ÅÙÐØ Ú Ö Ð Ò Î ØÓÖDavid A. SANTOS
[email protected] January 15, 2009 Version Ð ÙÐÙ× ree to photocopy and distribute Mathesis iuvenes tentare rerum quaelibet ardua semitasque non usitatas pandere docet. ii Copyright c 2007 David Anthony SANTOS. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”. Contents Preface 1 Vectors and Parametric Curves 1.1 Points and Vectors on the Plane . . . . . . . . 1.2 Scalar Product on the Plane . . . . . . . . . . 1.3 Linear Independence . . . . . . . . . . . . . . 1.4 Geometric Transformations in two dimensions 1.5 Determinants in two dimensions . . . . . . . 1.6 Parametric Curves on the Plane . . . . . . . . 1.7 Vectors in Space . . . . . . . . . . . . . . . . 1.8 Cross Product . . . . . . . . . . . . . . . . . . 1.9 Matrices in three dimensions . . . . . . . . . 1.10 Determinants in three dimensions . . . . . . . 1.11 Some Solid Geometry . . . . . . . . . . . . . . 1.12 Cavalieri, and the Pappus-Guldin Rules . . . . 1.13 Dihedral Angles and Platonic Solids . . . . . . 1.14 Spherical Trigonometry . . . . . . . . . . . . 1.15 Canonical Surfaces . . . . . . . . . . . . . . . 1.16 Parametric Curves in Space . . . . . . . . . . 1.17 Multidimensional Vectors . . . . . . . . . . . 2 Differentiation 2.1 Some Topology . . . . . . . . . . . . 2.2 Multivariable Functions . . . . . . . 2.3 Limits . . . . . . . . . . . . . . . . . 2.4 Definition of the Derivative . . . . . . 2.5 The Jacobi Matrix . . . . . . . . . . 2.6 Gradients and Directional Derivatives 2.7 Levi-Civitta and Einstein . . . . . . . 2.8 Extrema . . . . . . . . . . . . . . . . 2.9 Lagrange Multipliers . . . . . . . . . 3 Integration 3.1 Differential Forms . . . . . 3.2 Zero-Manifolds . . . . . . . 3.3 One-Manifolds . . . . . . . 3.4 Closed and Exact Forms . . 3.5 Two-Manifolds . . . . . . . 3.6 Change of Variables . . . . 3.7 Change to Polar Coordinates 3.8 Three-Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v 1 1 11 15 17 25 31 41 50 58 62 64 68 70 74 79 87 90 97 97 99 100 104 107 116 122 124 129 133 . 133 . 136 . 137 . 141 . 146 . 155 . 162 . 168 iii iv 3.9 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 172 3.10 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 3.11 Green’s, Stokes’, and Gauss’ Theorems . . . . . . . . . . . . . . . . 179 A Answers and Hints 186 Answers and Hints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 GNU Free Documentation License 1. APPLICABILITY AND DEFINITIONS . . . . . . 2. VERBATIM COPYING . . . . . . . . . . . . . . 3. COPYING IN QUANTITY . . . . . . . . . . . . 4. MODIFICATIONS . . . . . . . . . . . . . . . . 5. COMBINING DOCUMENTS . . . . . . . . . . . 6. COLLECTIONS OF DOCUMENTS . . . . . . . 7. AGGREGATION WITH INDEPENDENT WORKS 8. TRANSLATION . . . . . . . . . . . . . . . . . 9. TERMINATION . . . . . . . . . . . . . . . . . 10. FUTURE REVISIONS OF THIS LICENSE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 . 250 . 250 . 250 . 251 . 251 . 252 . 252 . 252 . 252 . 252 Preface These notes started during the Spring of 2003. They are meant to be a gentle introduction to multivariable and vector calculus. Throughout these notes I use Maple™ version 10 commands in order to illustrate some points of the theory. I would appreciate any comments, suggestions, corrections, etc., which can be addressed to the email below. David A. SANTOS
[email protected] v 1. We say that r is the tail − of the bi-point [r. 2). at r′ . Here the first coordinate x stipulates the location on the horizontal axis and the second coordinate y stipulates the location on the vertical axis. The bi-point [r. 5). Vector) A scalar α ∈ R is simply a real number. −4). Bi-point. Given two points r and r′ in R2 the directed line segment with departure point r and arrival point r′ is called the bi-point r. r ]. −1). The zero 0 → − vector 0 = indicates no movement in either direction. b2 = (5. 0 Notice that infinitely many different choices of departure and arrival points may give the same vector. Some authors use the terminology “fixed vector” instead of “bi-point. r = (x.1 Vectors and Parametric Curves 1. A point r ∈ R2 is an ordered pair of real numbers. the point (0. r′ ] and that r′ is its head. that is. See figure 1. a2 = (3. y) in the horizon′ y −y tal axis and of y ′ − y units from the current position in the vertical axis. 0) b = (2.” 1 .2 for an example. −6). r′1 and is denoted by [r. 2 Example Consider the points a1 = (1. Point. Those interested in more formal treatments can profit by reading [BlRo] or [Lan]. we associate to it the vector −′ → x′ − x rr = stipulating a movement of x′ − x units from (x. 0). 0) by O = (0. 1 Definition (Scalar. 1 b1 = (3. r′ ]. y) with x ∈ R and y ∈ R.1 Points and Vectors on the Plane W e start with a na¨ve introduction to some linear algebra necessary for the ı course. A vector → ∈ R2 is a codification a ′ of movement of a bi-point: given the bi-point [r. O = (0. with the arrow tip. We will always denote the origin. See figure 1. r′ ] can be thus interpreted as an arrow starting at r and finishing. 6. y) Points and Vectors on the Plane r x axis a2 Figure 1. 1. that is. consider the set of integers Z. Though the bi-points [a1 . in example 2. .2: A bi-point in R2 . −1. −6. b2 ] and [O. 3Z+1 = {. 2.1: A point in R2 . 4. −4. 5.y axis r′ r = (x. a 1 Figure 1. In more technical language. they represent the same vector. . Analogously. b1 ] ∈ . . that is. have the same direction. As an simple example of an equivalence class. and point in the same sense are equivalent. the bi-point [a1 . −2. . . −5. According to their remainder upon division by 3. 3Z+2 = {. . all bi-points that have the same length. −6 −6 Free to photocopy and distribute 2 . −3. b] are in different locations on the plane. . .}. The equivalence class 3Z comprises the integers divisible by 3.}. 3. as 3−1 = 5−3 = 2−0 = 2 −6 . 7. −4 − 2 −1 − 5 −6 − 0 The instructions given by the vector are all the same: start at the point. [a1 . 0. . and the name of this equivalence is a vector. a vector is an equivalence class of bi-points. See figure 1. O b2 b1 b Figure 1. . . each integer belongs to one of the three sets 3Z = {. −18 ∈ 3Z. . b1 ]. . and for example. 8.3. go two units right and six units down. . . b1 ] belongs to the 2 2 equivalence class . .3: Example 2. [a2 . . .}. − v 0 ← → → 5 Definition We say that [a. b]. both their heads lie on the same half-plane made by the line perpendicular to then at their tails.Chapter 1 − → 3 Definition The vector Oa that corresponds to the point a ∈ R2 is called the position vector of the point a. ← → 4 Definition Let a = b be points on the plane and let ab be the line passing through a and b. that is. b] and so we write [b.4: Direction of a bi-point Figure 1. Two bi-points are parallel if the lines containing them are parallel. π[ that the line ab makes with the positive x-axis (horizontal axis).4. ab = −ba. a] has the opposite sense of [a.6: Bi-points with opposite sense. The direction of a vector − → = → is the direction of any of its bi-point representatives. b] is the direction of the line ← → L. 3 . the angle θ ∈ [0. The sense of a vector is the sense of any of its bi-point representatives. B A θ A B D C D A B C Figure 1. b] is simply the distance between a and b and it is denoted by ||[a. They have opposite sense if they have the same direction and if when translating one so as to its tail is over the other’s tail. say that the bi-points [a. a] = −[a.5: Bi-points with the same sense. . w] if ab = ← We zw. See figure 1. Figure 1. − → − → Similarly we write. their heads lie on different half-planes made by the line perpendicular to them at their tails. b]|| = Free to photocopy and distribute (a1 − b1 )2 + (a2 − b2 )2 . if bi-point representatives of them are parallel.5 and 1. 6 Definition The Euclidean length or norm of bi-point [a. The direction of the bi-point [a. Two vectors are parallel. See figures 1. w] have the same sense if they have the same direction and if when translating one so as to its tail is over the other’s tail. b] and [z. when measured counterclockwise. Bi-point [b.6 . b] has the same direction as [z. 9. Free to photocopy and distribute 4 (1. It is easy to see that vector addition is commutative and associative. proceed as follows. It is clear that the norm of a vector satisfies the following properties: ¬¬− ¬¬ 1. Find a bi-point representative of → having its tail at the tip u v →. and that the additive inverse of → a →. − − u v u2 v2 u2 + v2 − −2→ u → − u B → − v C 1→ − u 2 → − u (1. 1. Figure 1. and 1. The norm of a vector is the norm of any of its bi-point representatives. a ¬¬→¬¬ → − − a 2. that → − − the vector 0 acts as an additive identity. − − − − 8 Definition If → and → are two vectors in R2 their vector sum → + → is defined u v u v by the coordinatewise addition → + → = u1 + v1 = u1 + v1 . In particular.11. 1. and (iii) its sense. ¬¬→¬¬ ≥ 0.10. A vector is completely determined by three things: (i) its norm. To add two vectors geometrically. then we have Chasles’ Rule: − → − → − → AB + BC = AC.7.8: Scalar multiplication of vectors. (ii) its direction.Points and Vectors on the Plane A bi-point is said to have unit length if it has norm 1. a ¬¬− ¬¬ 1 − 7 Example The vector → = √ has norm ¬¬→¬¬ = v v 2 √ √ 12 + ( 2)2 = 3.1) A →+→ − − u v Figure 1. See figures 1. if → = − and → = − − − − whose tip is that of the bi-point for v u AB v BC. The sum → + → is the vector whose tail is that of the bi-point for → and − − − − of u u v u → → →. Draw a bi-point − is − a − − representative of →. ¬¬− ¬¬ = 0 ⇐⇒ → = 0 .2) .7: Addition of Vectors. 11: Difference → − − − − 10 Definition Let → = 0 . The affine u u u u1 − line with direction vector → = u and passing through a is the set of points on u2 the plane ´ − a + R→ = u x y ∈ R : x = a1 + tu1 .9: Commutativity Figure 1.Chapter 1 − 9 Definition If α ∈ R and → ∈ R2 we define scalar multiplication of a vector and a a scalar by the coordinatewise multiplication a1 αa1 − α→ = α a = . then x − a1 (x − a1 ) u2 u2 = t =⇒ y = a2 + u2 = x + a2 − a1 . a2 αa2 (1. as x is constant. u1 u1 u1 u1 Free to photocopy and distribute 5 − → w − → − v µ − → w → − w . → − → − − − α(→ + b ) = α→ + α b a a − − − (α + β)→ = α→ + β → a a a − − 1→ = → a a − − (αβ)→ = α(β →) a a − → v − → v → − w + → − → − v v + → − w → − w → − u + → − v → − u − → v − → − → +w − → + v) (u − →) w − → (v + − →+ u → − v → − v +→ − w − → w Figure 1. t∈R See figure 1. If u1 = 0.10: Associativity Figure 1. If u1 = 0.3) It is easy to see that vector addition and scalar multiplication satisfies the following properties. Put R→ = {λ→ : λ ∈ R} and let a ∈ R2 . the affine line defined above is vertical. . 2 y = a2 + tu2 .12. every Cartesian line is also an affine line and one may take the vector 1 − − as its direction vector.Points and Vectors on the Plane that is. if y = that is. Solution: → − √ ¬¬− ¬¬ v ◮ Since ¬¬→¬¬ = 3. u2 u1 . Conversely. 1 − 11 Example Find a vector of length 3. Hence. then x y = 1 m t+ 0 k . parallel to → = √ but in the opposite v 2 sense. the 0 is parallel to every vector. It also follows that two vectors → and → are parallel u v m − − − − if and only if the affine lines R→ and R→ are parallel. → → if there u v u v → = λ→.12: Parametric equation of a line on the plane. the vector ¬¬→¬¬ has unit norm and v ¬¬− ¬¬ v − has the same direction and sense as →. and so the vector sought is v √ → − v 3 1 − 3 √ . the affine line is the Cartesian line with slope mx + k is the equation of a Cartesian line. −3 ¬¬→¬¬ = − √ √ = ¬¬− ¬¬ v 3 2 − 6 ◭ Free to photocopy and distribute 6 → − u − r = a + t→ u . v v a Figure 1. − − exists a scalar λ ∈ R such that u v → − → − − − Because 0 = 0→ for any vector →. ◭ 14 Example Consider a △ABC. − → − → 13 Example Given a pentagon ABCDE. CMC = 2 Free to photocopy and distribute 7 . and Solution: ◮ The desired equation is plainly x y ◭ Some plane geometry results can be easily proved by means of vectors. = 1 −1 +t 2 −3 =⇒ x = 1 + 2t. half its length. − → BC − → − → = BA + AC − → − → = −AB + AC −− −→ −− −→ = −2AMC + 2AMB −− −→ −− −→ = 2MC A + 2AMB −− −→ −− −→ = 2(MC A + AMB ) −−→ −− = 2MC MB . B] and [C. Solution: ◮ Utilising Chasles’ Rule several times: − → − → − → − → − → − → → − 0 = AA = AB + BC + CD + DE + EA. −− −→ − → −− −→ − → 2AMC = AB and 2AMB = AC. Demonstrate that 1 − → − → −− −→ CA + CB . t ∈ R. A]. let MC be the midpoint of [A. We will demonstrate that BC = 2MC MB . determine the vector sum AB + BC + − → − → − → CD + DE + EA. We have. y = −1 − 3t. Solution: ◮ Let the midpoints of [A. be MC and MB . − → −−→ −− respectively. Therefore. ◭ 15 Example In △ABC. Here are some examples. . B]. as we were to shew. Demonstrate that the line segment joining the midpoints of two sides is parallel to the third side and it is in fact.Chapter 1 1 −1 12 Example Find the parametric equation of the line passing through in the direction of the vector 2 −3 . whence 2 − b = 0. −−→ −− −− −→ Since △ABC is non-degenerate. See figure 1. MA ] and [B. the lines AMA and ←→ − − → BMB are not parallel. −−→ −− − → 2MA MB = BA − → − → = BG + GA −− −→ −− −→ = bGMB − aGMA −− −→ −−→ −− −− −→ = bGMA + bMA MB − aGMA . The point of concurrency G is called the barycentre or centroid of the triangle. MB ] of the non-degenerate △ABC intersect at the point G.Points and Vectors on the Plane −− −→ −− −→ Solution: ◮ As AMC = MC B. − → − → CA + CB = = = −− −→ −− −→ −− −→ −− −→ CMC + MC A + CMC + MC B −− −→ −− −→ −− −→ 2CMC − AMC + MC B −− −→ 2CMC . Free to photocopy and distribute 8 . ◭ 16 Example If the medians [A.13.13. ◭ and thus − → −− −→ BG = 2GMB . From example 14. Therefore. AG and −− −→ − → −− −→ GMA are parallel and hence there is a scalar a such that AG = aGMA . See figure 1. there is a scalar b such that BG = bGMB . − → −− −→ In the same fashion. b − a = 0. =⇒ a = b = 2. 17 Example The medians of a non-degenerate triangle △ABC are concurrent. ←− −→ Solution: ◮ Since the triangle is non-degenerate. from where the result follows. and hence meet at a point G. −−→ −− −− −→ (2 − b)MA MB = (b − a)GMA . MA MB and GMA are not parallel. demonstrate that − → −− −→ AG = 2GMA . we have. 1. Prove that − → − → − → AC + BD = 2BC. We must shew that the line ←→ − ←→ − ←→ − CMC also passes through G. Problem 1. y. Solution: ◮ Let G be as in example 16.1.13: Example 17. Q]|| : Free to photocopy and distribute 9 .4 Let A. y = 2 − t.3 (Varignon’s Theorem) Use vector algebra in order to prove that in any quadrilateral ABCD. C].1. E is the midpoint of [B. and then demonstrate that for any arbitrary point M on the plane −→ − −→ − − → MA + 3MB = 4MI and −→ −→ − − − → 3MA + MB = 4MJ. 3 Problem 1. z] such that ||[x.C MB G A MC MA Chapter 1 B Figure 1. Problem 1. demonstrating the result. whose sides do not intersect.1. − → −− −→ AG = 2GMA .6 Let x. By the aforementioned example.2 ABCD is a parallelogram. B be two points on the plane. − − → −→ = GB + BG′ −− −→ −− −→ = −2GMB + 2G′ MB −→ −− −→ − − = 2(MB G + G′ MB ) −→ − = 2G′ G. Let Q be a point on side [x. Construct two points I and J Problem 1. C] and F is the midpoint of [D. Problem 1. ◭ Homework Problem 1. the quadrilateral formed by the midpoints of the sides is a parallelogram. −→ − −− −→ CG′ = 2G′ MC .1. Let the line CMC and BMB meet in G′ . −→ − GG′ −→ − −− −→ BG′ = 2G′ MB . − → → 1− JA = − JB. Therefore −→ − −→ − −→ − −→ − → − → − GG′ = −2GG′ =⇒ 3GG′ = 0 =⇒ GG′ = 0 =⇒ G = G′ .5 Find the Cartesian equation corresponding to the line with parametric equation x = −1 + t.1.1 Is there is any truth to the statement “a vector is that which has magnitude and direction”? such that − → − → IA = −3IB. It follows that − → −− −→ BG = 2GMB . z be points on the plane with x = y and consider △xyz. 19). [A]. β with α + β = 1 such that → = α→ + β → − − − za zx zy. 1. Find the sum of the vectors. Problem → − a → − c → − b Figure 1.8. The point a belongs to the line segment [x.8 A circle is divided into three. P]|| : ||[P. [E]. n such − → − → − → − → that TP = lTx + m Ty + nTz.16: 1. [D]. Find rational numbers α and β such − → − → − → that TQ = αTx + β Tz. → − a → − a → − c → − a → − d → − b Figure 1. 