -t. 0) = · f unc t Ion au (x. B. 3104*.I) ant cos----l .Problems on Fourier's Method Sec. I kart 2 -L. dX=n1k A~=T J lI x (l-x)sinI 2 l ' o if k is odd. where a =. x (I-x) and the 0 ==. From familiar formulas (Ch. in sines on1 y. VII I. a string. 3105*.. to determine the coefficients All and Bk it is necessary to expand in a Fourier series. Sec. x = 0 and x = I. In longitudinal vibrations of a thin homogeneous and rectilinear rod. the equation iJ2 u dt2 = 2 a au ax 2 2 . A string of length 1 = 100 em and attached at its ends. x=O and x=l. At the initial instant t = 0.2'.3°) we have r4h knx 32h . the displacenlent u = U (x. the points of a straight string 0 < x < I receive a velocity ~~' = 1. Find the form of the strin~ at titne t if 1he ends of the string x=O and x=l are fixed (see Problem 3103). attached at its ends. at 0) . Determine the longiturlinal vibrations of an elastic horizontal rod of length 1 = 100 cm fixed at the end x = 0 and pulled back at the end x = 100 by Iii = 1 CIU.!!(E is Young's modulus and Q is the density of the Q rod). At the initial time t = 0. whose axis coincides with the x-axis.ernline the shape of the string at any time t.-dx=O. the function u (x.1)1 sin I • n=o 3103*. . For a rectilinear homogeneous rod whose axis coincides with the x-axis. where a is a constant. Determine the tenlperature distribution for any tinle t in a rod of length 100 cm if we know the initial tenlperature distribution U (x t 0) = 0. in the absence of sources of heat. the temperature u = u (x. t) in a cross-section with abscissa x at time t.01 x (IOO-x).· DiDerential Equations 366 [Ch. 9 3107*. satisfies the equation of heat conduction au at =a ! au 2 ax! . which satisfies the inequality IA-al~L\. Number of correct decimals. If the absolute error of an approximate nUlnber a does not exceed a unit of the last decimal place (such. then the number a has n correct deci mal places in the narrow meaning of the word. In particular. if it 1 ( 1 ' n-l is known that 6E. 3°. the latter (if it is the final result of calculations) is ordinarily rounded off so that all the remaining digits are correct in the narrow or broad sense. And conversely. . is called the limiting relative error of the approximate number a. when n > I we can take. Absolute error. (1) is called the limIting absolute error. 2 (k + I) 10 ) . we often take the number 6=~ for the lilniting a relative error. Since in actual practice A =::::a. are numbers resulting from measurements made to a definite accuracy). The number ~. ~ n .. which satisfies the ineq uality I A-at ~ (2) A 1Iilrl::a. The number ti.. The exact nUlnber A is located within the litnits a-&~A~a+& or. Operations on Approximate Numbers to.. By the relative error of an approximate nUlnber a replacing an exact number A (A > 0) we understand the ratio of the absolute error of the number a to the exact nurnber A. If there is a larger number of significant digits in the approxinlate number. 2°. for the limIting relative error. A=a± 6. The ab~olute error of an approximate number a which replaces the exact number A is the absolute value of the difference bet\\'een them. for example. the number 6=2~(I~r-1 · where k is the first significant digit of the number a.Chapter X APPROXIMATE CALCULATIONS Sec. We say that a positive approximate number a written in the fornl of a decimal expansion has n correct decimal places in a narrow sense if the absolute error of this number does not exceed one half unit of the nth decinlal place. the number a (/0) definitely has n correct decimals in the narrow meaning if f> E. then it is said that all decimal places of this approximate number are correct in a broad sense. more briefly. Relative error.. 1. In this case. if need be.3·4 12 = 104. Powers and roots of approximate numbers.12= 104 2 ± (J. we find that the limiting relative error of the product is + 1 6=2.1(8)+21 4=250.8. Therefore. The limiting relative error of a product and a quotient of approximate numhers is equal 10 the sum of the linliting relative errors of these numbers Proceeding froln this and applying the rule for the number of correct decilTaals (3°).21 +14. Addition and subtraction of approximate numbers. 7°.. all summands should be put into the form of that summand which has the smallest number of decimal places. If lill. they should be rounrled off and one or two reserve digits ~houl d be retained. Hence.n'erence is not amenable to simple counting. Then be guided by the foregoing rule of addition while retaining the appropriate extra digits in the sunl up to the end of the calculations. Multiplication and division of approximate numbers.3. it is always advisable to avoid subtracting close approximate numbers and to trttnsform the given expression. In subtracting the a~proximate numbers 6 135 and 6. 4°. in the sum of a ~mal1 number of approximate numbers (all decimal places of which are correct). The results of intermediate calculations may contain one or two reserve digits. Whence the number of correct decimals of the product is three and the result.itive terms bes between the least and greatest relative errors of these terms.12 = 104. ~5 3·4. 215. Example 1.368 Approximate Calculations [Ch. we get the dIfference 004.. 10 Hencetorth. The relative error of a d. Then add the resulting numbers as exact numbers.131 to four correct decimal places. if it is final. Calculating the error of the result of various operations on approxi- mate numbers. We note that the exanlples of this sectiCJn are. 5°. Particularly unfavourable in this sense is the ditference of two close nunlbers. we shall assume that all digits in the initial data are correct (if not otherwise stated) in the narrow sense.2°.003.25.236.• flail are the limiting absolute errors of the appro- .182+21. The relative error of a sunl of poc:.4==215. should be written as follows: 25.20.3·4. . as a rule. The limiting relative ° ~ 0. The product of the approximate numbers 25.. so that this undesir~!-Ie operation is omitted. only correct digits (at least in the broad sense). not one of the decimals of the difference is correct. and round off the sum by one decimal place If we have to add approximate numbers that have not been rounded off.2l1)+14. + 4.004 =4=0. or more correctly. Therefore..00\ ~ 0/'01 I error is B 0. and for this reason the answers to them are given as approximate numbers with only correct drcimals.!. in order to have. The limitIng relative error of the mth power of an approxi mate number a is equal to the m-fold linliting relative error of this number The limiting relative error of the mth root of an approximate number a is the . The limiting absolute error ot an algebraic sum of sevE'ral numbers is equal to the sum of the limiting absolute errors of these numbers. Assuming that all dEcimals of the factors are correct.th m part of the limiting relative error of the number a.01 I 01 ~0. and in each summand a reserve digit should be retained. we retain In the answer only a definite nunlber of decimals Example 3. the results of final calculataons. Example 2. 6°. .1· f. .093·2.1 20+2--20.Sec.0.3 and 4. R' tan a. Arplying the formulas of 7° for the quantities L\S or f..303 = 2. ~} • ~= . • an) .0976. that is.bb)' ~ 2 8~) = ~.3+ V4_4).4 are correct to one decimal place. . Example 5. Establishing admissible errors of approximate numbers for a given error in the result of operations on them. the absolute error we can be sure of the first decimal place.4 is equal to is then equal to b(~a +~ . I = I :1. ."4 = ~S= a+ ~ We have 2.20 l+U =2604=::::0. the problem thus posed is indeterm inate.S given us and conSIdering all particular differentials Ia~ I~ak I I or the quantities a~ ~. . hvo decimal places will be correct. L\a n . Let us first compute the limiting absolute error L\S in the general form: 8=ln (a+ vb). Now let liS do the calculations with one reserve decinlal: log (10. To what degree of accuracy should we measure the radius . may be evaluated approximately from the formula ~S = I:~ I~al + ···+I:!n IM n• The limiting relative error S is then equal to 6S = ~. In (10 3+ yU)~ 1. I) Operations on Approximate Numbers xirnate numbers all •• _ • 369 an' then the limiting absolute error ~S of the result S === f (at~ . 1 ( 1 1 1) 1 ( I) 13 L\S=IO. Hence. of the argul11ents of a function it is not advantageous to use the principle of equal effects. since the relative error of the approximate number ¥4. Evaluate S = In (10. And we ~('t the answer: 2 52 8°. we calculate the admissible errors L\a" .4.3+2. V4. The volume of a "cylindrical segment".. the approximate numbers 10..·+I:!n It'tr = = dln'l da L\a l l + ·. Thus.2.''a ~olid cut off a circular cylinder by a plane passing through the diameter of the base (equal to 2R) at an angle a to the base.· + Idln'l dan tian • Example 4.0. Thus.ai equal. • " of the approximate numbers at • ..517.. Solution.. strictly speaking. + ·.1 =12. is computed from thp formula ab~olute V= . It should be pointed out that sOlnetimes when calculating the adlnissible errore.• an' ••• that enter into the operations (the principle of equal effects). since the latter nlay make delllands that are practically unfulfilable In these cases it is advisable to nlake a reasonable redistribution of errors (if thi~ is possible) so that the overall total error does not exceed a specified quantity.. we leave 2.1. ~a=M~ .3 + y44) ~ log 12 4 -== 1.093.005. Determine the number of correct (in the narrow sense) decimals and write the approxinlate numbers: a) 48. Find the product of the approximate numbers.8 for an accuracy of 2%.0 +3. 3111.10 + 0.22 cm (accurate to 0.Approximate Calculations 370 [Ch. b) 25. which are correct to the indicated decimals: a) 3.72-11. 3114. R and a. 3110. c) 3.49· 8. which are correct to the indicated decimals: a) 148. Find the difference of the areas of two squares whose measured sides are 15.1 +2. Add the approximate numbers. 3113*.9360 for. we ensure the desired accuracy In the ans\ver to I % if we measure the radius to 1 mm and the angle of inclination a to 9'. b) 0. Measurements yielded the following approximate numbers that are correct in the broad meaning to the nunlber of decimal places indicated: a) 12°07'14". b) 38.7. an accuracy of 1 ?1l. b) 1. Compute their absolute and relative errors. c) 38.72 + 0. l\R and L\a are the limiting absol ute errors of the quantities V. b) 29.5 cm. 3109.102 + 41.215 kg.5. 3112.22-34. which are correct to the indicated decimals: a) 25. If l\V.6.2.1· 1. Compute the absolute and relative errors of the following approximate numbers which are correct in the narrow sense to the decimal places indicated: a) 241.28 cm and 15. then the limiting relative error of the volume V that we are calculating is 3L\R 2L\a 1 6=R+sin2a~100 · We assume 3L\R R 1 2L\a 1 ~ 200 and sin 2a. 600 ~ 600 = 1 mIn.. c) 62. 3108. ~ 200· Whence R 60 cm l\R E. b) 14.21.124.09.743.05 mm). c) 0. Indicate the possi ble limits of the results.25.02· 16.49 +3.5.386 + 0.871. Subtract the approximate numbers. 10 R:::::: 60 cm and the angle of inclination a so that the volume of the cylindrical segment is found to an accuracy up to 1%? Solution.14. A sin 2a 1 d· 9' ua ~ 400 <400 ra lan~ • Thus. . c) 34.1-63..035.361 for an accuracy of 1%. c) 592. 02 and 4. The radii of the bases and the generatrix of a truncated cone are R=23.715.01 cm.1 cm. the lateral surface of a truncated cone whose bas~ radii are 2 m and 1 m.4158 2 .Sec. 3121.96 m (accurate to 1 cm).01 cm).144 : 1. to within 1%. Use these data to compute the total surface of the truncated cone. b) V65.4 cm + 0.64 cm±O.1 cm. r-=17. while the relative error in weighing is 0. Find the quotient of the approximate numbers. c) 216:4. 3116. and the generatrix is 5 m (approximately).01.2.OI cm. The legs of a right triangle are 12. 3118.684 : 5.3 cm (accurate to I mm).14. 3125. 3122.31 cln±O.8 cm ± 0.2 1 . 1] Operations on Approximate Numbers 371 3115. 3124. The sides of a rectangle are 4.10 cm and 25. Compute the current if the electromotive force is equal to 221 volts ± 1 volt and the resistance is 809 ohms ± I ohm. 3123. b) 65. Evaluate the absolute and relative errors of the result.032. 3117. which are correct to the indicated decimals: a) 5.2.21 cm (accurate to 0.4 gmt The relative error in measuring the lengths is 0.1. To \\yhat degree of accuracy can we determine the second leg and the adjacent acute angle? Find their values. Compute the tangent of the angle opposite the first leg.5 2 • 3119. in order to obtain its oscillation period with a relative error of 0. 1= = 10. The hypotenuse of a right triangle is 15. The period of oscillation of a pendulum of length l is equal to T T=21t V- g . 1t == 3. It is required to nleasure. where g is the acceleration of gravity. Find the area.21 em + 0. The side of a square is 45.OI cm./81. Calculate the specific \veight of aluminiulTI if an aluminium cylinder of diameter 2 cm and altitude II em weigh5» 93. c) 1. Compute the area of the rectangle. one of the legs is 6. To what degree of accuracy do we have to measure the length of the pendulum. c) J. 3120. whose period is close to 2 sec. b) 0.5%? How accurate must the numbers 1t and g be taken? 3126.001. To what degree of . Compute the values of the roots (the radicands are correct to the indicated decimals): a) Y2. Compute the indicated powers of the approximate numbers (the bases are correct to the indicated decimals): a) 0. linear Interpolation. if f (x) ha') a continuous derivative f (n+J) (x\ on the interval la. and the quantities d and b are known to an accuracy of 1%. To determine Young's tnodulus for the bending of a rod of rectangular cross-section we use the formula I liP £=4· d3bs' where I is the rod length. Let Xo. Interpolation qf Functions to. q(q-l) .12yO + .. formula (1) i~ exact In tile general case.5 cm? Sec.50A) . then the error of formula (1) is R ()n X - "n. for n ~ 2.n's fOrlllula we obtain: for n= 1. the tabular values Yi (l =0.. s is the sag.. the {ollowing approximate formula is more convenient: .. qupdratlc Interpolation. band d are the basis and altitude of the cross-section of the rod. accorciingly..!lyo. 2. L\2yo = L\y.. it IS advisable first to set up a table of finite differences.1yo+ q (q2~ I) . n). i=O. Xl' ••• . Newton's interpolation formula.I. which includes the points x o.. the ro1ynomial (1) takes on. Xl' ••• . n) and x. YtI art' the correspon din~ values of the function y Then the valup of the function y for an intermediate value of the argument x is approximately given by Newton's interpolation formula Y=Yo+q. . . . 10 Approximate Calculations 872 accuracy do we have to measure the radii and the generatrix and to how many decimal places do we have to take the number n? 3127. \\ hen x=Xj (t=O.1 ~u. and P the load. + q (q-I). .[Ch. the difference of which h=~xj (L\Xj=Xj+l-xi. •. b). As particular cases of l'\ewtc. provided that the load P is known to 0. ·n~q-n + I) . x n and x. To simplify the use of Newton's fonnula. . 1.k. 1.Yo. (1) are successIve fini te dU'erences of the furction y. then L\nYi = const and ~n+IYi=-O and. .=0 (2) whE're ~ is some intermediate value between xi (I =0. n .. Xn be the tabular values of an argument.. (q-i+l) If ~'Yo_ . For practlcal use. To wha t degree of accuracy do we have to measure the length land the sag s so that the error E should not exceed 5.. n).1) is constan t (table Inter val) and YOt Yl' ... I ~ 50 em. S ~ 2. Y... If Y=f (x) is a polynomial of degree n.1nyo• x-X where q = -h-o and ~Yo = YI. hence. 1. . USln!! the table I 2 2 2 4 2 6 I II=- .0. the difterences 6. ._ 21 I!J. 11=60'.695 approximate the root of the equation Sinh x = 5. I 1562 I --14 1548 1 Here.'l yo should be constant to wi thin the given places of decimals Example 1. we have sin 26°15' =0. (-0.Interpolation 0/ Functions Sec. for q we take the common value (to the given accuracyl) of two successive approximations qC1n) '='lcm+l).009 1 229 6..01562+ 4 ) ' . Find sIn 26"15' using the tabular data sin 260 =0. sin 28° -:. we will have: IR21~ f(f.45399.220 . it is alsc pOSSible.00014)=0.46947. o1 I 26° 26°1fi'-26° AI/. 1.457 5.nl I ~ 1. then it is best (0 choose it so that the difference ~n+ lyO::::: 0 w.43837.thin the "mits of the given accuracy. Appl ying formula (1) and uSing the first horizontal line of the table.43837+ f . Using Newton's formula.44229. -lh. to find the correspolldlng value of the argument x (inverse interpolation). 0 42837 0 45399 0 46947 27(1 28° 2 I I Y. q= 60' 4 . To do this.466 1. ). Solution. 2.333~4·10.h. all the decimals of sin 20°15' are corr~ct. sin 27° ~ 0. Here. then I y. first deterrnlne the corresponding value q by the method of successIve approx.I )({-2)(3t)3 =12d·5rl.nlatlon.Yo (I = 0. from a ~iven intermediate value of the function y. 2) 373 If the number n may be any number.('!--l ~. Let us evaluate the error R2 USing formula (2) and taking into account that if y = Sill x. We set up the table X.lnl1 J 4. 0. Whence x=xo+q. in other words.!. 7 1 I _I 180 31 Thus. putting q'O) =Y-Yo tiyo and (I) ( ( I ) 1) A2 q(l+l)==q(O)_L:L-=-. . 0. Exanlple 2. A'II. log 2 = 0. (0) + q(O) (l_q(O» • A2 yO =0 538 + 0.462 • 0.565·0.009 =1. log 4 = 0.. 3129. I..2 +0. which for x=xi takes on given values Yi (i =0.009 = 0..IOx l + 2x· -t.3x for integral values of x lying in the range I ~ x ~ 10. and log 4. 2°.602. 3.538.000.113=2.435. Taking Yo = 4.565. Lagrange's interpolation formula.l ) + + + Yn· 3128. (Xk-Xn) Yk ••• (X-X o) (X-Xl)" . log 3 = 0.1) .(X-Xk_l) (X-Xk+l)" . 3130*.2+0.6. log 5 = 0.699.538+0. Utilizing the constancy of fourth-order differences. . Given the table log 1 = 0.477.i. n).. (X-Xl~ (X-X z\ .457 0.(~-Xn) )YI (X O-x1 (XO-x 1 ••• Xo-X n X1-X O XI -X 2 ••• X1 -Xn (x-x o) (X-Xl)" . log3.027=0. II. .565. 5.. 1.301. (X-X n ) ••• (Xk-Xo) (Xk-Xl)" •(Xk.(X-Xn .0. we have 5-4.565.538 +0 .543 -0 538.-l for the values x= 1. \X-Xn ) ) Yo «X-Xo~ ~X-X2)) · . In the general case.. Use 1inear interpolation to compute the numbers: log 1. 7.. Make sure that all the finite differences of order 3 are equal. (Xn-X o) (Xn-X J ) ••• (Xn-X n . Set up a 3131. 1.Xk+l)' .220 = (0)_ q .2 =2. Given a ta hie of the values of x and y: x y I I 2 3 10 I I 3 I 15 4 5 6 12 9 5 I tabl~ of the finite differences of the function y..313. a polynomial of degree n.027 = 0. .457. ~:~:=0.5. is given by the Lagrange interpolation formula JJ = + + . (I) _ q -q - ~Yo' 2 2 qIZ)=0. set up a ta ble of differences of the function y = x 4 .Xk-I)(Xk. Set up a table of differences of the function y=x s _ -5x 2 +x.009-· . log 2.538+ 0.7. We can thus take x= 2.0.'874 Approximate Calculations (Ch 10 Solution. 9. . 1736. Given the table sin 13°=0. . Fill in the table by computing (with Newton's formula.2250. 3133. Form Newton's interpolation polynomial for a function' represented by the table x y I I 2 3 I I 6 I 8 10 I 11 I 27 I 50 83 4 Find y for x=5. sin 12° = 0.2588.2079.2419. 3136. 2] 375 Interpolation of Functions 3132. Form Newton's interpolation polynomial for a function represented by the table I 40 85 3134*.1908. sin 14°=0.5. sin 10°=0.Sec. sin 11°=0.. Experiment has yielded the contraction of a spring (x 111IU) as a function of the load (P kg) carried by the spring: I I 10 P I 49 I 105 x 5 15 20 25 30 35 40 172 253 352 473 619 793 Find the load that yields a contraction of the spring by 14 rom. sin 15° = 0. For what x will y=20? 3135. A function is given by the table x -2 I Y 25 -8 I I 2 4 1-151 -23 Form Lagrange's interpolation polynomial and find the value of y for x=O. for n = 2) the values of the sine every half degree. Given a table of the quantities x and y x 0 131415 -3 I 25 1 129 y I 381 \A>mpute the values of y for x = 0. b) by Lagrange's formula. we should continue to calculate the approximations CIt • • • • until the decitnals retained in the araswer cease to change (in accord with the specified de~ree of accuracy!). where the function f (x) is continuous on [a. Sec. that is. of the rql1ation f (>. The succeeding approximations are constructed in the same manner. In approximating the root. that is. establishing the intervals (as small as possible) within which lies one and only one root of equation (1).\) is defined and continuous on an interval [a. It is sometimes conveni(:'nt to replace the given equation with an equivalent equation q> (x):. Generally sreaking. it is advisable to use millimetre paper and construct a graph of the function y = I (x). then to evaluate the absolute error of the approximate root c2 . The abscissas of the r-oints of intersection of the graph with the x-axis are the roo~s of the equation f (x) =0. . then on [a.Approximate Calculations 376 [Ch. The rule of proportionate parts (chord method).-:: 1f (~).•) converges to the root ~. then by repla( ing the curve y = f (x) by a chord passing through the points [a.) = O. 2) computing the roots to a given degree of accuracy If a function I (. This root will definitely be the only one if I' (x) > 0 or f' (x) < 0 when a <x < b. f (a)l and [b. The sequence of numbers en (n = I. we obtain the first approximation of the root f (a) C1 = Q - I (b)-I (a) (b-a)..' (b) < 0. b] and f(a). . If the function f (.. b) thrre is at least one root ~ of equaticn (1).oints of intersection of the graohs l/ = {fJ (x) and y == ¢ (x). Computing the Real Roots of Equations 10 • Establishing initial approximations of roots. 3. If on an interval (a. Then the roots of the equation are found as the abscis~as of f. 2°. we apply formula (2) to that one of the intervals (a. (2) To obtain a second approximation c2 . b].5 and for x = 2: a) by means of linear interpolation. b] there i~ a unique root . I (b)). 10 3137. b). 2.\) has a nonzero continuous derivative f' (x) on the interval [a. take one or two reserve decimals Thi~ is a general remark. The approximation of the roots of a given equation (1 ) f (x) =0 consists of two stages: 1) separating the roots. for intermediate calculations. cll or [Cl' b) at the ends of which th(:' function f (x) has values of oppcsHe sign. .. x. f (a) (a) > 0. where the number A¢:. For practical purposes it is 1110re convenient to use the simpler formulas xo=a. 3] Computing the Real Roots of Equations 377 en' we can make use of the formula P:-c 1~lf(cn)1 where J. X n ~ b (n = 1. . (4) where' <P' (x) I ~ . (3') where a = t ~a)' which yield the same accuracy as formulas (3). Let the given equation be reduced to the form x==q>(x). If f' (x) :1= 0 and f" (x) :A 0 for a ~x ~ b.•• . 0 is chosen so that the function d!! [X-AI (x» = I-Af' (x) . ••• according to the following law: X1=q> (x o)' x 2 =q> (Xl)' .. we replace the latter with an equivalent equation x=x-Af (x)..=xll _ t -af(x n .. 1. Proceeding from the initial value xo' which belongs . b). (3) and (3') we should put xo=b.. Iterative Inethod. the sequence tonic and lim xn=~..= lim x n n ~ IX) is the only root of equation (4) on the interval [a. 2.t . The evaluation of the absolute error of the nth approximation to xn is given by the formula Therefore. xn=x . If f (b) /" (b) > 0. bJ.t = ~ min a~.. ) to the root ~ of an equation f (x) =0 are computed froln the formulas f (x( n . xu=q> (X"_l)' (5) If a <.1 = min I f' a~x~b (x) I. 1 2..t<. xn are succes- sive ap proximat ions to the root .to the interval [a. X n (n = 1. ). we butld a sequence of numbers Xl' X 2 . then the 1itnit .t ) ~n=l. where I (a) f (b) < 0. 2. then the limiting absolute error for xn will be . 2.t' n I· 3°. ). ) is nl0no- n-+e:tJ To eval uate the errors we can use the formula where f.-!' r' n t Xn - l Under the given assumptions. that is.. 4°.. if xn and x. < 1 (r is constant) for a ~ x ~ b..81 • -r In order to transform equation f (x) =0 to (4). . 2. ) (3) xo=a.+t coincIde to within e. Newton's method (method of tangents).b f I' (x) J.. .t) ) ( rz..Sec. then the successive approximations X n (n = 0.=. then in fornlula'.. . .. 5.5 as one of the suitable val ues of A. Example 2.001. We write the equivax lent equation X=X-A (2x-ln x-4) and take 0. this number is close to the root of the equation .449.458:::::: 2.5. is. 2.5]. Reduce the equation 2x-ln x-4 =0 to the form (4) for the initial approximation to the root Xo = 2.448.448..!-) I _ =0.Approximate Calculations . we can take '~2. of the preceeding equation that lies bet\veen 2 and 3. and all the decimals of this apprOXimate number will be correct in the narrow sense.450. 2 I x4 =2 +2 In 2. f(x}=2x-lnx-4.378 [Ch.450:::::: 2.5 (2x-lnx-4) I or I x=2+2"lnx.5 . Computing the root by thJ.!-6:::::: 0.458.21.II-A ( 2 _. close to 1. xt I =2+2" In 2..4. f'(X)=2-1. by virtue of the remark made above. we can put I-At' (x o) =0]. x x-2. the roof. 1 lp(X)=2+{lnx and lp'(X)=2 x. 10 should be small in absolute value in the neighbourhood of the point Xo (for example.5::::::2. Here.. . I xa=2+ In 2. The initial equation is reduced to the form x=x-0. Considering that all approximations to xn lie in the interval [2. And so ~ ~ 2 45 (we can stop here since the third decimal place has 'become fixed) Let us nQw \evaluate the error. Hence. Here. we get f=max I lp' (x) '=2. We make use of the result of Example 1. the limiting absolute error in the approximation to x.6.45.0012::::: 0.~. Example 1. to two decimal places. iterative method. Compute. the exact root ~ 0. that is. Solution.. ~= 1~~~1 Thus. I x2 =2 +"2 In 2. putting Xo = 2. We carry out the calculations using formulas (5) with one reserve decimal.448 ~ 2.4=0. of the equation lies within the limits 2 447 < ~ < 2. We olnit the evaluation of the error. graphically. 4481-ln 2. )-1 =0. 5°.3-ln3-4)=24592. On the interval 2 ~ x ~ 3 we have: f (3) 1" (3) > O.4477 -4) =2 4475. 4481-0. Yo) = O.=2. l)) of this system x = X o' Y = Yo' This initial approximation may be obtained. + { + + q> (x o.6(2. since _the third decimal place does not chan~e any more.f~ (Xl' + ~ Icp. for example. y) =0 and q> (XI Y) ::::::0 and by determining the coordinates of the points of intersection of these curves.45.Computing the Real Roots of Equations Sec. Yo) +o. (Xl' YI) =0. the conditions l' (x) > 0 and ()f We take Q= 1 ( 2. f' (X)=2-~..4481-4) = 2. Let us suppose that the functional determinant 1= a(I. f(x)=2x-lnx-4.6. . linear equa lions I (x o' Yo) aof~ (xo• Yo) ~of~ (x o. y) =0.oq>: (x o.6 (2 ·2. We carry out the calculations using formulas (3') with two reserve decimals: x 1 =3-0. The second approximation is obtained in the very same way: X 2 =X 1 where a p ~I +U l . q» a (x. f (2) f (3) < 0.6 (2·2 4592-ln 2 4592 -4) = 2 4481.In 2. and let there be an initial approxinlation to one of the solutions (~. Yo) + ~oq>~ (x o. Similarly \ve obtain the third and succeeding approximations. 8) 379' Calculating the root by Newton's method. x 4 =2. 4477. are fulfilled. a) Newton's method. Hence. = 2. y) does not vanish near the initial approximation x=xo' y=Yo' Then by Newton's method the first approxitnate solution to the system (6) has the form XI = X o + uo' Yl = Yo ~o' where ao' ~o are the solu tion of the syst~ln of two. XI =2. Y2=Yl + Pl' are the solution of the systeln of linear equations f (Xl' { YI) +alf~ (Xl' YI) q> (x I' !II) + UI cp~ (x I' Y I) + P. The answer is: the root.4477 -0.3 3° for Xo =3 r (xl> 0. At this stage we stop the calculations. . by plotting (in the sallle Cartesian Loordinate systenl) the curves f (x. Yo) = O.4592-0. y) =0. Here. x. Let it be required to calculate the real roots of a system of two equations in two unknowns (to a glven degree of accuracy): f (x. YI) = O.6 (2· 2 4477 . (6) { q> (x. The case of a system of two equations. Yo) + ~q>~ (. p. Yn ) (n === I. { vt (x. y)l~r<l. In other words. y) -+. . by transforming this 'iystem to an equivalent one x= F (x. which is equivalent to (6) provided that y) = 0. lJo) Condition (8) will h(' ohserved in such a choice of parameters a. 110)' xz=F (XI' YI)' Yz=<1) (Xl' YI)' X a = F (x z. ~ as approxImate solutions o the systeln of equations 1 1 1 + at: (xo' at~ (x o' vf~ (x o' +. Y) + ~q> (x. ~ such that the partial derivatives of the functions F (~. to the solution .to' Yo) + yt~1 (xo. 110) = O. then lhn xn ==~. n -+ o7J n lim y" = 1'). !J) do not vary very rapidly in the neighbourhood of the initial approximation (x o. y == Y + vI (x. y) ==: F (x. y). fJ on the ilSSUlllption that the partial derivatives of the functions t (x. y) + 6q> (x.55. = 0. I~: ~ I :. Yo) + ~q>~1 (x o. \Ve consider the system of equations at (x. ~. lJ) will Ul equal or close to zero In the Initial arprOXlIllatlun. We can also apply the Iterative method to solving the system o( equations (6). y). Reduce to the form (7) the sYliitem of equations x2+y2_1=O. -+ \XJ The following technique IS advIsed for transforming the system of equal tions (6) to (7) with condition (8) observed.. y) -:nd <D (x. what IS the sanle thing. y) and q> (. !lo>' Example 3.y)I+I<D~(x. y. 2. ). y. { x3 -li=0 . we find u.given the initial approximation to the root xo=0. { y=cD (x.of (6). YI = cD (x o. IF~(x. Yo).. Un) belong to U. Yo =0. WhlCh converges to the solution 0' the system (7) or.8.Approximate Calculat tons 380 (Ch 10 b) Iterative method. . ~. y). is constructed according to the following law: XI = F (x o. <}) (x. y. y) (7) and assuming that IF~(x. Y)f+I<D:J(x.:. The sequence of approximations (x n . y) + ~(P (x. y)l~r< 1 (8) in some two-dimensional neighbourhood U of the initial approximation (xo' Yo). Yo) = 0. Y2)' Ya-= cD (x 2' Y2)' If all (x n.(~tp (x. x= X Choose the parameters a.l: o.. which neighbourhood also contains the exact solution (s. 11) of the system.\. y) -= 0. Yo) + ~CPII (x o' Yo) = 0. y).~cp~ (xo. Rewrite it in the form + at (x. q> (x. ~-O. l.3(x2+y2-1)-0. 3146. Then the system of equations x=x-O. x·lnx-·14=0.1. ~ ::::::-0. 3] I (x. "" (x o.1.6a+l.6y -t.5. 3153. 3143. I~ (x o. y and l> choose the solution of the 1+1. V:::::::-0. .92.92~=0. (xl_y). l+l. + ~ (x l - y).logx=-. 3147. 92~ = 0. 