2.1.18: 1.Points and Vectors on the Plane ||[Q. Problem → − b → − c → − c → − b Figure 1.1.1. z be points on the plane with x = y. Find rational numbers l.8. The point a belongs to the interior of the triangle △xyz if and only if there exists scalars α > 0. y.1. Let T be an arbitrary point on the plane. [B].1. 3. β ≥ 0 with α + β = 1 such that → = α→ + β → − − − za zx zy.8. Problem → − a → − d → − c → − b Figure 1.8. y] if and only if there exists scalars α ≥ 0. The point a belongs to the line ← if xy and only if there exists scalars α. Assume that the divisions start or stop at the centre of the circle. Problem Free to photocopy and distribute 10 . z] such that ||[y. → − c → − d → − e [F]. [C]. Problem Figure 1.15: 1. Demonstrate that → 1. four equal. β > 0 with α + β < 1 such that → = α→ + β → − − − za zx zy.19: 1. or six equal parts (figures 1.7 Let x.8. Q]|| = 7 : 2. as suggested in the figures.1.1. z]|| = 3 : 4 and let P be a point on [y.17: 1. Problem 1.14: 1. Problem → − b → − d → − a → − f Figure 1. m.1.8.17 through 1. Problem 1. 2. Free to photocopy and distribute 11 . x y x y x y SP3 Commutativity SP4 SP5 SP6 → •→ = → •→ − − − − x y y x → •→ ≥ 0 − − x x − →•→ = 0 ⇐⇒ → = → − − − x x x 0 Ô ¬¬→¬¬ − − ¬¬− ¬¬ = →•→ x x x →•(→ + →) = →•→ + →•→ − − − − − − − x y z x y x z (1. inner x y product) is defined and denoted by →•→ = x y + x y .2 Scalar Product on the Plane We will now define an operation between two plane vectors that will provide a further tool to examine the geometry on the plane. − − 18 Definition Let → ∈ R2 and → ∈ R2 .8) (1.7) (1.Chapter 1 1. α ∈ R. x y z x z SP2 Scalar Homogeneity − − − − − − (α→)•→ = →•(α→) = α(→•→). then →• b = 1 · 3 + 2 · 4 = 11. a 19 Example If → = a 2 4 The following properties of the scalar product are easy to deduce from the definition. SP1 Bilinearity − − − − − − − y z (→ + →)•→ = →•→ + →•→.4) (1.6) (1.9) → → − b −− a → − b → − a Figure 1. − − x y 1 1 2 2 1 3 → − − − → − and b = .5) (1. Their scalar product (dot product.20: Theorem 21. − is orthogonal to b . If → a → − − → ⊥ →. → − It follows that the vector 0 is simultaneously parallel and perpendicular to any vector! − 23 Definition Two vectors are said to be orthogonal if they are perpendicular. a a → − − 21 Theorem Let → and b be vectors in R2 .4) through (1. b ). − Equality occurs if and only if → a Free to photocopy and distribute → − b. u π − − → Putting (→. 12 .9) we have → − − → − → − − − − − → || b − →||2 = ||→||2 + || b ||2 − 2||→|||| b || cos (→. we write a b ¬¬→¬¬ ¬¬− ¬¬ → − → − ¬¬− ¬¬ − − 24 Definition If → ⊥ b and ¬¬→¬¬ = ¬¬ b ¬¬ = 1 we say that → and b are ora a a thonormal.Scalar Product on the Plane → − − 20 Definition Given vectors → and b . and (1. Since | cos θ| ≤ 1 we also have 25 Corollary (Cauchy-Bunyakovsky-Schwarz Inequality) ¬ ¬ ¬¬→¬¬¬¬→¬¬ ¬→•→¬ − − − ¬¬− ¬¬ ¬ a b ¬ ≤ ¬¬ a ¬¬¬¬ b ¬¬. 2 22 Corollary Two vectors in R2 are perpendicular if and only if their dot product is 0. as the angle between the affine lines R→ and R b . a a a as we wanted to shew. →). a − → − − → − denoted by (→. using Al-Kashi’s Law of Cosines on the length of the vectors. b ) a a a a → − − → − − → − → − − − − − → ⇐⇒ ( b − →)•( b − →) = ||→||2 + || b ||2 − 2||→|||| b || cos (→. − − − − − a b a b a b Proof: From figure 1. Then a − →•→ = ||→||||→|| cos (→. b ) = a in Theorem 21 we obtain the following corollary.20. b ) a a a a a a → − − → − → − − − → − − − − → ⇐⇒ || b ||2 − 2→• b + ||→||2 = ||→||2 + || b ||2 − 2||→|||| b || cos (→. b ) ∈ [0. b ) a a a a a →→ − − − → − → − − − → − − − − − → ⇐⇒ b • b − 2→• b + →•→ = ||→||2 + || b ||2 − 2||→|||| b || cos (→. π]. we define the (convex) angle between them. b ) a a a a a − → − − − → − − → ⇐⇒ →• b = ||→|||| b || cos (→. Proof: → − − ||→ + b ||2 a → − − → − − = (→ + b )•(→ + b ) a a − − − − − − → → → = →•→ + 2→• b + b • b a a a → − → − − − ≤ ||→||2 + 2||→|||| b || + || b ||2 a a → 2 − − = (||→|| + || b ||) .◭ 2 ¬¬ ¬¬→¬¬ ¬¬→¬¬ →¬¬ − ¬¬ ¬¬→ ¬¬− ¬¬ − − ¬¬ a + b ¬¬ ≤ ¬¬ a ¬¬ + ¬¬ b ¬¬. y. b = .10) 26 Example Let a. the CBS Inequality takes the form a2 b2 |a1 b1 + a2 b2 | ≤ (a2 + a2 )1/2 (b2 + b2 )1/2 . u 28 Example Let x.− = c . Minimise a2 + b2 subject to the constraint a + b = 1. 13 Free to photocopy and distribute . Prove that √ 2(x + y + z) ≤ x x2 + y 2 + y2 + z2 + Ô z 2 + x2 . ¬¬− ¬¬ ¬¬→¬¬ ¬¬− ¬¬ ¬¬− ¬¬ + ¬¬→¬¬ + ¬¬→¬¬ = a c ¬¬ b ¬¬ x2 + y 2 + y2 + z2 + Ô z 2 + x2 . z be positive real numbers. Solution: ◮ By the CBS Inequality. a from where the desired result follows. ¬¬ ¬¬ ¬¬ ¬¬ ¬¬ − ¬¬→ − + → + →¬¬ = ¬¬ x + y + z − ¬¬ b c ¬¬ a ¬¬ ¬¬ x + y + z Also. 1 2 1 2 (1.Chapter 1 a1 b1 → − − If → = a and b = . 1 = |a · 1 + b · 1| ≤ (a2 + b2 )1/2 (12 + 12 )1/2 =⇒ a2 + b2 ≥ Equality occurs if and only if a b =λ 1 1 1 2 . a = b = λ. Then y z x ¬¬ ¬¬ √ ¬¬ ¬¬ = 2(x + y + z). . In such case. and so equality is achieved for a = b = 27 Corollary (Triangle Inequality) 1 . b be positive real numbers. − Solution: ◮ Put → = a y → z → − . B′ . = ′ ′. ◭ We now use vectors to prove a classical theorem of Euclidean geometry. ← − → Since AA′ Free to photocopy and distribute 14 . − or more routinely. On the one hand. C be distinct points of D .22: Corollary to Thales’ Theorem. We will denote this distance by AB − .Scalar Product on the Plane and the assertion follows by the triangle inequality ¬¬ ¬¬→¬¬ ¬¬→¬¬ ¬¬→¬¬ → →¬¬ − ¬¬→ ¬¬− ¬¬ − − ¬¬ − − ¬¬ a + b + c ¬¬ ≤ ¬¬ a ¬¬ + ¬¬ b ¬¬ + ¬¬ c ¬¬. u u → is patent. C′ be distinct points of D′ . B. and A′ . A = B. by AB.21: Thales’ Theorem. −→ − − − −→ − − −→ − − → −→ − → → BB′ = BA + AA′ + A′ B′ = (A′ B′ − AB) + AA′ .2. there is a scalar λ ∈ R such that −′ → − −→ − − → A B′ = AB + λAA′ . because they are unit vectors in the same direction. −′ → − −′ → − − → − → AB AC A B′ A C′ = . A′ = B′ . ← → ← → B = B′ . Observe that AB = −BA. A = A′ . Then ← ′ − → AA D D′ ← ′ → AC A′ C ′ CC ⇐⇒ = ′ ′. u − → →. Figure 1. C = C′ . AB AB D′ D C A B A′ B′ A B C A′ B′ C′ Figure 1. AB AC A′ B ′ AC On the other hand. ← ′ → BB . Proof: Refer to figure 1. Let ← → ← → A. Let AA′ BB′ . then λ is the directed distance or algebraic measure of the line − If AB = λ u − → segment [AB] with respect to the vector →. by Chasles’ Rule. − 29 Definition Let A and B be points on the plane and let → be a unit vector. if the vector u ← ← → → 30 Theorem (Thales’ Theorem) Let D y D′ be two distinct lines on the plane. AB AB proving the theorem. → and →. BB ⇐⇒ CA CA′ 1. then either x2 = 0 or y1 = 0. and a fortiori x → →. Then ← ′ − → AA ← ′ → CB CB ′ = . B. what x1 x2 does it mean x1 y2 = x2 y1 ? If none of these are zero then = = λ. − y 2 j 15 Free to photocopy and distribute . x 1 y2 − x 2 y1 The above expressions for a and b make sense only if x1 y2 = x2 y1 . points on line D′ . (See figure 1. x 1 y2 − x 2 y1 v2 = ax2 + by2 x1 v2 − x2 v1 b= . Operating formally. → = 0 .2.u From the preceding theorem. that is. b such that x y C → = a→ + b→ ? − − − v x y The answer can be promptly obtained algebraically. be points on line D . since all vectors are parallel to the zero vector.) ← → ← → 31 Corollary Let D and D′ are distinct lines. − x 2 j − → = y →. Under which condix y → on the plane as a linear combination − tions can we write an arbitrary vector v − − of → and →. → = a→ + b→ − − − v x y ⇐⇒ ⇐⇒ v1 = ax1 + by1 . say. we immediately gather the following corollary. v1 y2 − v2 y1 a= . when can we find scalars a. say. the equality above reveals that ← ′ ← ′ − → → AC A′ C ′ AA CC ⇐⇒ − ′ ′ = 0. intersecting in the unique point C.3 Linear Independence − − onsider now two arbitrary vectors in R2 . A′ B ′ AB AB ← − → ← → As the line AA′ is not parallel to the line AB. In the second case we − − x y have − → = x →. and y1 y2 to x1 y1 − − =λ ⇐⇒ → →. x y x2 y2 → − − If x1 = 0.Chapter 1 Assembling these results. and A′ . In the first case. B ′ . −→ − − − − → −→ − → CC′ = CA + AA′ + A′ C′ − −→ − AC − A′ C ′ − → −→ → = − · AB + AA′ + ′ ′ AB + λAA′ AB AB − A′ C ′ AC − A′ C ′ −→ → = − AB + 1 + λ ′ ′ AA′ . But. ← → ← → Let A. since one a b we can express is not a scalar multiple of the other. . . . The fundamental parallelo− − gram of the the vectors → and → is the set x y − − {a→ + b→ : a ∈ [0. and the respective tiling of the plane by various translationsµ it. for b ∈ [0. x y ´ . coloured in 0 1 brown. − − 34 Definition Given two linearly independent vectors → and → consider bi-point x y representatives of them with the tails at the origin.23 shews the fundamental parallelogram of Free to photocopy and distribute 16 1 1 µ . . b ∈ [0. then we − independent from vector y say that they are linearly dependent. − − − v x y a ∈ R. In essence 0 0 1 1 then. 1]. Given an arbitrary vector it as a linear combination of these vectors as follows: a b 1 0 1 1 = (a − b) +b . For a ∈ [0. linear independence of two vectors on the plane means that we may obtain every vector on the plane as a linear combination of these two vectors and hence Figure 1. In this last case we say that → is linearly x y x →. → and →. In the same manner. 1]. − − − Consider now two linearly independent vectors → and →. Observe of ´ 0 1 1 2 that the vertices of this parallelogram are . 1]. b→ is parallel to → y y →. The linear combination a→ + b→ is also a − − − and traverses the whole length of y x y vector on the plane. b∈R − − − if and only if → is not parallel to →. an arbitrary vector → can be x y v written as the linear combination → = a→ + b→. We have demonstrated − y − − − 32 Theorem Given two vectors in R2 . 1 0 1 1 33 Example The vectors and are clearly linearly independent.Linear Independence → − − and so both vectors are parallel to j and hence → x the following theorem. 1]}. a→ x y x → and traverses the whole length of →: from its tip (when a = 1) − − is parallel to x x − − to its tail (when a = 0). →. If two vectors are not linearly independent. Free to photocopy and distribute 17 W . will be certain construct called matrices which we will study in the next section.3. Find a formula for the distance from t to L. Problem 1. Problem 1. é −1 Problem 1. 1 1 1.2 Write an arbitrary vector æ é a on the plane.4 Prove that two non-zero perpendicular vectors in R2 must be linearly independent.3 Consider the line with Cartesian equation L : ax + by = c. and draw their fundamental parallelogram.4 Geometric Transformations in two dimensions e now are interested in the following fundamental functions of sets (figures) on the plane: translations. Let t be a point not on L. Figure 1.23: Tiling and the fundamental parallelogram.3. Homework Problem 1.3. where not both of a. scalings (stretching or shrinking) reflexions about the axes. b are zero.1 Prove that æ é 1 æ and 1 1 are linearly independent. It will turn out that a handy tool for investigating all of these (with the exception of translations).Chapter 1 cover the whole plane by all these linear combinations. and rotations about the origin. as a linear combination b æ é æ é 1 −1 of the vectors and .3. that is.5y Figure 1. → − v Figure 1. to a copy of itself a given amount of units from where it was. it does not distort it shape or re-orient it). 18 . by x y 2x 0.b (r) = .Geometric Transformations in two dimensions First observe what is meant by a function F : R2 → R2 . but nevertheless an important one is the following. 37 Definition A function Sa. then T→1 ◦ T→2 = T→2 ◦ T→1 . Figure 1. where a > 0. I(x) = x is called the identity transformation. if − − − − − − T→1 . Observe that the identity transformation leaves a point untouched. b > 0 are real numbers. 35 Example The function I : R2 → R2 . It is clear that the composition of any two translations commutes. where → is a fixed vector on the plane.24 for an example. − 36 Definition A function T→ : R2 → R2 is said to be a translation if it is of the v − − − form T→ (x) = x + →. v v v v 1 v 2 v 1 v 2 v 2 1 2 1 and − − − − − − − − v v v (T→2 ◦ T→1 )(a) = T→2 (T→1 (a)) = T→1 (a) + →2 = a + →1 + →2 . v v v v v from where the commutativity claim is deduced.5 Free to photocopy and distribute = . T→2 : R2 → R2 are translations. and the output is also a point on the plane.0.25: A scaling. A rather uninteresting example. Then v v v − − − − − − − − (T→ ◦ T→ )(a) = T→ (T→ (a)) = T→ (a) + → = a + → + → . We start with the simplest of these functions. v v v A translation simply shifts an object on the plane rigidly (that is.b : R2 → R2 is said to be a scaling if it is of the form ax Sa. See figure 1. This means that the input of the function is a point of the plane.24: A translation. For let v v v v v v − − − T1 (a) = a + →1 and T→2 (a) = a + →2 .25 shews the scaling S2. b′ )(r) = Sa. but − (S2. Its reflexion about the y-axis is magenta (on the second quadrant).b′ ◦ Sa.b ◦ Sa′ .1 ◦ T→ ) i −1 0 = S2. however. then Sa. Then i a2 − (T→ ◦ S2.b′ (Sa.b′ = Sa′ . For consider 2a1 → − − the translation T→ (a) = a + i and the scaling S2.1 ) i a′ x b′ y ax by = a(a′ x) b(b′ y) a′ (ax) b′ (by) . −1 0 − = T→ i S −1 0 − = T→ i −2 0 = −1 0 .b and (Sa′ .b .Chapter 1 It is clear that the composition of any two scalings commutes. Its reflexion about the x-axis is cyan (on the fourth quadrant).1 − T→ i −1 0 = S2.b′ (r)) = Sa. Its reflexion about the origin is blue (on the third quadrant). that is.b )(r) = Sa′ .b′ : R2 → R2 are scalings.b (Sa′ .b′ ◦ Sa. 38 Definition A function RH : R2 → R2 is said to be a reflexion about the y-axis or −x horizontal reflexion if it is of the form RH (r) = .b .1 (a) = . The original object (in the first quadrant) is yellow. Translations and scalings do not necessarily commute.b (r)) = Sa′ . A function RV : R2 → R2 y Free to photocopy and distribute 19 .26: Reflexions. For (Sa. = .b ◦ Sa′ . Sa′ . y axis x axis Figure 1. if Sa.1 0 0 = 0 0 .b′ from where the commutativity claim is deduced. We may now formulate the definition of a rotation.27. ρ cos α ρ sin α θ α x-axis Figure 1. y = ρ sin α as in figure 1. and ρ sin(α + θ) = ρ sin α cos θ + ρ sin θ cos α = y cos θ + x sin θ.27: Rotation by an angle θ in the levogyrate (counterclockwise) sense from the x-axis. We now define rotations. This definition will be somewhat harder than the others.4. Some reflexions appear in figure 1.Geometric Transformations in two dimensions is said to be a reflexion about the x-axis or vertical reflexion if it is of the form x .4. we land on the new point x′ with x′ = ρ cos(α + θ) and y ′ = ρ sin(α + θ). But ρ cos(α + θ) = ρ cos θ cos α − ρ sin θ sin α = x cos θ − y sin θ. 2π[. so let us develop some ancillary results.26. If we rotate it. A few short computations establish various commutativity properties among reflexions. A function RO : R2 → R2 is said to be a reflexion about origin if RV (r) = −y it is of the form RH (r) = −x −y . Hence the point x y is mapped to the point y-axis ρ cos(α + θ) ρ sin(α + θ) x cos θ − y sin θ x sin θ + y cos θ . by an angle θ. in the levogyrate sense. Here ρ = x2 + y 2 and α ∈ [0. See problem 1. translations. Consider a point r with polar coordinates x = ρ cos α. and scalings. Free to photocopy and distribute 20 . reflexions. It is easy to prove that scalings.Chapter 1 39 Definition A function Rθ : R2 → R2 is said to be a levogyrate rotation about the x cos θ − y sin θ . it is verified that L(a + b) = L(a) + L(b). where for the linear transformation L in the definition we may take I : R2 → R2 . then → − → − → − → − L(r) = L(x i + y j ) = xL( i ) + yL( j ). Various properties of the composition of rotations with other plane transformations are explored in problems 1. 0 1 43 Example (Scaling Matrices) Let a > 0.b is . L(λa) = λL(a). 2 columns) array whose columns are (in this order) 1 0 L and L .5 and 1. b on the plane and every scalar λ.6. For 0 b Sa. and rotations. The matrix of a 0 the scaling transformation Sa. and thus a linear transformation from R2 to R2 is solely determined by the values → − → − L( i ) and L( j ). reflexions. We will now introduce a way to codify these values. 40 Definition A function L : R2 → R2 is said to be a linear transformation from R2 to R2 if for all points a. If L : R2 → R2 is a linear transformation. origin by the angle θ measured from the positive x-axis if Rθ (r) = x sin θ + y cos θ Here ρ = x2 + y 2 . We now codify some properties shared by scalings.4. v It is easy to see that translations are then affine transformations. 41 Definition A function A : R2 → R2 is said to be an affine transformation from R2 to R2 if there exists a linear transformation L : R2 → R2 and a fixed vector → ∈ R2 such that for all points x ∈ R2 it is verified that − v − A(x) = L(x) + →. 42 Definition Let L : R2 → R2 be a linear transformation. We have seen that scalings.4. but not so translations. The matrix AL associated to L is the 2 × 2.b 1 0 = a·1 b·0 = a 0 21 Free to photocopy and distribute . and rotations are linear transformations from R2 to R2 . b > 0 be a real numbers. (2 rows. the identity transformation I(x) = x. reflexions and rotations are linear transformations. Geometric Transformations in two dimensions and Sa. Rθ is sin θ cos θ 46 Example (Identity Matrix) The matrix for the identity linear transformation Id : 1 0 R2 → R2 . Then the compo- Proof: We need to find (L ◦ L′ ) and (L ◦ L′ ) 0 1 . Id(x) = x is I2 = . 45 Example (Rotating Matrices) It is easy to verify that the matrix for a rotation cos θ − sin θ . r t s u a c b d and . N(x) = O is 02 = . and the matrix for the transformation RO is −1 0 0 −1 . We are now interested in how to codify the matrix of a composition of linear transformations L1 ◦ L1 in terms of their individual matrices. 44 Example (Reflexion Matrices) It is easy to verify that the matrix for the trans−1 0 1 0 formation RH is . that the matrix for the transformation RV is 0 1 0 −1 .7 we know that the composition of two linear transformations is also linear.b 0 1 = 0·1 b·1 = 0 b . 22 Free to photocopy and distribute . 0 1 47 Example (Zero Matrix) The matrix for the null linear transformation N : R2 → 0 0 R2 .4. 0 0 From problem 1. 48 Theorem Let L : R2 → R2 have the matrix representation AL = let L′ : R2 → R2 have the matrix representation AL′ = sition L ◦ L′ has matrix representation ar + bt as + bu cr + dt cs + du 1 0 . N = 1 −2 2 1 . and λ ∈ R = a+r c+t b+s d+u . 1 0 −1 1 .u The above motivates the following definition. 2M = 2 0 −2 2 . a b r s u c d t be a scalar. so is matrix multiplication. 1 0 = L L′ 1 0 =L r t → − → − = rL( i )+tL( j ) = r a c +t b d = ar + bt cr + dt . MN = 3 −2 1 1 . neither is matrix multiplication. Free to photocopy and distribute 23 . Since the composition of functions is not necessarily commutative. Since the composition of functions is associative. We define matrix multiplication as AB = a c b d r t s u = ar + bt cr + dt as + bu cs + du . 50 Example Let M = Then M +N = 2 −2 1 2 . We define scalar multiplication of a matrix as λA = λ a c b d = λa λc λb λd .Chapter 1 We have (L◦L′ ) and (L◦L′ ) 0 1 = L L′ 0 1 =L s u a b → − → − = sL( i )+uL( j ) = s +u c d ar + bt cr + dt as + bu cs + du . whence we conclude that the matrix of L ◦ L′ is we wanted to shew. as = as + bu cs + du . We define matrix addition as A+B = a c b d + r t s u 49 Definition Let A = and B = be two 2 × 2 matrices. is a fortiori. a c b d Solution: ◮ Let be the desired matrix.28: Example 51. We now assume. Figure 1.29. coordinate points with integer coordinates. d = 1. Then since a c b d 0 0 0 0 = . 0 Using these values. Figure 1. a c b d 1 1 = 1 3 =⇒ a+b c+d = 1 3 =⇒ b = −1. Assume in each case that the vertices of the figures are lattice points. Then the point a c b d 1 0 = 2 2 =⇒ a c = 2 2 =⇒ a = c = 2.29: Example 51. that each vertex of the square is transformed in the same order. counterclockwise. And so the desired matrix is 2 2 ◭ 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 −1 1 . that is. transformed to itself. to each vertex of the rectangle. Homework Free to photocopy and distribute 24 .Geometric Transformations in two dimensions 51 Example Find a 2 × 2 matrix that will transform the square in figure 1.28 into the parallelogram in figure 1. with0 out loss of generality. −1 3 2 1 0 0 1 2 Figure 1.4. Prove that the composition of a reflexion and a translation is not necessarily commutative.4. triangle in figure 1.4. and so the composition of two rotations on the plane is commutative.8 Find the effects of the reflexions RH . Problem 1.1 . R− π .Chapter 1 æ Problem 1. and 1.4.30.9 Find all matrices A M2×2 (R) such that A2 = 02 ∈ Problem 1.4.4.4.7 Let L : R2 → R2 and L′ : R2 → R2 be linear transformations. Problem 1.30. 1. B = .4. Problem 1. Problem 1.4. R π .4. and S2. Problem 1.10 Find the image of the figure below (consisting of two circles and a æ é 1 1 triangle) under the matrix . RV .31: Problem 1.10. Prove that the composition of a reflexion and a rotation is not necessarily commutative. and R− π on the 2 4 2 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 Figure 1. find a and b. S1. B 3 = Problem 1. 5 4 −2 (A + B)2 = A2 + 2AB + B 2 . Prove that their composition L ◦ L′ is also a linear transformation.2 . 1.4.4.3 Find the effects of the reflexions R π .4 Prove that the composition of two reflexions is commutative. Prove that the composition of a rotation and a translation is not necessarily commutative. Determine the the effects of the following scaling transformations on the triangle: S2.30. C = .2 Consider △ABC with A = −1 0 2 . as in fig2 −2 1 ure 1. It seems reasonable to require that this area determination agrees 25 W Free to photocopy and distribute .4.6 Prove that the composition of a scaling and a rotation is not necessarily commutative.2.4.5 Determinants in two dimensions e now desire to define a way of determining areas of plane figures on the plane. Problem 1.30: Problems 1.1 If A = 1 2 æ a 1 b é and é −1 .3.2 .8. and RO on the triangle in figure 1. Prove that the composition of a reflexion and a scaling is commutative. Problem 1.5 Prove that the composition of two rotations on the plane Rθ and Rθ ′ satisfies Rθ ◦ Rθ ′ = Rθ+θ ′ = Rθ ′ ◦ Rθ . = ad − bc. y4 ). listed in counterclockwise order. r2 = (x2 . a b c d 52 Definition The determinant of the 2 × 2 matrix a b c d is det = ad − bc.33). r− 2 1r y2 − y1 Free to photocopy and distribute −→ = x4 − x1 . This motivates the following definition. This quadrilateral is spanned by the vectors −→ = x2 − x1 .34. the area of a parallelogram should be as we learn in elementary geometry and the area of a unit square should be 1. Figure 1. y2 ). From figures (1. y1 ). as in figure 1.Determinants in two dimensions with common formulæ of areas of plane figures. c d b d a b a c Figure 1. r3 = (x3 . y3 ). r− 4 1r y4 − y1 26 . the area of a parallelogram spanned by and c d is a b c d a b . Consider now a simple quadrilateral with vertices r1 = (x1 . in particular. D .33: (a + c)(b + d) − 2 · ab 2 −2 · c(2b+d) 2 = ad − bc.32) and (1.32: Area of a parallelogram. r4 = (x4 . → = (x2 . → − →). r1 r2 r2 r3 r3 r4 r4 r1 1 1 4 2 4 2 3 2 4 3 We conclude that the area of a quadrilateral with vertices (x1 . r2 r1 r4 r1 Similarly. →) + D(→. →) − D(→. r1 r2 r2 r r r3 r3 r1 4 This is equivalent to 1 − − − − − − ¡ 1 − − − − − − − − − − D(→. its area is given by A = det x2 − x1 y2 − y1 x4 − x1 y4 − y1 − − − − = D(→ − →. →) − D(→.35. →) = 0. →) r3 r2 r4 r3 r3 r3 →. →) + D(→. r1 r2 → = (x . → − →) + D(→ − →. creating a parallelogram. y2 ). →) + 2D(→.11. noticing that the quadrilateral is also spanned by −→ = x4 − x3 . (x3 . is 1 → → − − − − − − ¡ D(− . y1 ). which. − ) + D(→. r− 4 3r y4 − y3 its area is also given by A = det x4 − x3 y4 − y3 x2 − x3 y2 − y3 − − − − = D(→ − →.Chapter 1 and hence. r1 r2 r2 r r r3 r3 r1 r2 r3 Free to photocopy and distribute 27 .36. y ). listed in counterclockwise order is 1 2 det x1 y1 x2 y2 + det x2 y2 x3 y3 + det x3 y3 x4 y4 + det x4 y4 x1 y1 . → − →). →) − D(→. reflect it about − r3 3 3 one of its sides. →))¡ + 1 D(→. →) − D(→. by virtue of 1. →) − D(→. →) + D(→. →) − D(→. →) + D(→. y2 ). r− 2 3r y2 − y3 Using the properties derived in Theorem ??. y1 ). (1. (x4 . → − →) r2 r1 r4 r1 r4 r3 r2 r3 →. (x2 . →) + D(→. →) r1 r2 r2 r3 r3 r1 r1 r2 r2 r r r3 r3 r1 r2 r3 2 4 We will prove that − − − − − − − − − − D(→. →) − D(→. →) + D(→. →) + D(→. y4 ). →) − − − 1 + 2 D( r4 r2 2 4 2 1 1 4 1 1 − − − − − − ¡ D(→. The area of the triangle is now half the area of the parallelogram. →) + D(→. →) . we see that A = = = = 1 2 1 2 1 2 1 2 − − − − − − − − ¡ D(→ − →. →) + D(→. →)¡ − − − − − − − − − − − − D( r r r r r r r r r r r r − − − − − − − − ¡ D(→. →) − D(→. y3 ).11) − − Similarly. →) + 2D(→. →) + D(→. →) − D(→. →) − D(→. →)¡ − − − − − − − − D( r r r r r r r r 2 4 2 →. to find the area of a triangle of vertices → = (x1 . →) + D(→. →) − D(→. →) − D(→. r4 r3 r2 r3 −→ = x2 − x3 . as in figure 1. as in figure 1. →) . listed in counterclockwise order. We have already proved the cases n = 3 and n = 4 in (1.12) In general. Proof: The proof is by induction on n. →) + 2D(→. Thus − − − − means r r3 r2 r1 − − − − − − − − − − − − − − − − D(→. →) − − − − = D( r r r r 3 2 2 3 − − − − − − − = D(→. is 1 2 det x1 y1 x2 y2 + det x2 y2 x3 y3 + det x3 y3 x1 y1 . →) = D(→. creating n − 2 triangles. y2 ). (x3 . we appeal once again to the bi-linearity properties derived in Theorem − − − − ??. Then its area is given by 1 2 det x1 y1 x2 y2 + det x2 y2 x3 y3 + · · · + det xn−1 yn−1 xn yn + det xn yn x1 y1 . yn ) be the vertices of a simple (non-crossing) polygon. →) + D(→. →) r3 r2 r2 r3 r2 r3 →. →) − − − − − − −D( r r r r r r 3 2 1 3 3 1 2 3 = 0. If P is not convex. whose vertices (x1 . Consider now a simple polygon P with n vertices. . (1. →) − D(→. then there must be a vertex Free to photocopy and distribute 28 . → − →) + D(→ + → − →. that since we have a parallelogram. →) + D(→. If P is convex then we may take any vertex and draw a line to the other vertices. y2 ). which r r3 r2 r1 → = → + → − →. listed in counterclockwise order. . → − → = → − →. . →) − D(→. we have the following theorem.12) and (1.Determinants in two dimensions To do this. and observe. →) − − +2D( r r as claimed. →) − D(→. →) + 2D(→. →) − D(→. We have proved then that the area of a triangle. y3 ) are listed in counterclockwise order. . (x2 . respectively.11). 53 Theorem (Surveyor’s Theorem) Let (x1 . − − − − − − = D(→ + →. y1 ). → + → − →) + 2 r1 r2 r2 r r r3 r3 r1 r2 r3 r1 r2 r2 r3 r2 r1 → + → − →. →) − − − − − − = D( r3 r2 r2 r3 r2 r3 →. (xn . triangulating the polygon. (x2 . → r1 r2 r3 r3 r2 r1 r →. y1 ). → − →) + 2D(→. →) − D(→. y1 ) Figure 1. y3 ) (x . the terms det xi yi xj yj + det xj yj xi yi = 0. but having opposite orientations. Applying (1. These two sub-polygons are either both convex or at least one is not convex. we repeat the argument. y4 ) r (x. For 1 2 → − − example. . Moreover. that has a reflex angle. the polygon can be triangulated in such a way that all triangles inherit the positive orientation of the original polygon but each neighbouring pair of triangles have opposite orientations. → = a and b = and a matrix A := 2 1 Free to photocopy and distribute 29 det x1 y1 x2 y2 + det x2 y2 x3 y3 + · · · + det xn−1 yn−1 xn yn + det xn yn x1 y1 . y3 ) r3 r2 Chapter 1 (x2 . y2 ) r1 (x3 . r1 (x1 .36: Area of a triangle.34: Area of a quadrilateral. creating a diagonal. disappear from the sum and we are simply left with 1 2 u We may use the software Maple™ in order to speed up computations with vectors. where the sum is over each oriented edge. y1 ) r2 (x2 . y ) 1 1 r3 Figure 1. Eventually. y2 ) Figure 1. This diagonal divides the polygon into two sub-polygons.35: Area of a triangle. A ray produced from this vertex must hit another vertex.12) we obtain that the area is det xi yi xj yj . y2 ) (x3 . In the latter case. Since each diagonal occurs twice. Most of the commands we will need are in the linalg package. let us define two vectors. we end up triangulating the polygon. since the number of vertices is infinite.(x3 . y3 ) r3 r4 (x4 . otherwise the polygon would have infinite area. finding another diagonal and creating a new sub-polygon. y) r2 (x2 . r1 (x1 . 7 Find the Cartesian equa−1 tion of all lines L′ passing through 2 π and making an angle of radians with the 6 Cartesian line L : x + y = 1. angle(a.5. ] 5 5 4 arccos 4 5 2 4 > > dotprod(a.6 Consider two lines on the plane L1 and L2 with Cartesian equations L1 : y = m 1 x + b1 and L2 : y = m 2 x + b1 . b := [1. v v a v Free to photocopy and distribute 30 . Let us compute their dot product. Problem 1.[3. 1] b:=vector([1.4 Let → . prove that − a b ¬¬− ¬¬2 ¬¬→¬¬2 → − ¬¬− ¬¬ − ||→ + b ||2 = ¬¬→¬¬ + ¬¬ b ¬¬ .2]. m 2 = 0. b be arbitrary real numbers. a Problem 1. Prove that (a2 + b2 )2 ≤ 2(a4 + b4 ).2]). . > A:=matrix([[1. −2 Homework − Problem 1.b). find a unit vector in the direction of →. (There must be either a colon or a semicolon at the end of each statement. − − − − − − u v u v u v 4 Problem 1.b).5.5. a a → − − then→ = b . Prove that − − u v → • → = 1 ||→ + →||2 − ||→ − →||2 ¡ . where m 1 = 0. Using Corollary 22. Prove that if − − − − − → ∀ → ∈ R2 .2 (Pythagorean Theorem) If − → ⊥ →. 2 Problem 1.5. √ √ 5 2 5 [ . The result will not display if a colon is chosen.1]). A := 1 3 > det(A).5 (Polarisation Identity) Let →. a := [2. → • → = → • b .5. b be fixed vectors a 2 in R .4]]).1 æ Findé all vectors → ∈ R2 a √ −3 − such that → ⊥ a and ||a|| = 13.3 Let a.5. 2] normalize(a). Problem 1.Determinants in two dimensions 1 2 − . → be vectors in R2 .) > > > > with(linalg): a:=vector([2. a 3 4 and the angle between the vectors. prove that L1 ⊥ L2 ⇐⇒ m 1 m 2 = −1. − − → Problem 1.5. If there exist t1 = t2 such that r(t1 ) = r(t2 ) = p. Our interest is not in graphing curves. with r(t) = .38: x = 2 y = 2t/10 sin t. The curve is simple if its has no multiple points. be vectors on the v w − → = → . then p is a multiple point of the curve. and a theory akin to the one studied for Cartesian equations in a first Calculus course has been developed. →. but in obtaining suitable parametrisations of simple Cartesian curves. 1 + t2 1 − t2 . y = cos 6t. − − − − a v w ¬¬→¬¬2 w ¬¬− ¬¬ w 1. We mention in passing however that Maple Free to photocopy and distribute 31 . A closed curve whose only multiple points are its endpoints is called a Jordan curve.40: x = (1 + cos t)/2. that is. and such that r([a. y = (sin t)(1 + cos t)/2.39: x = y= t − t3 . r(a) is the initial point of the curve and r(b) its terminal point. with w 0 → → − − → = → − v • w → is perpendicular to → . The trace of the curve r is the set of all images of r. Figure 1. t/10 cos t. b] ⊆ R. 1 + t2 Figure 1. Figure 1.5. Prove that the vector − plane.8 Let →.6 Parametric Curves on the Plane x(t) y(t) 54 Definition Let [a. Graphing parametric equations is a difficult art. Γ. b] → R2 . A curve is closed if its initial point and its final point coincide.Chapter 1 − − Problem 1. b]) = Γ. Figure 1. A parametric curve representation r of a curve Γ is a function r : [a.37: x = sin 2t. the curve is traversed from left to right. > plot([(1+cos(t))/2. y = t..cos(6*t). That is. and so this parametrisation only agrees with the curve y = x2 when −1 ≤ x ≤ 1... If x = t and y = t2 . x decreases. If x = t and y = t. √ √ 2.. These parametrisations may differ in features. We will give various parametrisations for portions of this curve. y : t → b(t) and an interval I such that the graphs of f (x.. y) = 0.y=-5. π].x=-5. As t increases. Our main focus of attention will be the following.. we want to find functions x : t → a(t).x=-5.10]. Free to photocopy and distribute 32 . t ∈ [0.5). t∈R works for the whole curve. √ √ 3.x=-5.. +∞[ gives the half of the curve for which x ≥ 0. we wish to find suitable parametrisations for them. Given a Cartesian curve with equation f (x. and hence the parametrisation √ x = − t. > plot([2ˆ(t/10)*cos(t). 