3138.1a-~=0. Solution. Yo) = 1. x4-~-05x-I. (I a. y) =xl q>~ (xo' Yo) = 1.. Write down the system (that is equivalent to the initial one) { a (x 2 + y2_1) -t. xl -5x--t 0. use Ne\vton's rnethod to compute the real roots of the equations to two decimal places: 3141.ly-~==O.Computing the Real Roots of Equations Sec. x l -2x-5=O. and by means of the rule of proportional parts compute thenl to two decinlal places. Proceeding fronl the graphically found initial approximations. 1 3144. x'-3x+ 1 =0. Find graphically the initial approximations and compute the real roots of the equations and systerns to two decinlals: 3148. x 4 +Xl .2 x ==4x.x -I 1===0. 3140.2x-Inx-4=0.1 =0. x Utilizing the graphically found initial approximations. has the fornl (7). q>~ (x o' Yo) = . Xl + 3x-0. 3142.4. 3150.5 (x 2 + y2_1) + 0 4 (xl_y).3. 3152. 6 ~ 0. { y == y -0. 1. Yo) condition (8) will be fulfilled. Here. e. 3139. x 3 .3(xl-y).55=0. Yo) = 1. x'-x-2=O. we put a which is equivalent to the initial system.2x-2 = O. y) = X2 + y!-l. ~ 1#= 0) y. x l -4x --I == O. x l -2x 2 + 3x-·5 = O.5 = O. .~ (x 3 _y) =0.3. 3149. 4x == cos x. system of equations { ~.6. 1. 4x-7sin x = o. use the iterative method to C01l1pute the real roots of the equations to two decirna I places: 3145. 3151. and in a sufficiently sInall neIghbourhood of the pOInt (x o. Isolate the real roots of the equations by trial and error. l> V(x2+y2-1)+~(xl_y)=O in the form x=x+a(xl+yl-l)+~ y = y +V (x 2 + y2_1) For suitable numerical values of a. 381 y.1. 382 [Ch. 10 Approximate Calculations 3154. XX +2x-6 = O. 3155. eX + e- 1x -4 = O. 3156. 2 3157. { x +y-4=O, y-Iogx-l =0. {xxl_y=O. + -1=O, l y2 3158. Compute to three decimals the smallest positive root of the equation tan x = x. 3159. Compute the roots of the equation x· tanh x = 1 to four decimal places. Sec. 4. Numerical Integration of Functions 1°. Trapezoidal formula. For the approximate evaluation of the integral b ) f (x) dx a [I (x) is a function continuous on [a, b]] we divide the interval of integratiol1 [a, b] into n equal parts and choose the interval of calculations h = b-a . 11 Let xi=xo+ih (xo=a, xn=b, i=O, 1, 2, ... , n) be the abscissas of the parti tion points, and let Yi = f (xi) be the corresponding values of the integrand Y = f (x). Then the trapezoidal formula yields b Sf(x)dx=::h (Yo~Yn+YI+Y.+"'+Yn_l) (1) a with an absolute error of h2 Rn ~ 12 (b-a).M 1 , where M2=maxlr(x)1 when a<ax~b. To attain' the specified accuracy e when evaluating the integral, the interval h is found from the inequality 12e h 2 ~(b-a) M 2 2 t ) • That is, h must be of the order of Ye. The value of h obtained is rounded off to the smaller value so that b-a -h-=n should be an integer; this is what gives us the number of partitions n. Having established hand n from (1), we compute the integral by taking the values of the integrand with one or two reserve deciJnal places. 2°. Simpson's formula (parabolic formula). If n is an even number, then in the notation of 1° Simpson's formula b Sf(x)dx=::~ ((Yo+Yn)+4(YI +Y.+ ... +Yn-I)+ G +2(Y2+Y,+ •.• +Yn_2») (3) Sec. 4) Numerical Integration 01 Functions 383 holds with an absolute error of h4 Rn ~ 180 (b-a) M" (4) where M 4 =max I t lV (x) I when a~x<b. To ensure the specified accuracy e when evaluating the integral, the interval of calculations h is determined from the inequality hC 180 (b -a) M 4 <; 8. That is, the interval h is of the order (5) ve. The number h is rounded off to the smaller value so that n = b-a is an even integer. n Remark. Since, generally speaking, it is difficult to determine the interval h and the nunlber n associated with it from the inequalities (2) and (5), in practical work h is determined in the form of a rough estimate. Then, after the result is ohtained, the number n is doubled; that is, h is halved. If the new result coincides with the earlier one to the number of decimal places that we retain, then the calculations are stopped, otherwise the procedure is repeated, etc. For an approximate calculation of the absolute error R of Simpson's quadrature formula (3), use can also be made of the Runge princIple, according to which where 11 and ~ are the result~ of calculations from formula (3) with interval h and II = 2h, respectivel y. 3160. Under the action of a variable force F directed along the x-axis, a material point is made to move along the x-axis froln x = 0 to x = 4. Approximate the work A of a force F if a table is given of the values of its tTIodulus F: x 1 0.0 I 0.5 11.0 11.5 1 2 .0 1 2.5 1 3.0 F 11.50 10. 75 1 0.50 I 0. 75 11.50 1 2 .75 1 4.50 I 3.5 4.0 I 6.75 10.00 Carry out the calculations by the trapezoidal formula and by the Simpson formula. I 3161. Approximate ~ (3x l -4x)dx by the trapezoidal formula o putting n = 10. Evaluate this integral exactly and find the absolute and relative errors of the result. Establish the upper limit L\ of absolute error in calculating [or n = 10, utilizing the error formula given in the text. 384 Approximate Calculations [Ch. 10 1 3162. Using the Simpson formula, calculate 5:~Xl to fOUf o decimal places, taking n = 10. Establish the upper limit d of absolute error, using the error formula given in the text. Calculate the following definite integrals to two decimals: 1 3163. 2 5l+x· dx 3168. 0 0 n I 3164. 51+x dx 2 3169. • 5co;x dx. 2 I 51+x,· dx 3170. I 0 n I 3166. ~ xlogxdx. 1 5l+xx dx. 2 3171. cos 0 I 3167. 5SI: x dx. 0 0 3165. 5SI~X dx. 5IO~x dy. 1 3172. ~ e- X2 dx. I 0 3173. Evaluate to two decimal places the improper integral 5 ~XX2 by applying the substitution x = +. Verifl the calculations U) 1 by applying Simps~~'s is chosen liO that formula to the integral 5 l~XI<{. 10- 2 • 51 ~\2' where b I b 3174. A plane figure bounded by a half-wave of the sine curve y = sin x and the x-axis is in rotation about the x-axis. Using the Sitnpson formula, calculate the volume ot the sol id of rotation to two decilnal places. 3175*. Using Simpson's formula, calculate to two decimal . x2 y2 places the length ot an arc of the ellIpse T+ (0.6222)2= 1 situated in the first quadrant. Sec. 5. Numerical Integration of Ordinary Differential Equations to. A method of su(cessive approximation (Picard's nlcthod). Let there be given a first-order differential equation y'=/(x, y) subject to the initial condition y=Yo when x=x o. (}) Sec. 5] Numerical Integration of Ordinary DiOerential Equations 385 The solution y (x) of (1), which satisfies the given initial condition, can, generally speaking, be represented in the form y (x) = lim Yi (x) 1-+ (2) r:D where the suCCeStHVe approximations Yi (x) are determined from the fonn ulas Yo (x) = Yo, x YI (x) =Yo+ ~f (x, YI_I (x)) dx Xo (i = If the right side f (x, 0, 1, 2, ... ). y) is defined and continuous in the neighbo~lrhood R{lx-xoJ~a, IY-Yo'~b} and sa t isfies, in this neighbourhood, the Lipschitz cond itlon (L. is constant), then the process of successive approximation (2) definitely converges in the interval Ix-xol~h, where h = m~n ( a, ~) and M = m;x I f (x, y) I. And the error here is _ n Rn-,y(x)-Yn(x)I~/v'L I X-X o In+l (n+I)1 ' If I x-xo ' ~ h. The method of successive approximation (Picard's method) is also applicable, with slight nl0difications, to normal systems of differential equations. Differential equations of higher orders rnay be written in the fonn of systems of differential equations. 2°. The Runge-Kutta method. Let it be required, on a given Interval X o ~ x ~ X, to find the solution Y (x) of (1) to a specified degree of accuracy 8. To do this, we choose the interval of calculations h=X-xo by dividing the interval [x o, Xl into n equal parts so that h4 xi are determined from the formula xl=Xo+ih (t < 8. n The partition points =0, 1, 2, ... , n). By the Runge-Kutta method, the corresponding values Yi=Y (x;) of the desired function are successively computed fron] the formulas Yi+1 =Yi+ ~Yi. ~Y'-.!. +2k(i)+2k(t)+k(i» '-6 (k(i) 1 I • .., 13-1900 38.::.6 .-,.;A:..:.!:..p!:....pr:....:o:.....x--.im_at_e_C_a_lc_u_la_t_io_n_s [ C_"'h_._10 where (3) To check the correct choice of the interval h it is advisable to verify the quantity e_ - k(i)-k(t) I 2 a k~t)-k~t) I • e The fracti'Jn should amount to a few hundredths, otherwise h has to be reduced. The Runge-Kutta method is accurate to the order of hI. A rough estiIn ate of the error of the Runge-Kutta method on the given interval [x o, Xl Illay be obtained by proceeding from the Runge principle: R = I Y2rn--Ym 15 1 ' Ym where n = 2rn, Y2tn and are the results of calculations using the schenle (3) \vith interval h and interval 2h. The Runge-Kutta method is also applicable for solving SYStClllS of dlffe~ rential equations y'=f(x, y, z), z'-=cp(x, y, z) (4) with given initial conditions Y==Yo, z=zo when x==x o. 3°. Milne's method. To solve (1) by the Mtlne nzethod, subject to the initial condi\ions Y=Yo when x=xo, we in SOlne way find the successIve values YI = Y (XI)' Y2 = Y (x 2 ), Y, == Y (x a) of the desired function y (x) [for instance, one can expand the solution Y (X) in a series (Ch. IX, Sec. 17) or find these values by the lTlethod of successi ve approximation, or by using the Runge-Kutta Inethod, and so forth]. The approximations and for the following values of Yi (i =4, 5, ... , n) are successively found frool the formulas Yi Yi Yl=Yi-.+ ~h (2/; __ -/; __ +21;_.), } h _ &:: Yi=Yi-2+ "'here fi= f (xi, Yi) and 3" (Ii +4Ii-l + Ii-I)' fl= f (xi, Yl). 1 (5) To check we calculate the quantity 1- =:7 1 ei=29 YI-Yi • (6) Sec ..5] Numerical Integration of Ordinary Differential Equations 387 If 8i does not exceed the unit of the last decimal 10- m retained in the answer for Y (x). then for Yi we take and calculate the next value Yi+t, repeating the pr8cess. But If 8i > 10-tn, then one has to start from the beg1nning and reduce the interval of calculations. The Illagnitude of the initial interval is detennined approXinlately fronl the inequality hoi < lo-m. For the case of a solution of the system (4), the Milne formulas are written separately for the functions y (x) and z (x). The order of calculations remains the same. Example 1. Given a differential equation y' =y-x \vith the initial condition y (0)= 1.5. Calculate to two decinlal places the value of the solution of this equation when the ar~ument is x -= 1.5. Carry out the calculations by a combined Runge-Klltta and Milne method. Solution. We choose the initial Interval It frOll) the condl tion hoi < 0.01. To avoid involved writing, let us take h-=0.25. Then the entire interval of integration frool x=O to x= 1.5 is divided into six equal parts of length 0.25 by means of points xi (i == 0, 1, 2, 3, 4, 5, 6); we denote by Yi and the corresponding values of the solution y and the derivative y'. We calculate the first three values of y (not counting the initial one) by the Runge-Kutta ruethod [frotH fonnulas (3)]; the remaining three values - Y4' Ys, Y6 - we calculate by the Milne I11ethod [from formulas (5)] The value of Ye will obviously be the answer to the problem. We carry out the calculations with two reserve decilnals according to a definite schenle consisting of h\'o sequential Tables I and 2. At the end of Table 2 we obtain the answer. Calculating the valueYl' Here, f(x, y)=-X+IJ, xo==-O, Uo=1.5 !Ii y; ~ (k~O) + 2k~O) -I- 2k~O) + k~O» = h = 0.25. !J.Yo = = k~O) == f (x o, Yo) h == k~O)=f ( xo+~. h +2 ' k~O)=f ( x o k~O) == f (xo-t-/z, ~ (0.3750+2·0.3906+2·0.3926+0.4106) =0.3920; (- 0 -1- 1.5000)0.25 -= 0.3730; k(O) Yo+-+) h=(-0.125+ 1.5000+-0.1875) 0.25= 0.3906; k(O) ) Yo++ h=(-O 125+1.5000+0.1953)0.25=0.3926; Yo+ k~O» h = (- 0.25 + 1.5000 + 0.3926) 0.25 = 0.4106; YI =Yo+ ~Yo== 1.5000+0.3920== 1.8920 (the first three decilnals In thts approximate number are guaranteed). Let us check: 8= I k~O) _k~O) I 10.3906-0.3926 1 20 k~O)_k~o) = 1 0 .3750-0.39061 "'='156=0.13. By this criterion, the interval h that we chose was rather rough. Sim ilarly we calculate the values Y2 and Y•. The results are tabulated in Table 1. 13* 388 [Ch. 10 ApproxImate Calculations Table 1. Calculating .vI' YJ' .vI by the Runge-Kutta Method. h=0.25 f(x, y)=-x+y; I t (Xi+: ' , Value of i 1 2 3 Value of i == f (xi, 1.5000 1.8920 2.3243 2.8084 0 0.25 0.50 o 75 0 Yi~ YI xl t( Xi+i ' k(i) k~i) ) a k(i) kIll) J Yi) 1.5000 1 .6420 1 .8243 2.0584 + f (xi h, Yt-j- k~i» , k(i) Y'+T 0.3750 0.4105 0.4561 0.5146 1.5625 1.7223 1.9273 2.1907 0.3906 0.4306 0.4818 0.5477 k(i) 4 ~Yi Yl+1 0.4106 0.4562 0.5148 0.5900 0.3920 0.4323 0.4841 0.5506 Yi+T 0 1 .5703 1.7323 1 .9402 2.2073 ] 2 3 1.6426 1.8251 2.0593 2.3602 0.3926 0.4331 0.4850 0.5518 1 .89-20 2.3243 2.8084 3.3590 Calculating the value of .P4. We have: f (x, y) = - x+ y, h=O.25, x4 = 1; Yo= 1.5000, YI = 1.8920, Y2 = 2.3243, Y. = 2.8084; I y~ = 1.5000, Y; = 1.6420, y~ = 1.8243, y; = 2.0584. Applying formulas (5), we find Y4=Yo " + 4h 3 (2Y -Ya + 2y.) = 1 = 1.5000 + 4.0 25 (2·1.6420-1.8243 if, = f (x it) = 4, == 3 1 h -/ " Y'=Y2 +"3 (Y4 +4y. s, +2.2.0584) = 3.3588; + 3.3588 = 2.3588; + Y = 2.3243+ 0.25 3 (2.3588+4.2.0584 + 1.8243)=3.3590; I ) I~Y: 1_,3.3588-3.3590 , _ 0.0002 7 10-' < 1 0 001. 29 29 - 29:::::' 1"'.' hence, there is no need to reconsider the Interval of calculations. Sec. 5] Numerical Integration of Ordinary DiOerential Equations 389 We obtain Y4 = ~ = 3.3590 (in this approximate nu'mber the first three decimals are guaranteed). Similarly we calculate the values of y, and Yo' The results are given in Table 2. Thus, we finally have y (1.5) = 4.74. 4°. Adams' method. To solve (1) by the Adams method on the basis of the initial data y (x o) = Yo we in some way find the following three values of the desired function y (x): (these three values may b~ obtained. for instance. by expanding y (x) in a power senes (Ch IX, Sec. 16). or they may be found by the method of suc· cessive approximation (1°). or by applying the Runge-Kutta nlcthod (2°) and so forth]. With the help of the nurnbers xO' x.. x 2 • Xa and Yo. Yl' Y2. Ya we calculate qo' qt. q2' qat where qo = IzY~ -= hf (x o• Yo). q2 = h Y~ === hf (XI' Y2)' qt = hy; = hf (XI' YI)' qa = hy::= hI (x a• Yale We thrn form a diagonal table of the finite differences of q: li.y= x Xo XI I ~!lo IY\ I ~YI x2 1 Y2! XI x4 ~Y2 II/I I ~Ya Y41 ~Y4 1 XII Y. y) q=y'h aq=Qn+l-q", li. 2 q= =li.qU+I-li.q,l li.1q=- =li. 2q,.+ .-li.'Jq" I 110 1/ =f (x, ;; l/n+ I -lIn " I XII Y.I ~y, I qo f (XI' I q, f I q" f a, I qa t (X Y4) I q4 f (XI' UI) I q. f (X Ot Yo) ~qo t:tlqo ~3ql) YI) ~ql /).2 QI ,\'q. (x 2 • Y2) ~q2 ~2q2 ~3q2 Ya) ~qa &2qa (X 4t I tiq4 [Ch. 10) Approximate Calcu lat ions 390 The Adam5 method consists in continuing the diagonal table of differences with the aid of the Adanzs formula A_I A u y ,,--qn+2 uqn_l + 125 A2 u 3 qn-2+"8 AI u qn-I' (7) Thus, utilizing the numbers qa, 6,q2' t1 2qh &a qo situated diagonally in Ihe difference table, we calculate, by means of formula (7) and putting n::.:.- 3 in It, L\YI=ql+-} L\ql+~ L\lql + ~ L\'qo. After finding L\YI. we calculate Y4= Ya + t1Ya· And when we know X4 and U,a, we calculate q4= hf (x 4 , Y4)' introduce Y4' &Ya and q4 into the difference table and then fill into it the finite differences 6,qa' 6,"q2' !J.3 q1J which are situated (together with q4) along a new diagonal parallel to the fIrst one. Then, utilizing the nunlbers of the new diagonal, we use formula (8) (putting n=4 in it) to calculate ~Y4' Y5 and q, and obtain the next diagonal: qa, t1q., ~Iqa, 6,aq2' Using this diagonal we calculate the value of y, of the desired solution y (x), and so forth. The Adams formula (7) for calculating fiy proceeds from the assumption that the third finite differences 6,3 q are constant. Accordingly, the quanti tv h of the initial interval of calculations is determined frol11 the inequality h 4 < lo-m [if we wish to obtain the value of y (x) to an accuracy of lO-m]. In this sense the AdalTIS formula (7) is equivalent to the fonnulas of Milne (5) and Runge-Kutta (3). Evaluation of the error for the Adams nlethod is complicated and for practical purposes is useless, since in the general case it yields results with considerable excess. In actual practice, we follo\v the course of the third finite differences, choosing the interval h so stnall that the adjacent differences ~3qi and 6,3 qi + 1 differ by not more than one or t\\70 units of the ~iven decil11al place (not counting reserve desimals). To increase the accuracy of the result, Adams' fOrlnula may be extended by terms containin~ fourth and higher differences of q, in which case there is an increase in the number of first values of the function y that are needed when we first fill in the table. We shall not here give the Adams fonnula for higher aceuracy. Example 2. Using the combined Runge-Kutta and Adanls nlethod, calcu· late to two decinlal places (when x = 1.5) the value of the solution of the differential equation y' = y-x wi th the Initial condi tion y (0) == 1.5 (see Exanl pIe I). Solution. We use the values Yl' Y2' 1JI that we obtained in the solution of Example 1. Their calculation i~ given in Table 1. We calculate the suhsequent values Y4' Ys, y, by the Adams method (see Tables 3 and 4). The answer to the problerTI is U, = 4.74. For solving system (4), the Adams formula (7) and the calculation scheme shown in Table 3 are applied separately for both functions Y (x) and z (x). Find three successive approximations to the solutions of the differential equations and systems indicated below. 3176. y'=XI_t-yl; y(O)=O. 3177. y'=x+y+z, z'=y-z; y(O)=l, z(O)=-2. 3178. y" = - y; Y (0) = 0, y' (0) = 1. 3588 I 2./1.50 IIIIIIIIIIIIIIIIIIIIIIJIIIIIIIIIIIII 4.00 111111111111111111111111111111111111 3. (ItalIcised fIgures are Input data) + f (x.9950 j 6 11.der j~ll I Do not reconsider I y (1. Calculating J.5000 Y.32431 /111111111111111111111111/111111111111111111111111111111111111111111111111111111111111111111111111111111 11/11111111111111/11111111111111111 IIIJIIIIIIIIIII 1111111111111111111/11 111111l1li111/1 111111/111111/1111111111111/1111111111111/11111//111/1//1/11//1///11/1111/1111/111/11111 25 1. I.=! (x.x h. IJI.'4' Y5' )'6 by the Milne Method. y) == .) o I I 10. 7447 j 3.. 7450 I Do not reconsi. Yi £1.10Answer: 5 13. h = 0.9950 I 2. y.251111111111111 111/1/1/111111111111111 3.25.3590 IDo not reconsider 4 5 11.0584 11/1/1/111111111111111111[1111111111111111111111111 1111111111111111111 I 1111111111111111111111111111111/11111111111111111111111111111111111 i .> 500 ° /1.7406 j ~ 1. °1 1.2402 1 4.5°12..3590 I ~7·1O-5 13.-- U. Y. U: =/ (Xi' Y.5) =4.7402 I ~ 10.6420 1111111111l1li1111111111111111111111111111111111111111111/11111111111111//111111111111 I/JIIIIIIIIIIIIIIIII 2 /°.) i =y. 11. : Value ot I X.9947 1 2.892°1 1. following Indications of formula (6).3590 I 2.3588 I 3.74 .80841 2.75/2.Table 2.4. Reconsider interval of calcula t 10 ns.5 3.8243 111/11111111111[1111111111/11111111 i a /°. Y: =/ (X. q t-1 t21\2Qi -.0964 511 251 3.3588 0.ve by the Adams Method.0054 0. =- y) .25 (Italicised figures are input data) Table 3.0069 0. li.6420 0. qi= .9944 0. + y.8243 I 0.0293 0.!.5000 I I I I I I I I I I I I I 0.2 6.vI' f (x.4105 0.8920 111111111111111111 1. 0 Q.0089 0.6861 0. Basic Table for Calculatlni Y4' .7444 0.6861 01 "Il 0 I t .. gil ~ >- 111111111111111111 t . 3q.0047 I 0.0018 0.7450 .0166 10.5897 0.2q.0584 0.751 2. 6. 2q'_2 . =f ~ Yi= (xl.5146 I 0.7450 I 2.&.+ ~ l\'qi-I .25 f t .9 S 12 dY.=qi+ .8084 0.3750 I 0 0355 o.0376 0..!.h /i:lq.q. 3 0. Y.0213 I I I I I Table 4 Auxi liary Table for Calculating by the Adams Method t1y.Yl Yi t. . 6. 6.0456 210 501 2.3243 111111111111111111 t .I x 6. !..6356 5 0. l\qi_l+ Value of I q. h = 0.5146 0.6356 1 2.5000 I 0.0585 3/0.0101 I 0 0028 I 0..5504 2.0751 411 001 3.4561 0.5504 4 o 5897 0.0129 I 0 0037 I 0.3588 0.0011 0.0014 0.0482 0.. z' =y-z. Compute y when x=0. :It. = . 3182. . x'-'l when t=O. Conlpute y and z when x = 0.Approximating Fourier Coefficients ISec. x 3181. y(O)=l. 3186. Y-== 1 when x=O.2 when x = o. y == 1 when x == O. y=l. Compute y when x = 1. y'=z+l. y=2. Applying a combined Runge-Kutta and Milne method or Runge-Kutta and Adams method. Cornpute y when x = I. 1. 3183. ~2t~+i-coS2t=0. z' = x -t. y' = x 2 -1. 3187. 3185. 12) be the values of the function y = f (x) at equidistant poi nts x n = of the interval lU. Y3Y"-i-1=0.5. 6] 393 Putting the interval h = 0.5.5 (O~x~I). Compute y when x = 1. = -x +2y-+ z.2n). Approximating Fourier Coefficients Twelve-ordinate scheme. Us ua Us Us U.. y=2. z(O)=1 (O~x~l). use the Runge-Kutta method to calculate approximately the solutions of the given differential equations and systems for the indicated intervals: 3179. y'=y-x. Compute y and z when x == 0. .5. z=-I \vhen x=O. {y' {y' 3189.I when x=O. Ya Y7 Sums (~) Uo Us U 2 I I Uo U1 Differences (l\) u2 Ua U.3y-z. y"=2-y. 3180. Us U. calculate to two decimal places the solutions to the differential equations and systems indicated below for the indicated values of the argument. and Yo . Compute y when x = 0. 6. x=O.Yn We set up the tables: I YO Yl Y2 Ya Y4 Ys Y. y' = JL_ y 2.y. y = 2. Let Yn =f (x n ) (n = 0. VI V 2 Va V. y(0)=1. Find x (rt) and x' (n). Sec. Z = . y'=O when x=l. y = 1 when x = O.. 3188. Yn YIO Y. y' = x + y. y'= .2y + 3z.5. y (I) = 1 (I ~ x ~ 2). Sums Dillerences Sums Differences . 3184.2.5. z'=y-x. y' =2y-3. 5t 2 • 6a 2 =50 -5.23 u 138 70 20 - 6 v 4 20 . Using the 12-ordinate scheme. = 3. b.3 cos2x-8.27 . Example.866t l -l.8. 12 14 4 Solution. + Ga.9.7.8 sin 3x). 2.2n) by the .Approximate Calculations 394 [Ch 10 The Fourier coefficients an' bn (n = 0.0 a. • (1) ya 1 1 where 0.8. 3) of the function y = f (x) may be determined approximately from the formulas: 000 = 50 +S 1+ 52 + 5.=0. 60 1 = to +0. a l = 24. n=J Other schemes are also used.19 28 41 27 38 70 20 .9 sin x) + (10.5 (51-51)' 6a l = t o-t 2 • 6b l = 0. b2 =-8.=:::: 1.0.9. I. Co n sequen t I y• f (x) ~ 4. a2 = 10.4 sin 2x) + + (3.8 cos 3x +0. 3. 1+ 2).8660 2 6b 2 = 0.+0. find the Fourier polynomials for the following functions defined in the interval (O.9 cos x + 13.866 = -2. Calculations are simplified by the use of patterns.24 .20 33 u _ 18 _ 19 _ 13 I b t I v 0 't 18 18 I 276 414 28 I-21-37 33 45 28 From formulas (1) we have ao=.10 .13 . +L (an cos nx + bn sin nx). We set up the tables: y 138 38 12 32 4 14 48 . a l bl =13.9..4. Find the Fourier polynomial for the function y= f (x) (0~x~2n) represented by the table Yo UI 38 38 Y.20 20 56 51 89 7 .866 ('t 't 6b a =0'1.30 • We have a /(x) ~ a.8+ (24.50'1 + 0. o = 4.72 Ys=7.18 Ya= 6. .273 Ys=O.495 Y8=O.357 YI = 3. Yo=O Ya=9.134 Ys =0. 3190.191 YII-==0. = 300 Yol == 0 Y" = .60 YI -== 6.0.Approximating Fourier Coefficients Sec.42 Ye =5.22 YII = 3. evaluate the first several Fourier coefficients for the follo\ving functions: a) f (x) = 2n1 b) f (x) = 1 2 (Xl . . Yo =-7200 Ya==4300 Y6=7400 yg=7600 y.370 Ye =-0.540 YIO = .3nx 2 -t. 6] 395 tables of their values that correspond to the equid istant val ues of the argument.68 Yti = 8. (0 ~ x ~ 2n).714 Ya=I.5200 Ys = 3850 Yl l = 250 3191.67 3192.97 Y7 == 6.2250 Y10 = 4500 Y2 = 700 Ys = .88 Y2 == 9.042 Y4 = 0. Yo-=::2.68 Ys == 8.81 y.788 Y7 = 0.2n 2 x) n 2 (x-n)2 (0 ~ x ~ 2n).767 3193.437 Y2=2. Using the 12-ordinate scheme. x"'. 5 I 0. 20.lal+l-bl=lal+lbl... -2<x<2. ). x > 1.la-bl+lbl..e. l<.¥2 .24. Utilize the identity 28. T ='1:. 27. I a-b I ~ I a I-I b I· Besides. It should be 2+X-X2~0. -3~x~1. c if O<. y=b if periodic c<x~a. 14.. <p (x) = 2x4-5x 2 -10. 4. d) odd.b) y=O when x=-l and x=2.. r . b) periodic. Ixl~ V2.I I +l!.C. I+x!.x~IOO. x~2. 11.. Solution. 17. d) T='Jt. Thus. or x 2-x-2<. f(x)="6x -6"x+l.oo<x<.Y3and x= Y3...V3 and O<x< V3. e.. y>O when x>-I.. ~ lxI-I. y>O when . f (x) = {If (x) +f (-x)] +ilf (x)-f (-x)]. 3t..0. . 29. x-2~0. 11 4· 38. m=Qll1+q212+Q.=I. 10.11.. . I. 37. Hence. y<O when -00 <x<-l and 2<x<+00. la-bl=la+(-b)I<. '1' (x) = .x~-1. + 6x. 0. 2). a) y=O when x=-I.y>Owhen -1<x<2. m=q1x when 0 . 2<x<+oo.. S=bx-~ if C < x". a) -I<:x<+oo.-2). 6. 3. -1~x~2· or x+l~O. 12. Whence either x+l~O.13. f(x)=-3"x+"3. Since a=(a-b)+b. d) y=O when x=O. 1\. b) x=O.3. 16. ±I. -2<x~0.r ..4 . 0. a) -00 <x~. 5.. y<O when . O~x~l. y=~x. (-2.4. e) nonperiodic. b)-oo <x<+oo. y<O when x<-I.i. 1 2(x+fxl).. then lal<. 13 I 8. 21. a) -2<x<4. 9. -1<:x~2. -1~x~2.. 0. c) -1 < x < 0. 0. a) Even.a-bl~ la I-I bl and I a-b 1=1 b-a I~ I b I-I a I. (X+2)2. m= =Q1 l l+q!(x-l l ) when 11 < XE. y<O whenO<x<1 39. T= 2~. b) odd.(x-l1-l2) . T=j n.ANSWERS Chapter I 1. (2. ±2. . that is.. 22. c) even.x. k11~x~krc+2(k=0. -1<x<l... cp [¢ (x)]==2 2X . Solution. 7 .V3 < x < 0 and V3<x< + 00.-oo<x~-I. b) x < . 23. x. (-00. l+x2 . 11 -2".. a) Periodic.- . a) x=2 (y-3) (-co<y<+oo). 18. Whence . -6. +00).6. ~ 1 I . 28.VU+1 l-l~y<+oo). 0. Hin!. -24.a. c) y>O when -oo<x<-1-oo. 'I' [q> (x)]=2 X2 30. i. c) periodic. x-2~0. V2~x<+00. but this is impossible. x=. (x+l)(x-2)~0. 11 "2. d) x > O. S=2~X2 C if O.vhen 11+1 2 < X~ll +1 2+l.3x' e) odd. b) x=l"Y+l and x=.7. 1 y>Owhen-oo<x<-l and l<x<+ 00.. 15. e) y=O when x= 1. 1 n 19.x<:c. I/f l+x!.. 84. Fig. Hint.65. The function is even For x>O we determine the points at which 1) y=O. 72. See Appendix VI.. 25. 104. 81. 18. Hint. Hint. Hint. See Appendix VI.'2a and Yo={4ac-b 2)/4a. See Appendix VI. See Appendix VI. Fig. 51. 67. Fig. 10. 41. u=311 . 85. 12. Completing the square in the quadratic trinomial we will have y=yo+a(x-x o)2 where xo=-b. Hint. See . See Appendix VI. 11. Hint. cos l x =} (l +cos 2x). 8. Hint. 61·). Fig.Answers e) x= =i V -I-ys tan y ( (-oo<y<+oo). a) y=-cosx 2 . Hint. Fig. 23. See Appendix VI. Fig. See Appendix VI. Fig. Hint. Fig. 102. Fig. Hint. e) x= x=y when -co<yo. Hint. 92. v=~. Hint. Fig. Fig. See Appendix VI. Whence the desired graph is a parabola y=ax' displaced along the x-axis by Xo and along the y-axis by yo. 14. Hint. See Appendix VI. Hint. Fig. Fig. 120. See Appendix VI. Fig. 133. 80. 101. 140. u=tanv. 31. 3. 89. 32. 99. 138. The desired graph is the sum of the graphs Yl = x and Y2 = sin x. See Appendix VI. 107. u=cosx.. Hint. -00 The graph is a hyperbola y= m . 18. 10 83. Hint. A=lO. Fig. Fig. Hint. See Appendix VI. 1. 12. Fig. See Appendix VI. -~ <y<~). Hint. See Appendix VI. See Appendix VI. c) y=-i-when -co<x<O and y=x when OQ<+co. Hint. b) y=2u . and 3) y=-1. See Appen· dix VI.. See Appendix VI. b) y= =arctan ¥logx.Hlnt. See Appendix VI. Hint.: ) . 87. See Append 1X VI. Hint. Hint. 9. 73. x=¥"y when u=2x-5. Fig. 7. d) y=arc sinu. Fig. 19 78. See Appendix VI. Fig. Hi n t. Taking the integral part. Fig. 141. 118. 94. y --+ 1. 71. Hint. Hint. See Appendix VI. Fig. See Appendix VI. See Appendix VI. 17. Putting a=A cos q> and b=-A sin q>. Hint. Hint. .In our case. 121.927. 17. 132. Hint. b) y=log(10-10X ). The desired graph is the sine curve Y = 5 sin 2x with amplitude 5 and period n displaced rightwards along the x-axis by the quantity 1 { . 13. Hint. 82. Fig. Fig. <x< 1. q>=O. Hint.Hint. and y=O 43. 134. v=-x l • 42. 2. 75. 33. 9. Hint. Appendix VI. 4.. Hint. Hint. 53. 139. 61. Hint. a) y=sln1x. 58. 103. we will have y=A sin (x-q» where A= ¥al+b l and q>=aretan( . e) U=2(x!-I) if Ixl<. a) y=u 10 40. The period of the function is T = 2njn. 2) y=l. Fig.) (Cf. ¥n'. d) 397 x=2·IOY (-oo<y<+oo). 119. See Appendix VI. we have y=i - ~ I (x+ .lxl~ ¥21t. 28. Hi nt. See Appendix VI. Fig. See Appendix VI.l. . O<y<+oo. 5. 105. See Appendix VI. Hint. 15. See Appendix VI. Hint. Hint. 93. shifted along the x-axis by x o and along x the y-axis by Yo' 62. The desired graph is the product of the graphs Yl = x and y" = sin x. 30. if Ixl> I. Fig. See Appendix VI.O. Fig 11. When x~ -1-00. Hint. See Appendix VI. Fig. Fig. Hint. Fig. 13. 6. 46. See Appendix VI. Fig.90.. c) y=logu. Hint. 40. O. b) x 1 =-3. Hint. See Appendix VI. 198. 27. 192. we will have r=e'~ (see Appendix VI. 00. (N). =3. 5rc x2 = 4"' 167. O. =2. 32) cos q> F = 32 161. 00. Y2~-2. 22 150. x2 =10. 1. Fig. 187. Hint.:::3. 184. b) 2x > N when 7 170. 29. Y-2 Y'=T. O. 145. -1. Be ab . Y2=- ~2. Yl =5. 3. It is now easy to construct the desired curve froln the points. we have x=. Fig. 186. Fig. O. e ~=5 <x< ~ (8 Y2:::=29. Fig. I I -1 -2 -8 -1 -3 I 1I 4 27 9 Constructing the points (x. 144. 177. See AppendIx VI. 179.8. d) xl~-3. 27. Passing to polar coordinates r=Vx 2+y2 and tan rp=1!. we get the desired curve (see Appendix VI. n 1 whenx=T' 164. Fig. n~4. 2. Hint. + I. e) X= 1. 176. ± a.6. we wIll have r = 3~in ~c~s all' SlIl <p (see Appendix VI. y=O. we get y= ± Y25 . 72. 0 x2~-2. 178. a-I 3a l 2 • 1 197. 159. the parameter t cannot be laid off geometrically!) 142. Y2=-3. Y2=2. Passing to polar coordinates x=r cos <p. (Here. 157. 00. Fig. Yl=2. 173. 156. assigning to the ordinate!. Y4~-1. a) x1 =2' x 2 =2. O. 2"1 196. b) x=O. 2. c) 2.1. 1. Use the formula 12 + 22 + n2 =6"' n (n -1-1) (2n + 1). 1). Hint. 193. c) N= 9. f) x=O.6. xa==2.. Ya::::: 1.Ya~3. . a) 0. 1 +. O. x2~3. 185.1. -1 =N.. b) 0 002. 163. 188. YI~-3. n >~­ < a) 0 02. 180. 1 3. 1.. 19.50. 169.9. 143. Fig.:.37. Fig. . y) obtained. y) of the sought-for curve.6x(10-x). Solving the equation for x. I y 0 4 9 . :rt e) x'=T' b) n> 10.=-2.4. 00. a) x 1 =2. 153. 18t. c) \ f (x) . ± Hint. 175. 151. Solving the equation for y. and y=r sin cp. x4 =:. Fig 32) x 160. See Appendix VI. d) x=0. 1 2". -2. arbl trary values (y> 0) and calculating the abscissa x froln the formula <*) Bear in nlind that log y ~ . c) N=999. when N 3 -2' < -N when I x \ > X (N). c) x1 =1. b) 1.. d) 3 4".. 165. a) :8' 166. 162.398 Answers Fornl a table of values= o 2 3 x 0 8 27 . See Appendix VI. c) Xl ==2. n~32."2 . 28. n> a) 171. Whence we get the points (x. > 172. 21.. y) corresponding to the abscissas x==O. 168. Fig. 190. y. 183. a) log x x> X (N).86.00 as y ~ O. 191.5. Ymax=2 Y=2 Ymax=15 when x=5. 199.68. 182. 10 log y-y <*). See Appendix VI. b) N=99. x 2 =-2.x!. x.0002. 189. c) 0. 7). 174. the points (x. 1. YJ. 3x.:::::2. 195. 194. 1.x. 30 .7. I t is sufficient to construct i.. ab slnx . See Appendix VI. x 2 =5. See Appendix VI. 2 1 (n _m 2 )... In a = 2 ] 1 that lim (cos x) x = eO == 1. 214. 2 n' 1 Y3' 230. In a. . ) =-2lim x--+o 1 ]_2~in2f . 216. 205. 2. 