55 Example Consider the parabola with Cartesian equation y = x2 . b(t)) = 0. the curve is traversed from left to right. This works for every t ∈ R.37 through 1. t ∈ [0. > with(plots): > plot([sin(2*t).2].5.5.t=-2.. then clearly y = t2 = x2 . Notice that as t increases. then again y = cos2 t = (cos t)2 = x2 . y) = 0 and f (a(t). This works only for t ≥ 0. 1. y = t2 .5). according to the choice of functions and the choice of intervals.2*Pi]. Similarly. t ∈ I coincide. For t ∈ [0.t=0. For example.y=-5. This works only for t ≥ 0. Both x and y are periodic with period 2π.40 follow.Parametric Curves on the Plane has excellent capabilities for graphing parametric equations. If x = cos t and y = cos2 t. and so the curve is traversed from right to left. and hence the parametrisation √ x = t. the commands to graph the various curves in figures 1.5..5).2ˆ(t/10)*sin(t). and hence the parametrisation x = t.(t-tˆ3)/(1+tˆ2).2*Pi]. 4.t=0.y=-5. y = t.x=-5. As t increases. then again y = t = ( t)2 = x2 . > plot([(1-tˆ2)/(1+tˆ2).5. if x = − t and y = t. the cosine decreases from 1 to −1 and so the curve is traversed from right to left in this interval. +∞[ gives the half of the curve for which x ≤ 0..t=-20. then again y = t = (− t)2 = x2 ...5).y=-5.sin(t)*(1+cos(t))/2. −2). Figure 1. √ Figure 1. 1]. π]. y = cos2 t. The first parametrisation must satisfy that as t traverses the values in the interval [0. tan2 θ − sec2 θ = 1. +∞[. y = t2 . (sin 4πt. sin t) traverses the unit circle once. y = −2 + 3 sin t. 56 Example Give two distinct parametrisations of the ellipse 1. Solution: ◮ What formula do we know where a sum of two squares equals 1? We use a trigonometric substitution. y = t. cosh2 θ − sinh2 θ = 1. clockwise. one starts at the point (3. one starts at the point (3. The identities cos2 θ + sin2 θ = 1. finishing at (3.42: x = t ∈ [0. The second parametrisation must satisfy that as t traverses the interval [0.” Observe that for t ∈ [0. notice that as t traverses the interval [0. traverses the ellipse once counterclockwise. the point (cos t. cos 4πt) traverses the unit interval twice. t ∈ [0. 2π]. 2π] Then is the desired first parametrisation. 2 and y+2 = sin t =⇒ y = −2 + 3 sin t. 3 x = 1 + 2 cos t. −2).41: x = t. (x − 1)2 4 + (y + 2)2 9 = 1. t ∈ R. traverses the ellipse twice clockwise. Put x−1 = cos t =⇒ x = 1 + 2 cos t. Figure 1. 2. a sort of “polar coordinates. For the second parametrisation.44: x = cos t.43: x = − t.Chapter 1 Figure 1. +∞[. t ∈ [0. t ∈ [0. −2). and returns to (3. y = t. are often useful when parametrising quadratic curves. starting at (1. 2π]. √ t. but Free to photocopy and distribute 33 . 0) and ending there. −2). 1]. Parametric Curves on the Plane begins and ends at the point (0, 1). To begin at the point (1, 0) we must π π make a shift: sin 4πt + , cos 4πt + will start at (1, 0) and 2 2 travel clockwise twice, as t traverses [0; 1]. Hence we may take x = 1 + 2 sin 4πt + π 2 , y = −2 + 3 cos 4πt + π 2 , t ∈ [0; 1] as our parametrisation. ◭ Some classic curves can be described by mechanical means, as the curves drawn by a spirograph. We will consider one such curve. O′ φ B P A θ O Figure 1.45: Construction of the hypocycloid. Figure 1.46: Hypocycloid with R = 5, ρ = 1. Figure 1.47: Hypocycloid with R = 3, ρ = 2. 57 Example A hypocycloid is a curve traced out by a fixed point P on a circle C of radius ρ as C rolls on the inside of a circle with centre at O and radius R. If the R initial position of P is , and θ is the angle, measured counterclockwise, that 0 a ray starting at O and passing through the centre of C makes with the x-axis, shew that a parametrisation of the hypocycloid is x = (R − ρ) cos θ + ρ cos y = (R − ρ) sin θ − ρ sin Free to photocopy and distribute (R − ρ)θ ρ (R − ρ)θ ρ , . 34 Chapter 1 Solution: ◮ Suppose that starting from θ = 0, the centre O ′ of the small circle moves counterclockwise inside the larger circle by an angle θ, and the point P = (x, y) moves clockwise an angle φ. The arc length travelled by the centre of the small circle is (R − ρ)θ radians. At the same time the point P has rotated ρφ radians, and so (R − ρ)θ = ρφ. See figure 1.45, where O ′ B is parallel to the x-axis. π Let A be the projection of P on the x-axis. Then ∠OAP = ∠OP O′ = , 2 π ′ ∠OO P = π − φ − θ, ∠P OA = − φ, and OP = (R − ρ) sin(π − φ − θ). 2 Hence x = (OP ) cos ∠P OA = (R − ρ) sin(π − φ − θ) cos( y = (R − ρ) sin(π − φ − θ) sin( Now x = = = = = (R − ρ) sin(π − φ − θ) cos( π 2 − φ) π 2 − φ). π 2 − φ), (R − ρ) sin(φ + θ) sin φ (R − ρ) (cos θ − cos(2φ + θ)) 2 (R − ρ) (cos θ + cos(2φ + θ)) (R − ρ) cos θ − 2 (R − ρ) cos θ − (R − ρ)(cos(θ + φ) cos φ). ρ OO′ =− ρ R−ρ and cos φ = Also, cos(θ + φ) = − cos(π − θ − φ) = − cos (R − ρ)θ ρ and so x = (R−ρ) cos θ−(R−ρ)(cos(θ+φ) cos φ = (R−ρ) cos θ+ρ cos (R − ρ)θ ρ , as required. The identity for y is proved similarly. A particular example appears in figure 1.47. ◭ Free to photocopy and distribute 35 − d→ x Parametric Curves on the Plane → − x → + d→ − − x x Figure 1.49: Area enclosed by a simple closed curve Figure 1.48: Length of a curve. Given a curve Γ how can we find its length? The idea, as seen in figure 1.48 is to consider the projections dx, dy at each point. The length of the vector dr = is ||dr|| = Hence the length of Γ is given by ||dr|| = (dx)2 + (dy)2 . Γ dx dy (dx)2 + (dy)2 . (1.13) Γ Similarly, suppose that Γ is a simple closed curve in R2 . How do we find the (oriented) area of the region it encloses? The idea, borrowed from finding areas of polygons, is to split the region into triangles, each of area 1 2 det x x + dx y y + dy = 1 2 det x y dx dy = 1 2 (xdy − ydx), and to sum over the closed curve, obtaining a total oriented area of 1 2 Á Á det Γ x y dx dy = 1 2 Á Γ (xdy − ydx). (1.14) Here Γ denotes integration around the closed curve. Figure 1.50: Example 58. Figure 1.51: Example 59. Free to photocopy and distribute 36 Chapter 1 58 Example Let (A, B) ∈ R2 , A > 0, B > 0. Find a parametrisation of the ellipse Γ: (x, y) ∈ R2 : x2 A2 + y2 B2 =1 . Furthermore, find an integral expression for the perimeter of this ellipse and find the area it encloses. Solution: ◮ Consider the parametrisation Γ : [0; 2π] → R2 , with x y = A cos t B sin t . This is a parametrisation of the ellipse, for x2 A2 + y2 B2 = A2 cos2 t A2 + B 2 sin2 t B2 = cos2 t + sin2 t = 1. Notice that this parametrisation goes around once the ellipse counterclockwise. The perimeter of the ellipse is given by ¬¬ →¬¬ ¬¬d− ¬¬ = r 2π 0 A2 sin2 t + B 2 cos2 t dt. Γ The above integral is an elliptic integral, and we do not have a closed form for it (in terms of the elementary functions studied in Calculus I). We will have better luck with the area of the ellipse, which is given by 1 2 Á Γ (xdy − ydx) = = = 1 2 1 2 1 2 Á (A cos t d(B sin t) − B sin t d(A cos t)) 2π 0 2π (AB cos2 t + AB sin2 t)) dt AB dt . 0 = πAB. ◭ 59 Example Find a parametric representation for the astroid Γ : (x, y) ∈ R2 : x2/3 + y 2/3 = 1 , in figure 1.51. Find the perimeter of the astroid and the area it encloses. Solution: ◮ Take x y Free to photocopy and distribute = cos3 t sin3 t 37 Parametric Curves on the Plane with t ∈ [0; 2π]. Then x2/3 + y 2/3 = cos2 t + sin2 t = 1. The perimeter of the astroid is 2π Γ ||dr|| = = = 9 cos4 t sin2 t + 9 sin4 t cos2 t dt 3| sin t cos t| dt 0 2π 0 2π 0 π/2 0 3 2 | sin 2t| dt sin 2t dt = 6 = 6. The area of the astroid is given by 1 2 Á Γ (xdy − ydx) = = = = = = 1 2 1 2 3 2 3 8 3 Á (cos3 t d(sin3 t) − sin3 t d(cos3 t)) 2π 0 2π 0 2π 0 (3 cos4 t sin2 t + 3 sin4 t cos2 t)) dt (sin t cos t)2 dt (sin 2t)2 dt (1 − cos 4t) dt 2π 0 16 3π 8 . We can use Maple™ (at least version 10) to calculate the above integrals. For example, if (x, y) = (cos3 t, sin3 t), to compute the arc length we use the path integral command and to compute the area, we use the line −y/2 integral command with the vector field . x/2 > with(Student[VectorCalculus]): > PathInt( 1, [x,y]=Path( <(cos(t))ˆ3,(sin(t))ˆ3>,0..2*Pi)); > LineInt( VectorField(<-y/2,x/2>), Path( <(cos(t))ˆ3,(sin(t))ˆ3>,0..2*Pi)); ◭ We include here for convenience, some Maple commands to compute various arc lengths and areas. 60 Example To obtain the arc length of the path in figure 1.52, we type > with(Student[VectorCalculus]): > PathInt(1, [x,y] = LineSegments( <0,0>, <1,1>, <1,2> ,<2,1>,<3,3>,<4,1>); To obtain the arc length of the path in figure 1.53, we type Free to photocopy and distribute 38 6. 1 −1 −1 1 Figure 1.3 Let a. c. y= . over Free to photocopy and distribute 39 . Pi/6.6.0. Problem 1.6. > Path( <(1+cos(t))*(cos(t))+1. d be strictly positive real constants.55: Problem 1. Find its Cartesian equation.x/2>). [x. x = at + b.54 > with(Student[VectorCalculus]): > LineInt( VectorField(<-y/2.5 Describe the trace of the parametric curve æ é x y æ = sin t é .2 Give an implicit Cartesian equation for the parametric representation t2 t3 x= .52: Line Path.4 Parametrise the curve y = π log cos x for 0 ≤ x ≤ . x = a sec t. t ∈ [0. 1 + t5 1 + t5 Problem 1. 6 5 5 4 3 2 1 1 2 3 4 5 Figure 1. t ∈] − π π . 1. y = 0 3.y] = Arc( Circle( <0. Pi/5 ) .6.Chapter 1 with(Student[VectorCalculus]): PathInt(1.6. x = cos t.6. > > 5 4 3 2 1 1 2 3 4 5 Figure 1. y(t) = 1 − t2 . Give a suitable parametrisation of this curve. y = b sinh t 4. b. angle ≤θ≤ . and find its length.7 Consider the graph given parametrically by x(t) = t3 + 1. 2 sin t + 1 Problem 1. [ 2 2 Problem 1.53: Arc of circle of π π radius 3. In each case give an implicit Cartesian equation for the parametric representation and describe the trace of the parametric curve. y = 2 + (1 + cos t)(sin t). y = ct + d 2.2*Pi)).55. x = a cosh t.(1+cos(t))*(sin(t))+2>. Problem 1.6. Problem 1. Then find its arc 3 length.6..0>. y = b tan t. 5 4 3 2 1 1 2 3 4 5 Figure 1. y(t) = t3 + 2t. Homework Problem 1. To obtain the area inside the curve in 1.6 Consider the plane curve √ √ defined implicitly by x + y = 1.1 A curve is represented parametrically by x(t) = t3 − 2t. The graph of the curve appears in figure 1.54: x = 1 + (1 + cos t)(cos t). 4π].6. Find the area under the graph. 3). Problem 1.57: Problem 1. Find the ity of magnitude V . Free to photocopy and distribute 40 .10 Find the area enclosed by the curve x(t) = sin3 t. Problem 1. t ∈ R\{−1}. The curve appears in figure 1.11 Let C be the curve in R defined by 2 2 Figure 1. known. known that the shell was fired from a is 1 + t3 1 + t3 gun on the ground with a muzzle velocwhich you may see in figure 1.8 Find the arc length of the curve given parametrically by x(t) = 3t2 .6. y(t) = (cos t)(1 + sin2 t).12 Let P be a point at a distance d from the centre of a circle of radius Problem 1. is called a cycloid.6. 3 1 1 t ∈ [− .6. Find a parametrisation of the cycloid. Problem 1. without slipping.6.10.15 A shell strikes an airplane flying at height h above the ground.58: Cycloid Problem 1.13 Find the arc length of the arc of the cycloid x = ρ(t − cos t).6. Problem 1.56: Problem 1. The curve traced out by P as the circle rolls along a straight line. It 3t 3t x(t) = . π]. 2π]. t ∈ [0. Figure 1. y = ρ(1 − cos t). t ∈ [0. Deduce that the gun is situated within a circle whose centre lies directly below the airplane and whose radius is 1 Ô V V 2 − 2gh .11. but the position of the gun and its angle of elevation are both unarea enclosed by the loop of this curve.14 Find the length of the parametric curve given by x = et cos t. Problem 1. y(t) = 2t3 for 0 ≤ t ≤ 1. + ]. y = et sin t. y(t) = .6.6. Find the length of this curve. 2 2 P d θ C ρ B ρ.56. and between the lines x = 1 and x = 2.9 Let C be the curve in R2 defined by x(t) = t2 .6. Problem 1.6. Find the equation of the locus of the vertex of the rolling parabola.Parametric Curves on the Plane the x axis.57.16 The parabola y 2 = −4px rolls without slipping around the parabola y 2 = 4px. g −1 −1 1 Figure 1. 2 y(t) = (2t + 1)3/2 . Problem 1.6.6. Free to photocopy and distribute 41 . y. 1. z x y Figure 1.59 we have pictured the point (2. 0 0 0 → − k = 0 . 61 Definition The 3-dimensional Cartesian Space is defined and denoted by In figure 1. we have a choice for the orientation of the the x and y-axis.Chapter 1 1. 0 and observe that 1 ¾ ¿ → − j = 1 . z ∈ R} . Let us explain. as in figure 1. 1 ¾ ¿ → − → − → − r = (x. 3). y.60. Most of what we did in R2 transfers to R3 without major complications. z) = x i + y j + z k . y ∈ R.7 Vectors in Space R3 = {r = (x.59: A point in space. We adopt a convention known as a right-handed coordinate system. In analogy to R2 we put ¾ ¿ → − i = 0 . z) : x ∈ R. Having oriented the z axis upwards. the dot product satisfies →• b = ¬¬→¬¬¬¬ b ¬¬ cos θ. By 42 Free to photocopy and distribute . π] a is the convex angle between the two vectors. where θ ∈ [0. z are positive. z be positive real numbers such that x2 + 4y 2 + 9z 2 = 27. Solution: ◮ Since x.63: ℓ1 ℓ3 are skew. ¬¬− ¬¬¬¬→¬¬ − ¬¬− ¬¬ − → a Just as in R2 . Maximise x + y + z. ℓ1 and → − − 62 Definition The dot product of two vectors → and b in R3 is a →•→ = a b + a b + a b . 63 Example Let x. |x + y + z| = x + y + z.→ − k → − k Vectors in Space → − j → − j → − i → − i Figure 1. Figure 1. − − a b 1 1 2 2 3 3 − The norm of a vector → in R3 is a Ô ¬¬→¬¬ − − ¬¬− ¬¬ = →•→ = a a a (a1 )2 + (a2 )2 + (a3 )2 . ℓ 2 Figure 1. y.61: Right Hand. The Cauchy-Schwarz-Bunyakovsky Inequality takes the form ¬¬− ¬¬¬¬→¬¬ − ¬¬− ¬¬ − → |→• b | ≤ ¬¬→¬¬¬¬ b ¬¬ =⇒ |a1b1 +a2 b2 +a3 b3 | ≤ (a2 +a2 +a2 )1/2 (b2 +b2 +b2 )1/2 . Left-handed ℓ1 Figure 1.60: Right-handed ℓ3 system.62: system. ℓ2 . y. a a 1 2 3 1 2 3 equality holding if an only if the vectors are parallel. z= √ 2 3 7 . Therefore for a maximum we take x= ◭ − 64 Definition Let a be a point in R3 and let → = v parametric line passing through a in the direction of √ 18 3 7 . 2 Equality occurs if and only if ¾ x 2y 3z ¿ ¾ 1 ¿ = λ 1/2 1/3 =⇒ x = λ. y = λ 4 . ¬ ¬ |x+y+z| = ¬x + 2y ¬ 1 2 + 3z 1 1 ¬ 1 ¬ ≤ (x2 +4y 2 +9z 2 )1/2 1 + + ¬ 3 4 9 ¬ 1/2 √ 7 = 27 6 √ 7 3 = . The → is the set − v − r ∈ R3 : r = a + t→ . y= √ 9 3 14 .Chapter 1 Cauchy’s Inequality. 43 . The desired equation is à à x y z ◭ Free to photocopy and distribute = 1 2 3 ¾ ¿ 3 3 +t 3 . v à 1 65 Example Find the parametric equation of the line passing through à 2 3 and −2 −1 0 .z = λ 9 =⇒ λ + 2 λ2 4 + λ2 9 = 27 =⇒ λ = ± √ 18 3 7 . Solution: ◮ The line follows the direction ¾ 1 − (−2) 3−0 ¿ ¾ ¿ 3 3 2 − (−1) = 3 . → − 0 be a vector in R3 . →.Vectors in Space Given two lines in space. that two vectors are coplanar if there exists bi-point representatives of the vector that lie on the same plane. This prompts the following theorem. See figure 1. Conversely. a c Any vector in R3 can be written as a linear combination of three linearly independent vectors in R3 . We will say. All the above gives the following result. b . they still span a plane. one of the following three situations might arise: (i) the lines intersect at a point. abusing language. (ii) the lines are parallel. a a → − − which is a line through the origin. Thus → − − − − → {s→ + t b : s ∈ R. any vector → − − a. again abusing language. b. they would span the whole plane R2 . passing through the origin. t ∈ R. We will say. Then every vector → of the v w u form → = a→ + b→.64. → in R3 be non-parallel vectors. → in R3 are said to be linearly independent if a c → − → − − − α→ + β b + γ → = 0 =⇒ α = β = γ = 0. − − − 66 Theorem Let →. v w See figure 1. t ∈ R} = {λ→ : λ ∈ R}. a We saw in the preceding chapter that if the vectors were on the plane. (iii) the lines are skew (non-parallel. b arbitrary scalars. and that x c are non-collinear points on the same plane and that r = y z Free to photocopy and distribute 44 is another . In the case at hand the vectors are in space. A plane is determined by three non-collinear points. − From the above theorem. → − − − − → Consider now two non-zero vectors → and b in R3 . that a vector is parallel to a specific plane or that it lies on a specific plane if there exists a bi-point representative of the vector that lies on the particular plane. is coplanar with both v w t → and → can be uniquely expressed in the form − − coplanar with both v w → − − − t = p→ + q → . without intersecting. then linear combinations of these three vectors may lie outside the → − − plane containing b . c . c − − → − 67 Theorem Three vectors →. one over the other. → b } a a is a plane passing through the origin. then the set a a → − − − {s→ + t b : s ∈ R. Suppose now that → and b are not parallel. If → b . if a vector → is not a linear combination of two other a → → − − vectors b . lying on different planes). − − − u v w − → and →.63. Suppose à a. Free to photocopy and distribute 45 . (u2 v3 − u3 v2 )(x − a1 ) = (u2 v3 − u3 v2 )(pu1 + qv1 ). and → are the position vectors of these points. b. Multiplying the first equation by u2 v3 − u3 v2 . − − − − − r a b a c a is the equation of a plane containing the three non-collinear points a. y − a2 = pu2 + qv2 . − also lies on the plane. are non-parallel. that there exist real numbers p. and the third by u1 v2 − u2 v1 . →= − n ¾ ¿ a b c → − v − − p→ + q → v w ax + by + cz = d → − x → − w Figure 1. − − 68 Theorem Let → and → be linearly independent vectors. The parametric equau v − − tion of a plane containing the point a. By Chasles’ Rule.65: Theorem 69. − − → − where →. and c. Since ax Lemma 66. we obtain. b. we have by → which are coplanar. and c are non-collinear. − → = → + p(→ − →) + q(→ − →). Figure 1. Since a. (u1 v2 − u2 v1 )(z − a3 ) = (u1 v2 − u2 v1 )(pu3 + qv3 ). z − a3 = pu3 + qv3 . the second by u3 v1 − u1 v3 . − − − − r a u v Componentwise this takes the form x − a1 = pu1 + qv1 . Thus we have the a c following theorem. q with → → = p− + q → − − ar ab ac.64: Theorem 66.Chapter 1 − → − arbitrary point on this plane. ab and → ac. (u3 v1 − u1 v3 )(y − a2 ) = (u3 v1 − u1 v3 )(pu2 + qv2 ). b . and parallel to the vectors → and → is u v given by → − → = p→ + q → . prov1 1 0 . Here a2 + b2 + c2 = 0. c are zero. b = u3 v1 − u1 v3 . Then x− y z d a á ¾ =s − b¿ +t ¾ a 1 0 − c¿ a 0 1 is a parametric equation for the plane. a − c a + b (0) + c (1) = 0. observe that if → and → are non-parallel vectors and → − u v r → = p→ + q → is the equation of the plane containing the point a and − − − a u v − − − parallel to the vectors → and →. (u2 v3 − u3 v2 )(x − a1 ) + (u3 v1 − u1 v3 )(y − a2 ) + (u1 v2 − u2 v1 )(z − a3 ) = 0. This gives the − − Since v u following theorem. ¾ ¾ ¿ and so is simultaneously perpendicular to − ¾ c¿ b¿ − a a and 0 . then if → is simultaneously perpendicu v n → and → then (→ − →)•→ = 0 for →•→ = 0 = →•→. The argument is similar if one of the other letters is non-zero and a = 0. the vector n b is normal c to the plane with Cartesian equation ax + by + cz = d. Proof: We have already proved the first statement. c = u1 v2 − u2 v1 . Put and a = u2 v3 − u3 v2 . We have a − a b c ing the second statement. 69 Theorem The equation of the plane in space can be written in the form ax + by + cz = d. c is non-zero. b. d = a1 (u2 v3 − u3 v2 ) + a2 (u3 v1 − u1 v3 ) + a3 (u1 v2 − u2 v1 ). − − − − − − − − − ular to u v r a n u n v n since at least one of a. For the second − − − statement. we may assume a = 0. which is the Cartesian equation of the plane. In this case we can see that d b c x = − y − z. that is. Moreover. not all of a. at ¾ ¿ a →= − least one of the coefficients is non-zero. b. a a a Put y = s and z = t. → is linearly independent from →. Now.Vectors in Space Adding gives.u Free to photocopy and distribute 46 b a + b (1) + c (0) = 0. Hence the Cartesian equation is x − y = 1. − n 73 Example Free to photocopy and distribute 47 . ◭ − − 72 Definition If → is perpendicular to plane Π1 and →′ is perpendicular to plane n n − Π2 . 71 Example Find both the parametric equation and the Cartesian equation of ¾ ¿ ¾ ¿ 1 1 the plane parallel to the vectors 1 and 1 and passing through the point à 1 . and ¾ ¿ 1 1 ¾ ¿ 1 0 =s 1 +t 1 . 0 0 −1 2 Solution: ◮ The desired parametric equation is à x y+1 z−2 This gives s = z − 2. the angle between the two planes is the angle between the two vectors → and n →′ .Chapter 1 à 1 70 Example The equation of the plane passing through the point ¾ −1 2 and normal to the vector −3 2 4 ¿ is −3(x − 1) + 2(y + 1) + 4(z − 2) = 0 =⇒ −3x + 2y + 4z = 3. t=y+1−s=y+1−z+2=y−z+3 x = s + t = z − 2 + y − z + 3 = y + 1. 67: The plane z = 1 − y. This appears in figure 1. 5.68: Solid bounded by the planes z = 1 − x and z = 1 − y in the first octant. the height of this pyramid is clearly 1. 2. then 1·0+0·1+1·1 1 π cos θ = √ =⇒ cos θ = =⇒ θ = . Figure 1.z z z Vectors in Space x y x y x y Figure 1. Figure 1. and the vector 1 1 1 is normal to the plane y + z = 1. 1. where A is area of the base of the pyramid and 3 h is its height. 3. If θ is the angle between these two vectors. Find the angle between the planes z = 1 − x and z = 1 − y. and 1 hence the volume required is . 2.66. This appears in figure 1. 4. This appears in figure 1.66: The plane z = 1 − x. 3 ◭ Homework Free to photocopy and distribute 48 . √ 2 + 12 · 2 + 12 2 3 1 1 4. Solution: ◮ 1. Now. The vector 0 is normal to the plane x + z = 1. Draw the intersection of the plane z = 1 − y with the first octant. ¾ ¿ 1 ¾ ¿ 0 3. Draw the solid S which results from the intersection of the planes z = 1−x and z = 1 − y with the first octant.68. 5. Find the volume of the solid S . Draw the intersection of the plane z = 1 − x with the first octant. Recall that the volume of a pyramid is given by the Ah formula V = . The resulting solid is a pyramid with square base of area A = 1 · 1 = 1.67. 12 Let a > 0. 2.7. Ǒ . S(a. (Vertex A is opposite to the side measuring a. c > 0 be the lengths of the sides of △ABC.7.10 Determine lines Ǒ Ǒ x 1 L1 : y z x L2 : y z = 1 1 0 = 0 1 whether the in R3 .4 Find the equation of plane Ǒ 1 containing the point −1 ing the line x = 2y = 3z. b. intersect. b > 0.5 (Putnam Exam 1984) Let A be a solid a × b × c rectangular brick in three dimensions. Prove that f (a. Problem 1.6 It is known that ¬¬→ ¬¬ = 3. Let B be the set of all points which are at distance at most 1 from some point of A (in particular.) Recall that by Heron’s Formula. a+b+c where s = is the semiperime2 ter of the triangle. c be arbitrary real numbers. − − − − a b b c c a Problem 1.7.7. a ¬¬ b ¬¬ ¬¬ ¬¬ → ¬¬ − ¬¬− Find ¬¬→ − b ¬¬. c) = S(a.2 Find theǑ equation of the 1 line passing through 2 3 in the direction ¾ of ¿ −2 0 −1 . etc.8 Find the equation of the plane perpendicular to the line ax = by = cz. Problem 1. c) is maximised when △ABC is a 2 + b2 + c 2 equilateral. z = 1−t.7.7 Find the equation of the line perpendicular to the plane ax + a2 y + a3 z = 0. b.7.7. ¬¬ c − → = →. −1 and contain- ¾ ¿ 2 +t 1 .Chapter 1 Problem 1. Express the volume of B as a polynomial in a. c. c) ¬¬− ¬¬ Problem 1. 0 point 0 1 Problem 1. y = −2t. Problem 1. b. Problem 1.7.7. where a > 0. c > 0. Problem 1. and find this maximum. the area of this triangle is = s(s − a)(s − b)(s − c).11 Let a. a ¬¬ ¬¬ ¬¬→ ¬¬ − − ¬¬→¬¬ − ¬¬ = 5 and that → + → + − a b ¬¬ b ¬¬ = 4.7. Prove that (a2 + b2 + c2 )2 ≤ 3(a4 + b4 + c4 ). a − → . b. → satisfy − b ¬¬ ¬a ¬ − ¬¬→ →¬¬ − ¬¬ a + b ¬¬ = 24. ¬¬→¬¬ = 19. Problem 1.9 Find the (shortest) distance from the point (1. Ǒ = 0 and passing through the point abc 1 1 1 Problem 1.1 Vectors ¬¬ ¬¬ ¬¬→¬¬ − ¬¬− ¬¬ = 13.7. b > 0. Find the angle between them. 1 Ǒ Ǒ ¾ 2 1 ¿ + t −1 . A ⊆ B). a = 0 and passing through the Free to photocopy and distribute 49 .3 Find the Ǒ equation of plane 1 containing the point 1 and perpendic1 ular to the line x = 1+t. Find − c 0 − − → •→ + → •→ + → •→ . 3) to the plane x − y + z = 1. b. Problem 1.7. 7. 2x + 3z ≤ 24. and let α ∈ R be a scalar.7. 3 1. z ≥ 0.15 Find the Cartesian equaǑ 1 tion of the plane passing through 0 0 . y.16 Prove that there do not exist three unit vectors in R3 such that the 2π angle between any two of them be > . Draw this plane and its 0 1 intersection with the first octant.17 Let (→ − →)•→ = 0 be a r a n plane passing through the point a and per− pendicular to vector →. 1 0 and 0 1 . Prove that √ √ √ √ x+y+ y+z+ z+x ≤ 6. y. − − − Problem 1. each normal to a face of the polyhedron. Prove that the sum of → − these vectors is 0 .7. 0 0 .18 (Putnam Exam 1980) Let S be the solid in three-dimensional space Ǒ x consisting of all points y satisfying the z following system of six conditions: x ≥ 0. If b is not a point n on the plane.8 Cross Product We now define the standard cross product in R3 as a product satisfying the following properties. Problem 1.Cross Product Problem 1.7. √ x+y+z Problem 1. and length equal to the area of the face. Problem 1. y ≥ 0. Problem 1. Ǒ 0 1 and 0 0 Ǒ . 2x + 4y + 3z ≤ 36.14 Let x.19 Given a polyhedron with n faces. Prove that x+y+z ≤ 2 x2 y2 z2 + + y+z z+x x+y .7.7. − − − 74 Definition Let →. → be vectors in R3 . x + y + z ≤ 11.13 Let x. Determine the number of vertices and the number of edges of S. The cross x y z product × is a closed binary operation satisfying − − − − Anti-commutativity: → × → = −(→ × →) x y y x Bilinearity: − − − − − − − (→ + →) × → = → × → + → × → and x z y x y z y → × (→ + →) = → × → + → × → − − − − − − − x z y x z x y → − − − →×→= 0 x x − − − − − − Scalar homogeneity: (α→) × → = → × (α→) = α(→ × →) x y x y x y Free to photocopy and distribute 50 .7. →. z be strictly positive numbers. ¬¬− ¬¬ n Problem 1. Find the volume of the tetrahedron with vertices at Ǒ Ǒ Ǒ Ǒ 0 1 0 0 0 0 . z be strictly positive numbers. then the distance from b to the plane is ¬ ¬ − − ¬ → → •→ ¬ − ¬( a − b ) n ¬ ¬¬→¬¬ . consider n vectors. − − x y 2 3 3 2 i 3 1 1 3 j 1 2 2 1 k Free to photocopy and distribute 51 . since → − → → − − → → − − → − i ×( i × j )= i × k =− j → → − − → − − → → − → − ( i × i )× j = 0 × j = 0. Operating as in example 75 we obtain ¾ − 76 Theorem Let → = x2 x x3 x1 ¿ ¾ − and → = y2 y y3 y1 ¿ be vectors in R3 . allows us to “get out” of that plane. k × i = j . given two non-parallel vectors on a plane.Chapter 1 Right-hand Rule: → → − − → → → − − − → → → − − − → − i × j = k. 75 Example Find ¾ ¿ ¾ ¿ 1 0 −3 0 2 × 1 . It follows that the cross product is an operation that. Solution: ◮ We have → − → − → − → − ( i −3k)×( j +2k) = = = → → − − → → − − → → − − → → − − i × j +2 i × k −3k × j −6k × k → − → − → − → − k −2 j +3 i +60 → − → → − − 3 i −2 j + k. ◭ but The cross product of vectors in R3 is not associative. j × k = i . Then − − − → × → = (x y − x y )→ + (x y − x y )→ + (x y − x y )→. ¾ ¿ Hence ¾ 1 0 −3 ¿ 0 2 ¾ 3 1 ¿ × 1 = −2 . Figure 1. we may obtain a third vector simultaneously perpendicular to two other vectors in space. the cross product of two x x y y x y vectors is simultaneously perpendicular to both original vectors. that is. obtaining.70: Area of a parallelogram → → − − → → − − → → − − → − Proof: Since i × i = j × j = k × k = 0 . 78 Theorem − → × (→ × →) = (→•→)→ − (→•→)→.×→ − y ¬ ¬¬¬→ ¬¬ − y ¬¬ → − x Cross Product θ − → x → − y ¬¬→¬¬ ¬¬− ¬¬ x Figure 1. Proof: We will only check the first assertion. proving the theorem. − − − − − − − − a b c a c b a b c Free to photocopy and distribute 52 ¬¬→¬¬¬¬→¬¬ ¬¬− ¬¬¬¬− ¬¬ sin θ x y . →•(→ × →) − − − x x y → − → − → − → − = (x1 i + x2 j + x3 k )•((x2 y3 − x3 y2 ) i → − → − +(x3 y1 − x1 y3 ) j + (x1 y2 − x2 y1 ) k ) = 0. the following theorem. we have. u Although the cross product is not associative. however.69: Theorem 79. = x 1 x 2 y3 − x 1 x 3 y2 + x 2 x 3 y1 − x 2 x 1 y3 + x 3 x 1 y2 − x 3 x 2 y1 completing the proof. we only worry about the mixed products. u Using the cross product. − − − − − − 77 Theorem → ⊥ (→ × →) and → ⊥ (→ × →). →×→ − − x y → − → − → − → − → − → − = (x1 i + x2 j + x3 k ) × (y1 i + y2 j + y3 k ) → → − − → → − − → → − − → → − − = x 1 y2 i × j + x 1 y3 i × k + x 2 y1 j × i + x 2 y3 j × k → → − − → → − − +x3 y1 k × i + x3 y2 k × j → → − − → → − − → → − − = (x1 y2 − y1 x2 ) i × j + (x2 y3 − x3 y2 ) j × k + (x3 y1 − x1 y3 ) k × i → − → − → − = (x1 y2 − y1 x2 ) k + (x2 y3 − x3 y2 ) i + (x3 y1 − x1 y3 ) j . the second verification is analogous. →). C y − − − 79 Theorem Let (→.72: Example 83. Then − y − − − − − − ||→ × →|| = ||→||||→|| sin (→.71: Theorem 82. →) ∈ [0.Chapter 1 Proof: − → × (→ × →) = − − a b c = = = → − → − → − → − (a1 i + a2 j + a3 k ) × ((b2 c3 − b3 c2 ) i + → − → − +(b3 c1 − b1 c3 ) j + (b1 c2 − b2 c1 ) k ) → − → − → − a1 (b3 c1 − b1 c3 ) k − a1 (b1 c2 − b2 c1 ) j − a2 (b2 c3 − b3 c2 ) k → − → − → − +a2 (b1 c2 − b2 c1 ) i + a3 (b2 c3 − b3 c2 ) j − a3 (b3 c1 − b1 c3 ) i → − → − → − (a1 c1 + a2 c2 + a3 c3 )(b1 i + b2 j + b3 i )+ → − → − → − (−a1 b1 − a2 b2 − a3 b3 )(c1 i + c2 j + c3 i ) → − → → − − − − − (→•→) b − (→• b ) c . u z P D′ − →×→ − a b A′ → − c D x B′ C′ N θ −b → A → − a M B Figure 1. a c a completing the proof. x y x y x y Free to photocopy and distribute 53 . Figure 1. π] be the convex angle between two vectors → and x y x →. See figure 1. →).Cross Product Proof: We have − − ||→ × →||2 x y = (x2 y3 − x3 y2 )2 + (x3 y1 − x1 y3 )2 + (x1 y2 − x2 y1 )2 2 2 2 = y 2 y3 − 2x2 y3 x3 y2 + z 2 y2 + z 2 y1 − 2x3 y1 x1 y3 + 2 2 2 +x2 y3 + x2 y2 − 2x1 y2 x2 y1 + y 2 y1 − − − − − − = ||→||2||→||2 − ||→||2 ||→||2 cos2 (→. The volume of the paral− − where θ is the angle between c a b lelepiped is thus ¬¬ →¬¬¬¬→¬¬ − ¬¬ − − ¬¬→ − → → − → − ¬¬ a × b ¬¬¬¬ c ¬¬ cos θ = ( a × b ) • c . →) x y x y x y →||2||→||2 sin2 (→. The area of the base of the parallelepiped is the → − − area of the¬ parallelogram determined by the vectors → and b .70. → satisfy → × → = 0 if and only if they x y x y are parallel. be linearly independent vectors in R3 . proving the theorem. 81 Corollary (Lagrange’s Identity) − − − − ||→ × →||2 = ||x||2 ||y||2 − (→•→)2 .71. which a ¬ ¬¬ ¬¬→¬¬ →¬¬ − ¬¬− has area ¬¬→ × b ¬¬. x y x y The following result mixes the dot and the cross product. →. The following corollaries easily follow from Theorem 79. a c c a c a In particular. The signed a c → − − − volume of the parallelepiped spanned by them is (→ × b ) • →. it is possible to “exchange” the cross and dot products: − − → • (→ × → ) = (→ × →) • → − − − − a b c a b c Free to photocopy and distribute 54 . it follows → − → − − → − − − − − − (→ × b ) • → = ( b × →) • → = (→ × → ) • b . b . − − → − 82 Theorem Let →. → − − − − − 80 Corollary Two non-zero vectors →. u 2 2 2 = (x2 + y 2 + z 2 )(y1 + y2 + y3 ) − (x1 y1 + x2 y2 + x3 y3 )2 − − − − = ||→||2||→||2 − (→•→)2 x y x y ¬¬− − ¬¬ Theorem 79 has the following geometric significance: ¬¬→ × →¬¬ is the area x y of the parallelogram formed when the tails of the vectors are joined. − − − − = || x y x y whence the theorem follows. The altitude of the parallelepiped is ¬¬− ¬¬ cos θ a c − → and → × →. a c Proof: See figure 1. u that Since we may have used any of the faces of the parallelepiped. D ′ (0. 1. 2 −→ − 3. 2. and hence the line AC′ has parametric à x y z = 2 0 0 +t ¾ −2 3 1 ¿ =⇒ x = 2 − 2t. 1). 1). 3. B(2. 0). D(0. We have AC′ = equation à −2 3 1 ¿ ← → . −6 −3 ¿ 2. Find the area of △AD ′ M . and M. D ′ . Suppose that a line through M is drawn cutting the line segment [AC ′ ] in ← − → ← → N and the line DD′ in P .Chapter 1 83 Example Consider the rectangular parallelepiped ABCDD ′C ′ B ′ A′ (figure 1. Free to photocopy and distribute 55 . 