11m ( .:. b) 1. lim (cos x) x 250. Solution. e. 3.59) Hint. 268. a) -1. 261. 1 -"4' 237. 255. S'Ince l'1m 2 x~o X 2 253. 270. 233. · 1n a • a-b.56' 204. 263. Hint.Answers 200. Put 262. 2' "9' 202.'e' 4x 4x Solution. 219.' x~o r e case e-&. 2' 236. a) -00. 269. 5 2"' 211. a) -1. 249. where a-+-O (see Exalnple 2. 1. 225.O = _ 2 lim x->o = '. 201. a) 0. 221. cos a. 223. b) 1.n = 1 "2 . O. O.!. 241.251:' = ex-+o rl(Sin ~ )2 X~2] = _ . 245.) I ( x 2 It X f) As x Iitn-=O. . 1. 222. it follows x -+0 in 4 the preceding (_ 2 2i) Si11 . 217.\ ---x2sin2 _ 2 [(Sin. a) 1. 2 4 3 · 1 1 3 203. 11. 251. b) O. where a ~ O. - 3 n' 244. 254. 267. a) 1. ..1 • 246. 220. 1. 266. ~ sin 2. 232. 10 log e. 265. b) . =e 2 1 4'.X2] =-2·1· x 2" . 264. 1. 206.2sin -i)x = ~i~ l(1 . 257. a -"3' 210. 1 2 1 e.. 1t. b) +00. a) 1. . 1 -3. 207. ~ e2 • ::= x-+o = ~i: r 1 .-i)2 -.2 sln i) 2 0( "2' 240. b) 1. 259. 242. a) • 2 . 252. 227.. 231. n. 224. ~.2' 2 n' 235. Utilize the identity a=e 260. a. 208.2" 4"' 239. 3 V x ~ 215. 247.1 • 248. O. 0 2 1 229.')8. a) 0. 226.X'-~o x b) I.sin a. X 2~fn:!. 238. I -2' 213. O. x-+o Since Iitn ( . 399 t I l Vx · 209. e. I. 3 · 243. eX. I. x-+o -~. = ~ follows that lim (cos X)i> =e Put eX -1 =a. 3. { . 234. a) -1. b) O. 1. In 2. r. lim [1-(I-cos x)) x lim 2sin l. a) -1. 256. 228. x-. b) 1.!(see a ). cos x. 12. I1m (cos x )X2 I X-.o 1. 212. b) 1. 218. b) l. Hint. where k is the proportionali ty factor (law of compound interest). the straight line y = x is the asyrnptote of the curve y = x2 + 1 . 278. x=-1 is a removable discontinuity. Qt = Qoekt • 288. 1) 0.l'~\~x\ 313.!. e 0 <e< ~93 1 N=~. \ x I> 100. 272.0005 290. 2n.00952). 1< x < n. }. 2) 3) 1. X. a) 1.. 6) 0 72 (0. x= .04139). then when lL\xl<x we have \Yx+L\x-Yxl= =1L\XI/(Yx+L\x+yx)~IL\xI/Yx. 297. 289. b) x ¢:. b) Ix-11<0. Hint. 'x- 1\< e 2 \vhen .y. (k= ± 1. b) 4. 0. . I x I> c) I x .9849).8730). 4) ~=0. ) are infinite discontinuities. y=O when x=O. d) 2. 317. 327. . 0. x= kn. x>O.. A=4. where Ii is an integer. 301. 318.985 (0. 8. a) 2.-+ 00. 5) 1. d) -}. c) 3. 315. Hint. 314. 2· 286 .01. e) {(O)=O.7480).. b) /(0)=. y=x when 279. c) 300. k =1. f) {(O)=l. 1 but if x=kn. d) {(0)=2. 3. X= 2 is a discontinuity of the second kind. 61 276. No 316.1 I < 0.043 (0. kn. 13".0309). Hint. Y=-T when x<O.Answer~ 400 271. Y=T + 00. 303. b) third. 7) 0. 319. b) a) 292. 3.I is a removable discontinuity. Hint. 326. 307.03 (1. n .0095 (0. XI O<. 296.12 (1. 322. y=O when x> I 273.05. e 2 -1 I x'+ 1 b=O..954(10.x~ y= 1 when c -'-1i. 321. then cos! x <1 and Y =0. \x-2\< 29S. c) 3. x= 2n. c) ~. where k is an integer 311.125). b) d) I x. b) x=O is a removable discontinuity. a) /(O)=n. Y=2 n when x==l. 1. a) i.. 1(0)=1.1623) 10. 450' 277.98(09804). 324... -1. y=x when O-c. 291. b) 6=0.167 (3. ± 2•.k ± ~ (k = 0. ). x=O is a discontinuity of the first kind. n. 325. > 1000.R.001. %= 1 is a point of dislontiiluity of the first kind..282. a) x i= ~ + kn. c) Ix-l \<0. -1. ± 1. Y=lxl· 274. a) 1. x=-l is a discon- . a) x==O is a discontinuity of the second kind. x=o is a discontinuity of the first kind. . +~ r.954). ± 2. e) 3. 275. i.. b) 2.005. Q\nl = Qo ( 1 287. ± 2.x< I.. x= kn (k='(). e 1 Yen + 1 2 4 l' AC l ab 280 e-l' 281. 295. c) 6 = 0..1. ± 1. n~ n=3' 285. b) { . Take advantage of the inequality Ilx+L\xl-... when x> O. c) /(0)=2. a) 1. yfO=Y9+1=3 VI +{.03 (I 0296). 15. x=2 is a removable discontinuity 320. x=O i~ a relnovable discontinuity.. 298. then cos1x=1 and y=1. ± 2. a) Second.309.875 (3. If x ¢ kn (k =0. ) are infinite discontinuities 323. Solution. a) \ x \ > 10. x=-2 is a discontinuity of the second kind. No 1. ) are infinite discontinuities. Take advantage of the If inrquality I cos (x+ L\x) -cos x I ~ I L\x I 310. ± 1. f (x + L\x) . 366. 15 cm/sec. c) 344.. 339. f' (0) = .!.f (x) • Lin (1+ £\x~o L\x L\x 352. x= 1 is a discontinuity of the first kind. a) 3x!.-::::::: - y~=! = . 330. + x L\x+ L\x). -1). 345. sec' x. a) -1. b) 3x 2 Ax a. 367. a) a~x. 2Hint. 338.o cos x cos (x d) --=. b) dT = lim L\t dt 6/-'+0 ~ T. x2 =3. x= 0 is a removable discontinuity. 334.L\x) -f (8) ~x I1x-. Use the results of Exa Inp Ie 3 Xo and Problem 365. 2. f' (1) ==: 0. 349. 2x L\x + (~X)I . Chapter II Va+h-Va. sin ~ = lim sin L\x X £\x -+0 L\x cos x cos (x ~x) £\x-+o L\x _1_ = sec 2 x.53.f (x) . dQ . 364. b) _ c) _1_. . x= 1 is a discontinuity of the first kind.. -1. f' (x) = Jim f (x + L\x) . 1560.000011. sin 2 x = lin1 AX~o + ~x) 359. 335.21. a) 0. a) 3.. O. \vhere Q is the quantity of substance at dt 355. b) 0. 0. 1. c) -h.0. b) x=k (k ~ 0 is integral) are points of discontinuity of the first kind. 341. b) the function is continuous.. a) I' (O) == linl ~x-+o V (L\X)2 Ax 1 11111 J _.01. -1 365.-!.-+0V Ax ± OO~ . 337.16.o 12 . No. 363.. x = 3 is a discontinuity of the first kind. b) litn L\m L\x 6x-+oL\x ~l =:::: -0. Hint.. Yx+L\x+Yx c) - Xl (x+ L\X)2' x2(x+ AX)2 ' 2x(2~~ 2x(2£\x_ 1).:::: lim ~Q. V 8-l-~x-V8 == 11m L\x time t. a) x=O is a discontinuity of the first kind. 21.!. b) dq> = lim L\q>. = Solution. a) c) - £\t -'0 L\t ~m. 3 2 3 (2.5. 2. y' = lim tan (x + ~x) . we have P (-x o) P (xo) < o. l' (8) = lim f (8 -t.tan x = lim ~x -+0 0. 360. 6 L\x X Urn 1 ~x . L\x e) ~X). a) L\q>. l\t dt £\t-+o L\t 353. 3 (A 2 + + + (L\x) x ox) ¥x+L\xd) f) In 348. For the given function the equation I' (x) = f (x) has the form 3x = Xl. 357. a) x=/tt (It is integral) are discontinuities of the first kind.. 7. -2~1 ~-O 238. . 329. cos 2 X x' 2 ¥t" ~ + Solution. 358.1. . since the function y = E (x) is discontinuous at x= 1.::: ~x.= 3. 30m/sec. 328.!. 332. 333. 346. 1. c) 2h+h 2 • 342. 2x + L\x .. Show that when Xo is sufficiently large.Answers 401 tinuity of the second kind. . x Yx.1.25. 100. b) 0.8. Solution. 356. Hint. a) 624. b) 0. X~i\X. 362. 1 _ . f' (xo) = .=O.249. c) x= k'Jt (k is integral) are discontinuities of the first kind.! . b) t. c) -1. tan q> . ~x-+o V(8+ L\x)I+2 V8+L\x +4 12 I' (2)=0. where T is the ~t temperature at time 354. a) L\T . a) _. b) -3. Ax-+o ~x 8+~x-8 lV (8 + ~X)2+ V = _ (8 + L\x) 8+ VB!] = lim 1 = . The function is continuous. x. 361. 351.1. 347. 350. where q> Is the angle of turn at time t. 399. 5 cos x-3 sin x..2.t h2 404 -3(xlnx+sinhxcoshx) 402. 2 bX I )2 '1r 1arc. xle X. eX (cos x-sin x). x In 2 x-sinh! x . 3x1 x1 + b (m+n ) tTn+n-1 . 5x"-x xe.. 390 . 398. 369. . 3 S1l1 X co~· X • 426. sin1x 424.(1):: 11m &x Ax-+o = V (Ax)e cos(2k+ 1 n+~x)l . 36k. V (a + 2 412. l • 416.2-15 cosl x sin x. 3x Inx. arc sin 1+2xarctanhx 410. x-yX2="farccoshx 2 2 -1 x Yx 12ab + 18b y.. x2 418. eX (arc sin x+ y_l__) . 6ax n · .. 400. slnhx+xcoshx. x x x xn x 2xcoshx-x2 sinhx . . 383. 391.. x.397.. 371 · -15x! .x 2) arc tan x y I-x! (1 + x 2) (arc tan X)2 419. y=XIX I =x l • 377. 376. 5 373· . 394. 382. 2x I 385.!. ~(ax+b)2 (l_x2)2 e c x 2 -1 -x 2 16x (3 + 2X )3. 2ax+b. 386. -110--.-+2x-2x l • &x 0 f'- i' + (2k I =_ I. lim (sin&xl =1. 422. 5x -12x!+2. 372 · rna t m - a 1 375. 387.~. rl-~ fl+~ . arcsinx+ y x • I-xl 5 x 392 · ex -xa. y'=O. 395.. . V -2x 2 -6x+ 25 1-4x 380.3x ----:v-. X 388. 3 + 1 n)= (2k -.. cotx-+.+-o I Ax~-o &x &x =00. cosh 2 x • 403.401. S111 8 x (7) xe x-I. sin I1 x + . Y 428. Y ( tlsin t. 389 · 2 + 1 n) = 2 370.374 . sin x 423. x-2 393 . 2 = lim I sin &x lim Ax-+-o = 1 Ax-. -2x 405. V Xi 38t. 413. an x . 396. V (2x-l)8 l-x 2 408. . Hint. 2 (1 -t. z 1l sin 2x (sin x-cos X)2 2a . r 1 406. -16cos ~ . -+-2-2. 407. (1-3cosx)a cos4 x 2 15sin x-l0 cosx 2 cos x + 3 si n x 1 425. - Vr Va2 -1. -e-x. C Ax-+-+ c) be-ad (e+dx)! 378. 379. 411. 384. Hint. (x 2 -5x+ 5)2 Xl (2x-l)2 1 4-2 y-)I .COsx+2sinx .1 427. l-tan 2 x+tanol x cos2 x . xarc t anx. 1 -x! In 2 x 2 2 In x 2 21n x l . 414. . -2" rr Q2+b 2 x 3 8 ~ 2 8 4b -fix l -3x-·.402 Answers b) V-1+&x-l= lim /. V . x=sin-It + 2 sin 2 x cot x sin a 2t +cos-2t. 2 V I-Xl 'V I + arc sin x · 1 3 (arc sin x)! . . 3"x l . 1-~ bx 415. 42t. 420. x (2Inx-l) . X l 488. .r y-a . (x+ 2)' 1+2 470... y' = (sin! 5x)' cos· . 448. (2x-5) X 432. 1+x 2xl0 2x (1 +x In 10). 452. i ){ = 15 sin! 5x cos 5x cos! i .-!. -2x l-x 2 21n x 451. . 487.3)5 Vat f 2 (eX --:t-r====:::::l:::.7 + sin! 5x (cos! ~ )' = x ·t4x-6 . x + ~ cos 2 f a sin 2 .. 1-1 f . . Solution. . 3t 2 sin 2t'. 3 cos x cos 2x. . rr======:::. 1.. xarccosx(1-x 2 ) /2 49t. -1 sin 2 = . 475. .r f 453. ~.: .X2 COS 2 435. tan x. 454. ~:'-l X .r--1 -2x 1.2)8 469.Answers 429.:=== 2 2 484.r-- 2x fIn x + I 3 sin!5x cos 5x 5 cos! ~ 7 x2 +x 2 · 462. 438.. Solution. r f (a -I· ~ 2)3 f (1 464.bA 'l )m-l 467.f sin! 5x cos i sin i .:::=:: .473 ' .• 444. )' · + x5 2 - 2x -t. 486. 2ubrnnx". _10xe. Yy 6Yy V(y+ ¥y)2 • + 2. 449. 1 Ya - 492. +x -ex sin 2t -t. 2x 2 -1 .r · 440. 478. f t2ay-y2)' f e +1 477. x 436. a x2 450. x-I V 2t x2 (1 + VYx X)3 · 463..J • (eX +5 sin x-4arc sin x) Y l-x 2 1 ) + 2 (f. sin'" x 1 arc sin x (2 Jrc cos x-arc sin x) .. 10 tan 5x sec! Sx.· 482. + S cos x) ¥1-x 2-4 -_--... -x. 480.r X1 · 474· · S1na X COS 2 X. 2 5 • 1-2Sxl arc sin 5x . 1 466. I-xi yf=X2 . (1_x 2 )5 (x.-l 1 +x ¥a -x•. a X cos(x'-5x+ 1 ) .2r . ( .. r 485.489.1 .2f t cos 2t In 2. ¥2x . + x) X (1 + x 2 ) arc tan x 455. 479. r-x + X ) .X2 In 5.e2X ----X 1 + q».-12 . 456.r 2 f 1 -4x 2 x cos 2x2 sin 3xl • 437..sin 4x+3 (x. (a_bx 'l )Tn+J 3x 2 + 2 (a -1. 2 .b + c) x + ab + be + ac 2 Y(x+a) (x+b) (x+c) sin 4 x cos 4 x . ..r x ..:. 447. . 4 cos 2x _(U-~) sin 2x . + eX +xex 1 x +x 2 Vxe x 2e V 3 (2e 430.459.490. 439.r . -. x cos Xl.r f x _2COS 1 (2x)' 1-(2x)2 2x - 5 In4 x . cot x log e. . ~ 403 a-3x 2 Ya-x • 468.::====::::::::. + 1)2 + -x. 458.sin! 5x 2 cos ..2)5 471. 442. 2 457.4) . 476. V(x-l)3 (x --1." (" ~ . 1 · 441.. 460. _2tS. sin' x · 2x In 2 X 433.. 446. 443. 483. (1 + 1n 2 1 x In x 1 -I. X 1 .2 .. 481. -a sin (ax + ~).. 2 (7t . O. V--a-2 (1 +x) • 465.1 (a 1. 2 445. . arc sin bx l yx. 2 V a x + 493.. 472. 461. x-x 2 x . . sin (2t 434. .. d) " . ea.. Y Y 520.554.. .y (I . 1 -1+.-2 tan . • Y eax.cosx r sin x xoi_1 1R 537 · 6' sIn h2 2x·cos h 2x. r x . e) f _ (0) 2 =(i . . cos bx 530. 509.x (I-x). f _ (0) = f + (0) = 0. 552. 546.I when x < 0. 512. x + 1 . 548. 538. 507.515. y' (0) does t exist. _ (x + 2) (5.. I+ x y tan x (I Y a 1+ 2ax+b . ~na 491. x ' mx + n x 2 -a 2 521. 3cot -In 3 ( \). 544. I ~nl ~ . x arc sinh x. We have y'=e.JL (I-x) x + 3x) + 3 (x + 1) (1 + 2x). X 527. a. arc sin 3 x /. sin 2xe.x COSI3x. and . x ~ l V' · Xl -2 I x-I 2x+11 Hint. e. (I _ Y Y2 sin In x. sin' x + 2 (1. 503.16x + 19 I y=5In(x-2)-3In(x+n. a) y' = I when x> 0. 505.Answers 404 494.XI (n-2x 2 In a). V 1 x 499. eax (acoslpx+ C x{I+ln x) 536.fn x • 1-2xcosa+xl 498. O. l' a +. f. . .-2 -6x 15a In 2 (ax + b) 517. x-3 f + (0) do not exist. 504. a cos ax cos bX. .Xl ol 1 -1 2 542. x it follows that or xy' =.I when no y -= x. 555.. (x + I)" (x + 3)5 y.. 1 Y l-ln x 497. 502. .arc cos 3x)}. (I + 2x) (I + 3x) + 2 (1 + x) (I 567. -2. 2 (3:~: ~:-ln + slnllax ) 529..F • 534. + (0) = c) f _ (0) = I.I 3 cos bx I x (1 + In 2 x) 1 53t. -I. 526. 556.x whenx ~ x >0. 514. 522. 611. 535 · . /2 6 tanh 2 2x (I-tanh! 2x) 540. a) · " . 545. xn-1a. . (3 _ 2xl ) 1n (3 _ 2x ) ' 519. 528. SUt y x! x +x2 -2 . .-~'~-.3x 5 3 3 . b) f. (x-I}(x-2)(x-3)' 516. a 2 -x cos x 2 501.. 2 x Ina x x x x -x-2 2 3x . Since e-x=JL. (2 I 49 ' I 550 f'() { . Solution. 2m p (2ma mx + b)p-Iamx In a. ax2 + bx + c 508. 4x ' /6 x 1 ..\2 + 19x + 20) • 568. 513. 2+ -4-' 557. 506. r 543. sln'xcosx' .a• 518.. f + (0) = 0. f ya "2+3' I =x . 2 coth 2x. I+2' x . t 2) [ 2arc sin IX In 2 3 1-9x 2 523. xl . I-x. 1 5+4sillx 500. 553. 498.x 2 arc sin x x · x 1 1-1 n 2 x 2 x 2 .x sin px. ax + b 511. Y 1 + In x -I1. 2 + ~ sinh ~x)~ 539. b sin ax sin bx. I --. x -4x + 2 Y x(x-I) (x-2)' 2 2 . f+ (0)=1. efJt (a cos pt-p sin Pt). 541. 525.x) 566.=. 524. 5 · Y x = _e. b) .(0) . . x arc tanh x 2 x r 1n x-I cos2x sin x l-x 2 547..{O)=-I. y810. I 2ax +x2 x 2 2 +a yf:t:Xi. x sin I + YCOs"'X In a). 558. 532. r2. y' = .0 561. 5x+6y- Y"2 6x-5y+21 =0.627. y=O. 576. ylnx-xx 2 2 arc tan 2 ~ 63° 26'. 638. x · · 3(x!-y2)+2xy-l+3 xy l+4 y'· · 10-3~(jsy · y cos! Y y 1. 1 -x cos2 612. I~XI + X-. 634. 2 Y'X-2. 36° 21'. x+2y-l=O for the point (-1. 589.-b t an t . · 1 t (2 .sin x tan x). . 600. 2x-y+3=0. xy 2 ' 2-cos x : c) xy 10 = ---:f. 581. 592. 7x-IOy+6=O. x+y-2=0. y-5=0. 613.. y=x2 -x+ 1. -12). 45°. 3 Vi' 580. since the equality is an Iden- b 00. (1+~r[ln(I+. cy + x V~ cx-Y¥X2 +y2 . a t < 0. y=-2-. 1 ~n x . 624. ~ In x ). 1). t + 1 .. .tan 3t.- . ( "8' 1 -16 I) · k=lT 631. t y=2x-2. (n+4)x+(n-4)y--4-=0. 14x-13y+ 12=0. 637. Yes. d) y=-=x-I. -"2. xX (l + In x). (-2. 583. 625. at t he pOln . 618. x I ( 1+ rio X (Sl~ x + cos x Inx). a) y=2x. (0. x2+2y . 1. c) 6x+2y-n=0. 632. a) 0. y=O. n2 IOx+7y-34=0. x[tnaretanx+(l+xltaretanx]' b) 40& . 3 (x-I) Vx 577. 15). -x I + x2+~· Y I I ~ x+y x 616. Yx' = ~ -11 when a w hen t > 0 . -1. -5· 602. t -1 -2t 582. x+2=0.626. b) x-2y-l=0. 571. 610. 13x+ 14y-41 =0. No. 620.!. 604 . 2x -2e at• 599. 3-x y= -2-. xXX xX( ~ + In x+ In'x ). (1. At the point (1. 623. y = y · 611. tan t.. + 5e! t +1 b 587.0): y= -x+2. -1 · 629. 1-2tl 2 ' 586. I). 2x+y-2=0. -3).575. 578. 2" t. 573. arc tan - e ~ 622. x 607 2y 2 _ l_y3 608 10 606.. (x+y). x yY x2 x(3x+2y) -.x2. (1. (cos x)&tn x (cos x In cos x . xx' +1(1 + 2 In x). 20). t (t l + I) 588.!L. 628. f+e • 615. Y= I -2. x-I =0. 597. 2x-6y+3:rt=-=O. 0): y=2x-6. 639.Answers 3x2 +5 3(x' +I) 569.~4 2 (x + 3) If • 572. x In y -y . XYy x-y 2 tity. y ay L 617. c) O. y=x-2. 591.0): I-x at the point (2.t I) 585. e) 2x+y-3=0. 574.y2 2 ' 609. =eY-I=x+y-I' 614.605. 636. x-2y+l=0 for the point (I. 93 590. 584. 5 · 594. 601.. 596. 45°. 3 2 a) (arctanx)X x~=3(I~X'). 1_/2. 635. 633. tan /. y=l-x. b) -.x. (3. V---xr (x-2)9(x 2 "':"7x+l) xl+I' (x + 2) 8 (x-I)(x-2)(x-3)V(x-I)I(X-3)c' 570. _Vu --2-· 603. -13=0.)+I~X]' 579. The surface area increases at a rate of 0 211 In 2 jsec. 2(1-x ) 671. elinlinate the parailleter t from the given systcnl.6. 1~2' point (1. ~. -9cln/sec 658. 3 (I +X2)1 Y(a l +x2 )3 I -j-x2 2 2x arc slnx 674 1 t x 679 y'" -_ 6. 686. The equation of the tangent is 22. 2 g . The velocity is v=5. we get the point (x o' Yo)' 643. the InaximUln absolute value of acceleration is aw . 3ctn/sec: 0. at a rate 01 0. --+ 'i/" . V o sin a-gt v:-2v ogtsina+g 2fl. the volume. Initial velocity. The mass of the rod is 360 g. 687.x2 ) 2 a a 681. The projections of velocity on the axes are V (~~r + ('~r . 647. 659. I). . The range is the abscissa of the point A (Fig. Sn=a. 2 cos 2x 2 670. St=a. initial acceleration: -am2 . velocity when x=o is -aro. N=2asin~. -0. 2. acceleration when x=o is O. the slope of the velocity vector is vocosa Hint. 2nx J . 656. and ~ The magm tude of the velocity is the velocity vector is directed along the tangent to the Diminishes with the velocity 0.COSal . 663. YI +431 n=a Yl +431 tan ~l=2n. the density at B is 60 gJcIn. 660. 0) and intersect at an angle ~ ~8°8' at arctan 648. 644. 4. Sn= (Po -.. 8 t =8.x 2 ( I .7. the density at A is 0. the 653. 2 arc tan x+~ .mjsec. The equation of the trajectory isy=x tan u2 t=2na 2 . yVI= -Msil12x 684. {) . the tangent crosses the x-axis at the point A (2x o' 0) and the y-axis at B (0. a) sin (x+ n ~ . ) ==4 3'20 1 .: . YV=(X~41)5' 682. y(n) =nlan. The law of motion of the point M 1 is x =a cos wt. a=O. -0.0511: mlJsec. To detennine the trajectory. 665. ~~~3). 17). 8 f'" (3 673. 40° 36'.0. 655. t=Va +Q:. . 654. 6 O.. The maximum absolute2 value of velocity is au>. 0. the velocity at time t is -aw sin wt.8 cln/sec.Answers 406 640. . 4 997... ({. b) 2" cos (2X +n ~ ) . the density at M is 5x gjcrn. the area. 56x6 + 210x4 • 668.2. T=2asin~tan~.%. at a rate of 40 cm 2 /sec 664. Hint. trajectory 66l.x2 • 2v:cos 2 a The range is v :sin 2a g The 1/ V velocity..=2. 685. n=~Va2+Q~. 1 arctanT' 31 2+2cp. ex2 (4xl -1. the acceleration at time t is -aU>2 cos rot. -x 672. Hence. Xo + 2YYo = 1.2). 15 em/sec. 1 L b) (_I)II+1 . 669. tanll=-q>o' 657. 1. . 2yo)' Finding the midpoint of AB. t=n=2Y2. The parabolas are tangent at the point (0. St==431 2a. 667.4 62.3. a) n! (l_x)-(tz+l). The diagonal Increases at a rate of -. Sn=asint. 689. The acceleration. ~ em/sec 666. 688.006. 8t=2asin2~tan~: 652. ) -2nx cos ( x+ f or (-l)n-l. 744.a 2 y s • 2y! + 2 d 2Y y d2x l i i I 1 -----ys-. In x dx. 739. · a x· e ne . '{ . 729. 0.009001 ~ dy -== 0.+T x (I 723. 4 n~ · 2 (O):::-(n-I)! y(h) b) 3 699 (n-/) n) -n (n-I) cos ( x+ (n -.) (-I )"-Z ] : c) (I-Xl) X + + g) + X cos (X + n. -mdx 722. 2.\ 1 b) 2"-'e.13. ) e (1 +x)n' . dx dx .753 . t: a cos sin X (I +3t 2 b) ). -1.03. a) 3a 2 x yS. 695. m · d1x _ -1" (x).. 752. c) 0. a/ • 749. 407 2n! f) . 2 ' h) (-l)n-J(n-l)!a" (ax b)" .57. c) e (_l)n-l (n-I)!.. 738.. 732. d) t-: It' 694. I dY=2700~0. b) 2t 2 +2. 734 ~+Ydx. 728. 2. For x=O. a) 0. n 4+0... (I +X)n-t 1 .2.l·. 712.n !. 256· 710..0019. c) 1. 2lnxX-3 (dx)!. 717. 3 · 2n sin (2x + 5 + 2nn) (dx) 11 . d) -0.it· 706.16. . 718. 384 (2 _ (dX)4 x)s· 754. c) 2 693.cos t) 701 _ 6 at (I + 3t t 2) 702 "t'" sint· · (sint+cost)s· · e .zx 2 (_I)" x Z 2n (-1)"-'x+ n (n 2.Answers (_3)n -'x. (2n-3) ('-(2 -I)]' ) (-I)n6 (n-4)! 2n + I X n . 714. _ e-xx ( 2 X (x - s ox + '" )( 6 d)1 x .1.00037. x=1 No. 2 d<p. .. x l'tdt IOx+8y -1+e2t.2. (I-xl) :I (l-x 2 ) 2 d 2 · Inx+-x---.485. . 1[ a dx --z-+ x a 725.2 . .0698. a) 0. 750. L\y = 0.)2 • 724. V70~4. .e _ xn _ I d) 2n x 691 I. 730.13. . 1. 707. + 2 cot· t · 703. 741.x. 740.[I' (X»)I ' d yl = [I' (X)]5 • 705. aSln +t I) X (d!Y) 1 dx 2 t=o = · · 2t 700 4e (2sin t .xr 2 cos x sin X) (x). L\S=2xL\X+(L\X)2. 16. I 3~ b) - 71 t.. Y17:::::::4. 748.38.009. d1x 3 [f" (X)]2_f' (x)f"'(x) p2 b4 dy2 . Y640==:::253.965. 713. x-y x-y -ye Ydx x y2. -x sin q> dY=-45 ~ 0. 742. b) 0. 733.85. ---. (l-x)'Z+I' 690 ) x x 2 n . e) 735.I51' n ['2x +( n -l)~J. 720.9.7x+5y dx. d) (-I)"+I.737.0436. 565 cnl'.25.2) n) . . c) 692.xe y 12 rrdx. dx2-=(I-y)l.2xe dx. 743.025~O 81. dS=2x!:J. t . a) 5.708. -Slnx 75t. dy 2=Y2· 709. 721. b) 1. VlO:::::=2.r--=· -x r a 2 -XZ -x 2 -2dx I + cos cp 726.93~ . d (I-Xl) = I when 1 and L\x=-a. -Ldx.(dX)2 -x (dX)2 d) 0.045. a) (I 4a sin4 2" a sin t l+t (l _t)I· 696 t -2e- · 697 (cos t + sin t)1 · -:-~t. a) • -I t . 727. 719. V200:::::::5. n dY=-f2~ -0. -VI-t a) 9t'. Y70~8. Y5~2. . b) 2e·a t. slnx=x-3f+51 -6fsin~2' where ~2=e2X. ( . -2). decreases.-X) a ( . 1. 1. (.00. We have t I a-+ x) -x= ( 1+- 6 ~=~. 00). 00. 789.. 0) and (2. -2). 795. (0. I 1 x 3 x +-8 a 2 ::::: 1+-2 a 2 • x Multipl) In:. 5. Xl 8 a2 1. / a+x:::::: 1 + ~ V a-x 2 1 2" 779. decr~ases 815. .' t: ~= I +e(X -I).. (0).w here + 2 (x3!..t x n :.. we will have: . . decreases. O. 2) and (2. I 2 + ax + 2ax n! 2" > I. 00).1nx=(x-I)-"2(x-I)2+ 758. (-2. t '='1' x x 1vhere ~1=eJx. 00). Solution. (0. 793. (0). in· creases. e Then. (. a. e 2 • 777. ( ~. 816.=4. Xl 5/ +~) b) - . increas~s. ( - 00 . 773. a for n = 1. since 757. x 2 Xl xn .=ex. 791. where . 820. excosa sin (x sin a +na)· (dx)n. (2) is a discontinuity of the function. 772. 805.crea!>es. increases. .we a ro\vcr~ o for a e 00 797. 817. 00). +(n_I)I+ii1e". Hint. The point x= 763. ea. 00). 781. . 796. 1-a Expanding both factors in powers of x. 810. O. 00). 768. (. 3 775. 3.. 2).1 0< 769· Xl · SlOX=X-3f + 4f x· sin . increases. 803. increases 821. 808. 800. 812.. 784. Find lim u -~ 0 _ Chapter III 811. 8( . 6=0. (-00. 783.00. (. 4). decreases . (.2). 1 -i-. Error: a) -16 0<6 < I. \. increases. No. (2. 1 782. (-00. 11: n < 1."5. I 780. 1). where S=~ (a-sin a) is the exact expression for the area bh 2 3 of the seglnent (R is the radius of the corresponding circle). -}-). (-1.<1. 806.Answers 408 755. 0).. increases. (2. 2). 813. get the same po 1ynomial ea ::::: I of ~.. 785. 1. 00). 2 788. i f' No. for n 807. decrease~. I 12 I I 804. 0<8. (1. 2 I a CI-X Xl +~) • In both cases ~ = 8x. 762. 770. co). I). (-2. . increases. a) S=g. (-2. 801. decreases. expanding e a in x 778 00 802 + 2a ~. decreases.<1. 818. decreases. -1)1 . -1.00. t I get: ( 1 + -X) I ::::: I +-I -x a 2 a \ve 5 81 (I X' (1 5 - -1 . (1. 14 n I 765. 1.:. 823. 1 -"3. (-00. I o < e < 1. 792. does not exist. 790: O. increases. 814. 8) and (8. 00). 1. increases. -I) and (1. 822. eX==I-1-x+2f+3f+ . 1). The error is less than v ( X ) I . 799.co. e <. 819.00.00. b) . 786. de- decreases. ()<O. (0. Ymax=-ya \vhen x=-2Y3. 850. No extremunl. Altitude of the cylinder. The altitude must be half the is equal to the diameter of the base radius of its base R vi. Ymax =-2 9 when x = 0.841. 833. 824. q::. (I.Va. Ymax = 16 when x = 3. . The rectangle must be a square with side cut-out square must be equal to base.whenx=12(k±})n.2. 863. 870. Ymax = 1296 when x = 6. b) nt = -1579 when x = -10. when x=O nnd x= 10. where i-. 866.. X=12(k±{)n. M =5 for x=5. Altitude of the cone. 830.. Ymax = lr- 2 837. ). Ymax=1 when x==O i) n. No extremum.23.Ymin~·-0. !lm. Ymin=3. Ymtn-=O when x= ±I."2 when x == 1. m=2"1 when x=(2k n + I) T. Ymin = 2 when x = 2 =-3 lrr 3 when when x=O 834. I). M = 3745 \vhen x=-I. R Y·2" R is the radius of the given sphere. M ~ 236 when x = 5. Ymln=1 va when x==2V[ r- 839. increases. No extremum when x = 2.05when x:::=::: 1. 83t. 861. Ymax= r 2 Ymax=~ V3" 840. Altitude of the cylinder. 00). increases f x=~. nz = 0 -e ±1. 3 when x=}ra 836. ~~. Ymin =0 when ~==O. The side of tllt: 862. x=12. decreases: 825. Srl1al1esl 1 value is m=. Ymin = . 854. 1 1 . 851.when x=-l.43. ±2. 4 1 when x=l 843. 409 . .Ymin~-O. 842. 00). Ymax == e2 when x = '2 844. a) and (a.Answers (2. Ymin = when x= e .... ±I. Ymln=O when x-=O. 867. Ymin=-2 1 r 3 \vhen (k=O.lx ~e2 e when x=2 847. Ymin == 1 wh~n e . ±2. No extremum. Ymin= 838. decreases.-} for x=-I.4. 848. x-= ( k- . p=-2. 856. Ym3x-=5 when x=12 kn. nl=-l \vhen x=-I. Ymln=O when x = 12. Ymin= x=6(2k+l)n when (k-=-=O. 832. (-00. greatest value. x= when 3 (k +{ n) Ymax=5coS~ when =-5cos'. ±1. M==l for krt x=2 (k=-=-O. 864. The side adjoining the wall must be twice the other side 865. M =. (-00.::. ~ F is the radius of the given sphere. when 828.. where 1< 869. 852. Ymin=O when x=O. ±2•. ). a) fn----6 when x == I. 849. A1=J1 \vhrn 853. 835. Ymax=O when X:=I.r- 2 x= . Ymax = (0.76 when x~0. Ymin=e when x= 1. Ymin = 0 4 1 x=O 845.0) and 827. Isosceles. Each of the terms must be equal to a 2 f.. <Xl). ). That whose altitude 868. M=27 when x=3. m-==O when x=l. 846. 879.V3') and (0. (4k + I) i ' (4k+3) i). 888.concave down (k=O. d x=y= Vi". concave . Altitude of the cone. cp = 1£. (-3. 877. V2a'lQ. 00). points of inflection M1 Ml I. (. • 884. (. (2ft + I) 0)' 897. 896. the abscis- ~as of the points of inflection are equal to x= kn. ± I. 882. (± ~3' 1 .\ve take the closest integer which is a divisor of N). 898. concave up. -6) and (0. concave up.. /2aQ V-q- . 00). 893. ~ R. 887. concave up. That whose altitude is twice the dianleter of the sphere. X= r - Y2· -. v3"). n = . 0 (0. floints of inflection M v3: 0) and 0 (0. 0) and (va. that h (I: -d: ):. 873. ±2. 874. -3). (± i. 2). . (6. The central angle of the sector is 2n Vf. 889.Answers 410 where R is the radius of the given sphere. The altitude of the cylindrical part must be zero. 2k1t). 872. ). the velocity imparted to the stationary sphere of (nass m l after inlpact with a sphere of mass m 2 moving with velocity [) is equal to Pmin = 2. 875.00. is. concave up. concave down. (2k + I) n). . a.. I 00.arc cos -k 885. cone 871. 0). M (2. points of inflection. = X 878. ). 2 of the battery is the physical meaning of the solution obtained is as follows: the internal resistance of the battery must be as close as possi ble to the 2 external resistance. (-6.. 6).00. (-6. «2k-l):rt. (4k +3) i. ±2 88t. no points of inflection. (2. concave down. concave up..+n1 v nZI n1 2 r . 0) and (6. a) and ~)- y~). VMm. concave down. (. 880. 00). y=d Vii p J + Vi /2 V 3"' -. / N R (If this number is not an integer or is not a divbor of r N. Y="3 h. concave up. concave do\vn.4M =aV · d V3. l . The angle is equal to the greatest of the numbers h arc tan -d . 895. concave up. that is. {r. 00). 894. b) X= 883. {). -{). point of inflection.00. ( .va. ±2. where R is the radius of the given sphere. the vessel should be in the shape of a hemisphere. Since the internal resistance n. V"ii)... 2 +2Y = 1. ±1. For a completely elastic impact of two spheres. (2kn. The sides of the rectangle are a V2' and b y~ where Xo Yo a and b are the respective selniaxes of the elli pse. 886. The coordinates of the vertices of the rectangle which lie on the parabola (. 12). Radius of the base of the where r is the radius of the base of the given cylinder. 892. 0).00. 891. (4k+5) i) . Hint.concave down (k=O. the crosssection of the channel is a semicircle. (0.. ( . 00). . 876. Asymptotes. . inflection ~) 918. 911. =f 2). 904. x == I. y=x-n. y==-2. y == . Ymin=-I (when x=O) 939. pOint of Inflection. 938. x-::O. Ymln= 4 when x=4. 0). 2) are end-points. asytnptote. x==o and y=3t 932. x= ±I. x ~ 3. left. y=-=2 910. 0) and 64 ) ± I. Ymin = 2 \\'hen x -:= 2.:--l. right 907. point of inflection. V V + ~. 27 T6 YmJx = -4 \vhen x=---l. of inflecti~l1. 934. 917. x=O. M I . Ymax= 1 \vhen x= ± Y3. y=-=O. (-3. 923. M 1 (0. y == l.). 925. End-polllt. x --= 0. 919. concave down. Ymin=-l \vhen x-:-::O. A(-V-~ 0). x --:: ± 2 and Y::= o. 941. :). 900.-". right 908. A (0. asymptote. -4) and B (8. Y == x. I) and M 2 (1. 905. 915. point of inflection. 00). Ymin=2 when x=O. 929. pOlnt~ of inflection. A (-3.-2 \vhen x =-= 0. 6 1 Ymax = 1 when points of lI1f1ection. 0 (0. 2) and R (4. aSy111ptotes. -I). points of inflection. YmJx= 1 \vhenx= 1. point of inflection. 912. (±l' -{-). aSylnptote. x-=. concave up.=0. Ymax-=4 when x=4.) (~ ii' - concave up. 0) is a point of inflection. M (0. -2). Ymin==O \vhen x==2. asymptote. 909. points of inflection M 1 2(± Y5. 906. Ynain = 3 when x= 1. x=O. Y==l. Ymax --. !lmax=1 \\'hen x--:4. down. left. y=:::O M point of inflection.0).4 ( V- x=O.41 ] Answers . . 0). x ==-= 1. Ymin ==-4 when x==2. nSYlll ptotes. y==O. Points of inflection. x. concave ~) and x == ± 2. MI . left. . asymptotes.4). ±~2. . 927.0. Ymax ==. 2 (± 1. aSyl1lptote. 0). point of inflection. Ymax==2 when x=3. Ymin=-4 when x=-4. 940. y=a. . x=±2 and y=O. 914. when Ylhin=O points of M I .125 . y==-I. 935. 0(0. YmlO=O \\'hen x=l. ri~ht. Ymin==4 \\hen x== I. x=2. Ymin= 4 when x=2. End-points. 0 (0. Ymax==O when x==O.00. POint of inflection. y-=2x-2. Ymln~4 \vhcn x==l. 937. point of inflection. y==x. 913. 4) are end-points. concave up. ( ± I. 00). 0). left. right. ( ~ea' 00). y== I. 924. O. f). 0 (0. M 1 (I. right.0) and M I . ( . lIlhax={ when >. Ymax=4 when x=-I. (5. y==l. -3) and (-I. 922. :e. Ymin-=-=-J when x:=. Ymin-= Y3 Vi - when x=Y3. 901. Ymax=O when x=-I. y = 0 926.x-I. x=O. 944. YmJx~2 when x=O. A (-8. left. 0). (0. y =. y==-x. left. y==x. 0). Y == 0 903. Ymax=8 when x=-2. \\Then x =--= "3. y~O. (-00. Ym.2). M 2 (-I. rIght. 921. M(¥3+2Y3. M I 2 (±1. as)lnptote. aSylnptotes.2 J-r 3 ) ( ±2Y3. y==O. V- V- 943. y == o. Ymax=2Y"2" \vhen x=2 933.0) and B(V~ 0). asyrnptote. M is a' point of inflection. x = 2 and y =--= 0). x= ±2. points of inflection are M1 (-3. Ymax= ¥2\vhen x=-I. x-=-l. 899. 1\1 (-' 2. 916. YmJx=-=-4 when x-=-l.iown. 0(0. POint aSyl1lptotes. 942. 2 (± 1. M a. 902. points of II1flectlOn. 0(0. 928.1X -== - 931. Points of inflection M I. 936. y=-x. x===O. 920.±2 and y -= O. Ymin-=-2 \vhen x=-2. asynlptotes. y=x+n. V2). M 1 (0.:=: 4 and Y == 0 930. aSy1l1ptote. aSylnptote x==O.-2 \vhen x-=-. Ymax=~ whenx=±:: x= ±n. 0(0. the function is not defined on the e I.. asymptotes.13) and 962.33). ¥). of inflection. l~aa) and M. Ymax= Y 2 when x=4+ 2kn interval points of inflection. M (12. M (2.. -0. Y=O. rn \vhen x=4 n+2kn. Periodic function with k . N k (arc cos ( .. ( -a. M 4 M 3a 2 a ye. + 958. points M9..70). MI .± I.39. . asymptote. 951. 954. M ( (e 2 .. 945. Ymax=O.0) when x=6..4e 952. Y= I. Asymptotes.37): + Y -.' - 4e2 asym ptote. 953... x=O. MI (-3a.57. 1. asymptote. points of inflection. 0]. points of Yrnax = inflection. asymptotes. points of inflection.54 e Ymin=O when x=O: point of inflection. ). 950. ( 8±2[2 . perIod 211:.71. Ymax=l 2 when x=±I.) 43 ~ n + 1m. M I . yi"' pOInt of Inflection. !/rnin=O when x=O. 956. O. Asymptote. when x . 4 (± 2 ~O. 955. ~).Ymax=!e when x= I. . 1 \vhen x=s--1:::::::-O. ~~0.. y=O (when x -+ (0) and y=. M 1 'On M a. Asymptotes. x=_!.. . x=4. Ymax=2 when V2 ~).. period 2n. x=. Ymin = - Y2 (k=O. Yrnln = 1 when x = ± y-2'.x (as x ~ . ).X=2 946. 0. function with 3 1r '4 Y 3 when Periodic 5 x ="'3 n + 2kn.74 when x=e =::::. Ymax=e2 when M I . and when x=-3.739: point of inflection.. - point 947. e (+-1~-0. when x=i: Ymax=l when .63. Odd periodic function with' period 21t. 2 (± 1. ±2.95). . (±l.(0). Ymln= (-3. n X= Mk ( Yrnln = - 3+ 2kn (k=O. points of inflection. ~ YIS). a 2 Ymln = . Ymin = y". y=O. 2nJ.rY 3 when ±I. . -1 0 (limiting end-point). Periodic function with period 2n. X= ±l of Inflection. x= ) • 1. 0) and 961. point VI2100 ). ±2. y=O.Answers 412 YmdX=- Y3 V2 M. 960. 0 as x . M) ~).asy m ptote.. x=O and Y =0.~ . points of inflection. 0.89. 957. Ymln=-2 when (±0. of IIlflection. On inter- -val [0. a.O e when x= when x = e. Yrnax="T=::::'O. + 2kn.. (3. M (e Mi • :::::: 14.. Points of inflection. ~). asymptote. ±I. e~).{.. Mk(kn. Ym3ll= 1 when I x=O: Ymln=0. el ). Y=? 948. the Interval [-no nl. asymptotes.86. 0). Ymln =0 when x=O. x=O. point of inflection. 959. rwhen 3r6 r6 3r 6 2 ( I) :6" . (. + 0(0. M. Ymax= 3: -I when x=-I. ± I. Al 2 (±2 83. ' .r. M k (k~. ). asymptotes.1 when x=n. Odd function. Ymln=I. M.Answers x ~ =2. -0.2 . V~). Periodic function with period 2'1:. Even Ymax = Ymln = (k=O. Ymax= lrwhen x=arc cos .50.86).35. x=O and y=x-ln2. ~ y~).. points of inflection.21. 0.. point of + inflection (centre of symmetry) (0. 0). 975.. points of ~ V2) (k=O. y=l. -0. r. points of inflection. _ 4~f). points of inflection. x= x =~ +kn. ±2. End-points. Ymin = - 3r 3 0. ± 2. . M. 4~7). Ymin=O when x=O. ±I. y = x 2~ (left) and Y = x (right). 00). Ymln = 1 +~ when x= I. Ymin=O when x=O (node). k~) lk =0.86). 2k~). M a (2. Ymax=--2-whenx=-"41t+2kn ~ n + kn 965. Point of inflection. asymptotes. 