0). 1). 3. C ′ (0. 1). 4. ← → 3.72) with vertices A(2. Find the Cartesian equation of the plane containing the points A. C(0. B ′ (2. 0. 0. Solution: ◮ 1. −6 −3 ¿ The equation of the plane is thus ¾ x−2 z−0 ¿ ¾ y − 0 • −1 = 0 =⇒ 3(x−2)+1(y)+6z = 0 =⇒ 3x+y+6z = 6. 0. 3. Let M be the midpoint of the line segment joining the vertices B and C. Find the parametric equation of MP. 0). −→ − AM = −1 3 0 ¿ ¾ − → −→ − − =⇒ AD′ × AM = −1 . Form the following vectors and find their cross product: ¾ −→ − AD′ = −2 0 1 ¿ ¾ . 0. 0). 3. z = t. The area of the triangle is ¬¬− → −→¬¬ − ¬¬ ¬¬ − ′ ¬¬AD × AM¬¬ 2 ¾ 1Ô 2 = 3 + 12 + 62 = 2 √ 46 . A′ (2. Find the parametric equation of the line AC′ . y = 3t. 0).4 Redo example 71. a). z = 2s.8.8. Solving the first two equations gives s = 1 3 . find the Cartesian equation of the plane parallel to the vectors ¾ ¿ 1 1 and ¾ ¿ 1 1 and 1 0 passing through the point (0. x y z = 1 3 0 +s −3 z′ −1 =⇒ x = 1−s y = 3−3s. Thus P = desired equation is à à 0 2 and the x y z ◭ = 1 3 0 ¾ +s −3 2 −1 ¿ =⇒ x = 1−s. by finding a normal to the plane. 2). 2a) in R3 .8. Putting this into à 0 the third equation we deduce z ′ = 2. Find a vector of unit length simultaneously perpendicu- Free to photocopy and distribute 56 . y = 3−3s. and (0. t = sz ′ . P = 0 z′ ← → The parametric equation of the line MP is thus à à ¾ ¿ for some real number z ′ > 0. 1. Since P is on the z-axis. ← → ← → Since N is on both MP and AC′ we must have 2 − 2t = 1 − s.t = 2 3 .5 Find the equation of the plane passing through the points (a. a a a → − − − Problem 1. (−a.6 Let a ∈ R. Homework Problem 1. − − i − j and a b j k b c Problem 1. −1.3 If b − → and → − → are a c a − − parallel and it is known that → × → = c a → → − − − − − − → × → = → + →.Cross Product à 0 4. z = sz ′ .2 Prove that → × → = 0 x x follows from the anti-commutativity of the cross product. 3t = 3 − 3s.1 Prove that → − → − → − − − − (→ − b ) × (→ + b ) = 2→ × b .8. find → × → . 1.8. 0.8. Problem 1. → − − − − Problem 1. Problem 1. that is. 15. Find the coordinates of the points P .8. Q.8.8. d be veca c 3 tors in R . Solve the equation − − → × (→ × →) = → × (→ × →). Find the equation of the plane Π. x x x − − → Problem 1. 6.12 The vectors →. c be vectors in R3 .73: Problem 1.14 Let → . 0. ¾ 0 ¿ ¾ ¿ 1 ¬¬ ¬¬ → − ¬¬ → ¬¬− 2. as shewn in figa 0 ure 1. a c − → − → 1. Find the parametric equation of the line LD E joining D and E. Prove the following vector identity. − a c a c a 0. − − − − a b b c c a → − − − Problem 1. ||x|| = 1. − − − − − − − → − → − − → → − → → − c (→ × b )•(→ × d ) = (→ •→)( b • d )−(→ • d )( b •→).8.Chapter 1 Problem 1. Prove that → − − → − xA − ( b × →) • (→ × d ) c a → → − − − − +(→ × → ) • ( b × d ) c a → − → − − − +(→ × b ) • (→ × d ) a c = S Q R B y 0. →. − − − − a x b b x a − → − − → − Problem 1. x y a y x a c − − − → − → Problem 1. 5. → are a c constant vectors.8. − Problem 1. 4. a c a c a z C P − − − → − → Problem 1. Free to photocopy and distribute 57 . R. 8. Solve the system of equations → − − − − − − − − 2→ + → × → = b . with a parameter t ∈ R..11 Assume → • ( b × →) = 0 a c and that − → = α→ + β → + γ →. Find the CP QRS. d . one of whose vertices is at − Problem 1.9 The vectors → . with a parameter s ∈ R. β.7 (Jacobi’s Identity) Let → . − − − x a b c → − − − Find α. 3→ + → × → = →. prove a c that − − → × → = → × → = → × →. 3). The plane Π intersects a 3 × 3 × 3 cube.8 Let → ∈ R3 . Prove that → → → → → → → → − − − − → − → ×( b × c )+ b ×(− ×− )+− ×(− × b ) = intersection forms a pentagon CP QRS. 4.8.13 Let → . 7. b . and S. 0). 3. 9. and γ in terms of → • ( b × →).8. − lar to → = v −a − and → = w a . as in the figure.10 If → + b + → = 0 . 0) and C(0. Figure 1. Find x → 2 − → − → − − − − ||→ × i || + ||→ × j ||2 + ||→ × k ||2 . This b .8. Find the parametric equation of the line LC A joining C and A. area of the pentagon − − → − Problem 1. the origin and that has three of its edges on a → → − − the coordinate axes.8. → . b . b . Find the intersection point between the lines LC A and LD E . 0.8. B(0. Find CA × CB. Find the area of △ABC.15 Consider the plane Π passing through the points A(6.73 below. Find ¬¬CA × CB¬¬. b are cona stant vectors. be veca c 3 tors in R . Prove that L is a linear transformation. Most of the material will flow like that for 2 × 2 matrices. T (j).Matrices in three dimensions 1. 85 Example Consider L : R3 → R3 . 84 Definition A linear transformation T : R3 → R3 is a function such that T (a + b) = T (a) + T (b). Then àà u1 + v1 L(u + v) = L u2 + v2 u3 + v3 = à (u1 + v1 ) − (u2 + v2 ) − (u3 + v3 ) (u1 + v1 ) + (u2 + v2 ) + (u3 + v3 ) 3 à u + v3 + v1 − v2 − v3 v1 + v2 + v3 3 àà à = u1 − u2 − u3 u1 + u2 + u3 u3 u1 àà v v1 +L v2 v3 = = L u2 u3 L(u) + L(v). for all points a. b in R3 and all scalars λ. Find the matrix corresponding to L under the standard basis. and T (k). Such a linear transformation has a 3 × 3 matrix representation whose columns are the vectors T (i). T (λa) = λT (a). Solution: ◮ Let α ∈ R and let u. 58 Free to photocopy and distribute . with àà à x1 L x2 x3 = x1 − x2 − x3 x1 + x2 + x3 x3 .9 Matrices in three dimensions We will briefly introduce 3 × 3 matrices. v be points in R3 . and matrix multiplication are defined for 3 × 3 matrices in a manner analogous to those operations for 2 × 2 matrices. Free to photocopy and distribute 59 . AB = k=1 aik bkj . and L 0 1 = −1 1 1 . à u2 u3 proving that L is a linear transformation. B be 3 × 3 matrices.Chapter 1 and also àà αu1 L(αu) = L αu2 αu3 = à α(u1 ) − α(u2 ) − α(u3 ) α(u1 ) + α(u2 ) + α(u3 ) αu3 à = α u1 − u2 − u3 u1 + u2 + u3 u3 u1 àà = = à αL αL(u). αA = [αaij ]. 86 Definition Let A. whence the desired matrix is ¾ 1 1 0 −1 1 0 −1 1 1 ¿ . à à à 1 We have L à 1 = 1 0 .L 0 1 0 = 0 0 −1 1 0 0 . scalar multiplication. Then we define 3 A + B = [aij + bij ]. ◭ Addition. ¿ a + 4b + 6c 2a + 5b 3a a + 4b a 2a 3a 6a 3b c BA = 2a + 5b 3a . Easy to find counterexamples should convince the reader that the product of two rotations in space does not necessarily commute. and B = a ¿ 0 . Free to photocopy and distribute 60 . 1 ¿ A rotation matrix about the y-axis by an angle θ in the counterclockwise sense is cos θ 0 sin θ 0 − sin θ 0 0 cos θ Ry (θ) = 1 .c = 0 . 6a 6b 6c 88 Definition A scaling matrix is one of the form ¾ a 0 0 0 b 0 0 c ¿ Sa. c > 0. AB = 9a 9b 4c .b. 89 Definition A rotation matrix about the z-axis by an angle θ in the counterclockwise sense is ¾ ¿ cos θ − sin θ 0 Rz (θ) = sin θ 0 ¾ cos θ 0 0 . where a > 0. then ¾ ¿ a 0 1+a 2+b 3+c 0 0 6+a 0 3 18 6 0 9 0 A+B = 4+a 5+b ¾ . b > 0. A rotation matrix about the x-axis by an angle θ in the counterclockwise sense is ¾ 1 0 0 cos θ sin θ 0 cos θ ¿ Rx (θ) = 0 − sin θ .Matrices in three dimensions ¾ 1 2 6 0 ¾ 3 0 ¿ ¾ a b b c 0 ¿ 87 Example If A = 4 5 0 . It is an easy exercise to prove that the product of two scaling matrices commutes. ¿ 3A = 12 ¾ 15 0 . a general formula for An ... Free to photocopy and distribute 61 . 1 . . ..1 Let A = 1 1 0 1 0 ¿ 0 .. 1 1 . m(a)m(−a) = I3 . 0 1 1 .. 2 0 0 1 If a. b are real numbers. Problem 1....2 Let A ∈ M3×3 (R) be given by Describe A2 and A3 in terms of n.9. . that An = 3n−1 A for n ∈ N. ¾ A= 1 1 1 1 1 1 ¿ 1 1 . m(a)m(b) = m(a + b).9..3 Consider the n × n matrix ¾ ¿ 1 1 1 1 . 0 1 1 . using induction. 0 1 0 . 0 . Find 1 1 1 A2 .. 2. 1 1 0 A= 0 . A3 and A4 . the 3 × 3 identity matrix. 0 1 1 . prove that 1. Problem 1.4 Let x be a real number. . Homework ¾ Problem 1.. A reflexion matrix about the y-axis is ¾ 1 0 0 0 0 1 ¿ Ry = 0 −1 A reflexion matrix about the z-axis is ¾ 0 . .Chapter 1 90 Definition A reflexion matrix about the x-axis is ¾ Rx = −1 0 0 0 1 0 0 1 ¿ 0 .. Problem 1..9.. .9. 1 Demonstrate.. . prove by induction. . Conjecture and. n ≥ 1. 1 0 0 1 0 0 0 −1 ¿ Rz = 0 . . and put ¾ ¿ 1 0 x 2 m(x) = −x 1 − x .. b . then D( a b c → → → − − − 3. b . for Free to photocopy and distribute 62 → − − − − (→ + →′ ) • ( b × →) a a c − − → • (→ × →) + →′ • (→ × →) − − − − a b c a b c → → − − → → − − →.15) D(→. the volume of the parallelepiped → − − − spanned by these vectors is → • ( b × →). b . D is linear in each of its arguments. → = c2 . − − − linearly dependent. are a c →. − → − − → − − − − → − − − − → − D(λ→. →) = det a2 b2 c2 = → • ( b × →).Determinants in three dimensions 1. b . b . D( i . →. a c to be ¾ ¿ a1 b1 c1 → − − − − − → − a c (1. c ) + D(→′ . Since thanks to Theorem 82. b . b = b2 . →) = (λ→)•( b ×→) = λ((→)•( b ×→)) = λD(→. a c a c a c a c The linearity on the second and third component can be established by using the distributive law of the cross product. j . det A. If the parallelepiped is flat then the volume is 0. c ). a c a3 b3 c3 We now establish that the properties of the determinant of a 3 × 3 as defined above are analogous to those of the determinant of 2 × 2 matrix defined in the preceding chapter. →). we define the determinant of A. Proof: − → − − → − − − 1. − − D( a a . b . and accords with the right-hand rule. and the 3 × 3 matrix A = a c ¾ a1 a2 a3 b1 b2 b3 c1 c3 ¿ a3 b3 c3 c2 .10 Determinants in three dimensions We now define the notion of determinant of a 3 × 3 matrix. 91 Theorem The determinant of a 3 × 3 matrix A as defined by (1. →) = a a c = = and if λ ∈ R. − − → − 2. b . k ) = 1. →) = 0. If D(→. Consider now the ¾ ¿ ¾ ¿ ¾ ¿ a1 b1 c1 → − − − vectors → = a2 . For example. in R3 . linearity of the first component a c a c follows by the distributive law for the dot product: − − − → − D(→ + →′ . →) = → • ( b × →). →. if →.15) satisfies the following properties: 1. that is. we have det A = 1 det 5 8 6 9 − 4 det 2 3 8 9 + 7 det 2 5 3 6 = 1(45 − 48) − 4(18 − 24) + 7(12 − 15) = 0. → → − − → − →→ − − 3. b . →) + D(→.18) b2 b3 c2 c3 − a2 det b1 b3 c1 c3 + a3 det b1 b2 c1 c2 a1 det . c ).19). u Observe that ¾ ¿ a1 a3 b1 b2 b3 c1 c2 c3 det a2 = − → • (→ × →) − − a b c (1.19) which reduces the computation of 3 × 3 determinants to 2 × 2 determinants. →. k ) = i • ( j × k ) = i • i = 1.Chapter 1 the second component we have. j . − − − − = D( a b c a and if λ ∈ R. →. →) = →•((λ b )×→) = λ(→•( b ×→)) = λD(→. then they lie on the same plane a c − − → − and the parallelepiped spanned by them is flat.16) = = = − − → − → • (b c − b c )→ + (b c − b c )→ + (b c − b c (1. and i • i = 1. b + b ′ . → → → − − − → − → → − − →→ − − D( i . λ b . ◭ Free to photocopy and distribute 63 = −3 + 24 − 21 . are linearly dependent. Since j × k = i . b . →) = → • (( b + b ′ ) × →) a c a c −′ → → → → − →•(b ×− + b ×−) − = a c c −′ − → • (→ × →) + → • (→ × →) − − − − = a b c a b c → → −′ − →. hence. b . (1.17) − a 2 3 3 2 i 3 1 1 3 j 1 2 2 1) k a1 (b2 c3 − b3 c2 ) + a2 (b3 c1 − b1 c3 ) + a3 (b1 c2 − b2 c1 ) (1. − − → → − − − → → − − − D(→. − → − − − − − → − − − → − − → − D(→. →).D(→. →) = a c 0. If →. ¾ 1 7 2 5 8 3 9 ¿ 92 Example Find det 4 6 . b . a c a c a c a c − − → − 2. Solution: ◮ Using (1. 0. or Student[VectorCalculus] to perform many of the vector operations. dotprod(a. An example follows with linalg.0. −6] 2 √ 50 25 Homework Problem 1. 1] b:=vector([-1.2 Prove that ¾ ¿ a b c det b c c a a b = 3abc − a3 − b3 − c3 . > > > > > > with(linalg): a:=vector([-2.b).1 Prove that ¾ ¿ 1 1 1 det a a2 b b2 c c2 = (b − c)(c − a)(a − b). M is the −→ − midpoint of edge [BB′ ] and N is the midpoint of edge [B′ C′ ]. 93 Example Cube ABCDD′ C′ B′ A in figure 1. Problem 1.b). b := [−1. we may use the Maple™ packages linalg.10.1]). a := [−2. 3. 1.b). −1.11 Some Solid Geometry In this section we examine some examples and prove some theorems of threedimensional geometry.74 has side of length a.3. LinearAlgebra.10.0]). arccos [−3.75: Example 93. Prove that AD′ −→ − MN and find the area of the quadrilateral MND′ A.Some Solid Geometry Again. C′ N N M C D′ Q P A Figure 1. Free to photocopy and distribute 64 . 0] crossprod(a. angle(a.74: Example 93. D′ A′ B′ D M A B Figure 1. 2 2 The area of the trapezoid is finally. ← → Draw line AD cutting plane P 2 in G.76. 94 Theorem (Thales’ Theorem) Of two lines are cut by three parallel planes. C. Given the lines AB and CD. M. and D intersects plane P 2 in the line GF. and hence MN BC′ by example 14. Similarly the plane containing ← → points A. From the figure Solution: √ ¬¬−→¬¬ ¬¬ 1 ¬¬− →¬¬ ¬¬−→¬¬ a 2 →¬¬ ¬¬ − ¬¬ ¬¬− ¬¬ ¬¬ − ′ ¬¬ ¬¬ − ¬¬ ′ . The aforementioned √ ¬¬ −→ − 1 ¬¬−→¬¬ a 2 − ¬¬ ¬¬−→¬¬ ¬¬ − ′ ¬¬ example also gives ¬¬MN¬¬ = ¬¬AD ¬¬ = . 2 8 3a √ · 2 2 ◭ Let us prove a three-dimensional version of Thales’ Theorem. ¬¬− →¬¬ ¬¬ − ¬¬ ¬¬D′ N¬¬ = Ö ¬¬− →¬¬2 ¬¬−→¬¬2 ¬¬ − ′ ¬¬ ¬¬ − ¬¬ ¬¬D′ C ¬¬ + ¬¬C′ N¬¬ = a2 + a2 4 = √ a 5 2 . Because − → −→ − − they are diagonals that belong to parallel faces of the cube. The plane containing points A. In consequence. The height of this trapezoid is thus ¬¬− ¬¬ ¬¬ →¬¬ ¬¬NQ¬¬ = 5a2 4 − a2 8 = 3a √ . by the Pythagorean Theorem.Chapter 1 ¬¬− →¬¬ √ ¬¬ − ¬¬ ◮ By the Pythagorean Theorem. M and N are the midpoints of the sides [B ′ B] and [B ′ C ′ ] of −→ − −→ − △B ′ C ′ B. ¬¬D Q¬¬ = ¬¬AP¬¬ = ¬¬AD ¬¬ − ¬¬MN¬¬ = 2 4 Also. we must prove AE EB = CF FD . ← → and D intersects plane P 2 in the line EG. This means that the four points A. D′ . AD′ 2 2 −→ − MN. ¬¬AD′ ¬¬ = a 2. Proof: that ← → ← → See figure 1. Since P 2 and P 3 Free to photocopy and distribute 65 .75). Now. their corresponding segments are proportional. N are all on the same √ √ a 2 ′ plane. Hence MND A is a trapezoid with bases of length a 2 and 2 (figure 1. B. √ à √ a 2 a 2+ 9a2 2 = . AD′ BC′ . 78: Example 95. EG (theorem 30). and so by Thales’ Theorem on the plane AE = AG . Figure 1.Some Solid Geometry ← → are parallel planes. = AG GD . u = CF FD . since P 1 and P 2 are parallel.77: Example 95. ′′ C ′′ A D N M B ′′ A C B A′ Figure 1. Free to photocopy and distribute 66 . ← → GF and P1 A C A′ D′ C′ B′ D M A N C B P2E G F P 3B D D′ C′ D ′′ B′ Figure 1. AC EB CF FD It follows that AE EB as needed to be shewn. ← → BD.76: Thales’ Theorem in 3D. GD ← → Similarly. ¬¬− →¬¬ ¬¬ − ¬¬ ′′ ¬¬B N¬¬ = xa. ¬¬−→¬¬ ¬¬ − ¬¬ First notice that ¬¬AB′ ¬¬ = ¬¬−→¬¬ √ ¬¬ − ′ ¬¬ ¬¬BC ¬¬ = a 2. by the Pythagorean Theorem. If ¬¬MN¬¬ = ¬¬AB¬¬.78. Since MN is parallel to face ABCD . The intersection of P with the cube produces a lamina A′′ B′′ C′′ D′′ . Put ¬¬−→¬¬ ¬¬− →¬¬ ¬¬−→¬¬ ¬¬−→¬¬ ¬¬ − ¬¬ ¬¬ − ′ ¬¬ ¬¬ − ′ ¬¬ ¬¬ − ¬¬ ¬¬AM¬¬ ¬¬MB ¬¬ ¬¬AB ¬¬ − ¬¬AM¬¬ ¬¬−→¬¬ = x =⇒ ¬¬−→¬¬ = ¬¬−→¬¬ = 1 − x.Chapter 1 95 Example In cube ABCDD′ C′ B′ A′ of edge of length a. There are two possible positions for the segment. ¬¬− → ¬¬ ¬¬− − ¬¬ ¬¬ − ′ ¬¬ ¬¬ − →¬¬ ¬¬MB ¬¬ ¬¬B′′ B′ ¬¬ ¬¬−→¬¬ = ¬¬−→¬¬ . ¬¬ − ′ ¬¬ ¬¬ − ′ ¬¬ 3 ¬¬AB ¬¬ ¬¬BC ¬¬ ¬¬ ¬¬ − ¬¬ →¬¬ ¬¬−→¬¬ ¬¬− ¬¬ ¬¬AM¬¬ ¬¬BN¬¬ 2 ¬¬−→¬¬ = ¬¬−→¬¬ = . ¬¬− →¬¬ ¬¬−→¬¬ ¬¬ − ′′ ¬¬ ¬¬ − ¬¬ ¬¬BB ¬¬ ¬¬AM¬¬ ¬¬−→¬¬ = ¬¬−→¬¬ . 3 3 . giving the solutions ¬¬ ¬¬ − ¬¬ →¬¬ ¬¬−→¬¬ ¬¬− ¬¬ ¬¬AM¬¬ ¬¬BN¬¬ 1 ¬¬−→¬¬ = ¬¬−→¬¬ = . − ′ ¬¬ ¬¬ ¬¬ − ′ ¬¬ 3 ¬¬AB ¬¬ ¬¬BC ¬¬ Solution: ◮ There is a unique plane parallel P to face ABCD and −→ − containing M. ¬¬ − ′ ¬¬ ¬¬ − ′ ¬¬ ¬¬AB ¬¬ ¬¬BB ¬¬ ¬¬− →¬¬ ¬¬− − ¬¬ ¬¬ − ′ ¬¬ ¬¬ − →¬¬ ¬¬MB ¬¬ ¬¬MB′′ ¬¬ ¬¬−→¬¬ = ¬¬− ¬¬ ¬¬ − ′ ¬¬ ¬¬ →¬¬ ¬¬AB ¬¬ ¬¬AB¬¬ =⇒ ¬¬− − ¬¬ ¬¬ − →¬¬ ¬¬MB′′ ¬¬ = (1 − x)a. ¬¬ − ′ ¬¬ ¬¬ − ′ ¬¬ ¬¬ − ′ ¬¬ ¬¬AB ¬¬ ¬¬AB ¬¬ ¬¬AB ¬¬ Now. ¬¬ − ′ ¬¬ ¬¬ − ′ ¬¬ ¬¬BB ¬¬ ¬¬AB ¬¬ ¬¬− →¬¬ ¬¬− →¬¬ ¬¬ − ¬¬ ¬¬ − ′′ ¬¬ ¬¬B′′ N¬¬ ¬¬BB ¬¬ ¬¬− →¬¬ = ¬¬−→¬¬ ¬¬ − ′ ¬¬ ¬¬ − ′ ¬¬ ¬¬B′ C ¬¬ ¬¬BB ¬¬ =⇒ √ ¬¬ 5 − ¬¬ ¬¬−→¬¬ Since ¬¬MN¬¬ = a. as in figure 1. find the ratios ¬¬−→¬¬ and ¬¬−→¬¬ . the points M and N are −→ − located on diagonals [AB′ ] and [BC′ ] such that MN¬ is parallel to¬ the¬face ABCD ¬ −→¬¬ ¬− ¬ ¬¬ − ¬¬ ¬¬ →¬¬ √ ¬¬ ¬¬−→¬¬ ¬¬ ¬¬AM¬¬ ¬¬BN¬¬ 5 ¬¬− ¬¬ → ¬¬ − ¬¬ of the cube. P also contains N. ¬¬ − ′ ¬¬ ¬¬ − ′ ¬¬ 3 ¬¬AB ¬¬ ¬¬BC ¬¬ ◭ Homework Free to photocopy and distribute 67 . 3 ¬¬ ¬¬− − ¬¬2 ¬¬− →¬¬2 5 2 − ¬¬ ¬¬−→¬¬2 ¬¬ − →¬¬ ¬¬ − ¬¬ ′′ ′′ a = (1−x)2 a2 +x2 a2 =⇒ x ∈ ¬¬MN¬¬ = ¬¬MB ¬¬ +¬¬B N¬¬ =⇒ 9 1 2 . as △B ′ AB ∼ △B ′ M B ′′ and △BC ′ B ′ ∼ △BN B ′′ . The corresponding ordinate for the ellipse and the circle are proportional. as the proof for the first is similar. Find the angle between the lines [M N ] and [BC]. 68 Free to photocopy and distribute .11.3 In cube ABCDD′ C′ B′ A′ of edge of length a. a b Solution: ◮ Consider the circle with equation x2 + y 2 = a2 . Prove − → − → that AC ⊥ BD. points M and N are the midpoints of the edges [AB] and [CD]. Problem 1. a > 0. −→ − − → −→ − 3. As x1 A(x)dx = c x1 A(x)dx the corresponding volumes must also be proportional.79. respectively. and hence. and the Pappus-Guldin Rules 96 Theorem (Cavalieri’s Principle) All planar regions with cross sections of proportional length at the same height have area in the same proportion.1 In a regular tetrahedron ¬¬ ¬¬ → ¬¬− ¬¬ with vertices A. x2 ]. Proof: We only provide the proof for the second statement. u 97 Example Use Cavalieri’s Principle in order to deduce that the area enclosed x2 y2 by the ellipse with equation 2 + = 1. Prove that MN ⊥ AB and MN ⊥ − → CD. 1. Cut any two such solids by horizontal planes that produce cross sections of area A(x) and cA(x). Problem 1. find the distance between the lines that contain the diagonals [A′ B] and [AC]. we know that x1 A(x)dx and x1 x2 cA(x)dx x2 give the volume of the portion of each solid cut by all horizontal planes as x runs over some interval [x1 . b > 0. ¬¬ ¬¬ ¬¬ ¬ ¬ ¬¬ ¬¬ → → → → ¬¬− ¬¬ ¬¬− ¬¬ ¬¬− ¬¬ ¬¬− ¬¬ ¬¬AB¬¬ = ¬¬BC¬¬. Area of the ellipse = b Area of the circle a b = πa2 a = πab. as in figure 1. B. 1. y= bÔ a a 2 − x2 . for y > 0. C.12 Cavalieri. Find the length of the segment [M N ]. By Cavalieri’s first principle. ¬¬AD¬¬ = ¬¬DC¬¬.11. at an arbitrary height x above a fixed base. Then.11. x2 x2 From elementary calculus. D and with ¬¬AB¬¬ = a. 2. where c > 0 is the constant of proportionality. is πab.2 ¬ In¬ a tetrahedron ¬¬ ABCD. y= Ô a 2 − x2 . the corresponding chords for the ellipse and the circle will be proportional. and the Pappus-Guldin Rules Problem 1.Cavalieri. All solids with cross sections of proportional areas at the same height have their volume in the same proportion. By the Pythagorean Theorem.81. who was so proud of it that he wanted a sphere inscribed in a cylinder on his tombstone. consider a punctured cylinder of base radius a and height a. We need to recall that the volume of a right circular cone πa2 h with base radius a and height h is . 3 Free to photocopy and distribute 69 . Volume of the hemisphere = Volume of the punctured cylinder πa3 = πa3 − 3 2πa3 = . By Cavalieri’s Principle. a Figure 1.81: Hemisphere. Figure 1. as in figure 1. as in figure 1. producing a circle of radius r. and so this circular slab has area πr 2 = π(a2 − x2 ). Now. 98 Example Use Cavalieri’s Principle in order to deduce that the volume of a 4 sphere with radius a is πa3 .Chapter 1 ◭ x a x a r a x Figure 1.79: Ellipse and circle. x2 + r 2 = a2 . with a cone of height a and base radius a cut from it.80. Cut a horizontal slice at height x. which agrees with a horizontal slab for the sphere at the same height. 3 Consider a hemisphere of radius a. 3 Solution: ◮ The following method is due to Archimedes.80: Punctured cylinder. A horizontal slab at height x is an annular region of area πa2 − πx2 . 12. Free to photocopy and distribute 70 . Also. 100 Example Since the centre of gravity of a circle is at its centre. and radius of gyration R (as in figure 1.82: A torus.13 Dihedral Angles and Platonic Solids 101 Definition When two half planes intersect in space they intersect on a line.Dihedral Angles and Platonic Solids 2πa3 3 4πa3 3 It follows that the volume of the sphere is 2 = . the surface area of the torus with the generating circle having radius r. 99 Theorem (Pappus-Guldin Rule) The area of the lateral surface of a solid of revolution is equal to the product of the length of the generating curve on the side of the axis of revolution and the length of the path described by the centre of gravity of the generating curve under a full revolution. The volume of a solid of revolution is equal to the product of the area of the generating plane on one side of the revolution axis and the length of the path described by the centre of gravity of the area under a full revolution about the axis. See figure 1.83. 1. The portion of space bounded by the half planes and the line is called the dihedral angle. The intersecting line is called the edge of the dihedral angle and each of the two half planes of the dihedral angle is called a face. r R Figure 1.1 Use the Pappus-Guldin Rule to find the lateral area and the volume of a right circular cone with base radius r and height h.82) is (2πr)(2πR) = 4π 2 rR. the volume of the solid torus is (πr 2 )(2πR) = 2π 2 r 2 R.◭ Essentially the same method of proof as Cavalieri’s Principle gives the next result. by the PappusGuldin Rule. Homework Problem 1. we use the term trihedral angle.84. In the trihedral angle of figure 1. Free to photocopy and distribute 71 . Hence the measure of a dihedral angle is the measure of any one of its rectilinear angles. The angles formed by adjacent edges are called face angles. In analogy to dihedral angles we now define polyhedral angles. The portion of the planes lying between consecutive edges are called the faces of the polyhedral angle. △V AB. notice that in any polyhedral angle. 102 Definition The rectilinear angle of a dihedral angle is the angle whose sides are perpendicular to the edge of the dihedral angle at the same point. 104 Theorem The sum of any two face angles of a trihedral angle is greater than the third face angle.83: Dihedral Angles. each on each of the faces. Figure 1.Chapter 1 Figure 1. All the rectilinear angles of a dihedral angle measure the same.84: Rectilinear of a Dihedral Angle. A polyhedral angle is said to be convex if the section made by a plane cutting all its edges forms a convex polygon. △V BC. any two adjacent faces form a dihedral angle.85. △V CA are faces. V is the vertex. The common point is called the vertex of the polyhedral angle. In the particular case of three planes. 103 Definition The opening of three or more planes that meet at a common point is called a polyhedral angle or solid angle. See figure 1. Also. Each of the intersecting lines of two consecutive planes is called an edge of the polyhedral angle. as wanted. Through any point D of the segment [V W ]. Let the faces be cut by a plane. An illustration can be seen in figure 1. Observe that △AV D ∼ AV B. . Ak Ak+1 Ak−1 Figure 1. then we are done. Hence ∠AV B + ∠BV C = > = ∠AV D + ∠BV C ∠AV D + ∠DV C ∠AV C.85: Trihedral Angle. by the triangle inequality in △ABC. A2 . Since we are assuming that ∠ZV X > XV Y . . Ak+1 are three consecutive vertices.86. Proof: Let the polyhedral angle have n faces and vertex V . X. Consider now the plane containing the line segment [AC] and the point B. u 105 Theorem The sum of the face angles of any convex polyhedral angle is less than 2π radians. where for convenience.V V A4 A X B Y Figure 1. Observe that the polygon A1 A2 · · · An is convex and that the sum of its interior angles is π(n − 2). This implies that ∠BV C > ∠DV C. D W V C Z Dihedral Angles and Platonic Solids A3 P A5 A2 A1 Figure 1. in ∠XV Y the line segment [V W ] such that ∠XV W = ∠XV Y . Hence AD = AB. Free to photocopy and distribute 72 . Take the point B ∈ [V Y ] so that V D = V B. Z. say. intersecting the edges at the points A1 .87: A. which proves that ∠XV Y + ∠Y V Z > ∠ZV X. We must demonstrate that ∠XV Y + ∠Y V Z > ∠ZV X. = Now. Ak . ∠ZV X > XV Y . AB + BC > CA.86: Polyhedral Angle. say. we may draw. An . .85. we have depicted only five edges. Proof: Consider figure 1. draw △ADC on the plane P containing the points V . so assume that. If ∠ZV X is smaller or equal to in size than either ∠XV Y or Y V Z. as in figure 1. let Ak−1 . V Ak Ak+1 + ∠V Ak+1 Ak + ∠Ak V Ak+1 = πn. Observe that as k ranges from 1 through n. This notation means that Ak−1 Ak Ak+1 represents any of the n triplets A1 A2 A3 . 1≤k≤n since this is summing the sum of the angles of the n triangles of the faces. 1≤k≤n being the sum of the interior angles of the polygon A1 A2 · · · An .Chapter 1 We would like to prove that ∠A1 V A2 + ∠A2 V A3 + ∠A3 V A4 + · · · + ∠An−1 V An + ∠An V A1 < 2π. ∠V Ak Ak−1 . An−2 An−1 An . Also. we let A0 = An . the sum ∠Ak−1 Ak Ak+1 = π(n − 2). that is. An A1 A2 . since one sum adds the angles in one direction and the other in the opposite direction. . An+1 = A1 . A2 A3 A4 . u Free to photocopy and distribute 73 . and ∠V Ak Ak+1 . . (V Ak Ak+1 + ∠V Ak+1 Ak ) 1≤k≤n (V Ak Ak+1 + ∠V Ak Ak−1 ) 1≤k≤n < πn − π(n − 2) as we needed to shew.87. For the same reason. A3 A4 A5 . . V Ak Ak−1 = 1≤k≤n 1≤k≤n ∠V Ak Ak+1 . Now. . By Theorem 104. Ak . But clearly V Ak Ak+1 = 1≤k≤n 1≤k≤n ∠V Ak+1 Ak . etc. Ak+1 be three consecutive vertices of of the polygon A1 A2 · · · An . Consider the trihedral angle with vertex Ak and whose face angles at Ak are ∠Ak−1 Ak Ak+1 . An−1 An A1 . Thus V Ak Ak−1 + ∠V Ak Ak+1 > 1≤k≤n 1≤k≤n ∠Ak−1 Ak Ak+1 = π(n − 2). An+2 = A2 . ∠V Ak Ak−1 + ∠V Ak Ak+1 > ∠Ak−1 Ak Ak+1 . Hence ∠Ak V Ak+1 1≤k≤n = πn − = πn − = 2π. 0. where V is the number of vertices. m 3 4 3 5 3 n 3 3 4 3 5 S 4 8 6 20 12 E 6 12 12 30 30 F 4 6 8 12 20 Name of regular Polyhedron. Tetrahedron or regular Pyramid. Since n ≥ 3 and m ≥ 3. That there are exactly five can be seen by explicit construction. Dodecahedron. Hexahedron or Cube. π] radians. Thus there are at most five Platonic solids. 1. Octahedron. φ ∈ [0. Icosahedron. y. m). Since OP = ρ sin φ we find x = ρ cos θ sin φ. Free to photocopy and distribute 74 .14 Spherical Trigonometry Consider a point B(x. Appealing to Euler’s Formula for polyhedrons. z) in Cartesian coordinates. and let its distance be ρ. z). let us say it is an angle of φ.93. Let θ be angle that this projection makes with the positive x-axis. and E is the number of edges of a polyhedron. Since each interior angle of this polygon π(n − 2) measures . We measure its inclination from the positive z-axis. as in figure 1. n m π(n − 2) n < 2π =⇒ m(n − 2) < 2n =⇒ (m − 2)(n − 2) < 4. We now project the line segment [OB] onto the xy-plane in order to find the polar coordinates of x and y. Observe that z = ρ cos φ. we must have in view of Theorem 105. Figures 1.88 through 1. 0) we draw a straight line to B(x. Suppose a regular polygon with n ≥ 3 sides is a face of a platonic solid with m ≥ 3 faces meeting at a corner. y. From O(0. y = ρ sin θ sin φ. we obtain the values in the following table.92 depict the Platonic solids. F is the number of faces. which states that V + F = E + 2.Spherical Trigonometry 106 Definition A Platonic solid is a polyhedron having congruent regular polygon as faces and having the same number of edges meeting at each corner. the above inequality only holds for five pairs (n. By convention. Figure 1. In the case where the two points are not diametrically opposite. y. We call the smaller arc the geodesic joining the two points. Figure 1. Here φ is the polar angle. z) in Cartesian coordinates. The endpoints of such a diameter are called the poles of the circle. The axis of any circle on a sphere is the diameter of the sphere which is normal to the plane containing the circle. measured from the positive z-axis.92: Icosahedron. A pole of a circle is equidistant from every point of the circumference of the circle. but this is not the case in a small circle. By the pole of a small circle. the intersection will be a circle. 108 Definition If a plane intersects with a sphere. a plane can be drawn. 109 Definition Given the centre of the sphere.88: Tetrahedron. we mean the closest pole to the plane containing the circle. we call it a great circle. The poles of a great circle are equally distant from the plane of the circle.91: Dodecahedron Figure 1.90: Octahedron. Spherical coordinates are extremely useful when considering regions which are symmetric about a point. If the two points are diametrically opposite then every plane containing the line Free to photocopy and distribute 75 . 107 Definition Given a point (x. measured from the positive x-axis. Otherwise we talk of a small circle. If this circle contains the centre of the sphere. and θ is the azimuthal angle. and any two points of the surface of the sphere. 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. Figure 1.Chapter 1 Figure 1. the great circle formed is split into a larger and a smaller arc by the two points. z = ρ cos φ. The radius of a great circle is the radius of the sphere. y = ρ sin θ sin φ. its spherical coordinates are given by x = ρ cos θ sin φ. This plane will be unique if and only if the points are not diametrically opposite.89: Cube or hexahedron. and the arcs formed are then of equal length. OB × OA . − → − − → → − → ∠B = ∠ OB × OC. ρ cos φ Free to photocopy and distribute ρ si n φ P − → − − → → − → ∠C = ∠ OC × OA. OA × OC . If A. In this case we take any such arc as a geodesic. 76 . B. − − → → ∠b = ∠ OC. but in lowercase. 110 Definition A spherical triangle is a triangle on the surface of a sphere all whose vertices are connected by geodesics. The three arcs of great circles which form a spherical triangle are called the sides of the spherical triangle. the angles formed by the arcs at the points where they meet are called the angles of the spherical triangle. OB . − − → → ∠c = ∠ OA. OC . B φ ρ θ Figure 1. ∠a. and three arc angles. ∠B.93: Spherical Coordinates. and A spherical triangle has then six angles: three vertex angles ∠A. ∠C. C are the vertices of a spherical triangle.Spherical Trigonometry forms with the sphere a great circle. OC × OB . OA . ∠c. it is customary to label the opposite arcs with the same letter name. − → − − → → − → ∠A = ∠ OA × OB. Observe that if O is the centre of the sphere then − − → → ∠a = ∠ OB. ∠b. x2 + y2 + z2 = ρ2 . so that Free to photocopy and distribute 77 . 112 Theorem Let I be the dihedral angle of two adjacent faces of a regular polyhedron. and let O be the centre and ρ be the radius of the sphere. OC. CE and DE will be perpendicular to AB. c. y2 . 2 2 2 2 Since x2 + y1 + z1 = ρ2 . In the plane containing CE and DE draw CO and DO at right angles to CE and DE respectively. z2 ). y2 = ρ sin θ2 sin φ2 . Then cos π I n sin = π . 2 sin m Proof: See figure 1. Then cos a cos b + sin a sin b cos C = cos c. ρ2 + ρ2 − 2ρ2 cos ∠ (AOB) . z1 = ρ cos θ1 . Let AB be the edge common to the two adjacent faces. u Figure 1.94: Theorem 112. e respectively. y1 . Proof: Consider a spherical triangle ABC with A(x1 . that is. cos a cos b + sin a sin b cos C = cos c. x2 = ρ sin θ2 cos φ2 . Therefore we obtain cos θ2 cos θ1 + sin θ2 sin θ1 cos(φ1 − φ2 ) = cos ∠ (AOB) .Chapter 1 111 Theorem Let △ABC be a spherical triangle. OE at a. and join CE and DE. we gather that 1 2 x1 x2 + y1 y2 + z1 z2 = ρ2 cos ∠ (AOB) . Then the following quantities give the square of the distance of the line segment [AB]: (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 . bisect AB at E. and meeting at O. By a rotation we may assume that the z-axis passes through C. we shall denote it by I. z2 = ρ cos θ2 . about O as centre describe a sphere meeting OA.94. say. In spherical coordinates this is. and the angle CED is the angle of inclination of the two adjacent faces. B(x2 . z1 ). C and D the centres of the faces. y1 = ρ sin θ1 sin φ1 . x1 = ρ sin θ1 cos φ1 . Then the angle ace = ACE = = . cae = = . the radii of the inscribed and circumscribed spheres of a regular polyhedron. Proof: Let the edge AB = a.Spherical Trigonometry cae forms a spherical triangle. it is perpendicular to the plane CED. that is cos π n = cos π − I 2 sin π m . Let m be the number of sides in each face of the polyhedron. Since AB is perpendicular to CE and DE. I 2 = a 2 cot π m tan π m I 2 . so that r is the radius of the inscribed sphere. From the right-angled triangle cae 2n n cos cae = cos cOe sin ace. and R is the radius of the circumscribed sphere. cot From the above formulæ we now easily find that the volume of the pyramid which r ma2 π has one face of the polyhedron for base and O for vertex is cot . respectively. R= a 2 tan I 2 tan π n . π n . and 3 4 m Free to photocopy and distribute 78 . hence the angle cea of the spherical triangle is a right angle. 