970. (arcsin ~2 when ±2. Y= ~ +n (when (as x-++oo). x=2n.when ~2. x x-. Ymax=I..x=4n+k~. -1 57) Ymax == I 57 when x=O (cusp). . Asymptotes. Even periOdIC function with period 2~. r 3 3r 3 when x=O. (arc cos arc cos ( 967. asymptotes.1t 4 . Periodic function with period n. Points of inflection. On the interval [0. ).. VI ~ 3 x=4+2k~.36.86). ±I. 968. x= M 1 (0. 0) .. -{ y~} M. -0. Odd function. 972. Ymin= . l x=arc cos Ymin = ..Ymin=-1 when n. 0). M k M s (4.r. 974. inflection. M 1 ~ (± 1. 971. . Even function. ~).51. (5. Even Ymln= ~ function.36.r. ). points of M 4 (3. .856 when x=-I. . per iodie function x=arcCOs ( - ~3> Ymln=O ~2. 966. .-oo) and Y=2 M (0. ±2. with 3 ±I. Mk(k~. inflection. M 1 M.31 (k=O. asymptotes. Odd function. point of inflection. 2k+1 asymptotes.. asymptotes. 969.:: ~.. 963. ~). 0). ± I. asymptote. Ymax = ~ -I + ?kn when n+ 1 +2kn when X= ~ n+kn. ).. period 2n On the interval 4 I . Ymax=-O. Odd function. points of inflection.asymptotes. 964.86). fJ=- ~ x-I (as x -i""-oo) and ~ y =2""x-1 (as x-++oo).0). M 1 Y~..54.285 when %=1.00) and (I.( n-arcsin [0.34).( ( .71 when x=4"1t. Ymax= 1 when M 2 (1. .. Ymax = 0 when x::. Litnlting points of graph (-I. -2( 4+k~. ~] Ymax = I \vhen 2 x==O. x = ~ 413 5· Ymln=-I when x=1t.when x = arc cos . x= . . 973. 5. Points of inflection. asynlptotes.Answers 414 976. x=k31. 2k31 + 31). xmin = . 988.81). O)~ e e =::::: 1. Limiting end-point. Ymax==+a (cusp) when 3Jt I xmin=-- e 731 (x==O).32 when x=±I.a when t = 3t (y == 0). ). end-point).n + 2kn . 985. 979. A(O. when A (+0. Ymax=. Periodic function wit h period 231. Y -i""-1.a (cusp) when Ymax = ° 331 t=+T (x=O). ±I. Le. x=O. asymptotes. ± 1.69 987.. ±I. Points of innection~ M (0. ( 531 4' T' T ~. ).) domain of is the set of intervals [( 2k .lTCSin---+2k31. M (0.57 as x -+ ° (linliting AI. Periodic function with period 2rt rt x==2kn (k==O.12) and M2 t4 ~5. x=O. 1. Ymln=O when x=O.. Periodic function with period 2n. (k asymptote.59)..e. Y=± V2 . where k is an integer. Ymln=( f)"e ::::::0.74). . 4. ). ±2. y = 1. monotonic increasing X= function. x==T+kn. ±2. . 980. 992. asymptote. x=o and y=O.::= I. Ymin==O "'hen t=O. 157).. I when t = . 982. . ± 2. t==2 Y2. ±2. period 2n.t when t = -I (x = 3) 989. M k ( . e')' 981.00). ). Domain of definition.81 (as x ~ -1.00) and y=4. ± I. ( - :::~. points of inflection. asymptote. y -+ x I as -+ +° Ynllll = I when 984. ±I.. asymptote. The ( 2k + -4.. . asymptotes.28. - V 2eva ) and when 1 = V 2.. xmin == -I \vhen t == 1 (y == 3).57. ). point of inflection. 977. asymptotes.when t=1 (x=e). 1. . End-points. I) and B (I. 983. M I (0.. == 0. To obtain the graph it is sufficient to vary t from to 2rt. ± I. . Periodic function with Ymax=e when x= ~ +2kn }IS-I') lrr5-1 (k=O. I 40).. where k = 0.0.44 when x == e ~ 2. I x-2kn 1< . ±2. xmax ==a when t = 0 (y = 0). ..3I. 31 ± T+2k31. Domain of definition. y=0. Asymptote. definition n]. I x == e ~ 0.58..(±1. Ymln= i when x= i.72. Y. )12' Y -2 eV -" eva). . Ymtn = . M k (2kn. The domain of definition of the function is the set of intervals (2k31.. Ymll1=1. Point of inflection. pOints of inflection ( a a) x=± 2y"2" .I (y=-e).. Ymin -:: . 991.. 31 when 1=4' 990. ±2. e 2 2 Y5-1 YS+l\ and N k ( -arcsin-2-+(2k+l)n. Xmin= I and Ymln = 1 when t=O (cusp). y=2x when t -++OJ. End-points.37.{ ) n. 978.21 (as x -+. x> 0. 0) (k ==0. 31 Ymax=O when x=2"+ 2kn (k==O. 1 986. asymptotes. points of e inflection when 1 = I. ds= y" ds=cosh~dx. '993. ds = V + = == 3a sin t cos t dt. V aI_xi a 415 p V p2+ .. aSHl /. sin (J.. ds =-= . -3).9X 4)3/2/ = I r V!+FI .-2 . ted for the sake of brevity. 1: a ).ds= 2 f 1 dcp.sina-==- Ya 2 . dt.2). 2 2 V' a · -dx. ds = 2a sin 4 2 2 _C X dx. a x + 2 2 b2x" -1. a b 6 1008. sina-==- cosa== bx . 2 ~ = cos ~. K B =(i2' 1009.:= sin t. a. nx 11 n-l' . 1000. (y.!£ y Sill . cosh ~ a ~. cos a=-. R = ~ a cos ~ t023.r . cos a =sin x a sin a= cos ~ . 1025. i ' 3) ( (1 -1. . sin p= cos 2q>. 999. 5 x.Answers X a dx.c2 x! y . 1 -1. Sin a-= . 1006. (2. ~ = cos ~.. ( 1003. T+ + (a +3b) Xl + abx· 2· 1l-1 abx4 1034. py2 = a ds-= 1004. sma= ~ sin a'-" tanh : . vertices. K---=36. (In a)2 dcp. 2t 1 .~ and _ (b 4x2 r R == I at I. V-_. cosa=_l_. 1036. y 1 -1- -t · K -= a 1010. 997. KA=b 2 . cos a = -cos t. (X-3)2+ ~ (X - 1020.where c=-ra"-b • 995. +a ab4y!)3/2 4 4 • 1018. 1013. + 1)21 · 1016.d<p. 2 + (bl") a == 1024. r--2 1 a = 1001. R= 1022. \vhere c2 ::::: a2 _b 2 • Chapter IV In the answers of this section the arbitrary additive constant C is olnit. 103t. RIc..!-. p)' (scmicubical parabola).st = = ~. 1027. (:' .. (aX) I + 4 C 3 . ds == . 1014. a a -x . K=13 V13' V 1 + (In a)2 Y2· both ~ dcp. cos ~= 1002. ( _1.. ya r d s=- 994 · y. (X+2)2+ . 998. 996. 2x '1m=::- 3 ' 2px. 1026. sin 1005. 2x 3 +4x 2 +3x. ds= V p2+ y 2 VI. 2 6x . R - I + (y -3)2 = 8. 1033. R = I 23 R =-:: (y2 4y t015. 7 a2x7 • 1032..7 · 1035. 101 I.cp2 q> 1 <p2 cos l . 1012. ds=y ~osa= 1 =Y a Va 4 ---. cosa= ~VXa.c2 x2 a V p2+y 2 dx.. .dcp. K-=3 :'2 _ 1n 2 2' at 1 Y2 ) 2 1 1 _I . 1019. I p I. R - 1017. + +q> ds = a cos sin p= 1007. cos ~ 1 a . ds == a 1 cp2 dq>. -2b f l-y. x x' 4"+3" +x 2 +2x +31n I x-II. 2a r ax.. 1 • 5 ~r3 __ 1073. 2 \x.( ~2 2 + 1 ~I In I a2 -x ll ) Y2 1n (2 1 .. arc arc 1046. Xl 2+2>:+ 4 -~ 1n I x +31· 1058.r.. 1037. Solution.coth x../3 . r - dx a 1062. In (x 1009. a-x a-x 1 h I . 1048*.rarc tan x JI -5' f 15 Ir 1 Ya-b 1067. 1064..rr 4+ Xl). Y21 Y2 · 1074.1 dx= Hint. a In cos I I· aC x =-a5~(a-X)=-alnla-xl+alnc==alnl-C-I. aX+~lnlax+~I. 2 '1r ( . 1049. 2 yx + . Hint. Dividing the numerator by the denominator.. we get 2 =1+2x+l' Whence =x+ ln I2x+ll· a + x. 5 + =5 5 2x 3 2x+l dx 3 11 -2x+41nI3+2xl. ~-In(x+ yX1 +2).~ x-a dx \ +1 1 · In 1 x+ll+ x 1059.. ~ r 2x+ r 7+8x l ).Answers 416 4 V/ -nx.2"+x+21nlx. alx-Sa Vx I - 5 X I I Vx ... V Xl + 1. b? t060. arc 10 9 - +7a I 7 4m 2x' . I. a) tan x-x..2xl 1044. r. Sx+I-J (x+l)2' 1061. is Vsarc sin x JI 7' 1072.2 f6 1n " V3+ Ix Xl t068. l 3x 3x 1040. .. 4n tan sin 2 Yx 2 . 5 r ax 2x 5Yx l Xl -3.. ba-a~ S 2 dx d (2x + 1) 2x+l=x+ 2x+l = x a 1054.~~ = Solution. x 3 Y35 arc tan /S V 7"- . + - X I 1053. x+lnI2x+ll· .r1042. S. 2x 2m 1041. . -3ij'" (a-bx). b) x. In(.: dx = 2 S + I) = y' Xl + l. Put tan1x= Sin =sec 2 x-l.4ax + 4x r ax. 4 9 - 1038. 3"ln l 3x -21... +1 · x ¥7". . 1057. Hint.7 . Sa a x dx = 1052.rf 10 x+ 1047. 1070. 7]-bllnla+bxl. a2 x+2ab 1n! x-al- 5(x+l)2= x dx S(x(x+l)2 + 1) . 1050. 2x 2n Yx 1 1043. II"" In . dx+ x2 1055... Ix YY7-2 V· Y21 x.rr-2 x- f In (x 2 arc 7 +2 tan 2 x 1r-' r 2 +4) +arctan ~ ./"_ 1045. yX 4xm + n +1 Yx 2m + 2n + 1 + Y7 + .rIn 4 . 1056.V VIOl.. X1 1 t063.1 1. Put tanh 2 x=1 -x.63V-x. f"' a+b+x Y 2 f a"-b 2 r a+b-x a-b 1069. 1 In . 1 -.e~ I' 1051.a)-cotxx Solution. b) x-tanhx... 1 .14 1066. --13. x +1 +1 = i 2 + lnlx -2 · t065. 1071. 3 I~ 2 In (2" + 3) t 102. 1085. In a arctan (a").1. 1110. 2" arc sin (ii". 1086. . I 4 4 a sin ax b1 x 2 + co t ax lIlt.107&.2 p 1 X2 2 In 7 7. I "2 . (a-b)X -2InlcosVXI. Inlsinxl.-3 In (3 +cos 3x). e1 +e. 3 t an 4 3· x "4 a· X ~Inx 1 27 SI1)4 6x 1 · ~. I ~ 2 5 V-X li15 · 1096. l 1125. 1108.. Y(a-be")' Hint. I ( 1.r-tr an . -lnlcosxl. a a xlnlslna~bl. {(lnltan~I+2sll1ax)- . l ax 1 Xl arc tan b· 1080. 1084.rx y 2 ~1l) Y"2 xcos (log x) S1 n 2( +-4- 1109.' ' ' . -tan(ax+b). } In I x' + Y x6 . ItIS. " 3 2 . 3)' 1107.~ 1097. - 1 _Y2Inltanx~I. 4 ~(ea +1)3. 1133 . '105. n' an x I. . .2}: 1103..hX -e x 1095.-2 f Ct S 2x. 2 sin r x. + 31n I x+ 1079.+i)l. Inltan x- axtwr.Ya 3X + n a~ n a n a In a W/ r aX --2 ea. 1075. 1077. . - f l 1 3" arC tan x'. · 1106.. -In lOX 1 x ="2 (I-cos 2~). I wr-- . alnltanil. l i b bX--x -2x. 2U In (a1x' + b2 ) + a 1082. . 1094. t4-1900 3CO~ 'x. i t X I sin (x + I). 32 11. hint 2 -2 tan (x).I119. (tan3x+cos\x).1 I· 1083. l 417 -i r r-------::::---- In (x+ YI 2X + X2)~ 1089. S Hint. 1126 · asin . . 1122. (aX bX) 1 Yx ( arc tan 3 1088. 1091.5 1n (xY5+ + I+ I 1 1 -8 In (1 ~ 1087. 1078.-x. b I V-2 cot x r 2- 1120. 1092. 1114. 4 J! 2 -4 + I 41n (2x 2 +3). -:31~4 4 2 . ----x. 1117.r r (1 +3 cos•x)•· cos (a + bx) . 09 In 11 . Inle"-II.r- Stll! i- 1100.4 - ttt3.5 1. ~lnltan(?. 1136. I I "2 cos (l-x ). Y5x 21 In lxi . l 1101. 2" ~ 3 """ 1 In 111 +e2b _e. a wr-r x2 -41 . -.. - hX . 1081. • 1 1093.' + 4x 2) e-"'''. • 1104. 1 1 1131. ee 1116. 1118. "2 .1I24. 1123. 1121.9"2 . Y(arc sm x)'. Put . . ~ 1130.Answers i Y5x -{-In (5xl +7). X 2 +1. arc sin et Sitl 2~. x- . ¥(arc tan 2(.5Inlsin~l· 1128. 1135. 1098. 1112.I. • 1129.l090. -~ + 1). eli +2x2 - 2 3-1. x 1099. . x I 2a cos 2ax. Hint. 5 1134. In cosh x. . 3 (Inlsec3x+tan3xl+~3). a SI n x 1158. x-In(l+e X). . -2n cos ( 2nt) r+q>o · 1181. I 2 x + 1n (14+ x ) +arctanx."43 V (1+lnx).1139·-2+4s1nh2~. X Y3 . Inlx+cosxl.'lill 2x 4 . I .Answers 418 1 a .12 xarc tan _ 1 x. Y21n tan 2 (2 Y2 cos .\ta~~:2' ( (-7i) . x 1 a' - 1150. • 1149. -2cot2x.2x 1172. S1 +dxcos x Ssin! x +dx2 cos---.1164.. 1155. 5cosh5x-5s1nh5x. 1147. (arc dx ==. 1165._. f a2 -b 2 a a V T 1178. 1138._ 1n (X V (5-x 2 1142. Y sec t.Y 1 -Xl.=- Hint. 4 x when x > 1189. 1170. tl43.In ~5 -I. In 1tanh x I. x sin x-. e sin 2X · '.alntan x x n ) 2a+T 2 2 2 -- -21n I cos Yx-II. 2 arcsin (x 2 )..1177. 1182. 1160. 4" )'. x +x-2In/x+1I·1151. In 1x 1+2 arc tan x. 1184. 5 j-Vlln(x-I-Y1-I-X'))" 1/.1176. 1154. 1157. (r 11 a) '2 - 13 V 2" arc tan .I). . . 5 . X lnl tanx+ Ytan l x-21.1n(eX+Ye2X-2)." I i /. 1174. 4 Y5 1 X 1146. I ( 1161. y 3 2 ). r X)2 f 5-sin2x I. . V\X + Y2 arc tan (x Y2">. In (sec x -I- ~2 arc tan ( 1187. 1188.4 (2X:+ I) 1)2 1 1 --2-· 1159. In tanh 2 arc tan e~. tanh % 1190 · -1. Y5' -X2 -"2 e 1148. 2 1175•• r 1 a. a-tanax-x. 1156. l I. l"il""a.lnltanaxl.3). 1168. eX 1 1 1152. eare tan %+ 1167. arc sin 1185.e. I 1 1n /2 + In x 2-1nx · 1180. I i I· tt40. tt41. 3Q1nlb-acot3xl.i ~ arc cos 2 1183.!. a I- x+ ~21n / :~ ~~ /.arcsln.-. 2 . 5 tt44. -}. r--2 ya arcstn--+ y 4-3x. i- Jt sinh (x'-I. In I sinh x I · 1145. -lnlsinx+cosxl· 1169.'2' 1173.. 1171.:. {-In tan 1166.la-+bb. "lry 2. --1sin x n x.tan %.ctanx. tan-·1163. 2 x 1 1137.. 1162. ~ 1179. Y3+ (2+3x 1.) . 33 • 1191. 2 2 Y2 arc cos - I st ~ ~r. 1153.. I 1 In I ~-4x + 1 . a) n 2 x -I. 1186. ax = Q (x) e ..JC (x 2 +5). See Problem 1218*.. may be used in place of the trigonometric substitution.. In the general form. in similar cases we shall some'times give an answer tha' is good for only a part of the domain of the integrand.. x Hint. r2. '1r-x2 a2 f - -!. YI Xl. In place of repeated integration by parts we can use the following method of undetermined coefficients: ~ re1xdx=(Ar+Bx+C)e'X or. 14* . 2 }(tr-2) 1198. 0 = 38 + 2A. 2 arc sin . arc cos 1. Pn(x)eaxdx=-.. 1210. x In x-x. 0 = 3C + 8.. x=-}. x2elX = (Ax 2 + Bx + C) 3eIX + (2Ax + 8)e1x • Cancelling out elX and equating the coefficients of identical powers of x. ~ (2~lt5)tt]. 1192. _e. (arc~tlx)l. 1200. x arc tan x-'2ln (l-t.. yeGs x. X ~ V(x+1)1-2VX+'1. ~2X+T-1 1. 1218. Put x=-t . 2 (V.5 1194. 1217. we get: 1 = 3A. 2 whence A="3. z x .) if x <0 *) x 1206. after differentiation. and arc cos ( - + ~XZ+TI. 1199. VeX + I. 1220. 5 1 8=-2 .Answers c)8~(5r-3)'. II+Vx 1 1202. 1197. e~x 2f (9x 2 1215. 1204. In +21 arc sinx. Solution.-2 +"21 arc sin x. if x x x Put In(sinx+Vl+sin'x).-2 4 . /' +1 Inx-ln2Inllnx+ 1196. r . x .. d) [(2Xi. e) 2x+I+1 +21n21. 3x + cn~ 3x .r . Hint.(COS'X-5)X l Hint. C=2j. Se~ *) Henceforward. -6x + 2).. 1212. V~ . The substitution x =!. r . 1201. x arc sin x+ 1216. 2arctan Vex-I. Note. 2 y 1213. 1211. -3e a (x'+9x 2 +54x+ 162). x2 . +2 yX~2Inll+Vii) • 1195. x In 2 + 1 2 xlf1i2· sin x-x cos x.x2 ).5)12 . where P n (x) is the given polynomial of degree nand Qn (x) is a p01 ynomial of degree n with undetermined coefficients 1219.rr-x. Inl 419 1205. j. 1208.-2 -'3 y2--x -:r y2-x • -a arc cos ~. J"X'i+1-1n 11 > 0. 1 _x x+1 -ex-. x Sl. - Hint. - 1214. y 1193. "2 y x -a -+l a '1r-I +2" In 'x+ y X 2 _ a2 1. -i Vl-x'+ 1203.-2 1207. -x cot x+ + In I sin x I • 1233 -~ + 1 It!-. 2e Xl 1232. + 2X: 5 cos 2x _ x c~ 2x 1221.x . x 1247. . Xl x I I I + x2 1241.f + (1 at a -x'+2 Utilize the identity 1.wehave Yal Xldx=XVa2-xl-S. 1237.9" . -. 1251. 2 10 · x l .1245.x (arc sin X)2+ 2 r l-x 2 X 1244.. 1240. [In(lnx)-l).- U= f ai-xI and dv=dx. l + SillS2x.[ . whence andv=x. r .• +b 2 X2 -cos (1nx)). x dx 1250. I-x !. and Qn (x) and Rn (x) are pol ynomials of degree n with undetermined coefficients (see Problem 1218*). 1224. I + V I· . Putting u = x and dv = (x 2 -t.. x -1246.1)2 ' COS2X-2Sin2x X( 5 S 2 1 x dx x we get du=dx and v= 2(x2 +1)' Whence (x 2+1)2 2(x 2 +1) dx x I I x 2(xl+l) 2(xl+l)+2arctanx+C.::x Vaa-x..g +2"-3x. x In (x+ VI +x2 ) _ VI +xl • 1230. x 2 -1 Xl 1 35 2 • y- x (Vx-l).f r x +2 . --x---x--X-.Answers 420 Problem 12IS*. 1225. 1252. 2 arc sin x-"4 X 2 2 X arcslnx+ ~ VI-xl. arc sin x tJ' Solution.. --2-(x1 +l).2x In x 2x.arc tan x-"2. ax (sin x +cos x In 3) · 1 + (In 3)1 e- eX (sin x--cns x) 2 1234 eax (a sin b"t-b cos bx) al . where Pn (x) is the given polynomial of degree n. It is also advisable to apply the method of undeter· mined coefficients in the form ~ Pn (x) cos px dx = Qn (x) cos px + Rn (x) sin px. 1227. xarcslOx-2x. -2. . X arc sin .. r .2x l . 1242. 2 1) · 4 2' . Put f . 1222._ ~ 1248 e- arc sin x +In .I sin x n an 2 · 1231.rf x.. 1236. x In x..1 [(xl+al)-x l ). 1228. 2a2 £iarctan(i+ +S + xl~al)' X Hint. Solution. . 1239. -2.rx +I X x 1 r x. aarctan3x-T8+162In(9x2+1). dU=-V XdX II Yal-xi Val-x' . 1229. -2 yrI-xx 1 I-xl x x tan 2 x+ In I cos 2x .1243. Xl x' In x 1 2 1223. . ygl_XI_s(al-xl>-alctx.::::~2dX= aI-xi Va 2 -x2 l l . 3-X2+ In 2 x 21n x 2 +3x ) Inx.4x 2 • + 1226.r l..2 (Xl + 1) + 2 arc fan x.-2-X 1 x(arctanx)2-xarctanx+"21n(l+x2).. 3 In x.I-S va -x dx+ a2S dx . 1- X . • 2 sin (x (I nx )_ 3 1238. 2x + I~X+ II sin 2x+ Hlnl. 2 X 249 £+xcos(2Inx)+2xSin(21nx) 1..lnx. 2 V -x In x-4 .1n I+x. 161 ~x+~lnl(2X_I/~~X+1)81 I I. W. 1265.y 2-x-x 2 + 1274.arc tan y 3 J V. 421 . .. 1261.1 1+ 1292.r . \x + 2x+ 1 .· - 1289. 1257. 1256.2 (x> V2). -2(X 2 _ 3X+2)2' 1291. 1 2 1258. Hint. 2 +~ +~ X In I x+ In SVa2-x2dx=x ¥a 2_-x2+a2arcsin : VA +x" I. -2 Vl-x-x2 -9arc sin 1266. i. 1255.I .» + 3 I V \. - ~'5 In ( x V5- 1275. i- 2 1n (x 2-4x+5)+4 arctan (x-2). 8 . x a .1285·I~X+ln IX~II. 1254. In ( 1279..b I.-ml arc tan x. I f -arCS1tl 1273. • 1287. arcsin Xy 5 (I-x) 1272. x++ln 1. +{ + VI +eX +e2 X (x-l)4 I X x+l. x- ) 1286... I 27 ~ In 1 ~-5 49 (X+2)+\i4J x+2· 1293.-lnl(X-l)\~+3)~1 1282• (x-I) ___9__ px+q V5x -2x+ I) • 1280. x-2'ln (x 2+ ~ x+3In/x-31-3In)x-21 1281.l .2 arc sin 2 -:.i VA Problem 1252*.Answers Consequently.f r -.I . . V Xl + 1278.-arcslD 2-:-:. In x+ ~ 1267. See 2 Vll 2 . xarc tan V3 · 1259.3 1 -4 In . 1264. r. I x +-1. 1 3-~1flX .X + 2)4 12 6 2. %+Inl 1288. Ix+ 2x-7 1262.-(X~2)i-X~2. arc tan arc sin ~ · 1263. x +2x+5). x-x2+'8arcS10(2x-l)o 2x+l +-98 arc S111 \i I. I arc sin x HInt. I _ .::1- 6~ In lx + 4x +5) + I~O X 2 . . 5x + In x 2 (X-4) l 2--V- 2 Ir.!. V5x l -2x+ I + 5 1+ 2 In (r-7x+ I. arcsin(2x-I). VIT' 3 Vx 2-4x+5. 5 2(x-a) 2(x+I)O 49(x-5) 1 1290. In 7 V3 5 ~5 + l ° ~X 1260. +x2+ -i V9-x + x. 1270. ~ 1269. 1271. Sl2 1n I X-31-2~ In I x. 9 2x+3 + 3x + 4) + y"7 arc tan ¥"7 . IIX . x + 3 In (x l -6x + 10) + 8 arc tan (x -3). /-2 6x-1 arctan · 1253·.SX + V'1-4 In x-Inl x. See Problem 1252*. 1268.r y x~+2x+5+2In(x+l+.4 . ~b In 0- 11. (X-4)51 + :3)7 ° -i 1284. In l-x 2 2x-1 -4- x +1 I ' r 1277.. 1 1283. 1276.~ x . -Inl cosx + 2+ Vcos l x+ 4CI. 1322 . 1323.l. z= v x+ I + x 1 x-I· 1325. 2' ln lx4. . 48(I+x2 )' + ..2+ 1 +-4.S6"/. x+2 5 2x+1 Inlx+II+3(X2+x+l)+3 Y3arctan ya- 1298. x 1295. I ~ I 12X4 + 1. In)xl.-4 In I x +x4-II--. x'-I ' XYa +arctan(x+I). . 1318._ r3 arc tan ~r-l r~ . /x 2. .. 1 21 (81nl xl +81-1n I xl t305. 2 6 3 2 V"X - V-r+ rX-2 V'2x lrVX~-I 1 -2 arc tan r I-x. r 3 arc tan 2 Y7 3 l~alX 1316. In xii + 1=(x7 +1)-x7 • -3 1320.71 In lx7+11. g (x-I). 1304.. _x 2 + x I I I 1301. 3x-17 I 15 In (x'+x+ 1). I)2+ x 2 arc tan I xarc tan x+2+ .4 (X 4 _1)+16 In x+l· 1303.. Put I 7 (X-I)7' 1314.~_ X 4r2 1 x2 + X + 1 I Tin xl-x+ 1 + 2 Y3 X 1296.5 +2Vx-3 Vx-6vx-31n II + Vx 1+ 6 arc tan vx. 2(1+x2)+ 1299. X5 _ I (JrXT~_1)2 1-. 3 +-4 + x- 4)3 1309. 2 (x2. 1311. -arc tan x.x+3 . 1 5 1 In I x I-SIn) x + 1 1+ 5 (xs + I)' 1312.2 IX-II 3 x 3 15x5 +40x s +33x -Sarctanx. 4 (x+ 1) (x2+ 1) +"2 ln I x+ II-Tin (Xl + 1)+4 arctan x. 2 (x-2) +2 In I x 1 Z2+ Z + 1 2 22+ 1 2z 3 1n (Z_V+Y3arctan Y3 +zl-l' + where V2x+3 2._~I.r 1 .. ~ (2InIXlx~II-~I-XI~I)' 4 X. V(ax+ b)Zl 13 x- 1 I I s Y x+l.4x +5) + 2 1n (x 2 -4x+5) + 2" arc tan(x-2). Y2 + 1 Y2 x Y"2 JI. arc tan x 2 · 2x-l 2(x2+2x+2)+ 1297.x .11- + I . x+1 1 X -2-· 1313.I)"+3(X 1315.' 1326. r lr-~- +. 1302. ---8.1300. 1317.y=-ln 2 r 5 2x + I + 5 +2Inl.Answers 422 1 6 In x-x+ . x -II· 1324. X[2 V(ax+b)S-5b I 4 (x-I)! Hint. 15 x-I +48arctanx. 1308. 2( 3" arc tan (x+ 1)-6 arc tanx 2Yx=I[(X-. 1 .r x2 -x+l+ In(2x-l+ 16 . t xarcan(x+).2 1 I +In x-I. 1319. 1310. 1 + 11)· 1306.- I .(x + 1)1 1 2 1294. "76 x V"/-x.arc tan I-xl' x! + x X In xl-x xarctan + .Y Y51' 1307.5x 5 + 3x 3 -X-- V V -x+2rlr-x-6In(l+ 11h/-)x. x x2_2x+2+21n(x2-2x+2)+3arctan(x-l). 1321.. where z=Vtanx.I 13 8 ( S 5 3 1 2.r1 . 3 32 1n tan _ .. (2x! - 1334. va arc tan 73 2z+ 1 + -S· 1337.3 . 5 sin "6 +3 sin"6.rcos X 2· s.!. 1364. 1369 sin 2ax • 4a V COSiO + x cos22b· l 1370 t cns <p · 2 sin ~2mt + cp) 1371 ~ + 4uJ. sin x. 1347. 8-32.. 2 Vtan x.4 .3 .. 1333..y x -x+l+ S lnx X (2x-I+2 . . - ~ V cos· x + ~ 1354. . f x 7 8+4x'+3x -x+l).3 . sin' x. -cot x .+10 · 1367. -'2'-- 1. 2 1338. r-z-- 1 1 2x-l . -4 In y 1+2x2 V where I (x" 4 1) Y 1 + Xl • 1335. r 1 19 2 1330. 1345. . 2 ~2>< .: ~:~I: +f In 1t ) 1358.2 tan2 4 cos "2 I i I+ I i + ~ ) I]· X [In tan 1355. - In tan ( sin 4x 16 cos4 4x + 3 sin 4x 32 cos2 4x cot 2 X 2-. ~ 1 + 21n tan 2 xl . 3x s Z == VlXi +. f 423 r-l-x.Answers +2 'If . 1365.xa) 2/. 16 x -:24 x +"6 x s) X .. 4 . tan x + .·ft' X 1342. ----00.21n I Sln x I · 1343.x' 1 Yx'-x+I).+ 1 + 1 V--x.-} arc tan V x-· + 1.8 arc sln-X • .-S-· 1350. r (1 3 ) . • 3 1336... I I+x. -2Y2 X x 1. ' x.1.. 1353. 2 . -Seas• x. 2 tt")s"3- 1362. . 4 I (2x +""4 I. 3 2 x tan! 3" + coP x 1361.-3. T .1n I sin x I· + s~::: .2 cos x. 1329. 4x4 + 8x! r x .. 3x sin 2x sin 4x 2 sin' x .r2 .4 tan 1 x· 1352. 1332.r-----==-.. 16-64+----:ta· 1346... 1339. 1340 st·n' x s·ln l x 1 x 1 1 " 2-:rcOS · -3---S-· 1 3· 4 -. -2 -.. 1368. . 1 TO In (z .3 1 X y 1 + x2 -16 1n (x+ y 1 + x2 ).. IS 4 . 1356.. cot a x cotS x tanS x 1 1349.-4. "2 tan' x+ + + -I 3 In I tan I 3 x ..2 1 .1)' z~+z+ 1 + 1 4 + 3x .. rZI Xln + l f 2+1_ .+1-1 1 1 -b. tan x "3 tans x + "'5 tanS x. Vx.3" cot' x+cot x +x. r _ arc tan~. 1357.r 5 . 1360. 1348.8" x (2 -t. T6x+i2sin6x+64 sin 12x+ 1 cotl x 2 1 144 sin a 6x.+ ~ · x sin 4x x sin 4x sin a 2x 5 1 1 1344. 1 x x x +tanla-3tan"3+31n cos "3 1359. 132. 2(x+l)2 r x +2x-"2arcSlnx+l· 1331. 1363.. . 1351.2 cot 2x. .. x -l~ V cos Xl sin 2\:1 4--8-· I i I· tan 1 5" tan Sx -x. _C 0 s8x+ Z2_ Z Y2+1 r 2 z2-1 16 cos 2x sin 25x sin 5x 3 5x x 3 x +-4-· 1366.. -cosx+3" cos s x - + 1)1. See Problem 1381. 1391. 2 2 x 1 tanh! x 1395. . 1392. . · 1401. -In Icosx-sinx. 1375. 5 .'2 I 1377. Hint.4 .Answers 424 Si '1 5x S1 f1 7x 1 I 1 tan~-2 I 2 x +2J+28· 1372. Hint.arc tan r2 1 x) (J-sln x) = 1380. 1373.. In tanh 2" +coshx. · See Problem 1381. Use the Identity 2 == lsinh x + cosh x).ll We 3 sin x put -x+tanx+secx. 1385. h . -5 See Problem 1381. 1400. arctan (I+tan tan2"-3 + 3cf's x Solution. 2 . tan x + 3 + V IJ X tant y an x HX-~lnI2sinx+ 2a-3p ==3.. 1 5-')in x -4 In -1.. In(1 +5i11 x). v--1 3 In 122 tan .- r3 ~15 arctan ( V~~n x ). I . sin~4 x • 1394.. In (cosh X)--2. Divide the numerator and denominator of the 1 (2-Std r 2-S111 \388. y-garctan 3 I or 2 ("~)l V5 arc tan eX . 2r2 tan~ 2 t x 3-S1n identity x · 1390.· + i I I 2 cotha x 1398. _ + sin: 4x ..cos x ==-1+ 1 Sl'l x-cos x . x-coth x .. .8. V2+ V . .. 1 In X 1384._In V2 2 . + 2 cos x ==a (2 sin x + 3cos x) + S x + 3.In 1387.Sill 2x 1 lr.13. 1383.I -SU1 x-cos h X (3 x tanh 2 +2) Y5 2 sInh x sinh 2x x . tan ~-5 ~ ~).arc tan Use the X 3tan~-1 2 .to 2x · 2 . Hint. .2' Hint. -2coth2x. +2 1 2 -cos x )2· 1386. 24cos6x-I6cos4x-acos2x. 1376. 1399. ~ + SIO~ 2x + sin: 4x • 1393. } arc tan ( ta~ '..-. . 4 1n tan 1374. 1397. 1396. I 51 2 r 2 2tan!--1 2 X --. We + P(2 ~1I) x + 3cos X)' Whence 12 5 ~=-13· a=T3' X (2 S2 ~i11 x X) fraction by ces2 x 1382.r.. 3a+ 2P=2 and. -x+ 21n x tan 2+1 . I-sin x +c 0 s x 2 cosh' x Use the IdentIty I + Sltl x .c.-. 3 Slf1 X + 2 Cf'S x 12 5 have 2s11lX+3cosxdx=I3Jdx-13X + 3J I~"~ v)' dx= -13 12 5 x -13 In 12 sinx+3cos x I. consequently.. arctan(tanhx). -2(1 Hint. In In tan ( i+i) \. l cosh x + Vcosh 2%).r.vG .- Hint. 1379.--.. - S111 x 2 r 3 1389. x+ c ~ " 1381. '. I· Hint.. -a--coshx. x-tan ~. 1402. Xl +x-Sln 2x+ 1+2 y xl+x I· 1409.. cos x + IO·9x + .-2 x-I '1r l "2 f 9+x -2"ln(x+ f 9+x ).r 2 y 2 y 1 + x2 -x 2 2 .r-In. r-X-3 . "2 (2 Sl1l2t 5 +l:( S 2~ Y21 1 VI + x +x Y . x 2 -6x-7- +} In (x-I + 2t+ I 1408. x . ~(lnl':x~:II-x~I)' 1437.• 1). i .. -2.. X 2 2 ' 4 X sin x n-2 1429.1n=(n_l)Cosn-lx+n_l/n-l... S4t) 4 ~ln 4X+Cf 1420 eX ( (.cos 6X-~ sin 6x ) • 2X e 1418.r . 110 =-e _1_ -x 1431. X V20x-25x l -3. I n = 2 (n - 1 I) at X [(XI+~2)II-1 + (2n. + In (I-x 2) +x2 ] -Xln(x+Vl+x 2 )+2x.5x -51: + { ) 1416.. 5 1n V x"+5· 1435.-2 9 . -21405. . r .x 2 X (x4-2x -I. X cos 3x Stfl 3x X Cf'S X SHI x 1417. 2 1&= . In=-xne-" +n1n_t. x-ln(2+e"+2."8 (2-sin2x-cos2x). 1 8c I &=8 4 -1-6-. 1410.r .. I a = 4~a X 3 xJ cos x sinn X n -1 . IX-2 V x-I 2 + 2 ~2 sin 6x + 2X 1415. eT X i...2 . l 14tt. e2x +x+I).81n I x-3+ Vx 2. 1433. x In 2 (x+ VI +x l )-2 VI +x'X (X2"-100.. 1427. I 1. ~ (XI~2+~aretan. F . 2 ' 2 +x 2 In(~+ r 2+x 2 ) .1n-"· ' 3x cos 3 sin 2x I 4 . 1434. X (3XI + 5a X [ 2a 2 (Xl + a2)2 + 2a arc tan a· 1428.r 1 1 +"2arctan(2x+I).. Yx 2-4-2In I x+ y.4 - .r ..+ 1 8 + .1 I . 9) arccos (5x-2)-1QO 5x+6 X 2 cosh x-cns x sinh x 2 ..6x-71. ] x I I . 1423. •r_arctan Vx Y2 1414. .6 .. 4Vx2-2x+5' . sin x 1424.. .Answers 425 . In = n + -n. r . 1406. 1 1413. F 1421. ~ rXl In ~ +. . 1 = 2~2 (x ~ a + ~ arctan ~ ). -"2+31nle"-I/+lfln(e"+2) 1422..cos x 5sin x -15cosxslnx-ISosx. i4 (2x + I) (8x 2 +8x + 17) Vx l + x+ 1+ 27 +128 In (2x+I+2 Yx +x+I). 21n x+2 -x+2-x+3 ~ 1436..=. . + y f Y x 2-2x+ 2) 1407.. 2x+ 10·9.2)· . 1426.2 - ir (XI l eX 1419.. 2 x-I 1412... + 10·9·8 8 l -1 X""l ~nx 2 ) I ( sin x 1 1'=2cos1x+-2 1n tan 1430.3-2x-x"+2arcsln -2-· 1404. V14 arc tan (x I X 1t ) 10 + 10x'+ 2+"4 : V2(x-l) Y7 1432. 1425..3) I 11_1] . x+l x . .. (x 2 1)1 +i In (x1 + x+ ~) + IX+31 1 I . -2-" x -2x+2+ x+l lr 1403. In V xl -2x+2-4 arc tan (x-I)..3 '+-3 tanx.t=4l· . 1 7 ' · "2 x sin x +cosx )-Slnx. 1464. Inltanx+2+ytan l x+4tanx+ll· 1474'-a x + . 3 V"2 (x 1456. tl. 3' -In Xl 4"- _- Vl-x- ..x~+1 I In -2 (V5-x-I)I-4In (1 + V5-x).3 3 IV 1 -3 In 1487.arc tan 1469.· 1468. r . "2 In I tanx + sec x 1..- 2x -t 2 y~ V 2 YtcotX)3 ... 1445.2(x+l) arc sin 1/-2.yF---2 I· 1-2 y x x x 2 2x-I + 3 . /..l. _(xt l ) V. (x 1455.~3x -3- 1470. . + -61 In (z +z + 1) . C0S 5t" 3 . 1-2x+ 1 1448.rtanx) ..4" I-xl . · 1 . {sin 2x.2 1 2x . r. 2 + 2x + 2) arc tan 1471..r_arctan . . 1461.I+2X+2-fln(x+I+Yx'+2x+2). 1 1 1475. a l +5111 2 ax). X ).i1n I x+ y' XZ -91."3 x tan 3x +9 In I cos 3x I · x sin 2x cos 2x elx ~-4---~-· 1477. xl+4x ="4 -X. arc tan (2 tan x+ 1). Y2X-: (2 . x l + 1 . 2 tanS x + -5... x -V~+l' V'(2x)'. 1443.40 sin 2 5x + 40 In . Tl 2x . . 10 2 (3 xarc tan ~\ I V3).1nlx+Jlx -1 x (xl-I' --I z X . 8+-4-+32.'Xln(xl + y 1+x 2 1452. f cosec x.x l! Y l+xl -2 1(4- 1 -8 In 1451. r. 1477. 1444. 1454..1 1 1 · 1458. Itan"25x I· 1 . . --3 In I z -11 I-x' + 1 1 2z + 1 1 + x' 5 .I · 1 x.. 6(xl-x+I)I+6xl-x+I+ x(3+2 Yi) 1 4 I 1440.1). 21 - 1 2x VI-X a . ' x 1) 1(1 1453. 5 -V cos x. :3 y 3 1442.X 8. tan x .Answers 426 I + I) 1 ( 2x -L x 1 1438. 1 2x + 1 + 2 (Xl + X + 1 2 1)' - 3 I x I 4 1 -"4 cot 4 x.Y2 x (tan I 1473.r-· .In x ..r. 1462. i + ~ ) +21n Icos ( ~ + ~ ) I· 1 2Y_ . Hint. . 4 ~an -cotx 3 cos 5x 1 3 X In (sin ax • 1463.2 'xarctan ( 1 . 10 I i) · 2 .1488.-x.r . tanl ( x - ~ (8x-l) (x-4x' +~ arc sin (8x-l).X arc sin ~-1 1450· X+4. In(x+{+YXI+X+l).r . . 1472.3x .r· 1441.6) tans x 1465.x+4' .arc tan where z= · 1459. 1448. xl 1476. 5 I2lcos! x. 4. y'(x x 1449..-20 SiU4 5x . r-2y x . (2X-I. -2 X f 3 3 x 3x Sill 2x sin 4x 1460. Inl tanxl-cotlx- 1457.1 1439. 1478 a1 eX3 • 1479.9- 2 + v·.. r. . 2) dl __l_ 5 I _ lr--.. 1503. lim sn= n I = I xdx=-2· S o I 1519. 156.2. 1518. 2x (coslnx+sinlnx). } . {Vx-xl+(x.. r . 1500.. eX -e. +~) =*(_1-1 +_1_2 I+nI+n- I+n- n+ -+<Xl I I + --2 +.+x + 2 Vxl _ 1).3 Vtex+l)I.In b· 1 09.+ -+ n n n nn n . X ( x2 _ 1 + _ +_ In 10 21n2 10 1494. _x_l_arctanx. 1) "2 ~= .I 1481. 1507. Divide the interval from x= I to x=5 on the x-axis into subintervals so that the abscissas of the points of division should form a geo1501.sln. 1517.. cos x I I 3 1512. -I t . In 2. 1 3x 1 5x I x I I I -sln---sin--. 1511. The sum sn = --I -+. 2 -X4 -Xl In a ' db ..lnj 1493... 1504. 2 2 4 4 I I + 2x 10. 1491.+ 2 cos 2· 1513. 3. 2S1n 1508.. X 2= X n =x(lqn. 1505.. + -n- may be regarded as the Integral sum of the function f(x)=x on the interval [0..~. Solution. b-a. y 1+ x 2 1495. 1515. sin x.. -2InIO X 1481 V(e x i_ ) .__1_. In4 1-2x· 1492.-8 • 2 y x x x 1516. In 2. voT -g -2" . 1510. 1486.t . Hint.)arCSin V"X. Hint. ~ [(X -2) arc tan (2x + 3) + ~ In (2x 1499. In!!.x = 2 sinh x. nletric progression: xn=-I.~- ~. Utilize the tormula r(Jq2.3. -xcothx+ 1 x I I eX -3 +lnlsinhxl. -2cosh y I-x.. lii""2. n l+cotxl-cotx. .Answers . 427 VI + 1480. xe -e . arc tan x-In (x. r e +1+ln V-- eX + I + 1 2 1 4 (x.+ Xl VI + xl). Slncosh2x. 2 2 · 3 2 10 2 + anx sinh 2 x . 1483. 1487..I) =2+"2+··· + n-I -2-=. l 5)~ + 6x + Chapter V T2 210 -1 1502. +Shlna=~ [cos~-cos(n+~)aJ. arctan-. n x. 2eX-4+4InleX-21.7" +I)1.· 1482.2 .. 1505..~ sin 5x ). Xl =xoq. Therefore. a Hint.1 ..} In I x-I 1. .arCSlnJ..1]. 1488. 2 '1r-xVex + I . x = nn (n= 1. See Problem Sil1a+Sin2a+ . 1485. 1489. 1497..2X X ln 1490. 1498. . Solution. y 1 +. 1514. . 3 X I ( -x2 cos5x+ 2 xstn5x+3xcos5x+ 5 5 + ~COS5X. + -n+n -= n+ may be regarded as the integral sum of . . The sum s" = 1 2 I (I 2 n. ). x 1x 1 • 2 I -iJ.• " I-cos x. 1506. X 1496. . q)= I/J 1 ~ f(x) dx+ ~ f (x) dx. 31. Converges. 1581. Diverges 157J.2-2 41 1585. 1552. 2+4· 9 V3 31 . 2 2 n 1 1563.Y3· 1533. 1539. 1572. 1546. 8. 1545. o 16 "3. 1582. 1529. Converges.' if p>l. I I p' if p< I. 1577. arc tan 1544. since lim f (x) x'-P= 1 = II o x~o 2 and lim (I-x) -q f (x) = I. 1523. 8 . In 2. I. p~ 1. 1 7 (k=l. 1541. ~ 1554. 16· 1532.91.· 1592. 3 = 33 "3 . e..2 o 1t 1584. 3 na2 1593. 1575...2 . B (p. sinh 211:. O.9 . diverges. tanh (In 3)-tanh (In 2)= ~ 2 3· 1537.321n 3.n y 3-.· 1540. Hint. t) l dt. "3. Diverges. 1. 1542. S dt. diverges. 9 1536. if p. 1578. Diverges. 1 ~ 7+2Y7 1 1t Sin 112.T' 1543. Hint. 1534. both integrals converge when I-p< 1 and I-q< I. 1. Diverges 1569. sinh 1= . r. Converges 1568. 1590.· y 1 + a2 n 1587.. 1521. In . 1583. 1558. In 3". n I ' (arc tan 5 I +t In a 2 In 2 8 QI) 1580. 1 l+YS 1535. if p:. 1531.. . 2 3 · 1525. 1559. I-cos 1. 1 . where f (x) =x P-' (l-x)q-. 7i (e. . 2 100 lim . Converges. 1538. p Xk=I+~n +1 .5' 1556. In a' 1561. X= (b-a) t +a. '2.4-21n3. Diverges. 1524.1] where the division poi nts have 1 the form 1520.~ ). n). 2. 3" 1550.. 2 n •1 . - verges. 1570. "4. 1557.2 Y2 Sy'tdt. ~3 + ~ · n 1 8+ '4. 35 15. % -+ 1 • 1 that is. 2. 1. 1562. 1548. 1 1560. 4 1588. S YI~Sl11lt' 1579. ••• .. + T1 ln3 1565. . ~+~ . 1576. 8"' 1564.:: l.1-. 2 1 9 7 n~ct> 4" . 4-n.1 1555. No.Answers 428 f (x) = I ~ x the function on the interval [0. wherE I xp-1e.. Converges. Diverges 1567. 3 2n Y3 · 1566.X. _1. 1589. 1551. the second when p i~ n 2 arbitrary. Diverges. 1527. The first integral converges when p>O. 1574. 1522. DIverges. In 8· 1528. when p>O and q>O. Diverges. Therefore. 1t 1/y 5 1586. I (x) = r (p) = ~ f (x) CIJ dx o + ~ f (x) dx. 1526.. arc tan 3-arc tan 2= 1 4 n 1 n 11: = arc tan 7· 1530. - 1 In 3' . 1553. Sn=51 dX+x == In2. 1547. 1571. Di· 1549. 3" In . 1632.::i.X and + ~ r(p+I)=[-xPe-xI:'+p ~ xP-'e-xdx=pr(p) (*) o tf P is a natural number.x dx = dv. Fig. 1633. 1642. 1. - 2 2 2 2 2<I<V5. 110 = 512 . 1610. In 2. 42f) 1 1602. 2 - e2 +3 t601.. 1625. (2+ ~ ). Solution. v = -e. e. (p-l)!(q-l)! 1 (m+l n + l \ . 1609. lOa' 1635.. 2al e. m2 1n 3. ~ • 11 I 2-3.. 1 1 3 . o we get: r (p+ 1)=p1 . 9<1<7' 1619. 1614'"2 ... 1631. 2n+'3 and 6n- ¥. Hint. 3MI • Hint. See Appendix VI. 1647. 1626. 2q .3·5 2. 1... Hint. 2. 1624. 1 2 =2(coshl-I)• e+e- . -8-· 1600. Whence du =px P . 8" nal . 1646. 12. i3 n <1<7 11 . tIJ 5 1604. 4 . Take account of the sign of the function.Answer" 1594.1636. Applying " o the fornlula of integration by parts.. l. n 31 2' 1599. 1638 .r- 1639.5 . 1645. 15n. we put x P = U. 1629. 1621. 8"' 1616. 1640. a- 3 s . Fig. a. a) Plus. 4 1620. r (p I) = xPe. 1 c) first.and "3 n + -. 27. 23. c) plus Hint. 1618. applying formula account that r ll ) p times and taking into ~ r(l)=~ e-xdx=l. (2k-I)~ 'f =2k b 1607 • 12k = 1. 1612. n2 O<I<a2.6 . ~ p2. (p -t-q-l)! . 1603. 3' 1613. 2 1634. : a l • 1643. 1644. 1650 t nab. al 1648. Fig. (2k+l).. 1615. 4 V3 32 4 3 n . See Appendix VI..ln= + lsano dd numer 128 6331 10 =-= :~ 15. 4 1627. 1637.r3 f 3)]. 2 B . b) second. 4 In 3.4. 1630. then. 2k 2' 1 n is an even nurn er.' . Sketch the ~raph of the integrand for values of the argument on the interval of integration 1611. 1. 4 ~ • 1641. I tk + 1 = 2 . ab [2 r 3-1n (2+ 32 3.X dx. ~ ~ </< ~. 1628. a2 ~ b 1605."2 (e1r +ll. 16 a · 1649. s= Hint. 1.4. The integrand increases monotonically. 6 . " 1608.1 dx. 2 arc Sill "3 · 1617.2 ' -2-)' Hint. . va 24.. a) First. 1623. See Appendix VI. 3 ~ HInt. a2 ~ b2 ' 1606. Put sin x=t. 2k l ' b 1. nat. b) minus. 1659. 1714. lt~IH. 171t. 17t6. 29. y== b· ± Va2 -xl • Taking the plus si~n. c) 31a • . 1722.Answers 430 1651. 1706. 2 31 • 16rra a 32 a 4 a 3 v X =2"' vv=231. V3) 1(31 . 1697. the parameter t varies within the limits 0 E:. b ) 6n aa. aJ31 4 (eI + 4 -e). See Appendix VI. 15 31a . 28. 1713. 1653. 27 (10 flO-I). 1668. &t'al . taking the minus sign. 1719. 1696. 1723. 30' 1686. 1690. we get the internal sur2 face of a torus. n.8a.1674. Here. :r 31ab · -1 88 31 4 31 • 1687. Hint. It: .· 1676. 2 (4 + In 3). e V2 + I) . "3 np . {(el+I). we get the external surface of a torus. 1656. a c) 6na (931I -16) . m · 1684. 8.lVI7'-I]. -5-' 1693. 1681. Hint.ln 2). i · 33 60 9 14 -8 F Ig. V2 a. 1679. 1671. Fig. 2a [Y2 + In (Y2 + I)]. aln. 1703· 38 31a. 1677. v'=7 2 1691. 22. = 1. n [V2 + In (I + Y2)J. where Vaa l 8= + 8 - b 2 2 e + -e a 64M 32 2 (eccentricity of ellipse).. 1658. (e-l) (e 2 +e+4). In(e + Vel-I). al. :: lthla. 4 na a• 2i 8 2 3Ul o.y'2) +It In 2 (V2 + I) • 1717. b e sin h a l Hint.Y2 + In (VI+ei-I)( 1666. 1661.nal Vpq. na YI +4n l + a 1673. (I-el)"'o' 1 '3+-2- 1665. 2a lrf +. 1670.-. -2 aT. a) 5nla'. Fig. nabh -3-' 10 31 • 1699. · 16 · 2"lt. 1682. See Appendix VI. 2n[V2+ln(V2+I)]. It (hi + 2ab) . T (el+e. n f 2. 3 It (V5 . ~ al . Fig. Fig.I + 4)=2 (2+ sinh 2). 1654. -a2 • Hint. 2) 2na l 1(b In 11 e. t ~+ 00 See Appendix VI. 1694. I6a. vx=T. 32 31aa. l l dix VI. I 663 1tp2 1662. V2 + In (I + ¥2). b) 1631' a2 . a. e2b -1 sinh b 1 2 3 · 1675. t7t5. InW-l+ a . 83 31 2• 1689. It: . In(2+ V3") 1672. VI +el . 1652. See Appendix VI. In (21t+ VI +4ltl) . 5 3 1721. HInt.• 1707. I 2nalh. 1701. h ( Ab +a B ) 3" AB + 2 +ab. 4n!ab Hint. ltabc. 1709. Vh Hint.1 (\5-16. 1667. na s v: + aVl+m 1 4 + In 3+V5 2 • 1683. Utilize the formula cosh 2 a-sinh 2 a. 2ltla·. 3 ~r- 1664. 6ltal . 16.a 2 ~ In ~. 30. :ra·. 1692. 1695. V5+1 na 2 31a 2 12 n 1718. 1704 • 1708. ltal.b =ln-. '3 16 1710. HInt. 1698. 1669.-b ) • 1678. 105 1702• 1705. 1) 231b l + 2nab arc sin e. 1720. nabh (hI'. 1685. nal. 1657. 1680. 4 I+¥). 1+ • ~r- 8 Pass to polar coordinates. For the loop. a) -3-. 1712. 4 (a:. ~6nal[5V5-8]. See Appen1655. 128 • 105 a. &tal. 0. +: a ).. r = Ya 2 _z 2 is the a n radius of a cross-section. 1746. ) is the aHi tude of the tube. The volume of each such elementary tube is dV = 2xtrh dr. 2na 2 • 1732. 1734. I b =3 a 'b. The mas~ of such an elementary layer dm = ynr 2 dz. Solution. 1726.Jla Whence z = 2 =. a . Y=r. 43t a2b Mb=y.a sin a 4 . x=3xt: 4b y -=3xt' - 9 - - - 5 x=Y=20' 1737. x=O~ a 2 + sinh 2 . where r is the radius of the tube (the distance to the axis of the cone).. We partition the cone into elementary cylindrical tubes para lie! to the axis of the cone. 1736. The mass of an elenlen tary layer (slice) is dmi=y1tQ2dz. 1731. Solution. 1739. x=1ta. x-== Y=O.. (y=I). /a=4"nab 3 . We have do = 2na dz. Q = ~ z. where dz is the al titude of a slice . 1728. 1725.a. At a dis-- tance of : altitude from the vertex of the cone. (a ) 0. J = I~ nR~Hv. R1 Solution. 2" . Yal+bl . 1~8nal. 1 1741. Ib=T1ta'b.Answers 1724. J=T5hb'.. h = H ( 1. Due to symmetry. 1745.i ty. 1729. ~ M Y ="2 r a 2 +b 2 • ab 2 M a =2. I = in (R:-R:). la=aab' . The Solution. where y is the density. y=O. concentric circles.. x=y=Sa. Solu- tion.: Zl z = I0 . x =-a. x=y=O. We partition the ring into elementary mass of each such R2 the moment of inertia J =2nS r l dr= element dm = y2Jtf dr and ~ n (R:-R:). X=1ta. 1~8nal. 1740. then the moment 01 . 1735.xtr 2h 3 dz = : h. where y is the denc. z is the distance M of the cutting plane from of the cone.' a 3 2 x=Y=a' 1730. I=na 8 • 1 1742. Divide the hemisphere into elementary spherical slices of area da by horizontal planes. Due to symmetry. Whence the vertex h n S. 1744. Mx=M Y =5 a2 . a' M x =M Y ="6. Y=3 a . To determine 2' we partition the hemisphere into elementary layers (slices) by planes parallel to the horizontal plane. We have: z = ~ (al o 2 -na 8 3 4 1 Zl) Z dz 3 =aa. a 2n ~ az dz . 1738. 2na l (2-Y2). (0. 2 is the distance of the cutting plane from the base of the hemisphere. MX=: 1727. Partition the cone into elements by planes parallel to the base. 1743.4a y 4 sinh 1 · 1733. 0. S =. 1= .--. The elementary force (force or gravity) is equal to the weight of water In the voluIne of a layer cf thlcknpss dx. the same volulne V-=2Jtx 2bf1. trom the ~ In (1 +~). Solution. dF = = ynR2 dx.a double cone obtained from rotatin f1 a trIangle about it~ bClc.( 1 . a) x=O. whl're x is the dIstance of of gravity 1752.2 2 where y is the density of the sphere. When h=«J we have ArrJ=mgR. and since the mass M = ~ naly. 1749.. is the radius of a tube. 22 -2 2 2 lowsthatJ=r. 1760. where y is the weight of unit voluml' of \vater.104 ergs.8. Hence. =_!n V o 1754. 2g 11= ~ In (II a bl). (I) v av 1 Hint. Iz is the altitude of the tnan(!le.R h) = mgh R . V==2Jtab. We partition the sphere into elementary cylindrical tubes.. 1756. Y:=3 It Hint. = kmM .a "1/ Y ': 1- a rldr =~ rra5y.Ma 1748. S=4n 211b. • --9 -4r b) x=y= lOP. h 1+1[ 1 . A-~'Jl4Y R·TM::::::079·JOI=079·IO' kgm.~ ) r'dr= yrr~. 1147. it fol. The coordinate axes are cho- sen so that the x-axis coincides with the diameter and the origin is the centre of the circle..Answers 432 R Inertia I=y S2nH (1. A=ynR'H. Then the moment of inertia I =4rray Jor . wht're x is the water level. 1761.e. ' where b is the base. The volume of the solid .H • where y is the density of the o 'Cone. c~ =]y R2H! base. the axis of which is the given diameter. 15 . '!'b1z 2 3 f. h=2a a is its altitude. it follows that kM =gR 2 • The R+h S ~~ (~ ~ $ought. By the GuldIn theo-1 - . 1758. b) x= ~ Solution. ynR~ (H -x) dx. A=l~ yRJHJ. 1753. A= x== h= ~ x ~ x It "111 wt. that IS. Solution. 1750. \\ hence =3 tiL' centre 175t. the elementary work of a force dA =-=. 1755. A= mgh . is equal to V -==-!. An elementary volume dV=2nrhdr.10 nz. MaJ. [btl-Ia-bt. Aoo=mgR.) In a~bIJ. 1159. a) x=Y=5 a .. 1. rem. ' where r is the distance from the centre R we have F=mg. c : : : .for work will have the form A = k d. where. The force acting h l+ R on a mass m is equal to F= k of the earth. Since for f= m~ . Solution. 1157. 2 C 21. K=~ R 2oo 2 =2. For an XI 2 Isothennal process. hold.><MR2W 2.000 kgtn. the Poisson law pull = Pov~. where {ronl the axis of do is an element of area. Q is the surface density 9 Q= M :rr..3. M -= -2-. o 1767.3·1O' kgm. Therefore. 99.LP £i2 .. the work x 5c. at a distance .Ll 3 gm 5 laT . 1776. Draw thex-axfs . M The frictional force of a 2JJ. The kinetic energy of a particle of the disk • Qr 2oo 2 mo 2 dK=T=-2dO. A = 800 n In 2 kgm. 633 1 hb"p kMm... bh P=T. Hint. The work pera formed by frictional forces on a ring in one complete revolution is S. Whence K=R2 MR2CJ)2 r 1 dr=--4- 3 1766. Hence Pov. For an adiabatic process. Solution. true..: ri ng of width dr. 1772.' is the distance of it rotation.!!.gf cnl.1-) = 1. 1775.Tr B . from the centre.2d. If a is the radius of the base of a shaft. Solution. R Moo 2 Moo25 dK =2JtR2 r2da .:= %. 1 (a 2_. where k ::::::: lA.-. ] C npa 0 = 8l11' 1777. the com pi ete work A = (i2 i.LP X dA = 43lJ. The amount of work required is equaf 2 .1 . 1771. is rd.8 cal. npa 8. A =3" nJ...LPa.8·10· x ergs. P =abVJlh.MR oo 2 1765. prJ =Pov o' The work perfornled in the expansion of a gas =eoel VI from volume Vo to volume VI Jrp dv = Pov o In VVIo • 1763.RI' Thus.H (the vertical component is directed upwards).. o 2 Solution. P=--6--~11..!. 1768.. 4 a 4nJ. + 1773. Solution.2) rdr== 1tp =2J. (a +2b) h 2 1769. 1762... performed moving charge el in from point XI to XI Is A =e'l x) (. a (a I) (k JS the a a C . then the pressure on uni t area of the support p ==. Q = v2nrdr = 21tp -4. is A = A:::::: 15. P = llR. to the reserve of kinettcenergy.10' T 1770.tPa. Consequently.Answers 433 The force of interaction of charges is F= eO~1 dynes. Va Solution. Q= a Sv a dY:Z::32 P abiLl 0 l • 15-1900 S 0 Hint. 1774..2 dr = 33tJ. K=20. gravitatIonal constant).. ' Pov~ dv ~ k-I Sv A= V (V [I _ k o ) k- VI J] _ Vo 4 1764.". d) strip contained bet~een the straight lines y=± I.f(x)= - Hint.= u-v =-'-2-· It remains to name the arguments u and v. 1788. y2_x2 2xy ' 1785. x <. R4 2xy x -y -2--2· 1786. including the boundaries. t 791. f(x. v)=-2-· y2_ x 2 2xy ' I and x+y=u. ) 5 f ( -2' 3 =-3.v y)=x -. whence dt =1.12 T RI: cal. z= VX 2+y2 Solution. x2 )=I+x-r. including its sides (-I ~x~ I. til x 1780. :~ = dv. Hint. e) a square formed by the segments of the straight lines x= ± 1 and y= ± I. Y ~ 0 and (2n+ I) n ~ x ~ (2n + 2) 1£. Represent the given function in tile form replace X2_ y 2 2xy ' f I( ~)= 2 by x. Solution. including _ the circle (x + y2 ~ 1). y. r+ Designate x=-2-' Y=-2-. a Mx =. i. Then f ( ~ ) = 1 +( ~ rand Z=X 1 + (~r = JI x"+y2. f (y) = = VI +y2. u-t. x and y. -1)==-2. In the identity x=l+f(VX" -I) put then x =. y ~-x when x<O). hence.ng the coordinate origin (. Solution. e. Mx = - a. including these lines (-I ~y~I). x-Y=v. f (u)= 2 =u +2u. . i) strips 2nn ~ x~ (2n + I) rt. s= ~ v. 1778. and the y-axis. Chapter VI 2 X2)X. perpendicular to 5 V2 it in the middle. Hint. b) bisector of quadrantal angle y = x. j) th~t part of the plane located above the . 1792. 1787.-.S~ (x-t)dt+ ~ x=x Jra dv =5. f (u) = u 2 + 2u. z = . z=x-I+ VY. 2 ~r 2 S="3(x+y) r 4z -l-3(x-y)2. do. and consequently. f(l. g) two strips x~2. t'. f(u. where n is an integer.Answers 434 along the large lower side of the rectangle. a) Single circle with centre at origin. f) part of the plane adjoining the x-axis and contained between the straight lines y:= ± x.I = u.x ~ y E:. u+v VI-I-x 2 x V(. x when x > 0. u-v xy. -2~y~2.. -1 <aU ~ I). 1790. f(x. f (y) = y 1 + y2. o 5(x-t) kt dt +Ax= ~ W-x 2 ). ~ [xt- ~]:+~ x= ~x (1-+) . o Use the J6ule-Lenz law.R I2. 1 1784. h) the ring contained between the circles x2 +y2=a 2 and x2+y2=2a2. Q =0. c) half- V 2 plane loeated above the straight line x+y=O (x+y > 0). 1783. u2 -uv Then 1789.= (u + 1)2 and. on the other hand. the acceleration time is t = 1779. When x= I we have the identity Vx . including these lines and excludi. -2-+ (U-V)I -2. 1781. 0. -2~y~2 and x~-2. y y I + y2 = 1· f ( t ). V=3(y2_ 1782. .c) 2~ d) e'l. az=_ ax y x2' az=~ ay 1805. m) that part of th~ plane located above the parabola y2=X and to the right of the y-axis. 1799. and when YI ~ 0. g) level lines are parap· olas y =.435 Answers par~bola Y = . Hint. d) a sphere of radius 1 with centre at the origin.¥X xy (x 2 + y 2)3!:. a) 0. b) a paraboloid of revolution. dx=-=3(x -ay). ax=(x+y)2' ay=-(X+y)2· 1803. two-sheet hyperbololds of revolution about the same axis. c) straight lines y-=CX (x =1= 0). Third. I) the entire xy-plane. 1. b) O. e) linlit does not exist. the gcneratrices of which are parallel to the straight linex+y:+-I=O. o. it follows that z does not have a lilnit as x -+ 0 and y -. one-sheet hyperboloids of revolution about the z-axis. f) the lateral surface of a quadrangular pyramid. b) all points of the straight line x==y (line of discontinuity). 0) since there is no lim z. for u < 0. b) First. Sinl Harly.. .Y -1. consider the variation of x and y along the straight lines y == kx and show that the given expression may tend to different lilnits. dz 2 az 2 az 2y az 2x 1801. 1795. . ox = - y 2' 1806 dz _ .y2 function z is discontinuous at the point (0. Indeed. a) Planes parallel to the plane x -t. ay=3(y -ax). including its faces. the level lines are straight lines parallel to the straight line x+y=O. the -t. b) hyper2 bolas xy = C (I C I ~ I). c) a cube bounded by the planes X:c:: ± I. Yx. the level lines are concentric circles with centre at the origin. \ve ~et the function <rl (x) = 2XYI x2 + y: . since for Yt :P 0 the denolninator x 2 + Y~ :P 0. Since these linliting values of the function z depend on the direction of cp. c) circles x + y2 = C2. then z -+ sin 2({). ax =yx Y - I • . 2. (PI (x) === O. 0) the family of concentric circles 21tk~x2+y2~n(2k+l) (k=O. ~= y2 ax (x 2 -1_ y2)3!2 ' y 1804 · x az ay = Yx2+ y2 (x + Y x 2+ y2)· az_ x ax. Puttin~ Y = YI == const. we ~et z == sin 2'Pt \vhence it is eviden t that if x ---+ 0 and y -+ 0 in such nlanner that q> = const (0 ~ q> ~ 231). d) the tines of discontinuity are the coordinate axes. with the exception of the coordinate origin.z --= 0. 1802. the level lines are the outlines of squares. e) a parabolic cylinder. y==O. In Item(b) pass to polar coordinates In ltcnls ~e) and ef).k) the entire xy-plane.. including the pOtnts of the y-axis and excluding the points of the parabola (x ~ 0. a) First octant (including boundary). sin q>l. c) for u > 0. b) concentric sphere~ \vith centre at origin. both families of surfaces are divided by the cone X 2+y2_ Z2=0 (u=O). 1797. CP2 (y) = 2x !J is everywhere continuous. d) second-order cone. 1796. a) Parabolasy=C-x2 (C>0). ax 2_ x2 y + y2' az x az__ y ay.x2 (x2 + Y > 0). Y. the level lines are parallel lines. a) Discontinuity at x=O. Continuous.Yx2_y2· 1 Y x2 + y2' az iJz _ ay- ay = x 2+y2· 1808. x: x~o l/-+O passing to polar coordinates (x= . the level lines are equilateral hyperbolas. which is continuous everywhere. 1793. y =~ ± 1 and z== ± 1. when x=x 1 = const. d) straight lines y = ax + C. c) a hyperbolic paraboloid. a) a plane. Sixth and Eighth octants (excludIng the boundary). y > Y x). the level lines are equilateral hyperbolas.y = . f) linlit does not exist. the function . including its surface 1794. 1798. c) line of discontinuity IS the circle X 2 +y2= 1. az 1807. h) the level lines are parabolas y = C I) the level Hnes are the circles C(t 2 +y2)=-2x. ). 1800 Hint. Froln the set of variables x. cos cp. dependin~ on the choice of k.Cx2 . n) the entire place except points of the straight lines X= 1 ~nd y =0. Check to see that the function is equal to zero over the entire x-axis and the entire y-axIs.1)=2' . 1853. °iJU -== yz (xy)Z-I..2y 2 du 1856. 1828.l. )COSX( t--. du 1845. 1834. dl =0.S1n x cos x co x . Hint. 1809. Y X ~~ 1810.. a) 1.273. df=4dx+dy.:3 u y =xY lnx.: 1848. iJh =2 (a+b). 1 z 1 '1/(2.1813. (dy cos a-dx sin a).065 cm.. 1826. dz= ~ ( dyX S111 J!. du= (x y +: ~) zdx+ X ( xy + : ) dz]. ~=_1!. Put the dilTerential of lke area of the sector equal to zero and find the differential of the radius froIn that.df=-+1 (dX-!. 1) = dx - y+ Y Xl +1y2 +Z2 (x dx + y dy + z dz). 1 az 1811. 1844. 1830. L\l =0. du = 25l5dz-3dx-4dy). iJau ==xz(xy)Z-I. 1849. iJb =2 h.2y· xy'l . uz OU iJu _ ' 1 '. 1820. dt = 2t In t tan t+ 2 ~(ttl+ l)tant+(t +1)lnt 1859 ~=-=O 1860 ~. fx (2. d~t = • 1858. 1815. (1. p=4. 1836. (x 2 +yl + Z2) .I) du . dz=3{x 2 -y)dx+3(y2_ X )d y. 1814. dz =y xY. iJy=xzxYlnz. da=. L\f=4L\x+L\y+2L\x2 . b) L\f-df=0. 1835. 2. ~ dx ) .dx+ t +x)/(l+ylnx)dy. ' . Accurate to 4 lnetres (more exactly.e X • ay X yx l ¥2x l . y z=arc tan -+<1' (x). df (1. dz=2xy 3dx+3x 2y 2dy. dz= sin 2xdx-sin 2ydy.= t In 2 t · 1857.~~.. du = yz dx +lx dy + xy dz. 1838.062.. tanY=T. I 1846. x y Y 1841. Hint. . 1839. 5)--.0)=2 . 1850.. (1. Q t x ( x ) Yu eot yy' 6.dY)J. 4. I) tana=4. Ix (1.00.062 cm.. Y iJu uy sin . - +y 1 x 3/ • 1821. a) t1f-df=8. c) 0. . 1833. iJa =2" h. i[j= 4. dz=O.(. 1851. b) 4.0)= 1. z=-2 x + 1 1 lnx+siny-'2. tan~=oo. Y • sin 1L oZ::z.1L Xl at!:::a .Answers 436 iJz .T t cos2 t · · dt· · dx .!. 2dy. tan r.2: y dZ).2y· I y I (Xl_yl)' OZ x. 1831. +2L\xl\Y+~XI~y..998. tan 1 Y=4· as 1 oS 1 iJS 1 1829. O)=f~(O.2.!. 1837.0)=2' f (I. dz= 2 1 (. g 1854. 0)=0. 75 tive to inner dimensions). 2 x 2+ y 2 (xdx+ydy). x 1843. ~ Cl11 3 (rela- em. n a -:.25 m). "'"cos JL xcoslL. ox Y2x I . iJx X X +a cot yi' ~u = (xy)Ztn(xy). 1855. and take advantage of the definition of partial derivatives. oz==xyzxy I. df (3. x2 1827. Be convinced that f~(O.JY I (xl_yl)' ~% = _ X ~ra_ cot 2y . ~l • g f 19 e' (t 1n t . r-1[ ( 1842. 1)=0.2 1/2)2 (xy 2dx-x 2y dy). 2) tana~oo. ox=yzxYlnz.dz= t8oW..2. 1847.~2) xzdy + (xY+f) In ~< = X2y:~ Z4 (y dx +x dy. ox = Yi 1812. . the area increases at a rate of 70 rnB/sec. 2 2 1891 1886. + 4xYl"uv (u. cos<p= cp === 83°37'. 1).a . _ _ 1889.ll2 4.¥· 2 " v) +x f vv (u.. _9~3. (u. . 1902. ~2 . = (x+ ~ ) f' (x y + ~). x2 1893.(x 2+ y)2 ' ax ay (Xl + y)2' a y2(x" + y)2 • a2u a2u a!z XI/ alz a2. y) ~ x 2 . . 1881. abcxy • az 2 (b x2 +a!y2)'/. 3 YIO. aal/z:=21 . 1.61 1 2 dz • ax 2 abcL/2 ~3-. we find that f. O):=n(n-l).. y. a) (2. 1864 ::=0. . z)+ 1865. 1878. +xrYf~(u. 1899.(b 2 x 2 +a 2 y 2)'/2' ay"=2 abex" 1892 a z _ 2 (y-x 2 ) .(b!x2 + a'y2)'/2 . 1861. 1884.2 . V).Ie 0). ax a y az ::z:a~yx y. v) + 4xyf uu (u. verify that f~ (x.1896. Y] Y =x [ q>'(x)lnx+-X . v X y 2u _ alu _~__ _ a"u_. d-= x x Y2' d-=-lx +x. y):z: -1. 1885./1/(0. x x az. ~+y x+y) f: 1: that for x =-= 0 and for any y." I -f-. perimeter increases at 1+2ta+3t~ y a rate of 2 m/sec. Sirnilarl y. 211 2 ] . 0). 1882. The y) <p' (X». . a"z 1_ . dz 1862 az . -a:z=2I u (u. 1879. 1. Y. 1 + t 2 + te • az 2 + 3j + 2k. • b) (0. . -a2 = . ~_ _ 2t . 0) ~ 1. a 1876. a-a = 3/ • 1894. 1875. f:y(O. 0) = -1. z)+<p' (x)f~ (x. 1). + C0S3 ~ + C(l5 V .( I ! 2 (when. . ay z:-2Yf u (U. Whence (0..3j.y. :~=Y(1-:2)f'(XY+~). 2. v). " " iJ2 z 2 " ':'1 1903. . y. 4"1 (51-3j). .\2 + y2 :. dz -slnx 1n sin x). 0) = m (m - 1). y) + "'~ (x.v)+yrYfv(u. y) -== . == fxx + 2f~l<PX +I Zl (<Px) 2 + fz<px. a-=-~+ · a-=U~ . v) " + 2 (Xl + y2) f" uv (u. v) +y I vl.=1.'x \0. 2 u (u. v)+4xYf uv (u. cosa=3". Z • 2 . = y [ . tan<p:::::8. 1897. in par'1 ticular. c) (7.1863.v). Y2 km/hr. 1873. (0. V)+4Xlfuu a2 z . 1867. cos~=-3' cosY="3' 1888. l ". f x (0. :. a x vy A -0 y vZ i~ -aZ a x -I.2- 1898. consequently. 1877.(u.<>V-I. 1874. I grad u 1=6. Hint. 0).Answers 437 az y . 13. v). ax=2xfu(u. z) ("'~ (x. O)==mn. a.. f"x (0.v)+ . 20 V5-2 68 C0S 1880. v) . v) 2 1904.0 . and (1. a-a-. _ .a' QXay. 0) -= 0 and. :i2=-a2= x y {2xy + y2) I X Y x. az . (0. a'z ax oy2= -A 2 y cos (xy) -2x sin (xy).. v). ax ay = 'v (u. 1895. 1887. a!u O~I " + 4y f"uu (u. 91. + f~ (x. v) + xYfC'v (u. ax 2. :~=f~(X.944. d ~~ ~-J '3-1 1-1 ""-0 Z2 .= O. USIng the rules of differen- tiation ~nd the definItion of a partial derivative. 1932. f. -d . 1922. a= -1. d"f (0. 0) = 'ldx" + 4dy" + 6dz 2 .)XY [(ylln" ex+ y )dx2+ \ y y x ey y II +2 (x y In ex In ey~+ In":") dx dy + (X"ln"!. . x' Y--3 +sin x+C. . The equation definlngy IS the x ay x ay ---a"y ' equation of a pair of straight lines.. X2+y2+ Z2_tx y + xy +xz + yz C. Write the condition of the total differential for the xy 1931. 1915.4 5 · 1942. y)=xq> (y)+'I' (y). u(x. . 1929.=-2'3. oZ=_I. 1930.-~) d y2 ] .w + y2e!Xf~v) dx" + 2 +2 (eYf~ +eXf~+xe2Yf:u+e x + y (1 +xy) f. 1921. x or -8. ddY =--L.'v (u. ax-y ~=x"-yz. ~= a bZx. 1916. v) dx" + 2abf. ~=6Y"-3xz-2. X dy)" + 2dx dy].. z= ~-+Y2+C. v) dx dy + b2f:v (u. df(l. d"y b4 day 3b 8x . ax a2 z ay b?z . 1928. + + expression Xdx+Ydy. --+ Y y Yx"+y"+C.438 Answers a"z 1905. ax" '" """ "..Z" + C y z x 1938. x"yz-3xy2Z -t+ 4x"y" 2x+ y+3lz +C.tV+ye2Xf~v) dx dy + + (xeYf~ +x1e"Yf: u +2xeX+Yf~v+ e"Xf.3 cos y dx dy" -l.. dz= ( . d'z = .4dx dy + ya X + 8dxdz+4dy dz. az=_c x. X (Yln~dx+xln!-dY). ox ay =2'.y 1 x x +In(x+y)+C. ~=_c:21J. ~+ 1L+~+ C. u(x. 1935.v) dy". az=zsinx-cosy. Oz I 1951. ddY = 1 Y" x or d"y (a + 1) (Xl + y") dx"= (ax-y)3 2 In''~l' 1944.. 2)=0. Hint. x y.y cos x dx 3 -3sinxdx2 dy-3cosydxdy"+xsinydy3. d"y 2y dxl=XZ . x-t..=0 1946.1 -xy -I. . xy +C. . 1941. :. 1927. v) dy".3 sin y dx" dy . fY x 1919. X [(y dx + 1918. 1937. nx XJJ -z" =xsiny-cosz cos x-y sin z (:t=1=3 1943. = dx 1948. o"z " " " " " I'" .. 1924. .+ c.. I" ay 2 =fuu (<I'y)" + 2fuv<py'l'y+fvv('I'y)"+fuq>yy+fv'l'yy· 1914. (~:. d"f(l.. 0. d'z = eX (cos y dx 3 . Yy)I' x + ay . '!1t.t=1=8 1947.. ""'" = fuu (q>x) " + 2fuv<l'x"'x + fvv ("'x) + fu<l'xx + 'v"'xx.. az 3 (xy -2") ax cos x. -d3 = . iJlz I " .- . d"u = 2 (x dy dz + y dz dx + 2 dx dy). d2 z=4cp" (t) (x dx+y dy)1 +2cp' (t)(dx' +dyl). 1936. c!lt.sin y dya). d z = (yeXf~ + e2Yf:u + 2yex +Y f. b= -1. 1949. 1926. 1925. d 2 z= (!. Yx 2+y2 -t. y ey y y 1920. A= -I.=f. d"z = a2f~u (u. . _ 1939. d2z=e XY X 1917. ay au -==... 1933. 1934. -2 In(x"+y2)+2arc tan-+C. 1923.. x3+ 2xy 2+ 3xz + y!-yz-2z + C. y)=<p(x)+'I'(y).. 5 dy 2.y sin z' 2 1950. 1945. 2) = = 6dx" + 2dx dy + 4. ax ay = Iuu CPxq>y+ fuv (<I'x'l'y +'I'x<l'y) + Ivv'l'x'i'u + fuq>xy + fv'i'xy. = dx _ JL . 1940. u=Sf(Z)dZ+C. dz dx = . 2v d 2 -(I+y)2 y. 1971.Answers 439 1953. a) 1965. . y+xy+ 3! .902. a b a b ab Y3 the FOlots x=Y3' Y=Y3 and x=-Y3' Y=-Y3. y==O.1 when x= 1.1 =0. ± 1. ab =... y= f 2. y) = yX in the neighbourhood of the point (2. Projection on the xz-plane: Hint. 1-~ -. 2011. . 0. 2013. b) 0.. f (x.r 2. 1994. + (l + ~)n :::::: I +! (ma +n~) +! 2 4 32 [(3m 2 4m) _ «2 _ 2mna~ + -4n) ~2]o 2006. u min=4 when y= I.. I).z min=-8 when x= Y2~'y=and when x= . 2ml n = . 2007. ~f(x. f(x.z) + k 2(y . 0).(u+P)_. 2 m in=0 when x=y=O.. 2000. at points of the circle (x-I)2+(y+2J 2 _. the functions: a) f (x.y)=2h+k+h2+2hk+h2k. y+k. 0).. 1998.. x+y+z= (2 1987 At the points (1. y =-1. y) = In the neighhourhood of the point (I.x .2002. + (x-I)+ (y+I)]' 2005 31 b) "' / (I +a)m V2 + (3n 1+[(. f (x + h.t -I) 2004. 2 2017. Apply Taylor's formula for Yx. b) f (x. zmax= 108 whenx=3. at the points (0.: +cy) k -t-.ah 2 + + 2bhk + ck 2 1997.2012. z = 1 2 (x . nonrigorous n18XilllUIlJ (2 = } ) at points of the circle x + y = I.0081.r. z) + 2 (h (x .2 max ==---.y)] + t (h.1) . No ex\remum.1. The line of tangency of the surface with the cylin- der projecting this surface on SOIne plane is a locus at which the tangent plane to the given surface is perpendIcular to the planf of the proJcclion 1996.y=-2. z+l)= = f (x. of which one has a maximum (zmax = 8) when x = 1. I+(y-I)+ 2 . zmin ==. the tangent planes are parallel to the xz-plane. y+ k) =ax 2+ 2bxy + cy2+ 2 (ax + by) h + 2 (b. 1999.:::Z2_ I =0 {Z:=+O! 1-0 x y -xy. x+4y+6z=±21 1986.y .r-at + + . y) = l-(x + 2)2+ 2 (x + 2) (y-l) + 3 (y_l)2.4 2 4 -p.(y-l)-B (X_I)I+ 10 (x-I) (y-l)-3 (y-I)! 2008. the other has a minimum (zm·n=-2Jwhenx=-l. Projection on the xy-plane: 3 Projection on the yz-plane: { IY=O 3x 2 \ 4+ z2 . 2019.1).!. z)=(x-l)!-t-(y-l)2+ + (z-I)I+2(x-I)(y-I)-(y-I)(z-I).V. k.!(U2 RI). 0) and (2.Answers 440 3x+4y+12z-169=O. = ± y a2 + b2 + O the xy-plane. yi 3 r 3 . f(x+h.1-4! · 2003. 0. There IS no extremum for x=y=O. 3"(2 y _ y' X +y2 x +6x y2+ y" 2001. Hint. 1991.4 z) +2 I (z . 2015. 2016. a b a b at the pOints x=Y3t y=-Y3 and x=-Yat Y=YS·2014'Zmax==1 when x=y=O. The functions mentioned in the answer are explicitly defined by the equalities . There are no points on the surface at which the tangent plane is parallel to 1983. t 2018. Hint. n .:25. zmax= Y3 whenx= I.-+ +tx-I)(y. to the yz-plane. 1985. 1). y. 2010. zmln == 0 when x= 1. y=O 2009. · + (y + I)] + [(x-I) -t- 2! (y+I)]2 + a) arctan I+a __ ~+. a) 1. The equation defines two functions.3 .x . 2 4 2 1 Umin=-a when x=-3"' Y=-3 %=1.y=2.I). x= ~ . each of these funcbons has a boundary ext-emum (z =3). y = -2. y. z= I. 2036. 2=6. zmax = for 1 x==Y=2' 2022. 3 3 p 2040. Major axis. Cube. y=2. 2b=2. b. Y == 2. I sosceles V . Y = 2. l= ~. Sides of the tnangle are -4 p. 2). . both functions have a boundary extremUlTI -4":: - i at the points of the curve 4x l -4y2-12x + 16y-33 = O. The problem reduces to finding the extrenlum of the function X 2 +y2 provided 5x 2 +8xy+5y 2=9. 2044. zmin ~ . One of the functions defined by the 2) for x =-= .Answers 441 z=3± ¥23-(x-l)2-(y+2)2 and consequently exist only inside and on the boundary of the circle (x-l)2 + (y + 2l =: 23. slTIallest value z=o for x===y-~O (boundary rTIlnlmum). Hint. 2026. 2"1 2V. 2 . 2033. 2a=6. z=.. 2043. Zmin-':-:-S for x=-l.2 . The dimensions of the abc .· V. 2035.:.y== 1 (internal minimum) and for x=O. a) Greatest value z=3 for x=O. /2 x=± V 3' Y= 3: smal1t'st value z=. umtn -=4 at the potnts (2. 1) (2. In. y==2.~a_. Greatest value z=--2. r 3 axes of the ellipsoid. y=O..42. z=2. smallest 3 V3 value n .. y'= X=:: -+. umax =9 for x=l. 2034.63 for x=2. 2025. and c are the semi- x=y=2~+ V 2V. 2020. 2037. This value is the least for the first function and is the greatest for the second. + 111 + lila •• 111 1 2 2042. llm.for x=Y=3 (Internal maximulll). y 3 ± :2 In x + nl 2x 2 + rnat a . a= Va-.3 for x=± V 3' 9n + 8+ ( 4 4 7) 7 (7 4) l y ~- y~. a) Greatest value z=---. +) .· Va. M ( . 2045. lJ m1l1 = =-9 f~r x==-I. 2 b) smallest value z=2 for x= 1. 2047. y=4.: c for x == y = 0 z -=-= ± c.bounnary minirn urn).lx --= a for x == ± a. (4 3"' 3' 3" : 3' 3' 3 . y=-2. I. y = I. 2041. 2V. zmin = 13 for x = 13' Y == 12 2024.1.V. 2039. 2032. Y= ± I. !/--= 0. 2027. 2038. V. y=-1 .2 (T 2 . 2) (I. The radius of the base ot the cylinder .lh+ m 2lJ2+ n1 aYa. minor axis. z=-2. 2 max =S for x=l. 2030. at the points of which both functions assume the value z =3.r for 3 r 3 .1. 2031. x= ± Vf ' 2046.2):--. umax=44/27 at the P011ltS 4) 4 3' 3' 3" . 2. Y3- b) greatest value z = 1 for x = ± I. Y= krt. ~+1L-t-~=3. Greatest value Z= 13 for x==2. 2V. y ==-1 (boundary maXllnUnl)~ snlallest value 1. zmJx = I3' 2 t- V2- 2 7n x == "8 for + ktt.1 for x=O. -4 p.Vtriangle.~_.for x =8" + kn.V2 3n Sn Y= 8" kn. 2028. y) of the ellipse from its centre (coordinate origin) IS equal to x 2 + y2. and -2 . the other has a equation has a rnaxim UIn (<'max mininlum (zmin == 1) for x = . y=-2 36 18 2023. y = z = 0.. 11= nl. umax=2. + nIl + nla parallelepi ped are . r 3 where a. 2021. The square of the distance of the point (x. eu be. 2. . umin =.=-2 for x-=. and a node if a < b. must lie between AI and B l . 2058. Y = 2 sin t.21 sin t -t + 2jcos t-I-3k.t1=-31. 0) IS a node. I) '1n aO :t(abC)=(~~bC )+(ad~c)+(ab~~) 2082. !' = 4j. 2076.ze· 2077. 2055. : ~. 2068. 2054. 2075.y= 2 _gx .0). whose equations. 2065. y2=:. 2057. 11 Y2730. Y= ± x. x == 2 cos t. tv=-2J . 2048. 2052. 0). -31. 2070. A pair of conjugate equilateral hyperbolas.2px.= ~1: ~. 2067. AIM =a tan a. If a === b < c.. w=-2icost-2jsint.1. ~) 1.t In 10 2073.:/. X o -l. Isolated point (0. Cusp of second kind (0. 0).2071.=l'!a. 5. 2056. 0lCOS a v 2 cos P 2051.1+ 2 r 2J.± R. x / a + + y'/. 1. Taenode (0. b) the discriminant curve y = 0 is the locus of cusps and of the envelopE of the fain i ly. then the curve does not have any singular points. d) the discriminant curve deconlposes into the straight hnes: x == 0 (locu~ 2 00 1 2 of nodes) and x=a(envelope). x = 3 cos t. th~ straight s~n~=~. 2066.=7. [or t=~. a) Straight line. (~.t1==--2. v= yf3 for any t. XY=~S. have the form xy = ± 2~ . 12 .2060. 2074.-~2= t~e point M. then A (a.Answers 442 is ~ V 2+ . 8 1 M =b tan p. 'D=2j+3k.-92079. at which SlD p v2 4 the ray passes frem one medium into the other. 2059. Origin is isolated point if a > b. its length is !. if a === b = c. 11 :1. Node (0. The problem reduces to finding the minimum v 1 cos a °2 cos p of the functien f(a. Node (0. ~)= __ a_+~providedthat a tan a+b tan~:=:c. 11 -1. r 3 (e -1). a=~. 42. Z = 3t (screw-line). provided that 11 +/ 2 +1. Obviously. then B (b. y-=.5' where R is the radius of the spkere. 'D=-2i+3k. 3)~ao+add~0. 0). 2062. !' = . none are equal. 0). y'9+4n 2 .5 I the altitude R V 2- . 2063. The isolated point (0. and c. 2081. Y = 4 sin t (ell ipse). The duration of motion cos a cos p of the ray is __ a_+ ~. b. 2069. b) parabola. 0). for t =0. 2061. 3 g 20: . line ( ~. w=--2-1-2 Y2J. 2080. 2053. tv =-21. 7 .r. -: Hint. then A (a. The channel must connect the point of the parabola with the point of 204ft. w== = -4}. 2083. fo 1t 3 Y2 . Cusp of first kind (0. c) the discriminant curve y == 0 is the locus of cusps and is not an en· velope.)=/~Rl'+/:R2+ I:R. If among the quantities a.2072. Node (0. r-3 y~r -. 2064. 8M =~. 2) ad~O.' Hint. if the axes of symmetry of the ellipses are taken as the coordinate axes. d) hyperbola.fort="2 . 2050. Find the minimum of the function f(/ 1 . 2084. rt t=4. c) ell ipse. 0). w-=2for any t for t=O. if a < b == c. a) The discri minant curve y=O is the locus Qf pOints of inflection and of the envelope of the givell family. AM =_a_. tv =. 0). it is a cusp of the first kind if a = b. 4t (t 2 + 1). 0) IS an isolated point. 0) is a cusp 2 of the first kind. Va 2u>2+11.tol x. x+U+ 4z.(tangent). (sin t + cos t)j + =-J2(Siut+Cost)l+(Slnt-cost)}). -1-. x-a cos t = y-a sin t =z-bt I Th e d·trec t·Ion cosines 0 f t 0 ..x-I y-I z-2 (tangent). -1-=-4-=12 (tangent). 2091. 2099.. 2101.10 =O (normal plane). v = I ro I..I . b sin t -beast a lnorma.0 .y-4 cipal normal). y = sin a cos rot. 0 t2 y-a-t ' • t2 tf. z:z x-I y-I z-I 31 =26= -22 (principal normal). xy-z V2-. f1 = cos a sin rot -. c) I = -2t t2 z-- x-. 'f= yet = (i-k). cos ('f-:'Z) = 2092 = -4i-f-5J-8k. x-2 y-4 z-8 x+2z-5=0 (rectifying plane). b) -1-= -1-= -4- (tangent).(pnnclpa I norma). . 2095. 'f= J-3.~ u-2 V3 = z-: 1 2 r a -2 3 (tangent). a) 4x-y-z-9=0. cos ~l = sin t.= -2}=~ 8) x-2 M 2 ( 4.= .acost . !J= · x-acost = -a sin t _y-asint __ z-bt (t t)o x-acost y-asint z-bt (b' I). x-I -6 = y-l z-1 -S-==-r . o a 0 0 w=g. a + b2 a2 + b2 f a +b The direction cosines of the principal normal are cos a l =cos t. -I-=--=t= . 2102.Answe. a) -2-= .Y-8y-z + a = 0 (osculat_ ing plane).«cos t . where 00= ~t is the angular speed of rotatien of the l :screw. 3 (tan~ent). t~ t3 Y2096. =. -1x-2 y+2 z-2 ing plane).y 2 . (binormal). x y 2 -z= 0 (normal plane). z = sin rot (circle). -t (prin- -+. 2088. 2x-z = 0 (normal plane). 2098.ro 2i cos a cos rot . ~). 2090. w=ro 2 • 2086. y+2 z-2 =-=1=-2- cos ~I= 1 Y2 ' MJ ({-.ro 2} sin a cos rot - + -w2ksinrot. t 3 +2t = I . b) 9x.oo} sin a sin rot rok cos rot.b angen. y2f' • 't R= 'II yi05 v 'Y=-}. 2 Va x+y-2 y3i=O (normal plane). -t2. -3"' 2 · 2097. cos '\'1 = O. x+y=O.== . (bJJ10rmV~~ cos a l = = z-T R _ Y2 I Yi".= .t . 6. cos ~ = ya cos . cos Y= ~r 2b l . tv = . 00 Ya + hI. wx=wy=O. c) b2 x:x-a 2 y:u + (a 2 _b 2 ) = a2 b2 (a 2 -b!). x+ +4y+12z-114=O (normal plane).18=0.wz=-g. v=VV~+Vy2 +(v x -gt)2.sin t) l + ~2(l+k).-2aoovoslnoot. -2i+k '11 = cos('Y~)=O. y-I = 0 (osculating plane).4 x-T z-2 t3 t" . cos t sin a sin t t the tangent are cos a = .i+4j+2k. 2094. x+y=O (osculatx-2 y+2 z-2 +1=-1-=-0- R x-2' R Y-7[ COSY2=Q. 2093 Y5 · k). 12x-6y+z-8=0 (oscutating plane).=O. (principal normal). 2089. X= cos a cos rot. 2100.6U+ 2z.I z-2 a x~. ~3.~ 443 rot 2085. . x=y. y) dy. b) R = Q = t=O. -I. 2x Y2 e.!L. x=3. 2121. 2115. y=2. b) J. x=3.4. o 1 = I Sdx Sf (x./'19 3 = 31 . na 2 2119. bx-z -= 0 (osculating plane).X V~ 5 . x=O. Y= O Itb l+b 2 + 3y + 19z-27 = b) K -- = T = 2a C~Shlt' av l a2 +bl 2111. 0 Sdy S f (x. x=2-y. 0 1 o 1 y)dy= IX . y=-6. 2109. . : x g' } (principal +bz = O. 2125. y=4. T. w-. ~.. y) dy. 0 y2 Sdy S f (x. p = y--. i + bk . . -j- -v f f(x. 4 1 I ~ I II I 2131.= Y14' wn =2 2117. T K=2. a) R = Q = (y~a)1 1 . y) dx= 0 f (x.t 2108.=O. tx+lt Sdx S f(x. y)dx. 9 11: 4". o I 1 Sdy Sf (x.t (P~.. 2103. 2106. 2128. . y) dx+ Sdy S f(x. 2.y 111- Sdx S o V 2- ~ x" 1 f (x.:~~)' . y=x 2 . x= I. 2118. 22 + e. y=x+2.bl + k = y . 0 2-X 1 I 0 1 Sdx Sf (x. 2124. y) dy+ . y=3.} (bmormal). 12' 2116. 2 J Sdy Sf(x. 2 I Sdy Sf(x. 2122. y=2x. 2 1 y) dx+ 7 J 2 Sdy Sf(x. 1 x=2.4. 2123. y) dy.Answers 444 (binormal). y=x. y) dy= Sdy -I IX 2 'V 2 - X2 S dx S f (x. Y= IO-x. x=-I. O. y=O. In 24 2120. w'1. y) dx. when w n =2. a) Y2. x y=O. x=3. ~ = y) dx= y)dx= 0 X 1 y. = 'V 2 normal). + _ YI . y) dy. y=xl . 2132. y) dx + S dy S o 1 VI _ yl - 1 1 J S dx Sf (x. 50. 2129. K=7 JI 14' t=l." ./19 JI 14· Chapter VII 2 25 2113. 2114. yEf T. y) dy+ 0 Sdx Sf(x. 4"3. a) K = When 2112. y=x+9. Y= Y25-x 2 . 2127. 0 y) dx = 0 I 2130. J-y o Sdx S f (x. x= I. 2107. 2126. --I-I 2 a {lx.x 2 {(x. S dy 0 VX - a 2" Y 1 ] 1X . . -Vy2-1 -I {(x. c) SdX J 2 1 - {(X.y2 Sdy 5 a I IQ f (x..x2 .1 J Yi"'=X2 -1 V4 _y2 1-~ t. aa a a 0 • Vu a . y)dy. y) dx= S dx {If S 0 x fa 5 {lX. y) dx~ a) a V. y)dy+ S dx S {(x. _ yl 1 . f(x.. y) dx +S dy S f(x. {(x. y) dx + S dy -] vJ=Yi ] . y) dy= S {(x.SdX S . y) dy +S tlxS O!!.V 4 .y)d¥+ Jl 4 V4 - - y) dx.i2 dy S {(x.y. y)dx.1/2 {(x. lJ)dll-= -. y) dx+ S dy V..lJ2 a - V.JI i . y)dx. - f(x.10 Sf (x. V CJ2 y2 S dx S . y) dx.a _ 1+ - xJ S 5dy 5 y + 2a a 5 e) S dy u Y 5dy o 0 1 o S f(x. Vi V y 2 tIL.(2 _ . _11 2 x2 S ! - V.~ .t'2 . 2137.2 _ V 4 . y) dy= S dy V CJ2 _ Sdx S {(x. y) dA. S dy 3.. 12 2138. y) dy-J- VJ:4iii 1/'J . - 2135.lP y).x2 -] y..l/~ S 1 - 2 'V e - 2 2134.V 9 V e . y) dy= SdY S {(x.y2 S VU~-Jay {(x.l Y 1 5dx 5{(x.y)dy+ S dx Ye-x 2 a J! 1 + ! _ Ve . y)dx+ S dy 1/4 •• IJ2 1/4 .x 2 .1 2 S dy S 1 f (x.y4 {(x. _ y) dx+ II'.J/J 0 ..gldy+ .Y2"'=1 S 1 0 x2 - - {(x.y) dy+ S dx S f(x.r 1 2133. y)dy= f (X.a + X2 11 2 y2 5 + S dy S f (x. y) dx+ 2 Va I 0 JL a 2 .Answers - V4="X2 1 1 445 V1=Xi - V~ . y) dy f 1 0 {(x. S dx S t SdX S 2 JI"5 • t JI e - V.SdY S _.X I I - V a2 . x.1.X2 S 2 - - 1 {(x.-:-i a 2136. d) x -J VI4'.. a l a +5dx 5f (x. 1 r 1 + . .. y)dx. S dx S {(x. . y) dx + t. y) dx.-:-xi _ . y)dy= S dy .1 b) {(x.a 2 . 1 {(x.y)dy+ lJ2 - lJ2 dy J o 1/ I - 5 5 5 S dy J {(x. x. 4. . 2118. T Bpcx:' K = 2. y) dx+l'"f dy YIY' o 1 + S dx o 1 Y2 - S ~ . -V:. y) dx= S dx S f(x. 12 9 4 . y=x+9. T· 2108. 0 0 y" 0 • 2- 1 y) dx = IX + I 2 X S f(x. 2117.t 3 (y + a)2 t = 0. .] 2 2 S dx S f (x. y) dx. y=O.. 2 2128.1. 50.. a) Y2.. 2120.a .y)dY+ yl . Y=3.Answers 444 (binormal).1. x=O. y=2x. =SdXSf(X. Chapter VII 2 2113. x=3. p= .= 0. 2127. 0 o 1 o I S dy S f (x. y=x2 . 25 1"24· x=3. 2103. 1 x=. y=-6. y=x+2. l' = Y= O 2111. b) 1 b) K = T wn rl+b -- + 3y + 19z-27 = O. dy = S dy -1 Ix l - X -{ + f (x. I · w-. 2132. x=2-y. when } (principal normal). y=x. y=2.JL 4 2 I 2 7 2 =SdySf(X. bX-J - O.y _ y X· 1 f (x. 0 (osculating plane). o I 2 SdX S f(x. 2125. x 2123. S . y.-j- -V f . x=3. x=l. x + bz = = 2a cosh1t· 2109. 1 1 0 0 X 1 2-y 1 S dy S f (x. \' = J. n 2115.4. y=4. 2x + Y2 e. y) dx= Sdx YIX'f(x. b) R = Q = When t= I. rl+b yEf 1(=+ 2106..y)dx. 2122. 2 _ ~ • f(x.t = 3 (pC + 2xl )' Q = .} (binormal). y=x l .14' e. yl. y) dx= S dx S f(x.y)dy= 2130. 2124.y)dy+SdX 0 0 y) dy. y) dy.y)dy. 2121. wn =2 -V~. S dy S f (x. 4 I Sdy S f(x. y) dy. 2 1 2131. 2114.Ir + bk 2 . 0 2X 1 . na2 T. 43". y=O. n 6. a) R = av l a2 +bi . x=2.. Y= Y25-x 2. . y)dx+SdySf(X. y)dX+SdY S f(x. 2112.2. 2129. 2126. 2116. a) K = 2107. y=10-x.:~' .-r bi + k2 .. =2. x=l. 2119. x=4. w~= . 2. g) dg + S . 2137. - 2 - I 5 dx 5 dx S 2 + JI 8 - 11a 2 . J x + Vt:4ij2 2 - S y" _ y2 _ f (x..a _ Va I f (x.lJ 2 11 2 a) y) dg = S dy S . y) dg + S tlxS 1(x..V. Vii'2=J - 1 r 1 + x2 _ S S dg 114. S dx S f (x. y) dx. g) dy = f(x. g) dy= S dg . JI i JI. 48 II 2136. c) S dx y~ - 0 1 - x2 {(x. g) dx. g) dx: 1 {(x. y) dx+ S dy _.1L a a 12 o f(x..'V4=X2 445 - y1. .. 2 2138. y) tU~ S .Answers Y~ -I 2133.-:-xi _ . ttL y) dx+ V.V al . y)dU. gl dg= SdY S f (x.V9 a f (x. S dy S f (x. y) dx.'V.'/~ I-V~ - 2 a 1/). Jli JI y2 .x 2 a S dy . lJ2 _ 1 + f(x.!!...lJ2 1 b) f (x. g) dx + dy 2ay Ya 2 . _ yl S 1 1 J - e) 5dy o y+ 2a a X I I - Y 0 0 f (x.y2 I 'lx. . g) dx + S dg 2 -1 f (x.x 2 - f (x. a S S .1 . --1-1 fa aa a a S f (x. - S dy _1/ 2 - 11.1/ 2 f(x. 0 2 S f (x. S dx S y2 -V'y2-l 1 S f (x.1 . y) dx. lJ)dll S 11 x -- - 5f (x.. g)dg+ y~ 2 -Y~ 1 x2 2 _ J - 2 S dg - 2134." .g)dg-J.g) dg+ V'" _y2 1 ~-r 1 'V" . S V . J dx .la f (x. y) dx. d) YY2="i fIX.1 1- S 'V It _ - 2 Ye-x 2 I t Y8="Xi 'V r J-Y1='Yi '1 2 S VI 1 dx J f (x.gldg+ t'2 J! I + x 2 +S x2 - S 1 2 _ y2 Sdg S f(x. g) dx+ S dg 11" - 5 5 5 - V" _. S dy VUl S Vu~ - a 0 • 1 la x . .. y) dy + S dx S f (x. S dy o ~ { 0 0 I f . y) dg -_ V. y) dy = S -I X a:: x~ 1/ 1 S dx x 0 VX 1 dy S '(x.E. y) dx.a 1/ 2 = 2135. y) dx = S dx S f (x. 5 v"5dx - + 5dX dx - - S 5 5 f(x. y)dx+ S dg y2 V a 2 ..!!. O. y) dx + S dy S f (x.!!. 0 f (x.!I)d~ + JI" _ y2 _ f (x. arc sin y J 5dy 5 2144. ~ a'V a 2 2 Jla. 2151.: S I(x. 2145. 2a 1(x. dlp o cos q> S rl (r 0 2 ) dr. -21- 2153. 1- 0 1 5 + I 1(x. 2160. o a 2140. y) dx. the preceding one by the substitution x= R (t-SiD t). 2147.16 150 3". 2150. where the last integral Is obtained from 0 . 2143.y)dx. r sin '1') dr+ o n 1 '2 sin q> +S dlp n T 5 0 . 3t 4 2159•. ~ln i· 2146. Vl-(X -2)2 1 2156. In 2 I Y2 P5 . 2158. y)dy. u 5dx S 1(x. Hint. . dy I(x. 4 R 80· "61 .y)dY+ . 0 5 5 'V -..446 Answers aY""'I a I a a S dy S1 (x. 21 5.!.:- 0 'V'- dxSI(X. (I - 4 b) 15n . i· y Y=f (t') 5 5 dx y dy= 0 cos t) 5 y dy. c ) 2 2 5 · XYdy=i· ~ a Y2a. 2 a) 1 2. 2161. -I Y"-. 5dx S 2154.:!. 2157. y 0 2 n ..2- 5 5 f(x. y) dx + 5 dy S 2139. y) dx+ Sdy S dy o f (X. S a+V a 2 _ y 2 ~ 0 4a ta J 'V 1 . y)dy+ 2142. y)dy. 2148.-::y. y) dx+ S dy S 1 (x. y) dx. 2 5 4. VI 0 f (x. _y2 JI"ia 2a I 1(x. 6. y) dy+ SdX 0 x 0 dx 0 RV"I Y"i'=X2 1/ R2 _ y2 . r/(rcos<p. 55 R 13t 5 R (I-cos t) dt o i-..x 2 o S dx S 2141. 0 231R ydxdy= (8) = ~ a. 2155. 8 ~nRI. o arc 2149. rsluijl)dr. 1 5 S rl cos q> o dq> (r cos '1'. a- a-l'a 2 a S ~ y) dx. 5 5 dcp n rf (r cos cp. u I=~ (~ du ~t f (U~V. sin cp uta'V~ rf (r cos cp. r sin q» dr. (~+ i ) 2166. u=-v. whence u=O (since 1. Hint.. ces 2 cp o an 2164. 0.. 2165. We define the limits u as functions of v: when x=O. 0 4. U = . 441 '31 1 n 4. the Jacobian is 1= u. I-V S dv S f (u - a uv.c1 . for y=px we find +a 2173. 2rr.. 2167.. The equation of the curve . r sin cp) dr + 5d cp S an - 0 rt (r cos cp.v) and y === uv... After change of variables. The limits of integration are c f\ l+fJ o. the equa-2-' tions of the sides of the square will be u=v. 32 Y2 -20) ~ ( ~_16 3 9 2 • 2170. + ~ d< + ~(~ dv t> 2 -V 5f II (U 1 v. Solution.. 0). 0 ~ . u-v=2. We 0 l+a have x = u (1. III E. whence v=-1 a .v i:. 2169. it follo\\s that llv=au (I-v). 2174. u (l-v)=O.. r E..]· Hint. sin cp 4. 2172. nab. -v of variation of v: Linlits since y=ax. 5f 2163. ab [(~. uv) u duo Solution. The Jacobian is 1= ubr. r sin cp) dr. an 4) 0 ~ S dcp S n S r dr+ (tan lp) dcp sin cp cp n aY~ 4. u2 V) dv + U2V) du + (u+V u2 V) dU. 2171. ai.. I.~)arc tan ~~l +::]. 31: a cos cp ! 5dcp 5 2168.. u+v=2.2 S r dr + Sf (tan cp) dcp S r dr. 2V) dV] = ~ ( _~ dv I 2 f (U ~ v.Answers 2162. 1 ~ sin cp +S f (tan cp) dcp -n" 0 C8. when x=c.. 5 .r2217. /~ a~ ak r arc tan bii 10 Ihe ftrst 5 SS dx dy = 4 Quadrant: (S) ~171 • •) 4~. b 1 . . b) Y. r= cosl cp. Hint.178 • · ~183. 2193. Change the variables x 5 x-2y = u. '34 real (r.}. .1=u. I 8a2 • 2220. 217 7. t 0 J& 2179 · n I H nt. It b2 a2 follows that hI cos 2 q> -Ji2 sin l q> ~ 0.z ~I_a. 2205. Vy abr dr. RI na 3 l 4 3 nabc . 2207.ahe upper limit..448 Answers r4 . o 17 2 8. 4 (m-n) RI. 2203..:: slna cp). Since r must be real. 2221. SdYSO-X) t!x= SdX 0 . 2 aya. + ~: )'.rf 3 -5). Sallfe alna-. 2187. -2-(2 r 2 -J). SdX S dt/. n.' b) -l<x<. 2197. Change 2206. Due to symmetry of the region of integration relative 4"1 of the entire integral. 34 nal (2 . 2213. i· 0 mt as a- l i b "3(b-a)(~-a). Pus to polar coordinates.. 2194. 6. ~ =v. 3xab 3 ). 2196. Pass to . 2218. r' (:: cosacp. {naa• 2184. 2182. Integrafe in the uz-plane. 3(~-a)lna.X)dy.. Sdy o . 2204.r- "3 2180. 2198. 18 . 4:t 3.-. Hint..a. -2.r-. 2201. "'3 nal (3 . 2186..2 1 (2 +4n) a.-1 ].-::. whence the lower limit for r will be 0 and Y :: . the (6 xy=u. 16 . whence for the first quadrantal angle ak we have tan q> ~ bh . 3 -I). 2208. t y 10 "3 a I . tll8.4Q. 2215.f 3 . JUI(l-'- tllnt. 0= f :rra 2 [( 2219. v = f 15. b+4.a2 • Hint. r-2 -1). -2-.. we can compute . _ ki sln2 cp S dcp Yg 2 b2 cP - COS2 0 V II t hi 48 V6 88 al abc "3. • ) 29. 2210. na' (a-~). variables lr-. 0 2 IOn.:: sin l cp. 4a l . 2195. 0 S dx+ S dy S dx.l. 32 "9 a3 • va2 V2 lr-2 . na' T· 2209. T· 2202.· 2199. 3M2 • Hint. 2211. 2216.rf 2 -1). 7a 120· 2181. 2185. (' n 1) 3\ 4+2 . confining ourselves 10 lhe axel. ~ Valbl+blcJ+clal Y2 2214. 105· 2200.2212. where k is the proportional ity factor. 2231. 449 aaa. 32 'Vax tJ 8 5 aC• 2234. Placi ng the coordinate origin at the vertex. l=~ ka 5 [7 Y 2 +3 In (Y2 + I)).x 2 -a l-X-y I-X dy (x. ~ a' 2222. 1- 5 5 dx G i. x=y is equal to d=V-2~ and is found by means of the normal equation of the straight line. . ~==5. y the answer.. 2236. The mOlllent of inertia i5 detennlned relative to the x-axis Passing to polar coordinates.polar coordinates. Hint. Ix=4 2232. 2224 31 5 2~ sin a 2 Y 6(4-31)· 2228. \ dx R 2241. Hint. Pass to polar coordinates. For the variables of infeiratlon take Land 1 II (see Problem 2156). . -Y r a -x 2 o l 0 0 Integrate by parts. 2240. 2nfJR a'b a2b~ 12-312 Pass to polar coordInates 2225. Hint. 161n 2-9 2235. y. y) from the straight lin. x=1fa. Hint. 10 =2' 2239. 2226 . y=O. x=3(4_n). 2230.) d•. Jl'RI_X:il dx b 5 __ -a-JI'a 2 -xl dy I (x. -3-. Hint. f (Je. 2229. and then change the variable x= a ~ 3 sin to lransform 31 (b Yb 2 +cl -a a l +c1+c l In b+ ~~) . we direcf the coordillate axes along the sides of the square. b) IX==64 (DC-dt ). 12nac. .) da .) d. 1/. -JI'ax Th~ distance of the point (x. (y+a)1 dg. 8a l arc tan Answers a ~ 0'=5' dx5! y a -x a dy i-Sa 51 arc sin 2 . r~~. Y52 Hint. 2T' 2227. 2242. x==3<l. the distance froln which is proportional to the density of the lamina. n n y =0. H 5 5 5f 5 5 S 0 0 ' dx dll -R -V'RI-x1 b _ _0 y (J G a2 . Hint. and a 2223.. 4 a + al+cl 2 . a) I O Z::: (D4_d 4 ). II. y=O.12. \\'e have n n a sec cp t I" = J J kr (r sin <p)! r dr d<p a cosec cp 2 +~ dq> ~ C nac 35 2238. 4 't' 2na 1 V2 3 . -4-· . z) dz. 3 3 SIr 22. 2270. Pass to cylindrical coordinates. 9 32 2h ga. the axis of the cone. 15 (31 2248. Solution.'VI 8 2244. for the plane of the base of the cylinder. x2 +/p 3t --25 dm 5na 2 'V 2QX - 2a 2249. Hint. n 2a 59 2250.'3 [1= la rdr 5 dh = 0 4 ( 60 -:rna. 2252. . The nlass of an element of volume dm=Qr 2 cos". "52 a. 4 226. The vertex of the cone is tal<en for the coordinate origin and its axis is the z-axis. The component of attraction. 8 2255. S dy . - rcabc. If we introduce spherical coordinates. "9 - a.16· =25 dx o X2 S dy la cos q> 2 't' o S s ( . and the equation of the plane of the base will be r =--/!:.. z= - 22 65. to spherical coordinates. ) T Hint. Hint. Sin '" From the symmetry it follows that the resulting stress is directed along the z-axis. 51£11 f (x. we have for points of the surface of the cone: r =f (11.2 S dz=2 5dq> S ~9 n. the xy-plane The moment of inertia is computed about the x-axis. and the square of the distance of the element r dq> dr dz fronl the x-axis is equal to r2 sin 2 <p+z2. y.2 xtabc' I nR. e u Ion.z). RI 2253. YI-x2 _y2 Yi'='Xi I 2243. a be (a+ b -I-c. n~~al (2h Z +3a2). r . the square of the distance of an elelnent r dip dr dz from the x-axis is equal to r a sin a <p +za.. 4 2263. 10. along the z-axis.. 0 S(2aCOSljl)4 dm=!nal. 97) -"6 · n Ra· 2254 · Y2 4xt -3-· 2245. = 0. z =0.!a cos cp 2a 4 ="3' 4) · n. 2268. by this element of unit mass lying at the pOint 0 is equal to k ~m r sin "I' = kQ sIn". Hint.b2 ) · 22 66 · 24 spherical coordinates. the equation of the lateral surface of the cone will be '" = ~2 - a. x = 0. r - 2 0 3t 0 =. rf 2 -27 f 3). The base of the cone is taken for the xy-plane. 2269. dcp d". I r'dr 2a 0 2267. 18 f 3 xt 2258. After passing to cylindrical coordinates."3 2261. y =0. 0 . + 12 4 bc. 2 na h 12 (3a 2 +4h 2 ). The filoment of inertia is computed about the x-axis. 2rtkQh(l-cosa). For the axis of the cylinder we t~ke the z-axis. where Q is the density. \vhere k is the proportionality factor and Q is the density. for the z-axis.Answers 450 Sdx S -x2 -1 5 2"1 In 2. Introduce ab (6c 2 -a 2 . Hint. Pass 0 aI (3jt-). dr. 15 2257. y x 8 r8 22 56 .. 480 1 2247. cos '1' d'1' dq> dr. Passing to cylindrical coordinates. xth -4-· 4 31 RI . 720 · :llal 8· 2246. 2262. 2251. 2271. 22 59. n 2 a.Answers n '2 . 0 0 0 2272. ex> 4 ynRI:. where the eliminated region is a circle 01 (8 ) radius e with c~ntre at the origin. arc cot > a If ° 2283.cos (rz) = .P R2 -t"' twice. 2284. 4a 2 HInt. b) _1_ for p p-a _. Con. a 2280. z) with origin at the centre of the sphere and with the z-axis passing through a material point whose mass we assume equal to m. We introduce cylindrical coordinates (Q. Converges.!.z dF = ..= 1 . arc tan--arctan-.. Hint. it follows that But since "3 kMm F=~. Hint. 2287. 2273. Pass to ° ~z • 2289. 2290. Eliminate from S the coordinate origin together with its e-neighbourhood.P P =. - • Se y2 -XJ 2 dy _e...~. a) 2276. (p> 0) 2278.n. ° 2286.. Solution. -2 In(l+a). DilJerentiate p8 ~ a m 2279.-Z)2 be the distance from the element of volume dv to the mass m. P (p> 0). We denote by ~ the distance of this point from the centre of the sphere. 1.:: M. The projection of this force on the z-axis is kmydv A E. n 4 1£. d) 2.=. Converges for a > 1.a 2n The resulting attraction is equal to 451 5 5 dlp h cosec 11' 5 d'ljl kQ sin 'Ijl cos 'Ijl dr.!.I..X3 .. In ~. c) 2277. that is. r r Whence R 231 F=-kmy5 d lp YRJ . consider II = ~ ~ In V Xl + yZ dx dy. The attractive force of the element of volume dv of the sphere and the material point m is directed along r and is numerically equal to -kym ~. x 2275.Q dq> dQ dz.. z~ RZ PI-I-' 1 2" ° 2288. where y= ~ is the r -1tR8 3 density of the sphere and dv = Q dcp dQ dz is the elelnent of volume. cp. -. Sur8-. 2291. ut: 1.0 2 XdY round the straight line g=x with a narrow strip and put S\~ Z (5) V (x-g) = . Solution. 2285. -3.ktny _. Let r= Y QI+(. we have 1 S eSr Inr dr= 5[r~ In r /.zJ 5 S(~-z)dz -R Qr~Q=kmy4n:RI~~. Pa5sing to polar coordinates.pl dt o > 0).!..!. m 2282.oerges. ~"it. n r J (p e. -~ eSrdr] dlp o dlp= 2n: (~-~ In e-f)· 0 Whence lim Ii =. polar coordinates. Use the parametric equations of a circle. 2313. S = ~ x ds.40 O. Solution.xl that connects the c 8 points (0. b) abc-I. with base. and with altitudes equal to the values ·of the integrand. O. Y a2+b 2 -- 2 f. a 2 2 231a. c) eX-Y(x+y)+C. 2335.. 2 . Inlx+yl+C.r aa b-b arc sin YQi=bi) . t)=1!-. Y7-.V 1 + b2• 2322. 2 + V"-:-2-+-4-n b-2). n: + b) xl -x2y xy2_ yl C. 2337. + x2+ 3xy-2y 2 C. . -nR2cos2a 2326. a 2 2 2.6). Hint. y) ds may be Interpreted geometrically as the area of a cylindrical sure face with generatrix parallel to the z-axis. 2 b) 0. 2338. ds= 11t ds= 1 dy=O ec s c n) 2334. Hint. f a l • Hint. 2331. 1 54 (56 for 2Cl6 a 15 a . a) 62. - + 2329. c) 5 ~. 2318. 2336. a 2306. (5) b) 2nn. then cos (X. 1 cos (x. 2301. 2300. Put . 2332. (ia. a) (i+ n 1~2) RI.!] ¥a -1).. d) 1+Y2. Therefore. Hint In Case (b). %2 ~ q> (x) dx + e) In (x+ y). 2302. Green's formula is not applicable. 1= t . 2320. 0) and (4. 2319. 2297. 2312. r27 (10 y 10-1). d) O. Y(at+bl)I' ~. 2303.S hence. V2. 2317. 2324. 2309. 28. ab (a l + ab + b 3 (a + b) 2295. b) 12. where S is the area bounded by the contour C. I f we consider that the direction of the tangent coincides with that of positive circulation of the con1our. 19 30 . with centre at the coordinate origin 2333. b) I. ~ ab 2 • 2316. c) a) 8. 2315. -2na 2 • d) -4. = SSy2 dx dy. a) 0. -4. C){+ln2. 6nat • 2339.Answers 452 x-e I == lim 8-+0 > "2 . l ab dy • V(X-y)2 J Y-52 + 3 .{ . nab. a) Y3. c) 21rb -2 kMmb 2310. where C is the arc OA of the parabola y= ~. 6-+oJ 0 3 I [ . 2 +b . 2327. 2311. 2328. 2314. 2304.2299. Hint. the contour of integra tion. -2na(a+b). Green's formula is used in the region between the contour C and a circle of sufficiently small radiuc. l y (n Ya2+4nb2+~ln 2na Ya 2 +b 2 • 2308. d) 2325. a)- 2307. 16 . In 2294. d) ~. 2323. -2 sin 2. Y2. 2305. (1 +411 2)1 -1 r x+~ 0 2293. O. In all cases 4. a) -20. (b + . YI+a 2_ YI_- . 2330. -2n. l ) • 2296. ~ f (x. ~ 1 + linl r dx (X-y)1 o a 1 Sdx Jr V dy arc tan 21Tb a . ~ nat.. n) = = cos (Y. f) XI liz +~""(Y)dY. e) 4. Y5m 1+ m a" Converges 2292. 2298. 2 -lla2• 2361. 2376. work. . y 2392. yo. O. rot r= 0. 2354. U =1:.r 2343. 2357. rr J J (x + y +z) dx dy dz. O.. ~l ~ r a2 + b2 +C2 k k 8 c) potential.. 2 + iJ 2 +OZ2. 2391... FR.0 (r) ex r. ~ 2. a) kllR2H (3R I +2H 2).1. where k is a proportionality factor. • iJ2U o"U 02U dlV grad U = iJ>. b) - 2358. r). rot (re) = rx.-=/gradU/whena=b=c. z=c. where m = ook 2387. where is a unit vector parallel to the axis of rotation. work. the level surfaces are planes perpendicular to iJU au a. a 00.r. When calculatine . mg (Z. 2340.-2 2 ). y= (R + r) sin t-r sin R +r t. 2372. 2 V3a" + b 2na2 2348. as 2359. ) na 2 b2 2363. b) y potential. div F=O at aU points except the origin. rr (cos a + cos p + JJ • · 2355. -al . where t is a para meter. 2 2 iJ U) Jr~·5 (dox2U + iJ2U oyl + OZ2 dxdydz. rot grad U = O. 2 4 nat 2351. -2- I 5" na. X2+y2=C~.. 411. 31tR 2 H. rot for I J. ~ 2345. 1 oP oR oQ iJz = ox' ox 2363. 2393. V5 + 1 V2 n (y-. 11 (R . r b) 2r. O. Circles.r l ). (5) + cos yl dS. b) ~nR2H (R2+2H2). ~: = ~~ .h 2353. c). c . 2347. (a l -b2 ). n (R + r) (R +2r). The flux is equal to -4nm. 6nR2 for R=r Hint. 2 2366.2 I' (r) (c. 2344. 4 3" nabc.r) (R . The equation of the hypocycloid is obtained from the equation of the corresponding epicycloid (see Problem 2341) by replacing r by . grad (er) the vector e. 2342. div!'::. r ' 2385. 3 M •.\J S 2 . a) div r=3. U =mgz. 12 2360. .2r). dlV 2 a=. 2U x= y=z. a) !.-. 2382. r 2386. \ grad U (A) 1= Y99 =3 I' (r) !. 2362. U=-T(X 2+y2+ Z2). 2341. I rot 'D = 2m. r d) 2378. r the radius of a sta tionary circle i drawn to the point of tangency. mg (Zl-ZI). a) 0.2 . iJP = iJy· 2361._ ) a. b) div (re) -= ~ .. ar= =-'-' 2 .f (r) + f (r). Z2=xy. cylinders. ens ('. c) div (f (r) c) = r = - = . 3a4 3 4 2352.. •• 2371.. The equation of an epicycloid is of the form x=(R+r)cost_ r cos R +r t. 2388. Hint.. 25 2 2 2349.~ . Spheres. . (V) (V) 2365.Answers 453" 2 y = tx. work. where t is the angle of turn 01 . 2383. grad U (A)=91-3j-3k. a) Potential. llR2 for r = ~ Hint. Cones. r . 2367. 2oon°. 10 5 5 2356. 2" (R 2. 2373. (I (r) e) = I' . 2350. 2346. 2380.:: 0. 2377. r) 2379· iJU or au -iJl =0 = e. . rr V d1C +dyy2dz+Z2' . Diverges.] 2 < all ri · ~ 1 +( ~ ) n n+l I • (n ~ 1 + ··. 2499. 2447. 2397. Diverges. (2n-l) (3n-2)· 1 n2' n +2 2405. converges. 2432. use the Ostrogradsky-Gauss theorem. 2434. 2476. 2420. Converges. Converges. Converges. 2454. 2425. 2419. 2439. Chapter VIII 1 1 11 2402. . c) diverges. Diverges. Converges. 2477. Diverges. 2422. Diverges. r. Converges. 2421. a) 2398. 2491.1 < 120' I R. 2484. Diverges. b) converges absolutely. 2488. Converges.. Converges. 2n-1 · 2401. Converges conditionally. 2404. 2410. 2431. Diverges. 2502. 2460. 2467. 2463. 2489. 2456. Converges. U = Srf (r) dr. No. Converges absolutely. Converges. Converges absolutely 2473. Diverges. 2430. Converges. 2409. Converges. Converges. 2482. Converges conditionally. 2433. 2478. 2436. Converges absolutely. 2437. In exaillpies (a) and (d) 00 ·consider the series ~ (a 2k-l +a2k) and in examples (b) and (c) investigate k=1 00 00 -separately the series ~ a2k-l and ~ a2 k' k=1 2485. Diverges. 2441. 2429.>O. L + L2n n=1 1 1 . 2426. 2423. Converges. 2450. 2494. Diverges.Answers 454 the flux. 2465. < 720' R4. Converges. Converges. R. 2462. 2471. 2452. 2500. Converges absolutely. 2481. 2458. 2394. 2496. 2490. 2461. Yes. Converges. Converges. 2475. Diverges. 2418. Yes. 2448. Converges. Diverges. Converges. (n 2n + 1)2 • 2406.7 2408. The remainder of the series may be evaluated by means of the sum of 1 a geometric progression exceeding this remainder: Rn = an [2 _1_ +(i ) (n + 1)1(It +2) +··. Diverges. Converges conditionally. I R4. Diverges. Diverges. 2n1hl • 2395. Converges. Converges absolutely. 2455. 2451. 2497. "-Diverges. 2459. Con- verges conditionally. Converges absolutely. Converges. -~R· r '2396. 2438. Converges. Converges absolutely. 2n · 2403. 2493. Diverges. 2492. (IJ 1 n=1 (2n-l).4. Diverges. 2445. Converges absolutely. Converges. a) Diverges. Converges. Diverges. 2424. . Diverges. 2400. 2474. Converges absolutely. 2480. d) converges conditionally. Converges. 1·3·5 1. (_I)n+l. R n < 2n+ 1 =2 n (2n 1) n! Hint. Diverges.. 2442. c) U=xy+xz+yz+C. 2427. polential. an . Converges. Hint. Diverges. 2466. 2449. n+ 1 n(-I) 2416.<O. 2440. Diverges. 2495. 2n-l· 1 2407. 2486. Converges conditionally. 2444. all+ 1 > I an 2470. Converges. Diverges. 3n + 2 . Converges. converges. No b) U =xyz +c. Diverges. 2417. n(n+l). 2453. Converges conditionally. Converges. Diverges. 2487. Converges. 2468. Converges k=1 absolutely. Converges. 2479. Converges. 2428. Hint. 2443. an · 1 + "2501. Diverges. 2435. 2472. 2457. Converges.] · )1 + . 2464. Con(IJ 1 (_I)n verges absolutely. 2446. . Converges absolutely for x > 1. Hint. 2512.-1. (f yn+z+<n+2) ({-yn+4+ . -oo<x<oo. and when x > 0 the series with general term ).. < x < ~ of x. \ve find the sunl of 2507... 2517. Rn R < (n+ n+2 I) (n+ 1)1 = __1 n RIo < I 3·10-'. x~l.. and cos nx does 110t tend to zero as n ---+ 00. -_ r n _ (!)2n _(-} 4 -t -n Fronl this \ve 1- r. 2524. diverges for x ~ O. _ (n+ 1)2+ (n+2)2+ . ± 2. x < -I.. 11 =0. = n + 1 ' R - < n I+ I) + n (n + (n + I)1(n + 2) + . 2519. diverges for x~O... ~x. 2508.. Converges absolutely for x :1= O.. For the given series it is easy to find the exact value of the remainder: Rn=fs (n+ ~~) ( } Solution. ± I... x ~-l. 999.)2n 4 · 2506. I) Ian I~ . x<l. 2504. 2514. converges..rn+Z + . 2513. 8==0 when n n-tx == O. - _(_I I) (I I) I n + 1.. 2. 2511. Solution.. COllvcrge~ absolutel y for x> 1. ).. Convergls ahsolutely for x> 0. -oo<X<():). > (n+ I) (n + 2) + (n +2) (n+3) + . at the remaining points it diverges.1X -~ 0 it would follow that cos 2nx -+. I I a'J~----1 2509.I < x <. 2) )x ~ 1 for x ~ 0. 2522.. 5. When I x I . converges conditionally for 0 < x ~ I. For these values converge. Rn=(n+ I) (} We multiply by (f r ~Rn=(n+l) rn yn-z. 2515.)2n+. diverges for x~1. .& 4 + .n + 3 + . diverges for x ~ 1.8::::1 when x>O.. 2510.!.. x~53' x<4"3' 2523. the necessary condition for convergence is violated when x ~ O. +(n+2) ({.455 Answers 2503. bol h the series L xfl and k=l ~ the series 1 L 2x 1l k k=l I.::= I .• Whence we obtain ~ 16 R" = n (~)2n 4 + (!. 2516. .. 3. ·=(iI · 2505. 2521.. 1 2 2520. x>3. 2518.)2n 4 + (!)2n+2 4 + (!. 99. Hint. S = I..n + ~ + n + 2. S=-I \vhen x<O. _ ( I - n+ ~) 15 16 find the above value of R n • Putting the series S -= ( ~ (. Converges absolutely for x> e. n+ I 1_ I < i < Ii· Rn I Solution. Converges absolutely when 2kn < x < < (2k + I) n (k == 0. converges conditionally for l<x~c.. x> I. Diverges everjwhere. x~-I. . x > I. thus.. since from cos . 2559. 2539.-Y2~X". sInce l'1m n~C7J )n 2 5 13 -"4 < x <"4 . = ~2 2587. but also by means of the d' Alembert test).' z 1< 1 2566.:n ~+ . l-x 2 2 x 2581. 0 ~ x < 4.Answers 456 -and when I 'x I ~ "2' the general term of the series does not tend to zero O<x<l.= l rI eye <x~3. X+~)= . "2 111 2584.. USing the d' Alelnbert test.. - .. _H_l)n.. 2580. I z 1< 1 2fi6:>. -2 2540. but also to investigate the convergence of the given series at the extremities of the interval of convergence. series 2552. but also to investigate the convergence of the given series at the extremities of the interval of convergence..I < x < I 2542. it is rosslble not onl y to find the interval of convergence. -2 < x < 4. 2 I I-X) + 1 ( arctanx-2"ln the series x- sum of 2586. 1 < x~3 2562. . 2576. 2541. 2563. 2558. 1 2578. x' l x x Va (Ixl<l). (l_x)I(lxl<I).2531. -I < x < I Solution. I z 1< { -In(l-x) (-I~x<l) 2577. d'Iverges. 2553. z=O 2569. 2529..]. 2525. arc tan x (f x I ~ 1). The diver- 2534. < x < 4. gence of the series for I x I ~ 1 is obvious (it is interesting. aX=l+I. Hint. 2 < x ~ 8. I 3' 2538.2536. xn~~na.. to note that the divergence of the series at the end-points of the interval of convergence x= ± I is detected not only with the aid of the necessary condition of convergence..2<x<4 2557.I 2533.1 z 1< 00. 2544. 2532.-e < x <e. 2547. x -= O. -3 ~ x < 3. -l<x<l < x < 1. . 00.00 < x < 00. 1 ~ x ~ 3 2549.1)2 (I xl < I).. -3~x~-1 . it is possible not only to find the interval of convergence.-1<x<1 2528.1 1--!. x = - 3 2551. however. 2564. 2555. 1 ~x < 5. sin ( [l+X_~_~+~:~_ . X= I V3. Hint. 2556. n. 3.. -4 2537. When I x I < 1 we have lim n -+ :t> l(n+l)IX~n+I)J 1= lim n! x n n -+ '" l(n+l)xnfn'~lim(n+I)lxln= lim n+~==O n -+ '" n -+ ~ ~ II (this equality is readily obtained by means of l'Hospital's rule). -l<x<L I I 2527. 2554. ±~ the e # 0 2560. 2~45.V2' 2530. 2583 (X_I)2(Ix l >I). 2~x~4. 1 tx I i-x (I x f < I) 2579. -e-3<x<e-3. I z-2t 1< 3 2567. -2~ x < 8. -2 < x< 0 2561. -2~x<2. (see Problem 2579) for -oo<x<oo. (1+x2)2(1xl<l) 2582. 2526.6 .2570. . -1<x<O. <x< 00 -3I < x< 2535. Hint.nn. 2546. -4 ~ x ~-2. USIng the Cauchy test. e I . ~ x ~ I Hint. 2588. 2548. . -I 2550. < x < 2. In(l+x) (-I<x~l). -I ~ x ~ 1. Consider the 3 +5"-.<x<I+J. For x=1 e ( 1+. . 2585.-1<x~1.. I z 1< V22568. 2543.3 -2~x~O.7< x< . (x.. . 2605..2 ·3"+2.. ~ =-~ ~ n=o - +{_I)n-I x.4.6 .21+28.s1n n 2:. 1·3·5 . n=o ~ X 2n + 1 2600.<-l)n(n_l)n (lxl~I). L L L + n=1 2·x 2 2·5x' l)n-l 2·5·8 .627 1- 2602. 00 ( 1+3 2) x n +1 < xa< n==o ~ n.. .x21l+1 (2n+I)' . xn n. When investi- use the theorem on integrating a power series 2x-3 3x-5 00 (X_W=-L(n+3)x".5 x' -1. I+~~ 2 (-oo<x<oo) 2597. X .3 nl X. n=1 -oo<x<oo.. 2594. Hint. L(-I)n-12n-Jxn Jxl< 1. L(-I)n gn + J n=o 1. X2n+1 QCI L(-J)"2n+1 ('x.. . ' n=2 00.4S+···+ 2.x sin a - 2589. 2598.4.J x2n 2 )' + . ' ~ n. + .4.n (-oo<x<oo).6 2ft 22n+1 (I xI < I) 2603 l-2 < x < 2) ~(-1)n+12n-l.¥I~)· 2n S 1 x' I · 3 X 11 1·3· 5 . 2593.. (-oo<x<oo).0 0 <x< \ n .31+ . (3tl-4) xn 2611.~ 2596.-oo<x<00. 2 x 2 2t 2S 4 25 x' x=2T-4f-l-m-··· X In(2+x)=ln2+--2 2591 00. eX'::=: I -1.n( x n - 1 1) · "2 < ..~).2. 2~ 211 + I (-3<x<3).t··. (211-1) X 2n + 1 x-f-"2·3+2.. n . .2 -2. xe-1X=x+ (n-l)l 2n (SJ 00 < x < 00 00 n=1 ~..:33.. 2610. .4.. 8--t. n 2L(-ll (n + 2) 32n. 2595.. n=o 1 x' 1·3 x' 1·3. 2592.~~.. n=2 2606.2 3-" gating the relnainder.. 2609. (_1)"1'\ (2n)1 (-oo<x<oo).3.4 Xl 2f cos a + 3f sin a +'4f cos a -f. . . .. 2n 2n+l+··· (I.¥ < 2 n=1 x. 1 t(-l)"-lnnl x" n=1 n=2 00 1 2n + 3n-. .6 . + (--23n-l.lx 1<1.5 . x.2590. -2<x~2. ~ n=o 2604 260).. -l-'(-l)"-' x' x. x2 -4x+3 ~ 2 x%n t 1 ~(2n+I)' n=o (-1)"(2x)2" (2n)' 11=1 (-oo<x<oo). f7J 24n-1 x2n go 1 2608.3~-I.. I-t-'''' (_I)"2"x " .11-25. 2n 2n+I+'" (lxl~I). x2 x Sln ..32. ~ \2n)1 2599.4S-···+(-1) 2.451 Answers cos (x + a) = cos a. + ~x~n+..22 +3. (2n -I) x1n 2. r ···-l-nr a+ (n -t-21) nl +. (2n I) x + 1 2607. . .2-1-22." 00 x+L..3...2 n + . x. 00). 2629. x I s 1. 2n (4n+l)n! x .1) ti! (lxl<oo).9. X2n + 1 + I) (2n + I)! L (_I)n X ~ n Q) x!n+1 L (0).ln2+L(-I)n-l(I+2. -2 ~ 1 2n +1 (I-X)2n+l 1 +x ..~ {X-4)8 _ 1·3·5 {X-4)4 + 4. -oo<h<oo). 2639. ) 2624. f(x+h)=5x ' -4x2 -3x+2+ +(15x 2 -8x-3) h+ (15x-4) h2 +5h l (-00 <x<oo.. L(-W+l~1 2619.. 2625. 2634.. + + 2. 2628.n-l_3. 2x5 2621.. 2631. 2n x-~ yn-l (2n-=-l)! (Ixl < 22n 2638.5 . (I xl".00 2618. x'-2x2 -5x-2= -78+59 (x +~4)-14 (x+ 4)2 +(X+4)' (-00 <x< (0). Hint.6 1 2' 4·6. x+x + ~ x·+ . + x '}/ . ..3 +15-·'· (1-2"+6"-'.2r x + .Answers 458 2612. L (n+ 1) (x+ I)n (-2 <x< 0).' X 2 X4 ) .5. (. ••• I 2+ n=l + \1 n. Proceed· of the elli pse x = a cos eft Y = b sin CPt com· l ing from the parametric equations pute the length of the elli pse and expand the expression obtained in a series of powers of 8. 2617.5 x +2 . 2637. n=o L (2.. L(-I)n(\-t! (-2-¥3<x<-2+¥3). 3 L (I +(2n)1 3 x2n 2n .8 28 (0~x~8)...2623. ( X-T yn-l (2n-l)! i.1+2"+24+··. 2633.3...l:(-W(X-W (Q<x<2). .. I)... n=l <x< ( . 1n 1 2630. 2616.fd (-I) (0 < x < 4n 1 n - ClO). (-00 <X<oo)..n) : X (2n < (-2 s 2x +U-+ .. e-' [ I +~ (x~12)n] (I x 1<00). n=1 < (-I x. x3 x+ 3 2620. n=o 2835. 2622.. I). .I xl n a:> X x n=o 26tS. 5x4 xl .n - ) (x+4)n n=o 2 Q) 2n (-6<x<-2). - 2).l ) C» "26t3. 1I x 1< 00). n=l n=o + ct) ct) L(-WIX) 2632..~ (x 244 )1 + + (_I)n-l I· 3. x+ (_I)n (2n -r. e (~+ ~+ ~+ . n=o n=1 S < < Y2).. (X n ) (0 <x".. 2 n=l ~2n-l) 4n+l + + 1. 2626.2).. (lxl<I).3. 1+"4 x L 4n+1 4n Q) I< <I x 2614. 2 ~(_l)n ( 4·6·8 . (2n-3) (x-4)n + • •• Ql 2636. -J.. 00). . th3t is. It is necessary to evaluate the reillainder by nlcans of a geo.. f(x-t-h. 2648.005. IX+ Y =arc tan x+arc tan y (for -~\y I x I ~ 1.I..2505 2658.X -I-XY+"2 -L c2 . (-oo<x<oo. ~(_I)II(X_y)3n_(x+y)2n ~ (-co<x<oo..'3!:::::: 0. 21 S (± n) = 2 xlJ + ••• (1 +.[(x-2)+(Y+2)] ~ It' x2 _y! x8 -3xy2 n=l 2669. 11=1 2660. e . 2655. x 1< 0.• I .39. .2651.~ C2 • - .. I-x+y 2 Hint. Make the substitution ~640.2652. -<Xl<y<co). 2671. I - 7 2645.J [ x + ( y . Two terms. i. 2644..4931. 2664. 2657.087. / x / < 2654. y-t-k) =ax2+2bxy+cy2_r-2 (ax-j-by) h+2(bx+cy) k+ah 2 + 1-2bh+ck 2 • 2666. I R I < 1\ .. (-oo<x<oo.. 2650. exceed 0.608 00 (2n ~ (211)! 2656. 2643.. 4..1-L~(-1)n I i. 2641.00 . 2670.~) 1zn 2 (2n)!· n=1 3' + . X 2n + 1 -1 + 2n+l 211:1 ~y < I). (-I~". 1.. r += 4590 t and expand In x in powers of t.~68. metric progression that exceeds this renlainder..lx-YI<1 . 2):=9h-21k+31z 2 -t-3Izk-12k 2 +h!t1' -2kl.523..' ~ +11 'I n (-1 ~ x < 1.92 2649. i.Answers ~ Hint. Two ternlS.Hint. S (0) = c\ 2 11: ~ 21l 1 n=o + 1 1 -t. Eight terms. I =-1+1 ( . {(1+". (lrc tan 2665. 2-j-k)-f(1. 0. 2646.. I x 1< 0 0. = r r +}( I :-x +~:~C ~x + .026. e. -!I ----(2n-l)! ~ 11=1 oJ) 2662. 2659. I y J ~ 1). 2667..001. -co<y<oo). }-23 '."':""').1-1-2 ~ (y-x)n.1-t-~(-1)" x-y) 0 621 18 2653. n=1 2647.22./ x 1<0 39.. 0. I + I: ~I I ~. 2642.. ( .. 99. 1+~. I-x-y+xy=(l-x) (i-g)... 2. x-if. ::::="2+2 I :-x -3-+~-5-:::::0.-+ 1+x+2 . sin (2/1 + l) x. 999..8 I R 1< 0.. x < <Xl ) . To prove that the error does not. tt=1 I: (-1)/1 JJ Hin:. n=o -1 ~y ~ 1). ~(_l)n-I I- 2)'2n-1 -oo<y<oo). Hint.y-x) tt=1 n (X) a geometric progression 2663. 2·(211)' 11=1 00 ( 2 X 2661. + ~:~:~:: :~~:=~~( I :-x +'" .. c1 + c2 _ 2 (CI-C2~ . . ~ :::::: I I (~r 1'3(~r... 0.2668. J x 1< 0. 0. Use -J-x-y .69. I R I < ~ < ~. -. S(±n)=coshan. 11 ~ n=1 (2n-l)" ..X n=1 n=1 X ~slnnx. 2 xt ~ t2a-l)1 ' xt ~ l "8.. 2674. n=1 nxt rt> rt>[( 2~ ~~ + 11 ~ n=1 (X) -)Inan e . S'l± n) =0.. a n S(±:n:)=coshan. b) ~+ 2 n=1 nn 2 \.Answers -460 b-an_2lb-a)~~COS(2n+l)x+(a+b '2672..n. .b2k+l= n=1 . ) _2IJI-COS- n xt SIn . n +- 'J -- n=1 _~ . n=o nl '3 + 4 2673.L(-1)"11"2. ~ n 4 b-a . . sin ax if a is an integer..I)n ea1t] n Jstn nx.J "(~ k 1 1. n ~ n=l 8 xt =2k-l-n(2k-l)' and b2k =-/i.1) 2680.) ... . ) ~. [1-( _. where b2k =(-1) -12k. 2676.bnSlnnx... S (± n) = nl.n . a +n2 b) .. (2n + 1)2 ~(_l)n-I sinnx. ~ ~ n=1 (_l)n-l n ~i+"n~. n' b) ~_.. 2683. ~bnsinnx. a) 211 I 1 where n=1 n=1 blk - 1n ~ (. 11.1)11 ] 2 sin axt X [ 2"a+ ~a2+nl(aCosnx-nsinnx) .X R=I :1" ClIO • X ~ (_l)n n n: if a is nonintegral.2..a 4' ~ b) ~.. )~ n • n=1 cos nx 2 L (-1 )nfi"l . 2675. ~t'1 (211 . n L. 2677. xt a. . b) 011 ~ " cos nx I) n 3+ 4 . ~ a-n R=1 2 sin axt [ 1 CD Cf'S-nx] . . cos ax If a Is an a n a-n ~ n=1 CD integer. ~ 2rt. S (± xt) =cos an. S (± n) =0.I)'ens nx a +n l 2 2684 •• a 2 . 2682.+ ~ (_l)n -a 2I IS nonlntegral. 6' xt 2 2 n=1 2) ~121 .1 x . nx. a n 2678. 'I) ~.. n=1 CD ~.! ~c0s(2n-l)x.If a .2679.. 2685. ) 3' c 2 2681 xt V3 • · a) 2~ stnnx.. 2 sinh axt no 2Si"ha:rr[2~+~(_Wa~('+sn:]..s (± n)= -2.1 sin 2 cos nx.~ crs 2t2'1-1) x 2606. 1t sin h anx CD n=1 1 ~ ( . (2n+ 1)2 • 11=1 4n b2k =-/i. o n 0 2 The same substitution as in Case (I). f( ..X] Ii. . n n=l 11=0 2697. )... ~ n ~ 2 n=o 4 1 2k+I-\ 2k 1.1 a) SIn n=l 11=1 where nL 4' 2( n=1 ~ 21 a) n IlJ'tX • SlI1 . fl nx S1Il -1- n. =.( -1)"-1 _ _ 2 .Answers 461 ~ ({ +i: SI::h cos nx).t ) leads to the equalities ~ _ i.11 =0 that 3t n 2) It wl\l readily be seen n 2 Sf (x) sin 2nx dx = ~ Sf (x) sin 2nx dx +. .. l . ~ n-I n Solution. n-I 1-~+2 ~ (-1)"-1 COlnx. ~ [~+ i: cl~:hr cos nx]. b) . Sf (x) sin 2nx dx. (2'1 1. 2689.. 2702...1) 3lX sin ...1) 1lX • 2695.I) nx oc 41 ~ cos .. 2. ~ n2 1"J ! n=o {2. l n I Sf (x) cos 2nx dx + 0 If we make the substitution t .(_I)n ') b) 00.t ) = f (~..: . 2892. )..t )... 2 ~ n 1 2691.1)2 .a») bn Sll17)' (2n . 1l=1 00 10 ~ nL 2698. 1'£ [i n=1 + 2 n~1 ~ (_l)n-I ens 21tX] 4n . +t ) = - (n = 0...' . a) 2 I. taking advantage of the assumed f(~ .. with account taken of the assumed Identity f (~ +.. bill 0./ . fl \' .J l2+ n2Jt2 (-1)" . I) alII = ~ Sf (x) cos 2nx dx = ~ n 0 + ~ Sf (x) cos 2nx dx. '" l (_1)n+l _ _ . = 0 (n = l.1 8 [ =n n2 nx cos - n n=1 ctJ _ i.l . sin 2nnx . identity in the second..-l 11 2 n blk +.1)' -16 .2701. ~ cos (2n +. [T f-2 l. 2694. 2 n2 ~ ~- (2/1 sinh 1 ~ ~~ ... a) ... lJ (_I)n "" 5 2699. I.1- cos (2n + 1) fiX 2703.!.J 1) 21~~1 n~: I b) 2700 Il'J (2'1 .. 2696.1 + 1)2 b) 4 n::! -S-- . -x in the first 11: Integ:al and t = x .. 1_ 1)2 bZII Icos n..t. 2690..".x -lttl SIIlIl. then. 2729. 2752.2759. y"-y'-2y=O. = In Cxt. 2764. Yes. xyy'(xy2+1)=1.-f.c. 2xy"+y'=O. 2760. 2745. In 14x +8y +51 + 8y-4x =C.C. 2719. 2769. ± x.£. 2707. (X-C)2_ y2=C2. 2730. Straight line y = Cx or hyperbola y=. y=O. 8x+2y+l=2tan(4x-I-C). InQ=-2-. b) no. 2740. y"+4y=O. 2716. 2775. By hypothesis ) ydx o Differentiating twice with respect to x. 2727. 2e 2 ~ x=y__ . Pencil of lines y = kx. 2728. y-2xy'=0. 2717.593 (exact value y=e).946 (exact value y= 1). y=xe2X • 2731. Y 2725. y"=O.yt-xt=C. 2750.2 -x . C 1 y=2 X2 -2C. x =Ce a .y= x ~ xydx o 3 -x--=4 x.826 (exact value . Fax 2 roily of similar ellipses 2X'-t.cos (J> 1 or y2=2Cx+C2. Hint. 2709. y=ax". 3y2_X!=2xyy'. 2779. 2732. y"-2y'+y=O. I C . 1. y=x In x 2770. 2757. 2724. 2778. x=O. 2710. 2706. 2738. 2 1 +y2=-1. 2726. 2743. Hint. No. 2739. y=XY1-: . 2762. y2 = } x. 2 = Ye (l-t-e x ). 2721.c+y-l)'=C(x-y+3).3x+y+2X Xln 1x+u-1J =C. x The seg- x ' ment of the tangent is equal to Yyt+(f. y =a+ I ~ax' 2746. 2771. 2720.= 2744. 2723. Family of hyperbolas X 2 _ y 2::=:. y= I. Q= arctan(x+y)=x+C. 2776. y2_ X 2=25. 2756. 2774. Yes. . 2777. (l+y'2)y"'-3y'y"2-=O. x x y= (x-2)!_y2=4. 2715. 2741. 2755.. y-xy'=O. 2747. IJ 2748.y 2=C • 2766. . + 2767. 2773.. y= 1 X X 2X =(f(-5e. 2749.2.ry 3) Y= 2742. tan Y =C (I-eX)I. x+2y+ +3InI2x+3y-7J=C._x=Ce y. Yes. 2772. xdx+ydy=O. we get a differential equation. s. x=O. y2~2px.. y == f 4-x 2 21n . 2714. 5x+lOy+C=3InllOx-5y-t-61. 2708. Yes. 2765.Answers 462 Chapter IX 2704._!-2 . . y==-cosx. 2722. a) Yes. 2761. y = C sinx. y=-xy' In~. 2751.780 [exact value y =3 (e-l)].+ge -4e ). 2 (X +y2)3(X+y)IC. . cottl/=tantx+C.E. 2705. xl = 1-2y. y"'=O. xy'-2y=0.--1nlcosq>I+C or Inlxlcos" <p -2y22 =C. 1 + 1/2 X2 +y2.4. y=. Yes.ix V y+ln1yl=C. 2718. 2754. 0. 2753.r 2-Y4-x 2 2763. y'=y.2758. Family of circles x 2 +(y_b)2=b 2 • 2768. (. x 2 :=: +C x y y 2. 2817. -x C 1 2793.. (X-y)2_Cy=O. Hint. there is r RO 1 y -lnx+ 1. 2816. (y 2797. · {x=eP+peP+c. By virtue of symmetry Ule sought- for mirror is a surface of revolution.463 Answers 2780. solution: __ c. The plane section is a parabola. - a x X + -by = I. 2791. cos x y'(3+CeC09X )=x 2795. Singular solution: y=e". I. r - I x VI+yl+cosy=C. If a tangent at any point M (x. arc Sill x) -1. 2814.\2_t. 2800.y 2-Cy+a 2=O. y 2=Cy·. y2+ p 2 JX=2P--p~ +C. 2805. 2810.2_ y2= O. y) forms an angle a. 2811. 2794. 2806. x y 2 2 X ( x- X 2_ (. x=y . there is no singular integral.2Cy) X Qeneral singular integral. y= 2786. 2792. y) of the curve. y 2p . Xl "3+xy2+X2=C. 2812. 2788. 2815. -2 x! 2 -x .x 2 + yl+X . y=--f---. \ y=p'+21np.t C 2 + ' XlJ= 2804. xl =C(2y+C). General integral ( f- Inte~ral y+C ) X + C ) =0. p y=~cosx ± ~-3Sil1X. 2803. Solution. singular integral X1_yl=0. y y x x .cos P + p + Sin~ular 2801. X=C y 2 _ . y= 2790. y2+ X +ay==O. tl .. 2820. 2781. But tan a. x Inlp-x/=C+--. 278li.. 2819. Use the fact that the area x ~y is equal to 2784. y = ~Cy2+a2. x4 3 yS "4-2"X 2y 2_1-2x+ 3 =C. In yp2+y2+arctan-=C. 2789. 2813. 2809. 2818 y=O. The coordinate origin is located in the source of light. +xy+y2=C. 2782. x=sinp+lnp. The equation is linear with dx =-2 (x r l . 2 4Y=X +pl. tan cp = y'. 2802. x=ln-. (x Sill Y + y cos y -sin y) c"(::::L C. The desired surface is a paraboloid of revolution. Hint.. II" X Xl 2808. -II (x 2 -t. 2799. X (x + C2. x2 ?807. { Y = P sin 191. forms with the x-axis an angle cp. y=Cx +x l .ex) = 1. y = p2eP . p-x 2821. and the segment connecting the origin with the point M (x. + ye l/=2 - + C)I =x'. the x-axis is the direction of the pencil of rays.2Cy) = 0.-::i-X4+~I' 2787. Inlxl-~=C. 2q98. --1-.=C. Paraboloid of revolution. then ta-n a = tan 2cp = I 2t~n ~ . X + '2I y2 :--::: C. t~x respect I eX ab-e fJ and d-. singular integral :. (2 y 2_ X ')I.an q> x y-yy'!= 2xy' and its solution is y2:= 2Cx +C2. The desired diiTerential equation is . generated by the desired surface bein~ cut by the xy-plane.. =JL .. General integral UI+C1=-=2Cx. Y 2+yl_2 arc tan-:t=C. a y In . y =Cx-x In I x dx.=CX 2. 2783. y2=xln-. + y I + CI. k) Bernoulli's equation in x. 2851. 2827.=1--+Ce y. y=xI(l+Ce X ). x Yx-+y2 __ =C. y=x arc sin (Cx). Y = 2px + V 1 + pl. 2 2 -- 2849.~ _ _ 2p 2p ( + f 1+p2). x=Ce-P-2p+2. 2856.P . xy=C(y-I). 2858. f) Clairaut's equation. 2arc tan -2-=lnCx. no singular solution 2828. 2836. I . 2865. 2855. 2843. y=--Cx+ 2829.1X + ~ sin x+ +:cosx. reduce to y =xy' ± y g'l. y= uv. xy = C 2831. 2857. 2824. y=uv. 2825 3 I Hint. x 2 . y = uv.. 2861. 2847. t y::. 2837. y= xl+C. singular a I x'-l I 1 solution. y=Cx+C.. y=Cx+ V -aC. y= =Cx+C In C. + equation from which x is defined as a function of p IS honlogeneous. In 1Y +21+2arctan. 2862. a) sin 1t. 2etX-eY-arc tany1 1 -'2 1n (1+y')=C. y=X~l (x+1nlxl+C). y'=Ce.1. n) Bernoulli's equation. py __ 333 c::C(p-l). yl = 4x.-slnx+Sinx:-l. xl+yC=Cyl. y=uv.Answers 464 2822. i) leads to eq uatIoll with variable~ separa ble. 2841. e) with variables separable. y=ax+CVl-xl • 2846. x = -In Ix. 1 1 -- x=. 2845. u =x+ y.(Cp 2 -p).:O. a) Honlogeneous. singular solution. 2826. m) linear. Y=21 Cx I + C2. differentiate with respect to x. The astroid x 2 /1+ y2JI=a 2 . y=xu. y = Cx + ~ . x4 =CecY -y'-4 yl-SY-32 2860. x= uv. x= In I p I-arc sin p + C. 2840. 2863. + C. Y=-"4. 2eX _1t=Cy l. Xl + yl = I. 2830. 2838. x 2835. x2 y=-CX+C2. y 2853. The differential 2823. b) linear in x. j) Lagrange's equation: differentiate with respect to x. 3y+ln (y+l)'=C. y = uv. xy(C- 1 2 b) x= y.} (siny+cosy).pZ+2.g ± 2x. x=yl(C-e-Y ). 2834. - 2842. V~ + In I x I =C.xeY l y y2=C. 2859.In(p+ . 2833. y= =C. h) Bernoulli's equation. A ci rcle and the family of its tangents. x'~CeY-y-2 2852. 2832. 1n 1 x)= 1. x=Ce slny -2a (1 + sin y). g) Lagrange's equation. · 1 _ y="6 (2Cp I p2). 2848. 2850. { y=C (1 +p) e.2868.eCY+1. 2854. 1) exact differential equation. VI +p2 1 C x=t+. (xy+C) (x 2y +C):. { y=p+ VI-pl. 2839. y-l x x2 2 +3x+y+ In [(X_3)IO I y-11 3 ] = c. x=CeY . y=t-(x+l). 2844.IJ=xeCx • 2864. d) Bernoulli's equation.+:=C. Y=4i. x=uv: c) linear in y. V. 2881.-b 2. 2889. tionality factor). y= =_-::e. 2904. 1/2_4C (C-. Equation 2908. Q = Qo 1t dd~ =kQ. Y=~(SiI1X+COSX).0) is a siugular pOint of the differential equation. Hint..x ). In one hour. C-x" .3~O) · .kp dll. ( (~r rJ dt. y=-=(Y2a ± YX)2.x .Y= x X+ y a2 +x2 2872. Hint. i = 2906. 2882.2 sec. dr sure at each level of a vertical CUIUlll11 of air IBay be consHlcrcd HS due solely to the pre~"1I1C uf the upper-lyIng layers Usc the la\v of Boyh~-. y -=-e". Use t h~ fact x Ilut the area IS equal to % Sydx and the arc leni:tll.e. 2880. 18. II' l-ry'2 1r. Hint. 2876. a propor- (t V~ ).kl • 2903. Use the fact th:lt th~ • re"ultant of the force oi ~ra\'lt~ and the centrIfugal force is nornlJI to thl' surface. U= V t )00. 2 y -t. Pass to polar coordinates. R' :L2W2 [(R sin wt- . 2884. c) x 2 + y2 =--= x~ x (0. / grn T as t m~~ =mg-ku 2909. Y = 21 (eX -). y=O. v ~ 1t ~ Q -= Q. S=(Pl-{W)kl. X~:. (h'-2h) dh = (2"I)!!' I. The pr<:s. b) y=Cx.2 4( 1 2 = 0. 2899. a) y=x.11lgular velocity -of rotatll)l1. 2887. (0. 2886. Hint. the poinf (0.C2 • 2895. 2885. Hint. x 2 'Y= C'I ew x x+-=C. y=-=x-l. EqUation ds-= 2901. IS 1 dp -. h) y2== 2px. 2890. b) no sol ution. s -:. w'-' 100 ( ~ y rpm. y2--l-16x=O.2871. y-=-=x. 2 ~ .= 0. to ~ I u 2 2896. Equation ~~ = k ( 16-1900 kl~. Hint. ~c­ cordll1~ to \\ 11Ich the dCllsIty IS proportional to the pressure. 2879.465 Answers ~~ . 2875. Equation =::: 35. '1 akll1~ the y-axl~ as the :tXIS of rotation and denotlIlI1: by (J) the . Hint. \\ e ~t't for the plane aXIal cross-section of the deSired surface the differential equation g ~~! ={t)~x. r--kcp 2892. 2894. 2893. r=Cea~. tanh 2910. I (y -Cx)· (y2_x 2 +C):. 2898. 2907. Hint.2. IS t =::: HInt. 2902. a 2 In(x+Ya 2 +x 2 )+C 2 1)3. 2900. 3y2_2~=O 2891. Y={-(2x 2 +2x+I).x -i-2x-2. a) y2==X.00 (k g. y 8 286. p2+4y2:-=Cy 3.2870. Y=-= Cx I + C2' Y==- 3 - 2" Vi \ 2x 2 • x' 2874. ~ 2897. 2883.0) is a singular pOint.0) is a singular point. +-x ~ 2867 . x 2 -1-(y-b)2=. 2905. Hyperbola !/ _x 2 -~ "C or circle x 2 1.-a-x).. 2888.-+ CII. Equation dQ =-kQ dll. where C is arbitrary. y2=-= --::l-e.J\LI!" utte.1 kg. The sought-for dtflefl\J1tlal equation _'k'WI-~xdx.y==CS1l1X-a. T=a-j-(To-a)e. y=2. a) (x -C)2 + y! = C!.I)QIl1671l.IJ .Ce .. .y2 --. 2878. • y= 2869.: 2" Hint. 2877. 4 2% of Ihe initial quantity Qo will decay in 100 years.!.. p -=e. 2873. . + x y2 _y2x _ y3===C. Id24' -+. X==Cl-~ • 2 2 X=C ty2+ y Iny+C 2 • 2936. Y = ~2' 2922. !J=-x-~-l. +-1.r. I 2 1 In \ y+C !I x==C 1 +ln y-C y-t-C \ • 2934. y=Cx2. y:-=. y=C. y == 2 x2 • 2944. 2932. Y-=="2 (x 2 + I). 2931.1n I x I.X I x + C2e y=C.x(x-C 1 )+C 2 . y= = ~ Clee. !I - 2 x 2 • 2 1 3 x 2940.Answers 466 R . 2939. 1 2929. 2941. Equation Ri + L dt = E sin rot. . X 1 l-C~x. y=Clec~x 2916. di Hint. Y =-= 2e • e e.Lro cos rot) + Lroe -. I 2 tion). 2914. y=x+ 1. y=C 1 +C 2 In/xl. y==(C1x-C:)e Ct +C 2 . y=(l+C~)lnlx+Cll-C1X+C2' 2919. y=ln/e 2x +C I !- -x+C 2. 2938. x 2 _1 2 (e 2 -1) 2933. y--=C. Y = (singular salu- (singular solution). 2912. 2913. x == -"2 2 (y -~. 2945. ± sin (C.x y2="e_I+I_e· 2942. s +C 2927.y=±YCtx-l-C~ 2917. 2) JI2 I 2943. y=2X2+C ~ =12+"2 + C1x In Ix I +C 2x+C•.t L ]. Y-== -:~-x- 8 Y . ± x) + C x-t. 2915. 2y 2_4x 2=:: 1. 2937.eX.1n I xl or y=2 (e2+ 1) e2 +1 I + 4 . Y == 2911. Y = c=(C 1ex +I)x+C 2• ±{ [x V C~ _x + C~ arc Sin~] + C2 x -+J e 2924.X=~ In II -:C "1 C2 . . Y= ~ r =aln\sinY~C21· (x-C I ) 2918." + I (In I x 1)2+ Cl ln Ixl +C2· 2920. 2925. Y---= 2930. '2935. Y= C.Ca. Y= 2923. =xlnlxl+C l x+C 2. 2 Y= 2926. =xl +3x. ~ 2928. 2921. 1+Cly2=(C2+~. e2 -1 I-x! -4. x)e·X-/-~(2x2+4X+3). y~-:. x 2C 2 s1n c. 1I:::.C) 1.e 3 '.2X (C l co~ 3t -1.45 seconds. d) eX (A cos x-I.x --t 2 (Cle 2 -1-C 2e 2 ) 2986. Y-= c) Yl~S. 11) yes 2969. a) t/.=a.EqUatiOnOfmotJon. 2968. C cos x -t-In/ sec x + tan x 1-1.m:. _ y=eXx.n "in 2x. cos 2x 16* 3004..C 2 Slll (~rJ' 11 V -kx. f) xe X [(Ax 2 -i. d) y"'-3y"-J-4t/-2y-=-=O 2970. . Y == C 1e -12X -1-C2e3.e-·\· + xe·~. c) A co~ ~\" -t-1J Sill 2\' +. 2983. =- t Cle J Ii + C 2e-·d -1/. 298R. Y ~ C . c) x 2 y"-2xy' +2y=O. + x l 10Xl) . Equation of nlotlon. Xincosh(t Hint_ X 2967. y -= cos 2x + ~ 3005. 11_ Cle x 1.sill 2x). m 2 s~2(sin(L-l-tcosU) s=/i 2966.. l Y=e:(clcosX~3+ .: sin 2x.tI2-=--=~' x x variation of paranlcters we find: C I :-:: ~ + A. y == -3 -f.(:~r· ::~=1i ds dt 2 =g(sina-l-tcosa).' -5" x -Jr-:.. 2972.Losx+C 2 sinx.B - 2975. sin 2x) -/- = C I cos x+ sin 2x. e) no.. a) xe (Ax:! + Bx -t.. b) A C()~ ~~ t-1. b) yes. y = (C 1 -J. f) no.4e x -1. y=C. Hint. II=(C I -! XSln2x) . !I =-= e. 2976.\ (Cle t J -. V -llX Y-= Il> 0.. k If < y-~CI cos O.Bx 2 +Xl.C sin rrx 2994.- Y3) . 2996.C 2 Sill x) 2981. y Y=C.E). +C 2 __ (C I-f-:!X C .'\: • i + C . 3001.. x +C 2 Hint. 2973. 2995. 2989.(CI-!-C~x)e-x. g t d) yes.B "in \).) l 11 Clcos~x-t-C2Sill-6-X . '~ 2 2997 • +x 3 -1. where H IS a The differential equation a constant hOrizontal tension. La\v of nl0tlon. Y:=-. y x == a cosha- • • II -== 0 2993.y=e If 2984. In 6. y= A + -I B Sl1l x. 2990. !I =. !J --=.C\"2e2X.B\ + C) cos 2x+ (Dx 2 + Ex t-F) X 2X 2992.elX".+c2e-x+~x+~(2COS2x-Sin2X). -f. q 2965.C 2e7X 0 +2 = C1e:!''\ -t-C 2e.l!x+C2e-2X-i(:~S1l12X+COS2X).3X _ ~e-x X lC. eX (C I cos x -1. 2991.\..e2.Bt -t. -t. a) No..ex +C2SiIlX-l-ixSinx. (sin x -/.) e -x -19 L 1 lj--- 2999. y--=CI-1-C2e~ 2979.Ax + -. Equation of motion. 300 dd:!~ =-10 v.461 Answers xcosh x-t-C +C 2. 3000. 2980.Hint."in x Inl cos \" I-x cos x. gt 2 !!. !I:. !I =. 3002. 3006. y = 1. Use Ule substitution Y=YIU.2. g) no.!I ~ e-·". 2978. . 2987. c) e·\:. 2974. =~(CISlnx-J-C2COSX). and VI -/.xe:!'. Particular so- x lutlons~ the honlof1:cneolls equation YI~=X' . C2 = By the 1l1cthod of the x:s 6 -t. b) y" -2y' + y -= 0.25 e I 2X TeX Y= 2 • 3003.•. 2998. (!)r. y=C1x+ x2 B In x.C2e-X~'2).< ).C:! 5111 3x) 2982.C). r.' (A \:! -1. yg ~).C 2x) eX + 2"1 cos x + x4" e X y=c.~=mg-k(~r. Y = A -t..\: 2977.y=-e 6 x -- 298t).=3x-5x 2 +2x 3 • 2971. Hint. Y = . + y = 0.:Cle-3~ j-C2. y=C 1 cos x+C I slnx+ +COSXln/cot (-i-+~ )/. then i x"=4-k (X. T=2n vi sec. +Cr 'VI +xeX sin x+e x cos x. sin pt. y:::::l =Cle-IX+czeU'_~ eX+ ~ (3cos2x+sln2x). 1 2X Y=g e + c: C1 cos 2x+ Casin 2x - f (3 sin 2x+2 COS2X)++. If x Is reckoned from the position of rest of the load. 3034.6. cos rot + C. 8036. · 3014. y=-=C 1 cosx+ + Clsin x-xcos x+sin x'VIn I sin x I. Transform the product of cosines to the sum of cosines. where X o Is the distance of the point of rest of the load from the Initial point of suspension of the spring. II=C 1e +C 2e e2X . a) y=C1ex +Cst. 3033. y= C1e'x + Cae4X-xe4X. 3013. 8031. y=(C 1 +C 2X) In I x I. 3041.6-9· 3019. 3035. therefore.x 'VI. y=C1e. g-900 cm. -= (k=l). 3025. 3018. ~ sin 3x. y=eX t~l cos x + xSinx-~cosx+514 (3x-l)e lX • 3026.Answers 468 A -8007.200 t·cos rot. x- 2gsin30t-60 Visin Vit . + C 1 sin x-2x cos x). 11= (C 1 +C 2x) e. Y=Cle-. k (xo-l) = 4.x+cr-i(2xl+x)e-'x+~x II=C1cosx+C 2 sinx+ 8030. x 4 Xl X cosx+"4slnx-acos3x+ Hint.tnltan. II = (C 1 cos 3x + C2 sin 3x) eX 3016. y=C 1 cosx+C 2 sinx+xsinx+cosxlnlcosxl 3037. 8042. + + 3023.y =C1cos3x+ 1 3 3 16 (2x 2-x)e3X • 3027.x +xe. 3012. y=C1+Cze-X+eX + 5 + 2XI-5x.y:::::a )( (2xl + 3x) eX. Hint. sin rot + -. y=Cl+C2e2X+2xe2X-2"x.--. y= C1e x + Cze-x + +C a sin3x+ )( (2-3x) + ~ (xa-x)eX. 3024. 2 1 + 37 (sin 3x + 6 cos 3x) + e~}x . 3029. y=(Cl+caX+x~elx+xtl. 3040.y=Cl+C2e2X-2xex-4x-4"xl 3028. I. ro -p 2) x= C1 cos rot + CI sin oot- A x Xl x· .x + 4 eX. where ~:a:4. y =C 1 +Cae"x + 4-4-"6 · 1 5 -8010. 4 dlx I Is the length of the spring af rest.y= ( C1+C1 X+ 1x1)X 1 xe.X+ (r +e-X)x yxarc tanr.dt l ~ d·x g ---k(x-y). y=C1+CiI'X-f6(COsx+3sinX)- x3 x 2 X X _X (4x+ 1). 3038. y=C1COSx+Caslnx+slnx. 3021. eX +xex i-(~a:')=2-k(X+2). g=981 cm/secl. 3022. 3015. 3011. Y= C1e. 3008. Equation of motion.4+"4· 3020. 1) x== C. 3009.x+cae-x+{x -=(CI+Clx+~)eIX.+X-II-I). b) y=C 1ex I +C~-x I +exi . 3017. 3032.ex -3xex -x-xl ..x sin x-cos x.y=C 1+C.x In I x . hence. y = e1e. . Y == Xl x . mdil=k(b-x)-k(b+x) .2X + C~e2X -20 (sin 2x 2 cos 2x). y:::eX(C 1 +C 2X+X 2). CI 1.. 3076. z=C 2 cnsx-C t sinx. • Y:. Y =. y =c C.x -I. 3072.C2 3073. -1.e-X+el c 2 cos-+x+C.C 2x) e-'Y: --1.--3 ) 2 x 3048. 3079.CI -1. 3080. 3045.4/3• 3074. 3043. Y = CI C 2x 3057. cos x + C2 sin x).C 2 sin (2111 x).x (C. • y=C. y=C.. c~e-'>.+C~x+Cael2·~. 469 b) r = .x+C 2 )e- I 2X • 2 . y =.XI+C~2 . C -I-{ x.(C a C4 t) e.~ lUt + V.C 2 X 2 -1. Y == C 1 cos (In x) + C 2 sin (In x). y = (C I -t..2) .e.Col sin x) 3051.. y 3075.C4~) eX. x Z= ~ e.~ (elllf _e.(C I 1. I 20 (C S -l-C t)c'Y: • I -l 3 (rtf C cOS-i-X-1 Ca<. -I. z=(C.C2 "< -!. y=-=Cl+C2X+Caex·V. C'l"(n-I). Y3 ) 3053.Cx 2 x (:~x -1.1". 3078. V"3 1 }/3) -1-x-2 y=-e-X+-e . 3054.+1~88(4CnS'lx-sin4X) 3063. .x)e- 2X .C Ie -x + C 2 C( IS x -t' C 3 "1 e~ -I- I~ \:~ -I. + + 3060. -I-C 2 1n (x -I- 1)1 tex -I.3067.3065 .x • 3058.wf ). y -.(C -I.[(C:-2C.tan x sin x +x sin . V2~).C2 In 2 1 x) ..Answers and x=ccos (t = ~ is (e :. t= e. y-=-e x tCI1-C2X+Cax!). yJ .C 2x) e t -t. y.I)'. y:-::: (C I -f.( C.Col sin a~.cos x In I cos t 1.: C IX -1. "4x . Y =-= CIXS X + _2 • 3070.e~ ( {X-~) .(C a -1. y==Clex-!-e 12r:!_L:3X3---l_~ r "I 2'\4_L~. cos x -f..C 2 sin x) e.I .2C ) sin xl. In (6+ y-- X( 3047.e~ + + (Ca-l.(C a -I C 4 x) e. 3069. 3071.C2 x) cos x -t. y === Cle QX -t.sin y=C. 3061 • y==C I -l-C 2 X_II 3062. +C 2 x 1.+C 2 X+C aX 2 -t 3064.C. !I =-= C J cos (21n x) -{-.t' +- + + + • 3056. y =e. 3059.mt ) Yi5).'(.: -f-C 4 e..a)r= Hint. y=C. The differential equation of motion y=C.3049. y=C. Y ==.6 ::~=gs. ~ CIX~ +.Ca ("('S at Col sin a~ 1ra 1 3055. y::= (C I + C 2 x) cos 2x 3052. y:=. 2 3077. ( J 3066.x lr 2 .Cax s . y=C1COSx+C 2 sinx.e-·o.(C 1-t.C2 cos x 1. 3046. Y= (C 1 -C 2 -C.\5_. (C I -t. 3044. y=x(lnx+ln x).+C~-x+Carr. U _::.C~X+~) eX. + Cs cO~J'K -1. y =-.~ C I -t.C2e-a:~ -1.~2 ( COS--X1--sin-x Y3 2 3068.eX (C I cos x -1.C 2x -1.CaX ) -I-{ x 2 - +Xl ' n x --. y=(x+ I)" IC. y=--CI-rC~e-x I -1.x (C.(C a + G\x) sin x. =ro 2r. ~3 ) . 3050.C 4 x) sin 2x (caCOSTX1-C4SlIl-2-x.':lT X • -~ 2 I }'F 2 -x 3 -5. Y :-.C3 sin x -1 sec x -1.) cos x.. X= v m cos a ( _. z = -(C + Cal e. lr z b) In ==C 2. m ~. t'Ive propor t'Ions.. x = lOe 2t -8e at -et + 6t -.C s C4 • 1 x z=. a) =C . m d.y (C )2' z .. 81 2 2X 2 1 - d-f (x -x-I). r z y x2+ y2 \ve find the first x~+y~ x 2 + y2 1 Whence 1n z == -2 111 (x y X +p2-:-:-: 2 2 + y 2) -1.' 3088.C -xV.\). flint.. Z == -9 (1 _e.C~ From the initial condition<. C 2 3084. 3090 . Applying till'" . Hint.2x. + + + + C 2e.x + :~ +i (31n 2 x+ In x. y==l.-y -r dy -t. 2 2 +Z2 == C2• Thus... =C 2 .t) .z=-=-2. 3085.X 2.. ¥3 t) t+ e--~(-Cs Va-c ¥:3" t+ C2 Y3-C 2 cos 2 2 n 2 t t 2t t 2t x= Cle.~ V'2 Cs cos x C" sin .. we h ave · 0 f derlva proper t les whence dx tlltegral In Then. 3 • Y3-I-C y~r 2 SID -2- s . y = C Ie ~ V 2x Vi. nlilarl y. Y =-=Cae. y=c t +C 2eU + . y = 14 (1-e.: = -kvy-mg for the initial conditions: when . y=(C 2 -2C l -2C 2 x) e. c) x+y+-z-==O.C 2 ={).2e -4-cosx-TslIlx-2e+x. y = .X )-2x (3 + 4e.C Ie .+ C e .x +5x-9. y = C I C2x+ 2 sin x. +y0) + z= 1-2C. z=C e 2 3083..I.X x . x y -I. z==(C 1 +C 2 x) e.X ) + x (5 + 4e. 2 Ie "-f (x!+x).=arc tan JL +C . x=l.!!0 k l-e m t) 'Y=/i2(kv m (_!. x+y dx dtj dz dx+dl/-I-dz --=----=-==--== 0 y-z z-x x.. \ve havl' y~ -~z = x (x~ x-y ) = y (x + y ) hence.InC 2 and. xdx-'-ydtj-t-zdz-=--Q and x 2 fy(z-x) z(x-y) 0 ' I.x -6x+ 14. ya C t+ -{. + (x dx )- x Y-Z z dz x dx II dy z dz Y dy . consequentl y. x-y 1L + CI' x . 3091. Integrating the horno1 x y x2 + y2 Yx geneous equation = arcX tan ~=~. y==C l x2 +-f-T8 (31n 2 x-2ln x). ya. C2 ==4.+ C2e~t.X ). uSing the properties of derivative proportions. x + y + z == C + 1• S. X2 +y2+ Z2==6. 2C 1 C1 * (X 2 +y2)y Z _ • _ 3087.Answers 470 t -~ ( C CosTt+Casln-3081.X ).C .+ C e . \VC will haveC 1 =O. (I O Y 2+y2. the integra I curves are the circles x +!I + z -:: C1. x=Cle +e 2 2 . 3086.Z2 .t = -kv g . z = -2C I -C 2 (2x + I) -3 sin x-2 cos x.J~--== 0 and.20e2t + 8e at + 3et -I. -C Z - 3082..T mg t ' o sina+mg) l-e m Solution. ---C 2 . x ex . CI =9..t ) . C x2 3089.d. 2.12t + 10.(C s Y = Ie e Y3-C Y3 t 2 2 cos -2- C2 t) . y~x..320 X -1. It (0. x -. · y - Il -.- x' ... 3102.. dt 0) • Hint.Answers t=o.3-(3!)q 1.I) ant . y= 5 _ 309. au (x..\ 2h 1 for O·::::x ~ :l" . U -= 21 -2 -f)- It . Use the conoitlOl1s: 11=1 iJu ~/ 211t -1- 0) _-=-0. U (1. Vt' =uosina. 0)= Jt ( . l11tut rt=1 u (0. thc series converges for -l~x<l.· Hint.k y. 00 3105.-.. 3100. U • rzr£x l1n) Sill . Hint. 11-== sin x . 1-22-t-22. x ~ x~ xG ~'I 3101. t)=O. x ==a cos ~r-' i.~r! t 4 .--= . k'2~l nzv 2 o x2 -:. The differential equations of motIon: 3093.16 X (Yo+{) e (X-I)_~ x+~ . 3106 · u ~ ~ n=o An Cos --2-[-(2n -1.. 0) -= 0. tlon~: Ll the (0. It (I. (211 + 1) nx wllcrc tile fn t 1 sIn 2t' coe 11Clf11l S / II =: . r2 x4 XS 3098.k"x. U~C 2 -1. I (/.· · · ••• x4 3097.. U t 6 -1.()! .••• .sin -1-. (x.: 3" 3 X y= dt.4+ . YIll ~ d!t +mg == (kv o sin f1 + mg) em. Usc the conditions: = 1. U"e the Illethod of undctcnnincd corfficH:nts.t . 0) --.+-~.3+a. t)-::-::o. III kV y JIm sin --:. k == V o cos ae til k t. . senes converges for the IlIl.:: 0). y.9 x x2 7 2 X 1 a -1.00. J 3092.t =_0_ _ 471 1 -7.a U ° (x.thod of UI1(ktrrnlinl'd coefflclents. L 1 (I-cos no-a (I-~ t 2 -t2! . Hint. y= -2-2t-x 2 • 3094. III 1. lj-~X-(1·f)2.8! 55 t 8 -. sene~ converges for- 00 <X< + CX). xo==Yo==O.7.< x 1J 2 < I• 00 . IUtX cos -l.\ 1-3. (x.2-i-(2!)2. Hint. 0) --= A' sin nx T' iJu dt 00 3104. = Integrating. . I for. l'Jse t 1h~ condi- ) t)-:--o. thc series conven~es for -00< <x< -1. 1 3! 3099.-.l .. t)=O. Ii (l.. l'XO=vocosa. Y --= ) - X S )·4 1· 4·7 -t.27 X II - x3 4 21 5 -1. .. t) =-= 0.SUI -l.+T:2-1-2. . we obtain v x k 0 . the . Biz U == 312 ~ ~ 1 • 11:rl 1t2 SlIt 2 ll:Tat.4~-2l .~ 1 • Y =. t) --= 0. T :rl X · 31 03 · u =-= A cos art -l-t S1I1 · H·Int.6! x 6 - ~}-.-:: k t' 3096 2 1 2" -f-"4 X -1. 112 ..P.-=- u" d2~ dt 2 =-= .:-=. 2 9 + 32 X -t. Hint. Use the 111ethod of undeternlloed coefficients. u (x. r In l.6~-1.. 15.7°/0 (on the principle of equal effects). Take the number 1t to three decimal places (on the principle of equal effects).01.44±0. c) 9.Answers 472 1 x = T S-1 2 .480. t) == 0.8+0.3 y /).001.1 In 2 • 3116.5. 3124. 3126.16°/0.3 enl. b) ~ I mm. b) two decimals. b) 4. c) the result of subtraction does not have any correct decimals.108 or 49.1. a) 1.7±0.5 or 18 47 ±0. 48. Chapter X 3108. 1. u = au (x. (2.2.01 x (laO-x). u (x.9±0. not a single decinlal place in the quotient may be con~id­ ered certain.01. a) 1. Measure the radii and the generatrix with relative error 1/300. a) 84. 3117. a) 29. t) == 0. 6 5 crn!. ---ax=0.1.01. c) 2. Hence. 3128.05±0. 3115.003. since the number lies between 47. Use the fornlula for increase in area of a square. 0.006±0. 31 I t.05.3±0. ~0.0±0.=---=26°15' ± ±35'. 3112. b) 0. 3121. 3114. c) the quotient nlay vary behveen 48 and 62.9±0. 0) iJt o.8±0.10 3 cm 2. 3110.0023°!o.2 cm.4 y ti 5 y 3 7 -2 -6 14 -23 2 10 5 -8 8 -9 3 15 -3 0 -1 4 12 -3 -I 5 9 -4 6 5 x . Relative error. ~ 0. a) 30.01)·103 cm 2 • 3120. and s to within 0. Hint. Hint. 3123.877 and 48.l. 3119.20/ 0 . ~0. ~0.45% . a) ~ 0. c) 43. 19. sin a=0.10 2. b) 43. a) two deciInals. b) 277.16°/0.0005. t) x u (0. 3118.006. u (x.26°/0.845. 4.1729.025±0. b) 1.005. c) ~ 0. The side is equal to 13. ~ 0.10 3 . The length of the pendululll should be measured to within 0.1). c) ~ I gm. c) one decinlal. b) 18.6-102 . y /).y &2y /). The last digit 111ay vary by unity.1295±0. 0) 1 3107. U (100. 0) = 0.10 3 . 2 7 ±O.2.3 cm 2. Use the conditions 21 Sill ° au (1. 0. 3127. 0.0002. ~ 1.021%~ b) ~ 0.3122.e · n=l Hint.648. t)=O. a 2 n:!j't2t CJ) 400 ' ~ I na ~ fi3 . take the nurnbers :t and q to three decinlals (on the principle of equal effects). c) 0. since the difference is equal to one hundredth with a possible absolute error of one hundredth. 31131&. Use the conditions: II (0. l\\easure the quantity 1 to within 0. 6.27 ampere 3125. (1.120±O. Absolute error. nnx -1""002 (I-cos n1t) Sin 100 . a) ~ I". 3109. For practi~" purpo"es there is sense in writing the result in the fonn (5.OOI6%. a) 0.102.2. (2n+I)1tx dx. 473 ·4nswers J129.\ .:. .1 (nl0ving -=.-------- I I Hint. After this the remaining part of the table is tilled HI by the operation of additlo.. after obtaining L\4 yo 24.- I .. y &y /).I. from right to left). .2 y flSy -4 -12 32 48 3 -16 20 80 4.Q fi 4 100 12R 48 7 104 228 176 9 332 40t 11 7:36 x •• 3130.. Compute the first live values of y and.\ :l IJ I . repeat the nUlllber 24 throughout the C01Un1l1 of fourth differences. 1 5:) 1. 0. 1t 2 fOlm the formula of the arc length to the form ~ VI-Ezcos z t·dt. 53 ~in 2x . Y2(X)==6 + 2 -x + 1. Z (0 5) = . 3161. 3143. + + + + +.211. Y= 1 \vhen x==O. 36 . 3169. b) Y(0.5) --:-:3 15.83. 3144.09.01. 3166. Yl(X)=X.:~I and 4 2. 1.:~·10-5. y = :6 +~~ XZ-~x+8.: . 0. 3193. 3141. where E .x + 1.0.3. 3188.1993.14. 4.271 cos 3x -I.15. Hint.851 cos x 0. 3181 .5)=-~. -=0. 3178. -0. 3171.83. Y (0.1. 3186.506.11. 0. 3 t 87. 3135. 3146. 1. 18.325 3138.490. 3145.2. 3136. Z ( 1) == 2 .13. x3 .]242 cos 3x-33 S111 3x. y== -0. y-==O 56. x~-~~ Xs + y=::::22 for x=5. + + + + . 3162.542 cos 2x 0. 3191..3 15. a) y (0. 0.414 cos x 0. 1.0.. -1.02. 1.1. The interpolating polynolnial is y =-= x 2 . 72 3182.100 sin 3x. 3175.79. a) 0.80.67. 0.75.72.14 SUI x-l.076 sin 2x 0. x~ -0.5°/o~ 6. 3184. 0. O..5. 3147.67. Yl (x)==2. 0. 0. 0. Y== 1 22.67. y=20 for x=::::5.lOx -t. 3172. x=0. 6 49-1 96 cos x -t.399. 3173. 3149.10. 2.664. Y (0. x (It) . Hint.5).389. 0. 3155. 11. 0. Y:::: 1.85 3170. 158 kgf (approxilllately).04 sin 3x.80 . 1 27 3148. 3150. z (0 5) --. 1. 3180.2165..660.0 1822. Y (1) == 3..491. b) 0.73 and O. 55.88. 1 29. o the eccentnclty of the ellipse. 1. 0.5) = 0 .165. Bv the trapezoidal formula. 7.x + 1.84.28.O.0.245 and 0 019. ± 1 1997 3160. 3163. 429+ 1739 cos x-l0:37 SUI x-6:321 ros 2x -t. 3183.79. 4 493. 3174.2334. 63 ' Y3 (x) -= ""3 -tx4 3x 2 + 63 +2079 + 59535· 3177.84. 3153. -1. 3151. 13 cos 3x 0. -0 995. When computing x for y=20 take Yo== 11. z3(x)-==T3 2 3 x3 x3 XS -2x2 +3x-2. 0 24.0.0. -1.31 and -0. Y2(X)=X-tf' Y3(x)~x-6+1~0· 31 79.1. 0. Make use of the parametric equation of the elli pse x ~ cos t. 3159.960 0. 161. 1 + x + XZ + x s• 3134.0. 3190.2503. 0.338 0.12x 3x x' 7x -"6-l-2.:. 1:38 3156. 3167.126:J Sill 2x.0. y(2)=1l. 3152. 3137. Yl (x) ==:3 'Y2 (x) x7 2x JJ x2 XIS XS x3 x7 x3 == 3" +. 3158. == 1. 6.2. Zt(x)=-:3x-2.022 sin 3x. . z2(x)-==1f-2x2+3x-2. 11 417. y(2)=-3.620 sin 2x 0. 0. Y (1) === 3 .86.56 x== 1. 3176.6222 sin t r'::11d tran~- 3140.056 cos 3x.0. 3133.0. b) 0. 16. 3.79. 3192.625: by Sinlpson's formula. 3157. Y -== 0.0 . -1. 0.93. 0 69 3164. 3168.111 cos 2x 1. 2.915 Sill x -I0. 0 87.6H Cos 2x -t-t. 3142. 3185. a) () 608 ~In x 1+ 0. Y (2) == 0 .005. 0. IS . !/:s(x) =-:-. 58~ x' (rt) --=0. 0 35.09. 3165. 72. 3154.3132.25.3068.Answers 474 3131.0. 3139. 3189.005.229. Xi-3~ Zeta -Z~ Eta .7182R 0.ry Delta-8b Rho-PQ Slgnla -~O' Tau.28318 :~ 0 g r 56571 .99167 :t 2:t j( "2n 4 1 rt V.16572 0.r" Phi -(Dcp Chi-Xx Psi .99130 -=lnl0 1 M 2.APPENDIX I.86n60 0.4342) t .81 2.'l'~' Oll1ega-Qw LaInbda-AA Mu-l\f~l Nu-Nv Epsilon-Ep.30258 O.6:)778 n2 9. Some Constants x Quantity 1 log x QU3nt It) I I I x I log x 0.89509 V-. 1 39561 0.II.49713 G.79818 I e el 0.3890fi 1.21715 0.46459 2.57080 0.Tt Upsl1on.24857 0.Bri Ganll11a .19612 Ve I.14-176 -- 0. Ornicron-Oo Pi-Iln Theta-eO II.~3G222 Vn 1..24188 0.77245 1.31831 t . e 14159 0.R6859 0.43429 1 radian arc 1 57°17'45" 0.78540 I . Greek Alphabet Iota-It Kappa-}(x Alpha-An Beta.01745 9.50285 1\1 = tog e 0.G48i2 O.36788 7. 472 4.6487 1.00 29.733 3.549 1.588 1.383 6.063 1.357 1.714 2.263 3.542 6.191 ~!.668 2.831 5.744 3.890 3.4 0.504 1.16 1.49 19.66 50.960 2.5 0.3 0.574 1.464 3.000 7.899 5.3 0.768 6.17 13.1632 1.222 4.626 1.357 2.410 2.9933 1.962 3.8G6 6.8329 0.4110 1 4351 1.342 1 378 1.145 1.250 6. Logarithms xl I x 1.606 3.87 59.448 3.289 2.196 5.239 1.832 6.458 1.351 2.909 1.440 1.7 140.409 1.6 117.5261 1.18 91.213 4.410 9.694 6.5 0.261 10.526 2.56 12.312 3.0 1.791 4.000 9.192 5.619 6.170 1 193 1.313 5.2 0.36 20.292 5.3083 1.265 1.769 1.719 7.775 7.000 4.303 3.520 2.25 12.0953 0.64 18.4 00 0.258 2.463 6.362 3.5686 1.243 7.208 4.0296 1 0647 1 0986 1.932 5.7419 0. I 0.916 6.500 2.196 5.400 2.09 79. Inverse Quantities.62 17.98e 7.331 1.162 3.270 3.732 1.477 5.4 0.1 0.324 lOOx 1.34 103.12 97.489 7.606 7.233 4.873 4.571 2.323 3.240 3.634 3.332 3.924 1.663 1.250 2.225 1.278 3.5 0.071 7.'()4 28.744 1.0 0.435 2.32 64.796 4.01 25.474 1.3 0.2 0.844 1.Q24 2.8755 0.732 1.119 1.51 85.420 3.9555 0.663 7.7885 0.871 1.6292 1.339 4.2809 1.037 3.183 1.240 3.243 4.489 1.~9 2.204 5.09 23.533 1.642 4.601 1.560 1.754 2.168 2.141 7.610 4.476 2.8 0.410 4.414 1.25 21.5306 o 5878 0.949 1.7 0.913 5.6 0.708 6.39 27.5 0.000 1.848 5.096 4.325 6.476 3.339 1.517 1.7 0.154 2 224 2.2 0.609 3.0 4 2.210 1.5041 1.651 1.687 1.944 6.1 0.302 3.000 5.200 5.690 1.4 0.568 5.4 0.426 1.990 8.127 6.675 1.760 6.032 1.290 5.96 13.659 3.844 2.123 4.65 54.9 157.88 46.859 8.2 0.6419 0.00 16.780 4.320 1.6677 1.145 2.840 8.817 1.294 3.4~55 0.000 2.306 7.557 3.164 6.633 6.114 7.025 3.449 1.928 7.481 6.5476 1.0 0.6 148.1939 1.3365 0.280 7.290 7.708 3.6 0.79 32.385 5.65 12.7 0.721 1.098 2.530 3.211 7.646 5.6854 .454 2.745 5.625 1.271 3.703 1.189 5.581 1.245 6.244 4.9 0.9163 0.756 3.3 0. Powers..884 7.403 6.89 11.610 10.556 1.141 3.250 4.547 1.0 132.6931 0. 4816 1. 214 2.1823 0.92 74.107 3.091 1.613 1.7 0.5 I I Iv: IVloxl V-IIOgX I Vx VlO x 1.192 5.236 2.04 24.208 3.041 8.4586 1.8 110.94 *39.429 5.3350 1.2624 0.391 3.nantlssas) 0000 0414 0792 1139 1461 1761 2041 2304 2553 2788 30rv 3222 3424 3617 3802 3979 4150 4314 4472 4624 4771 4914 5051 5185 5315 5441 5563 5682 5798 5911 6021 6128 6232 6335 6435 6532 6628 6721 6812 6902 6990 7076 7160 7243 7324 In x 0.840 5.093 8.684 3.503 3.000 9.910 6.897 1.6 125.0 0.643 1.260 1.304 1.81 17.82 15.698 1.280 2 302 2.286 3.2238 1. Roots.9 0.00 26.000 1.21 16.95 24.00 68.281 1.300 6.69 14.742 3.830 7.30 42.095 1.789 1.385 2.256 4. 1 0.560 2.000 1.548 7.197 2.557 6.442 1.639 1.749 5.2 0.16 22.540 5.216 1.840 6.333 3.000 4.518 1.975 2.1 0.833 1.4 0.2528 1.761 1.01 27.612 1.58 19.370 2.782 6.760 7.621 2.8 0.066 5.09 29.000 6.24 10.000 1.657 5. 3610 1.728 2.884 2.317 3.179 7.483 1.217 4.710 1.8 0.214 6.44 15.3 0.049 1.466 2.476 Appendix III.9 0.583 4.937 7.583 3.77 35.0000 0.6 0.3863 1.5892 1.690 4.227 4.368 7.714 1.175 3.375 4.6094 1.0 0.6 0.8 0.238 4.392 1.667 1.759 2 802 2.185 I X2 X3 / 1.348 2.037 6.1314 1.345 3.417 2.000 3.121 2.673 1.375 1.587 1.140 1.68 21.072 3.9 0.143 (t.083 6.000 1.429 7.301 1.099 5.047 7. 166 8865 2.9 51.983 4 273 I 992 4 291 2.765 3.2618 9 899 9868 2 2721 9.61 531.546 9395 2.24 551.1401 9 510 9345 2.2083 9.9 6.9 S.00 216.139 o 137 0.1 75.185 8.1518 9.109 9.1163 9.485 8.435 9243 2.143 0.950 10.881 2.5 79 21 705.198 1.3 7.510 2.747 9.871 1.470 2 490 2.29 389.106 9.100 t''J I x" I 30 25 166 4 31 36 175.888 8.849 9.:3 0.1041 9.434 7782 1.1~ 884.0541 9.120 o 119 0.00 1000.937 8.2513 9 865 ge23 2.1 42.89 571.1 6.095 4.062 8.0 81.1 7.04 941.165 9.135 0.0 38.550 2 569 2 588 2 608 2.975 4.045 8692 2.0281 9.004 8633 1.56 287 5 44 89 300 8 46 24 314 4 47 61 ~i28 5 49.84 474 6 62.1633 9.156 ~.807 3 893 1.547 2.344 2 025 4.116 0. 58~~ 9445 2.796 9731 2 2407 9.000 8.0669 9.098 3.44 2a8 3 39.4 88.7579 8 387 7709 1.147 4.627 2.41 49~3.103 9.175 0.000 4.8 5.118 0.922 4.944 9.739 2 757 2 775 2.5 8.041 4.957 4.44 681.126 8808 2.41 357.7 5.244 8976 2.0 82.118 4.595 2.0015 9.966 4.481 2.579 2.832 8.4 92.1 7:3.397 2 049 4.0 0.967 9956 2.803 1 776 3.702 2.327 2 017 4.246 8 307 8.8405 8 618 8062 1.00 729.847 3.0794 9.64 195.8 7.4 7.122 0.0149 9.933 9912 2.7047 8.786 3.309 2 008 4.958 1.1282 9.322 9085 2'.915 I 827 3.057 4.431 2.130 3.7 0.7 72.544 8 602 8.217 1.7 86.033 4 380 2.610 2.660 8.2 0.086 8751 2.0919 9 360 9138 ~.172 o 169 0.033 3.01 970.5 77.691 9590 2.681 7 746 7.830 9777 2.81 753.949 4.0 54 76 405.514 2.2925 10.857 4.473 9294 2.8083 8 527 7924 1.979 1.179 0.25 421. 0 64.7405 8.864 2.498 2.367 8 426 8.563 2.9 57.655 9542 2 1972 9.6 84.2 8.465 2.362 2.940 4.915 2 933 2.08~ 1.8563 8.243 7482 1 7228 8.36 830 6 90.102 1.Appendix 471 Continued x 5.110 4.398 9191 2. - .164 o 161 o 159 0.1 34 81 205.720 2.936 1.' 14 3.141 1.124 8. 4 68.4.707 8195 1.G 8.6 7.0 v"x 2.8 89 1+1 o 182 0.5 56 5.121 1.108 9.683 2.963 8573 1.866 4 021 I 876 4.00 512 0 6G.8718 8.3 6 4 6.9741 9.487 9.151 0.041 I 885 4 062 1.1748 9 619 9494 2.817 3.2300 9.146 3.69 658.4 36.0 7.2192 9.0 6.416 7 483 7 550 7 616 7.4 67.220 9.3 100.2 98.080 4.1 0.7 8.849 1.950 2.179 1.1861 9.798 9.000 1.283 9031 2.193 7404 1.205 8921 2.09 912 7 96.141 0.340 7634 1.898 2.2824 9.101 10.000 0000 2.147 o 145 0.642 IOgX lOOx (~a n- lOx t ls~as) In x 8.7 7.76 439 0 59 29 456 5 60.8 70.539 9.84 373.23 614.133 0.7918 8 481 7853 1.931 4.149 0.a27 9.7750 8.811 2.966 2 983 3.114 o 112 9.530 2.105 9.381 9.91a 4.56 592.5 7.104 9.662 8129 1.837 8388 I 9315 8 879 8451 I 9459 8 921 8513 1.3026 .64 778.1 8.2 56.8871 8 750 8261 I 9021 8 794 8:325 1 9169 8.837 3.274 9.797 3.5 66 6 7 68 69 7.236 1.103 4.154 .000 9 055 9.646 2 665 2.0 O.162 I IV-lv-/v-/ I VTOX x' 7.2 53.254 1.4 0.644 9.00 343 0 50.~~ 154 0.017 3.088 4.761 9685 2.160 1.102 9.775 8.96 636.082 3.133 4.23 274 6 43.448 2 072 4.899 9.9879 9.110 9.000 3.0412 9.69 250 0 40 96 262.3 8 4 8.895 4.49 804.125 4.065 4.793 2.167 0.810 7 874 7.000 1.828 2 846 2.695 9.115 0.2 7.626 2.7 94.0 37 21 227.5 0.O 8.6 0.826 1.291 7559 1.449 2.9601 8.8 0.110 9.904 4.132 o 1:\0 o 128 o 127 o 125 o 12~ 0.2 6.25 857.050 3 066 3.9 0.140 4.726 9638 2.8245 8 57:3 7993 1.592 9.43·1 9.6 32 49 185 2 33. III 9.718 8.531 2.414 2.345 2 366 2 387 2 408 2 429 2. 8660 0.5411 0. 244'~ 0.2126 0.1106 1.4887 0.475 2.3142 0.3584 0.7431 0.0698 1 2 3 4 5 6 7 8 41ln .3491 0.7071 9 tan I 90 89 88 87 86 85 84 83 82 RI 80 I 79 78 1.8910 0.5299 0.9925 0.9774 0.2079 0.0297 I .6283 0.1405 o 1584 I cot x 57 29 28.8829 0.5934 0.4281 1.3746 0.7002 0.4226 0.1908 0.0175 0.0175 0.8G72 0.271 3.7679 0.1222 0.2349 1.2915 1.GOO:3 o 64q4 rot x I 00 0.4067 0.6458 0.2793 0.0000 0.5592 0.:.9948 0.32G6 0.3840 0.1736 0.1504 1.5399 1.9994 0.9423 0.176~ 0.1918 1.5000 0.5878 0.2493 0. I ~61~ 0.1047 0.6643 0.0175 0.8480 0.904 2.2588 0.2618 0.9998 0.52:36 0.4040 0. 7~)86 0.9135 0.n647 1.8378 0.2756 0.0349 0.3764 1.8098 0.0872 0.8090 O.3090 1.3640 0.0875 0..0 I 0 x (fad lans) I 0.1228 0.4826 1.3839 0.8716 0.747 2.8901 0.2799 1.8727 0.0000 0.) 64 ():~ 62 61 60 59 58 57 56 55 54 53 52 S! 50 49 48 47 46 45 ~ ..994S 0.6018 0.145 2.145 4.6109 0..4189 0.9639 0.48:35 1.43 9.1396 0..0000 0. :~~~1 4.052:~ 0.0349 o .9848 0.8203 0.514 8.8391 0.6009 0.0724 1.3249 0.4245 0. ~~057 0.4538 .1170 1.9~O5 0.9511 O.7880 0.1745 0..4014 0.08 14 30 11.5708 1.144 7.81Q2 0.2679 0.356 2.~44:3 0.6249 I tan x I 0.0()9~ 0.961:3 O.3316 0.9903 0.9877 0.89R8 0.732 3.1219 0.0472 1.9703 0.2094 0.9599 0.~ 0.7330 0.55:33 1 5:~5q 1.8552 0.:~420 0.0355 1.OI2:l 0.487 3.7536 0.5760 0.605 2.3263 1. o 4384 0.011 3.64 19.4540 0.5774 0.-_---------_.6293 0.34..3090 0.4363 0.9744 0.9816 o 9781 0.7854 0.7660 0.0000 0.6561 0.2924 0.0349 0.1051 0.246 2.9250 0.6745 0.8693 0.0000 cos x 1.5446 0.6807 0.0524 O.4695 0.3270 1.7771 0.5543 0.9<)76 0. 37~8 .0698 0._- IV.0981 0.0821 l.5736 0.078 2.6632 0..7265 0.7193 0.671 5.2867 O.fiOl0 1.5150 0. 9.478 Appendix ---------------=-.9986 0.945[) o 9397 O.5317 0.4661 1.7547 0.R81 1.156t 0.5061 o 484S 0.8387 o 8290 0.1691 1.050 1.3907 0.0000 0.7505 0.1868 1.4-lR6 1 4:H2 1 41:37 I ~963 l. 5184 l.963 I. 956.906:3 1.5585 0.1944 0.9962 O.2419 0.2741 I 2566 I 2:392 I 2217 1.0996 1.\ I 0.7854 76 (rad~al") I 75 74 73 72 71 70 69 68 67 66 6.0524 0.7156 0.1519 I 1343 1.6820 0.1045 0.7314 0.5095 0.2967 0.7071 I ~in x I 1. Trigonometric Functions .~9 77 1.3665 0.2250 0.087~ 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 I I cos x x I 0.1392 0.1571 0.9076 0.9004 o 9325 0.703 4.115 6 314 5.4452 0.1920 0.2309 0.7813 0.6428 0.2041 1.6947 0.9657 1.6157 0.8029 0.2269 O.4712 0.4663 o 4877 0.6691 0.804 1 732 1.9272 I.t~6 0. ~~):~ 122:> IIn8 () 10n:~ o 0907 24 11 2 .0550 6.9093 -0.1146 0. :3:350 -0.1700 15 4 4817 1.5 33.6967 0.~7 11.86~~ -0 5048 o 8085 -0.0997 0.9a87 0.1 17 I H 1~ 2 () ~ 1 22 2 :~ • o :3329 () 22al ~.4108 1.0496 h 68fi9 n 974~ 1.3233 0.2913 0.0416 -0.1979 24.4794 0. 537n 14.0408 0.9854 0.201:3 0.9320 o 9636 0.1288 -0 2272 -0.0302 16.23H2 -0. I I sinh x I cosh x I tanh x I sin x I cos x 0.0000 1.4966 o 4493 0.8011 o 5155 -0.9991 -0.0000 0.9668 --0.8569 0.9':>:lO -0.5646 0.0764 12.8617 o 8834 () () () O~.9704 0.6685 0.9738 n 9463 :~ 7622 4 .9365 .:3499 I 4918 1 0000 o 9048 o 8187 0.9973 0.1309 7 ~~8~} I H 1662 ()~.7833 0.3012 o 2725 o 2·166 :>.4066 o 5211 o 6367 o 7586 o 8881 1.5431 1.9996 0.0450 0.6663 (1 6753 -G 7374 0.1215 12.0502 6 6(}47 7.1 I e-.1052 I 2214 1.8221 2.1154 0.J.6703 0.7648 0.6640 0.29n3 0.0811 0.6044 0.0050 I 0201 1.0334 10.3:374 1.2 3.~~ 0.0453 1.2552 1.5403 0.)3 2 4596 0.1 3.7408 0.1411 0.12:~ o ~)890 0.9370 5. 4i~l5 8 2f)27 9.03()9 0.2555 0007071 -0.0042 :l 3201 :~ 66 3 4 O:l52 0.9983 -0.3894 1.6442 0.5748 14.8 o 9. 718:~ 3.5426 16 05728\ 0.0000 0.2459 I~l.9468 o 9:)62 0.9940 3 4 20.3679 o 8415 12 1 :~ 1 ·t 2.9926 0.1.6456 2.0292 -0.U8ul 0.1075 3 4177 0.) 26 27 2.9354 0.9801 0.5325 27.0596 6.Appendix 479 V.4621 0.0677 11.0138 2.9041 O.98:37 0.5885 o 74.8797 16 4·146 18.1741 0.9709 2.8005 1.3524 2 5775 2 8283 :3.3356 o 0608 6. Exponential.2675 0.9866 0.9422 :~.7690 7.3 o 8912 o 8337 0.22.1853 1.4536 0.47:39 6.7163 0.8107 1.0000 0. ~640 0.O 3.9875 -0.0000 1.9654 10.7174 0. 4().9950 0.1919 9.9710 o 2019 I.085!l 22.6063 0.3045 0.1987 0.0372 5.4274 -0.9959 0.3508 -0.6981 1.9978 3.9967 0.12n3 2.5370 0.1974 0.3799 0.5985 -0.5569 0.0821 o 0743 o 06i2 2 3756 2.6216 0.8776 0.2682 :~ 6269 4 0219 4 4571 4.5095 1.0584 -0 1577 -0.0000 0.9422 0.5488 0.6 ·t.9211 o 5 o6 o 7 0.9982 2 9 :l.9051 o 9217 0.991U 0.4161 O.9950 0.8253 0. 1 6487 1.1002 0.0179 11.9553 0.1276 1.9917 0.4331 0.so o 7616 1.2 o :l I 0 1. Hyperbolic and Trigonometric Functions e"'C I x o 0 0.144:~ o. 9043 o 1827 () 1653 o 1·1<)6 q 1.9973 0.0998 0.) I:J.:17 -0.46G2 4 567<) S.4063 8.9641 0.0263 1.0498 0.1126 29.4 1.8 1~ 182.9900 -0.9757 D.1752 1.23G6 13.3624 0. 1 +x 2 • y o 6. y == Xl. Graph of a fractional function.480 Appendix VI. y=¥x. Cubic parabola. 1 y=-X2 · -1 0 1 5. Parabola. Some Curves (for Reference) y y o X 3. The witch of Agncsi.. Cubic parabola. Y:_-=x· "''"' y -1 0 x 1 4. 1. 7. Parabola (upper branch). Rectangular 2. y=x 2 • hyperbola. V x .:. 11= a -x. 1 y. 0 x . f x=t'l. Tangent curve and cotangent curve. rtf -f 9. X: - t-. Y :..481 Appendix y y o x Bb Sem icubical parabola. y . y=fanx and y=cotx. 10.:: t' a or I x~ is . Sa.-= SA:l x . y2=X' or ' \ y=t'. Si llC curve and cosi ne curve. Neile's parabola.]nd 'I = cos x. Appendix 11. Graphs of the inverse trigonometric functions II = arc sin x and y = arc cos x. Gra phs of the f unct ions y -= sec x and y = cosec A. y=arc sin :r x 12. "" . . .. Gra phs of the Inv~rse trigonoInetric functions y =.. Graphs of the exponential functions y=ex and y=e.x ... y -1 o 14..Appendix 483 y y=arc -----..- tan:l ------0------1[ 13........arc tan x and y = arc cot x. x .----'-. . x=a cos 3 t. Bernoull i's lenlniscate. a-x x y x o 26. r=a (1 +cos <pl. y 29. x=a (t -sin t). (x 2 _I.y2r~ = a 2 (x 2_ y2) or r2 == a2 cos 2<p. Strophoi d. 27. IIypocyclold (astroid). Cycloid. a+x y 2 =x2 -. 28. 24. { y=a (1.486 Appendix x x 25. Cardioid.\ a 2 + 11 a = a 3-.cos t). { y = a sin a t 2 or . Evolvent (invol ute) of the circle X = a (cos t + t sin t) t { Y =a (sin t -t cos l). .. r =a sin 3tp.. ea'l'.. 34. a r=-.._ . x 32..481 Appendix x tI--. Four-leafed rose. q> r =aq>. (-= 33.._~ X ..... Spiral of Archinledcs... Logdrithl11ic spiral.30." " ' _. . Th ree-Iea fed rose.. Hyperbolic spiral... r = a sin 2<p......... 31. 105. 105. 256 /\rea of a pI ane region 256 Area of a surf ace 166·168. 486 Asymptot~ 93 Ifft horizontal 94 l~ft inclined 94 right horizontal 93 right inclined 93 vertical 93 B Bending point 84 Bernoulli's equation 333 Bernoulli's lemniscate 155. 393. 297 Acceleration vector 236 Adams' formula 390 Adarns' method 389. 484 Catenoid 168 Cauchy's integral test 295 Cauchy's test 293. 63. 486 Catenary 104. 480 Broken-line Illethod Euler's 326 c Ca rdioi d 20. 65. 326 Cavalieri's "leInon" 165 Centrc of curvature 103 Change of variable 211-217 ina de fi nit e in t eg raI 146 in a double integral 252-254 in an indefinite Integral 113 Characteristic equation ~~56 Characteristic points 96 Chebyshev's conditions 127 Chord method 376 Circle 20. 104 of convergence 306 of curvature 103 osculating 103 Circulation of a vector 289 Cissoid 232 of Diocles 18. 105.\rea in polar coordinates 155. 394 Comparison test 143. Beta-function 146.Archimedes spiral of 20.480 Algebraic functions 48 Angle bet\veen two surfaces. 156. 150 B inorm al 238 Boundary conditions 363 • Branch of a hyperbol a 20. 66. 141 generalized 143 Approximate numbers 367 addition of 368 division of 368 rnul ti plicat ion of 368 powers of 368 roots of 368 subtraction of 368 . 243 Angle of contingence of second kind 243 Antiderivative 140. 49 486 . 385 J\rc length of a curve 158-161 J\rc length of a spacE' curve 234 . 487 . 295 Cauchy's theorcln 75. 392 Agnesi Witch of 18. 259 Argument 11 Astroid 20. 219 Angle of contingence 102.INDEX A Absolute error 367 Absolute value of a real nUlnber 11 Absolutely convergent series 296. 105. 256 Area in rectangular coordinates 153. 485 Clairaut's equation 339 Closed interval 11 Coefficients Fourier 318.Approxil11at ion successi ve 377. 293. 390. 294 Composite function 12. 480 Curl of a vector field 288 Curvature centre of 103 circle of 10:~ of a curve 102. integration of 202-204 Differential equation 322 homogeneous linear 349 inhotnogeneous linear 349 Differential equations first-order 324 forming 329 higher~rder 345 linear 349. 351 .484 sine 481 tangent 481 Cusp 230 Cycloid 105. 385 Conditions boundary 363 Chebyshev's 127 Dirichlet 318. 242 radius of 102 second 243 Curve cosine 481 cotangent 481 discriminant 232. 270 Convergent serie~ 293 Coordinates of centre of gravity 170 ~en('rali7ed polar 255 Correct deci n1 aI pI aces i 11 a hroad sense 367 Correct decilnal places in a narro\v sense 367 Cosine curve 481 Cotangent curve 481 Coupling equation 223 Cntical point of the second kind 92 en tical points 84 Cuhic parabola 17. 363 Conditional extrelTIUm 223·225 Conditionally (not absolutely) convergent series 296 Contingence angle of 102. 243 Continuity of functions 36 Continuous function 36 prop·erdes of 38 Convergence circle of 306 interval of 305 radius of 305 region of 304 unifonn 306 Convergent i111prOper integral 143. 106. 232. 21. 234 Gaussian 92 integral 322 logarithmic 484 489 probability 19.Index Coneave down 91 Concave up 91 Concavity direction of 91 Conchoid 232 Condition Li pschi t7. 485 Deternlinant functional 264 Determining coefficients first Inethod of 122 second method of 122 Diagonal table 389 Difference of two convergent series 298 Differential of a 11 arc 10 1. 105. 319 initial 323. 234 first-order 71 higher-order 198 pri nci pal properties of 72 second 198 second -order 72 total. 486 D D' Alembert's test 295 Decreasing function 83 Definite integral 138 Del 288 Dependent variable 11 Derivative 43 left-hand 44 logarithmic 55 nth 67 right -hand 44 second 66 Derivative of a function in a given direction 193 Derivative of functions represented parametrically 57 Derivative of an implicit functicn 57 Derivative of an inverse fUllction 57 Derivative of the second order 66 Deriv at ives of higher orders 66-69 one-si ded 43 table of 47 Descartes foliuln of 20. 234. Lagrange' s 339 Laplace's 289. 485 Direction of concavity 91 Direction field 325 Dirichlet conditions 318. 270 Di vergent seri es 293. 104. 21. 294 Domain 11 Domain of definition 11 Double integral 246 in curvilinear coordinatps 253 in polar coordinates 252 in rectangular coordinates 246 Double point 230 coupling 223 difTerenti al 322 Euler's 357 exact differential 335 first-order differential 324 homogeneous 330. 483 F E Elimination met hod of 359 Ellipse 18. 20. 83. 351. 485 Energy kinetic 174 Envelope equations of 232 of a family of plane curves 232 E picycloi d 283 Equal effects principle of 369 Equation Bernoulli '5 333 ch·aractrristic 356 Clairaut's 339 Factor integrating 335 Field direction fiel d 325 nonstati onary seal ar or vector 288 potenti al vector 289 scalar 288 solenoidal vector 289 Field (cont) stationary scalar or vector 288 vector 288 Field theory 288-292 First-order differential 71 First-order differential equations 324 Flow lines 288 FI ux of a vector field 288 Folium of Descartes 20. 232. 485 . 351. 319 function 40 series 295. 222 356 327.Index 490 Differential equations of higher powers first-order 337 Oifferenti als met hod of 343 of third and higher orders 72 Differentiating a composite function 47 Differentiation 43 of implicit functions 205-208 tabular 46 Diocles cissoid of 18. 328 326 335 55. EX\remal point 84 Extremum conditional 223-225 of a functi 011 83. 349 inhomogeneous 349. 291 linear 332 of a normal 60. 218 of a tangent 60 of a tangent plane 218 with variables separable Equivalent functions 33 Error absolute 367 limiting absolute 367 limiting relative 367 relative 367 Euler integral 146 Euler-Poisson inte~ral 272 Euler's broken-line nlethod Euler's equation 357 Even function 13 Evolute of a curve 103 Evolvent of a circle 486 Evolvent of a curve 104 Exact diITerential equation Exponential functions 49. 234 Divergence of a vector field 288 Divergent improper integral 143. 296 theorem 318 Discontinuity 37 of the first kind 37 infinite 38 removable 37 of the second kind 38 Discontinuous function 270 Discriminant 222 Dicriminant curve 232. 356 honl0geneous linear difTerential 332. reduction 130. 116. integration of 135 trigonometric 48 trigonometric. 281.\aclaurin's 77.' Higher-order partial derivative Hodograph of a vector 235 H0l110geneous equations 330. 483 hy perbol ic 49. transcendental. 297 HIgher-order dIfferential 198 l-ligher-order dIfferential equation'. 351. integrating 128. integration of 133 inverse 12 Functions (cont) inverse circular 48 inv('rse hyperbolic 49 inverse trigonometric 482. properties of decreasing 83 DIrichlet 40 dl'icont Inuous 270 ('veil 13 (If C1 function 12 inlpl1clt 12 increasi n~ 83 491 48 equivalent 33 ('\ ponenti al 49. 296. ~75 Newton's interpolation 372 ()strogradsky-Gauss 286-~88 parabul1c 382 Sln1p~on's Lagr~lnge 39~3. 319 Increasing function 83 Increment of an arguI11ent 42 Incren1ent of a function 42 Independent variClble 11 Indetenn inate forll1s evaluating 78. 483 Ii nearl y dependent 349 linearly independent 349 Han11ltonian operator 288 Harn10nic series 294. 49 contInuous 36 cont inUOll':>. :~8 223. 349 llyperbola 17. 129 Fundalnental system of solutions 349 382-384 Stokes' 285. 141. 282 Guldin's theorenls 171 H Illultiple-valued 11 periodic 14 11 vector 235 F11nctional detern1inant 264 Functional series 304 Functions al~ebraic logarit hm ic 49 . 282 Lagrange's 145 Lagrange's interpolation 374 Leibniz 67 t. l-Ion10geneous Iinear difTerent Ial equation 332. 79 .485 rectangular 480 l-Iyperbolic functions 49. 319 Four-leafed rose 487 Fraction proper rational 121 I'unctlon 11 COllI posite 12. 220 Nc\vton-Leibniz 140. 289 T a\' I or's 77. 270 Incoillplete Fourier series 318. 105. 225. Ilypocycloid 283. Fourier series 318. 224 sin~lc-vallled 394 G Gamma-function 146. 484 integration of 133 Hyperbolic spiral 20. 487 Hyperbolic substitutions 114. 286. 281.Index Force lines 288 FOrl11 Lagrange's 311 Forn1ula Adams' 390 Green's 276. 135 Fourier'" c(_'efficlents 318. 294 Gradient of a field 288 Gradient of a function 194. 18.20. 484 hyperbolic. 220 tra"'pezoidal 382 Forn1111clc. 150 Gaussian curve 92 General integral 322 General solution 359 General solution (of an equation) 323 General term 294 Generalized antiderivative 143 Generalized polar coordinates 255 Geometric progression 293. 486 345 197 356 133 I 111 pI i cit fun c t i on 12 Inlproper integral conver~ent 270 divergent 270 Improper Jllultiple integrals 269. 55.227 Green's forrnul a 276. 195 Graph of a function 12 Greatest value 85. 492 Index Infinite discontinuities 38 Infinitely large quantities 33 Infinitely small quantities 33 Infinites 33 Infinitesimals 33 of higher order 33 of order n 33 of the same order 33 Inflection points of 91 Inhomogeneous equation 349. 149 path of 273. 356 Inhomogeneous linear differential equation 349 Initial conditions 323. 280 region of 246-248 by substitution 113 Integration of differential equation by means of power series 361. 261 mass and static nlornents of a 260 moments of inertia of a 261 Least value 85 Left-hand derivative 44 Left horizontal asymptote 94 Left inclined asymptote 94 Leibniz rule 67. 362 Integration of functions numerical 382. 363 Integral 322 convergent improper 143 definite 138 divergent improper 143 double 246 Euler 146 Euler-Poisson 272 general 322 improper multiple 269. 82 . 380 J Jacobian 253. 224 Lagrange's interpolation fornlllla 374 Lagrange's theOrel11 75 Laplace equation 289. 264 K Kinetic energy 174 L Lagrange's equation 339 Lagrange's form 311 Lagrange's fornlula 145 Lagrange's function 223. 105. 486 Involute of a curve 104 Isoclines 325 Isolated point 230 . 351. 378. 106. Iterative method 377. 232 Bernoulli's 155. 117. 269 Leibniz test 296. 297 Lemniscate 20.. 483 Involute of a circle 20. 274. 270 line 273-278 part icul ar 322 probability 144 singular 337 surface 284-286 triple 262 Integral curve 322 Integral sum 138 Integrating factor 335 Integration basic rules of 107 under the differential sign 109 direct 107 by parts 116. 372 quadratic 372 Interpolation formula Lagrange's 374 Newton's 372 Interval of calculations 382 closed II of convergence 305 of monotonicity 83 I nterv al (cont) open 11 table interval 372 I nverse circular functions 48 I nverse functions 12 Inverse hyperbolic functions 49 Inverse interpolation 373 I nverse trigonometric functions 482. 486 Level surf aces 288 L'Hospital-Bernoulli rule 78. 383 Inte~ration of ordinary differential equation nurnerical 384-393 Integration of total difTerentials 202204 Integration of transcendental functions 135 Jnterpolation of functions 372-374 inverse 373 linear 13. 291 Laplace transformation 271 Laplacian operator 289 Lamina coordinates of the centre of gravity of a. 380 Milne's 386. 351 Linear equation 332 Linear interpolation 372 of a hinrtion 13 Linearly dependent functions 349 Linearly independent functions 349 Lines flow 288 force 288 vector 288 Li pschitz condition 385 Logarithlllic curve 484 Logarithnlic derivative 55 Logarithlllic functions 49 Logarithnlic spiral 20. 48] Node 230 Nonstationary scalar or vector field 288 Nonnal 217 to a curve 60 equations of 218 principal 238 Normal plClne 238 NtlInber Napier's 28 real 11 Number series 293 NUlllerical integration of functions 382. 106. 390. 390 Newton's 377. 392 chord lnethod 376 of differentials 343 of elinlinat ion 359 Method (cont) Euler's broken-line 326 iterative 377. 352 Minimum of a function 84. 278-281 Linear differential equations 349. 349. 28:3 of the first type 273. 125 Picard's 384. 275. 379 Ne\\·ton's serpentine 18 Niele's par abo! a 18.mation 38f. 487 M l\\aclaurin's formu] a 77. 150 j\\ean rate of change 42 ~\\et hod l\dall1s' 389. 275 Newton's interpolation formula 372 Newton's J11ethod 377. 222 MaxinlU111 point ~\ean value of a function 151 1\\ean-valuC' theorellls 75. 351 of variation of parameters 332. 222 Mininlum point 84 Mixed partial derivative 197 Moment of inertia 169 static 168 Monotonicity intervals of 8a Multiple-valued function 11 Multi pi icities root 121 N nth derivative 67 Nabla 288 Napier's number 28 Natural trlhedron ~3B Necessary condition for convergence 29:3 Necessary condition for an extrCJ11Unl 222 Newton trident of 18 Ne\vton-Leibniz fOrillula 140. 21. 387. 378. 379 Ostrogradsky 123.493 Index. 383 NUlnerical integration of ordinary differential equations 384-393 o One-sided derivatives 43 One-sided lilnits 22 Open interval 11 . Pascal's 158 Limit of a function 22 Limit on the left 22 Limit on the right 22 Limit of a sequence 22 Limiting absolute error 367 Lilniting relative error 367 Lilnits one-sided 22 Line straight 17. 20 Line Integral application of 276. 278 Line integral of the second type 274. 390 385. 234. 385 reduction 123 Runge-Kutta 385-387. Lima~on of successive approxi. 105. 274. 389 of tangents 377 of undetermined coefficients 121. 31:3 }'v\axiUll1I11 of a fUllction 84. 141. 220 Maclaurin's series 311. 277. 150 Reduction nlcthod 123 Regi on of convergence ~04 Region of integration 246-248 Relative error 367 Remainder 311 Renlainder of a series 293. 234 Niele's 18. 225 Polar subnormal 61 Polar subtangent 61 Potential (of a field) 289 Potential vector field 289 Po\vcr series 305 Principal normal 238 Principle of equal effects 369 Runge 383. 234. 105. 104. 119. 386 of superposition of solutions 353 Probability curve 19. 480. 484 Probability integral 144 Product of t\VO convergent series 298 Progression ~eometric 293. 294 Proper rational fraction 121 Proportionate parts rule of 376 Q Quadratic interpolation 372 Quadratic trinonlial 118. 385 Plane normal 238 osculating 238 rect ifyi ng 238 tangent 217 Point bending 84 critical (of the second kind) 92 of discontinuity 37 double 230 extrern aI 84 of inflection 91 isol ated 230 nlaximum 84 minimum 84 singular 230 stationary 196 of tangency 217 Points characteristic 96 critical 84 stationary 222. 280 Period of a function 14 Periodic function 14 Picard's met hod 384. 243 Radius of second curvature 243 Radius of tor~ion 243 Rate of change of a function 43 mean 42 Ratio (of a geornetric progreC\sion) 294 Real nunlbers J I Rectangular hyperbola 480 Rectifying plane 238 Reduction fornlulas 130. 20. 234. 105. 304 Remainder term 311 Removab!e discontinuity 37 Right -hand derivative 44 Right horizontal asymptote 93 Right inclined aSYlnptote 93 Rolle's theorem 75 Root multi plici ties 121 . 20. 485 cubic 17. 481 Parabolic formula 382 Parameters variation of 332.on 158 Path of integration 273. 349. 352 Parametric representation of a function 207 Partial derivative hirheg-order 197 "mixed" 197 second 197 Partial sum 293 Particular integral 322 Particular solution 339 Pascal's limac. 125 p Parabola 17. 135. Quantity infinitely large 33 infinitely sl11a11 33 12:~ R Radi liS of convergence 305 Radius of curvature 102. 274.Index 494 Operator Hamiltonian 288 Laplacian 289 Order of smallness 35 Orthagonal surfaces 219 Orthagonal trajectories 328 Osculating circle 103 Osculating plane 238 Ostrogradsky-Gauss formula 286-288 Ostrogradsky-Gauss theorem 291 Ostrogradsky method 123. 481 safety 234 semicubical 18. 133 trigonolTIctric 114. 487 hyperbolic 20. 296 dl vergent 293. 359 part icu Iar 339 Spiral of Archimedes 20. 105.495 Index Rose four-leafed 487 three-leafed 20.34. 389 Sufficient conditions (for an extrclnum) 222 Sunl Inte~ral 138 partial 293 of a seri es 293. 65. 106. 296.234. 220 . 2. 105. 269 I' I-Iospit aI-Bernoulli 78-82 of proport j onat c parts 376 Runge-Kutta nlethod 385-387. 313 Serpentine Newton's 18 Simpson's forrllula 382-384 Si oe curve 481 Single-valurd function 11 Singular integral 337 Singular point 230 Slope (of a tan~ent) 43 Stnallest value 225. 385 rl1 et hod of 384. 297 IIlC0l11plete Fourier :318. 227 Solenoidal vector field 289 Solution (of an equation) 322 general 323. 116. 66. 286. 319 functional 304 harnl0nic 2~}4. 4~1 Series absolutely convergent 296. 487 Static moment 168 Stationary point 196. 294 Fourier 318. 115. 289 S t ra i g h t 11 ne 17.486 Subnormal 61 polar 61 Substitutions hyperbolic 114. 222. 225 Stationary scalar or vector field 288 Stokes' formula 285. 487 logarithmic 20. 385. 304 of two convergent serres 298 Su perposlti on of sol uti ons principle of 353 Surfncc jnte~ral of the first type 284 Surface iI1t('~ral of the second type 284 Surface integrals 284-286 Surfaces level 288 orthogonal 219 T Table d I ago naI tab I e 389 of standard integrals 107 Table interval 372 Tabular difTerentiation 46 Tacnode 230 Tangency point of 217 Tangent 238 Tangent curve 481 Tangent plane 217 equation of 218 Tangents met h od of 377 Taylor's formula 77. 297 with cornplex ternlS 297 condrtionally (not absolutely) cOllvergent 296 convcrgent 293 Series (cont) Dirichlet 295. 133 Subtangent 61 polar 61 Successive approxitnation 377. 20 Strophoid 157. 313 number series 293 operations on 297 power 305 Taylor's 311. 390 I~unge principle 383. 20. 319 Maclaurin's 311. 487 Rotation (of a vector field) 288 Rule Leibniz 67. 21.232. 386 S Safety parabola 234 Scalar field 288 Scheme twelve-ordin~lte 393-39!i Second curvature 243 Second derivative 66 Second dJJTerent ial 198 Second-orde~ difl'crential 72 Second partIal derivative 197 Segment of the nonnal 61 Segnlent of the polar BOrIllal 61 Segment of the polar tangent 61 Segment of a straight line 20 Segrnent of the tangent 61 Senlicircle 20 Senlicubical parabola 18. 105. 225. 227 least 85 mean (of a function) 151. 487 Torsion 243 Tractrix 161 Trajec. 123 Triple integral 262 applications of 265. 156. 294 Leibniz 296.295 Cauchy's integral 295 com parison 143. 351 Uniform convergence 306 v Value greatest 85. 276.496 Index Taylor's series 311.tories orthogonal 328 Transcendental funct ions integrati on of 135 Transformation Laplace 271 Trapezoidal fornlula 382 Trident of Newton 18 Trigonometric ~unctions 48 integrating 128. 252 smallest 225. 271 . 268 change of variables in 263 computing volumes by means of 268 evaluating a 265 in rectangular coordinates 262 Trochoid 157 Twelve-ordinate schenle 393 ·395 U Undetermined coefficients method of 121. 352 Vector acceleration 236 of binormal 238 of principal nOrIllal 238 of tangent line 238 velocity 236 Vector field 288 Vector function 235 Vector lines 288 Velocity vector 236 Vertex of a curve 104 Vertical asynlptote 93 Vertices of a curve 104 Volume of a cylindroid 258 Volume of ~olids 161-166 W Weierstrass' test 306 Witch of Agnesi 18. 297 Weierstrass' 306 Theorem Cauchy's 75. 480 Work of a force 174. 150 Theory field 288-292 Three-leafed rose 20.arable an equation \vith 327 328 Variation of paranleters '332. 293. 326 Dirichlet's 318 Theorem (cont) Lagrange's 75 Ostrogradsky-Guuss 291 Rolle's 75 Theorems Guldin's 171 mean-value 75. 349. 227 Variable dependent 11 independent 11 Variables sep. 313 Term general 294 remainder 311 Test d' Alembert's 295 Cauchy's 293. 129 Trigonometric substitutions 114 115 133 ' t Trihedron natural 238 Trinorrlial quadratic 118. 119.