2 sin m 113 Theorem Let r and R be. therefore u 2 π cos I n sin = π . that is. 2m m and the angle cae is half one of the n equal angles formed on the sphere 2π π round a. let OC = r and OA = R. Then CE = AE cot ACE = a 2 cot π m . and I is the dihedral angle of any two faces. therefore the plane AOB which contains AB is perpendicular to the plane CED. Here a is the length of any edge of the polyhedron. r = CE tan CEO = CE tan also therefore u r = R cos aOc = R cot eca cot eac = R cot R = r tan π m tan π n = a 2 tan I 2 tan π n . n the number of the plane angles 2π π which form each solid angle. Then r= a 2 cot π m tan I 2 . a parabola.1 The four vertices of a regular tetrahedron are Ǒ Ǒ 1 −1/2 √ V1 = 0 . Shew that its volume is a3 . 0 −1/2 √ − 3/2 0 0 Problem 1. and that the raa dius of the inscribed sphere is . 6 Problem 1.4 Consider an octahedron whose edge measures a. Shew that its vol√ a3 ume is 15 + 7 5 . its surface area is 12 √ 5a2 3. its surface area is a2 3. Shew that its vol√ √ a3 2 . its surface area is 4 √ 3a2 25 + 10 5. 12 1. and that the radius of the inscribed √ √ a sphere is 5 3 + 15 .3 Consider a cube whose edge measures a.6 Consider an icosahedron whose edge measures a.15 Canonical Surfaces In this section we consider various surfaces that we shall periodically encounter in subsequent sections.14.14.5 Consider a dodecahedron whose edge measures a. a circle.. Shew that its vol√ √ a3 2 . V2 = 3/2 . 4 m therefore the volume of the polyhedron is cot Homework Problem 1.Chapter 1 mF ra2 12 π . Just like in one-variable Calculus it is important to identify the equation and the shape of a line. m ma2 π cot . We remark that in order to parametrise curves (“one-dimensional entities”) we needed one parameter.14. Shew that its vol√ 5a3 ume is 3 + 5 . 12 Problem 1. and therefore Furthermore. and that the radius of the inscribed sphere is a 4 10 + 22 1 . V3 = 0 √ 2 What is the cosine of the dihedral angle between any pair of faces of the tetrahedron? Problem 1.14. ume is 3 and √ that the radius of the inscribed sphere a 6 is . and that in order to parametrise surfaces we shall need to parameters.14. etc. and ume is 12 that the radius of the inscribed sphere is √ a 6 . its surface area is 2a2 3.2 Consider a tetrahedron whose edge measures a. 2 Problem 1. the area of one face of the polyhedron is 4 m mF a2 π the surface area of the polyhedron is cot . 5 Ǒ . it will become important for us to be able to identify certain families of often-occurring surfaces. We shall explore both their Cartesian and their parametric form. its surface area is 6a2 .14. V4 = 0 Ǒ . Free to photocopy and distribute 79 . qthe equation of the plane is x − x0 = pu1 + qv1 . not all ¾ ¿ a zero. b. The line ∆ is called the directrix of the cylinder. 114 Definition A surface S consisting of all lines parallel to a given line ∆ and passing through a given curve Γ is called a cylinder. Under these conditions. To recognise whether a given surface is a cylinder we look at its Cartesian equation. B) = 0. z − z0 = pu3 + qv3 . y = sin v. B = 0. 115 Example Figure 1. u ∈ R.Canonical Surfaces Let us start with the plane. then the curve is a cylinder. Figure 1. A parametrisation for this cylinder is the following: x = cos v. If it is of the form f (A. or z is missing.95 shews the cylinder with Cartesian equation x2 +y 2 = 1. One starts with the circle x2 + y 2 = 1 on the xy-plane and moves it up and down the z-axis. where A. whose directrix will be the axis of the missing coordinate. 2π]. c are real numbers. 80 Free to photocopy and distribute .95: Circular cylinder x2 + y2 = 1. z0 ) is a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0. y. In practice. B are secant planes.96: The parabolic cylinder z = y2 . z = u. − − If we know that the vectors → and → are on the plane (parallel to the plane) then u v with the parameters p. v ∈ [0. y − y0 = pu2 + qv2 . then the surface is a cylinder. Recall that if a. z z x y x y Figure 1. if one of the variables x. y0 . then the Cartesian equation of a plane with normal vector b and passing c through the point (x0 . the lines generating S will be parallel to the line of equation A = 0. .s=-10. r θ x y Figure 1.10.x=-1.sˆ2].98 shews the hyperbolic cylinder with Cartesian equation x2 − y 2 = 1.. The Maple™ commands to graph this surface are: > with(plots): > implicitplot3d(z=yˆ2. One starts with the parabola z = y 2 on the yz-plane and moves it up and down the x-axis. v ∈ R.10). The Maple™ commands to graph this surface are: Free to photocopy and distribute 81 . In general.x=-10..z=-10..numpoints=5001).10. z (r cos θ.10.s. u ∈ R.10.s=-10.. y = sinh v. z): x = r cos θ. z = v2 .t=-10.97.96 shews the parabolic cylinder with Cartesian equation z = y 2 . z) y = r sin θ.10. We refer to it as the method of cylindrical coordinates. The method of parametrisation utilised above for the cylinder is quite useful when doing parametrisations in space.t]..t=-10. 0) parallel to the z-axis to (x..97: Cylindrical Coordinates. y in the xy-plane. z = u.1.axes=boxed). r sin θ. z = z. y. y = v.z=-10.numpoints=5001).sin(s).10. u ∈ R. 116 Example Figure 1. v ∈ R.10.1. > plot3d([t. See figure 1. 117 Example Figure 1. and then lift (x.numpoints=5001.. We need a choice of sign for each of the portions.y=-1.. > plot3d([cos(s).Chapter 1 The Maple™ commands to graph this surface are: > with(plots): > implicitplot3d(xˆ2+yˆ2=1.. y.y=-10. One starts with the hyperbola x2 − y 2 on the xy-plane and moves it up and down the z-axis. We have used the fact that cosh2 v − sinh2 v = 1. we first find the polar coordinates of x. A parametrisation for this parabolic cylinder is the following: x = u. A parametrisation for this parabolic cylinder is the following: x = ± cosh v. ) = 0. as its equation can be put in the form −1 = 0.numpoints=5001. the surface S obtained by drawing rays from Ω and passing through Γ is called a cone. 0).100.10. B = 0.. z = 0. z) to the axis of rotation is the same as the distance of (x.x=-10. The axis of S is the line passing through the centre of Σ and perpendicular to the plane A. x 2 y 2 If the Cartesian equation of S can be put in the form f (A. where A is a plane and Σ is a sphere.Canonical Surfaces > > > > with(plots): implicitplot3d(xˆ2-yˆ2=1. Considering z z the planes x = 0. y = 0. C = 0. 118 Definition Given a point Ω ∈ R3 (called the apex) and a curve Γ (called the generating curve). We then say that ∆ is the axis of revolution. y. and its distance to the axis Ô of rotation would be |x| = 1 + 4z 2 . the distance of (x.sinh(s). The graph is shewn in figure 1.10.10. then the surface is of revolution.. s=-2.[cosh(s).2. 121 Example Find the equation of the surface of revolution generated by revolving the hyperbola x2 − 4z 2 = 1 about the z-axis. z) to Free to photocopy and distribute 82 . if the Cartesian equation of a surface can be put into the form A B .10. Solution: ◮ Let (x. + 120 Definition A surface S obtained by making a curve Γ turn around a line ∆ is called a surface of revolution. plot3d({[-cosh(s).y=-10. and its apex is given by A = 0.t]. C. f( In practice. then the C C surface is a cone.. Σ) = 0. it would be on the hyperbola.z=-10. y.t]}.numpoints=5001). B. If this point were on the xz plane. Anywhere else. are planes secant at exactly one point. z) be a point on S.sinh(s). The intersection of S with a half-plane bounded by ∆ is called a meridian.. 0.t=-10. where A.. the apex is located at (0. 119 Example The surface in R3 implicitly given by z 2 = x2 + y 2 is a cone.axes=boxed). y. If this point were on the yz plane. which is to say x2 + y 2 − 4z 2 = 1. forming a torus T . See figure 1. Find the equation of this torus. z) to the z-axis is the distance of this point to the point (x. Anywhere else. We must have Ô x2 +y 2 = (a+sgn(y−a) r 2 − z 2 )2 = a2 +2asgn(y−a) r 2 − z 2 +r 2 −z 2 . v ∈ [0. z). y. We must have x2 + y 2 = Ô 1 + 4z 2 . y. When x = 0. on the yz plane (a. a2 b c (0. This surface is called a hyperboloid of one sheet. z = u. Figure 1. z) be a point on T . 0.98: The hyperbolic cylinder x2 − y2 = 1.104. y 2 − 4z 2 = 1 is a hyperbola on the yz plane. sgn(t) = 1 if t > 0. x2 + y 2 . the distance from (x.100: Cone x2 y2 z2 + 2 = 2.Chapter 1 Figure 1. and the of the distance to the axis Ô of rotation would be y = a + sgn(y − a) r 2 − z 2 .99: The torus. z) : Ô Ô Ô 1 + 4u2 cos v. Figure 1. 2π]. A parametrisation for this hyperboloid is x= ◭ 122 Example The circle (y − a)2 + z 2 = r 2 . where sgn(t) (with sgn(t) = −1 if t < 0. y. and sgn(0) = 0) is the sign of t. x2 + y 2 = 1 is a circle on the xy plane. When y = 0. that is x2 + y 2 . Observe that when z = 0. r are positive real numbers) is revolved around the z-axis. u ∈ R. Solution: ◮ Let (x. Free to photocopy and distribute 83 . y= 1 + 4u2 sin v. it would be on the circle. x2 − 4z 2 = 1 is a hyperbola on the xz plane. c a b 123 Example The surface z = x2 +y 2 is called an elliptic paraboloid. (it could not be 0.103: Two-sheet hyx2 y2 z2 perboloid 2 = 2 + 2 +1. why?). or (x2 + y 2 + z 2 − (a2 + r 2 ))2 = 4a2 r 2 − 4a2 z 2 since (sgn(y − a))2 = 1. c > 0. z = y 2 is a parabola on the yz plane.Canonical Surfaces Rearranging x2 + y 2 + z 2 − a2 − r 2 = 2asgn(y − a) r 2 − z 2 .99. When y = 0. y = u sin v. Free to photocopy and distribute 84 . +∞[. z = r sin α.101: Paraboloid z = x2 y2 + 2. The following is a parametrisation of this paraboloid: √ √ x = u cos v. is of fourth degree.102: Figure 1. A parametrisation for the torus generated by revolving the circle (y − a)2 + z 2 = r 2 around the z-axis is x = a cos θ+r cos θ cos α. Ô ◭ z z z x y x y x y Figure 1. z = u. z = x2 is a parabola on the xz plane. y = a sin θ+r sin θ cos α. x2 + y 2 = c is a circle. See figure 1. with (θ.101. a2 b Figure Hyperbolic y2 x2 paraboloid z = 2 − 2 a b 1. When x = 0. (x2 + y 2 + z 2 )2 − 2(a2 + r 2 )(x2 + y 2 ) + 2(a2 − r 2 )z 2 + (a2 − r 2 )2 = 0. u ∈ [0. α) ∈ [−π. Rearranging again. 2π]. and its graph appears in figure 1. For fixed z = c. The equation of the torus thus. π]2 . v ∈ [0. The equation clearly requires that z ≥ 0. x2 + y 2 < 0 is impossible. The following is a parametrisation of this hyperbolic paraboloid: x = u.104: z2 x2 y2 = 2 + 2 − 1. When x = 0. When y = 0. y = u sin v. v ∈ [0. z = u2 − v 2 . then the x2 + y 2 = c2 + 1 are circles See figure 1. u ∈ R. and hence there is no graph when −1 < z < 1. 126 Example The surface z 2 = x2 + y 2 − 1 is called an hyperboloid of one sheet. z 2 − x2 = 1 is a hyperbola on the xz plane. u ∈ R. Figure 1. For x2 + y 2 < 1. When x = 0. z 2 < 0 is impossible. 2π]. z = u2 + 1.103. y = u sin v. z 2 − x2 = −1 is a hyperbola on the xz plane. When x = 0. u ∈ R.104. The following is a parametrisation for this hyperboloid of one sheet x= Ô u2 + 1 cos v. If z = 0. When y = 0. u ∈ R. See figure 1. 2π]. z 2 − y 2 = 1 is a hyperbola on the yz plane. y= Ô u2 + 1 sin v. 2π] and the following parametrises the bottom sheet. When z = c is a constant. then the x2 + y 2 = c2 − 1 are circles. v ∈ [0.102. y = v. See figure 1.105: Ellipsoid x2 y2 z2 + 2 + 2 = 1. The following is a parametrisation for the top sheet of this hyperboloid of two sheets x = u cos v. 2 c a b One-sheet hyperboloid Figure 1. v ∈ [0. z = −u2 − 1. 125 Example The surface z 2 = x2 + y 2 + 1 is called an hyperboloid of two sheets. v ∈ R. and hence there is no graph when x2 + y 2 < 1. x = u cos v. For z 2 − 1 < 0. When y = 0. When z = c is a constant c > 1. a2 b c Free to photocopy and distribute 85 . z 2 − y 2 = −1 is a hyperbola on the yz plane. z = u. z = x2 is a parabola on the xz plane. x2 − y 2 = 0 is a pair of lines in the xy plane. z = −y 2 is a parabola on the yz plane.Chapter 1 124 Example The surface z = x2 − y 2 is called a hyperbolic paraboloid or saddle. 15. 2π). 2 a b c a > b > c > 0. (u. 1).7 Shew that the surface in R3 implicitly defined by x4 + y 4 + z 4 − 4xyz(x + y + z) = 1 is a surface of revolution. b.When y = 0.5 Consider cap defined by the sur= ∈ Problem 1. = 1 is an ellipse on the yz plane. Problem 1. Problem 1. and find its axis of revolution. Problem 1. 2π].8 Shew that the surface S in R3 given implicitly by the equation 1 1 1 + + =1 x−y y−z z−x is a cylinder and find the direction of its directrix.15. v) (1. v) ∈ (0.4 Describe the face parametrised by ϕ(u. the radius of the largest circle which can lie on the ellipsoid x2 y2 z2 + 2 + 2 = 1. bu sin v. y = b sin θ sin φ.10 Demonstrate that √ S = {(x. and Cylindrical coordinates. Problem 1. z ≥ 1/ surface in R3 given implicitly by 2}.6 Demonstrate that the surface in R3 S : ex 2 z 2 − xy = 2z − 1 is a cone Problem 1. π].15. x 2 x 2 plane.15.1 Find the equation of the surface of revolution S generated by revolving the ellipse 4x2 + z 2 = 1 about the zaxis.15.15. v sin u.15. We may parametrise b2 the ellipsoid using spherical coordinates: x = a cos θ sin φ. b > 0. For z = 0. Parametrise S using Cartesian. When x = 0. Homework Problem 1.9 Shew that the surface S in R3 implicitly defined as xy + yz + zx + x + y + z + 1 = 0 is of revolution and find its axis.11 (Putnam Exam 1970) Determine. a > 0.15. z = c cos φ.Canonical Surfaces x2 a2 y2 b2 127 Example Let a. Spherical. +y 2 +z 2 − (x + z)e−2xz = 0. with proof. +∞) × (0. u2 ). Problem 1. v) → (v cos u. Free to photocopy and distribute 86 . θ ∈ [0. 2π) × (0. (u. implicitly defined. Problem 1.15.15. is a cylinder.2 Find the equation of the surface of revolution generated by revolving the line 3x + 4y = 1 about the y-axis . z) ∈ R3 : x2 +y 2 +z 2 = 1. φ ∈ [0. See figure 1. the spherical Problem 1. y. v) (au cos v. The surface z 2 + + c2 z 2 = 1 is called an ellipsoid. c be strictly positive real numbers. c2 + y 2 a2 + z 2 a2 + y 2 b2 1 is an ellipse on the xy c2 = 1 is an ellipse on the xz plane. au). Problem 1. a.3 Describe the surface parametrised by ϕ(u.15.105. x2 + =1 4 z Chapter 1 Problem 1. z x y Figure 1. The trace of the curve r is the set of all images of r.16 Parametric Curves in Space In analogy to curves on the plane. Find its equation. r(a) is the initial point of the curve and r(b) its terminal point.12. we now define curves in space. Free to photocopy and distribute 87 .106: Problem 1.15. x2 + y2 =1 4 Figure 1. A parametric curve representation r of a curve Γ is a function r : [a.15.y2 z = 2. b] ⊆ R. x z = 0. 128 Definition Let [a.107: Helix. its projection on the xy plane produces the ellipse 4x2 + y 2 = 1. b] → R3 .106 has the property that if it is cut by planes at z = ±2. r Γ .12 The hyperboloid of one sheet in figure 1. with à x(t) r(t) = y(t) z(t) and such that r([a. 4x2 + y 2 = 1 y z = −2. 1. its projection on the xy plane produces the ellipse y2 x2 + = 1. and if it is cut by a plane at 4 z = 0. b]) = Γ. that is. A curve is closed if its initial point and its final point coincide. The length of the curve is ¬¬ →¬¬ ¬¬d− ¬¬. Γ. and acts as a source light projecting a shadow of R onto the xyplane.Parametric Curves in Space 129 Example The trace of → − → − → − r(t) = i cos t + j sin t + k t is known as a cylindrical helix.sin(t). Find the area of this shadow. y = sin t. 2 From the equation of the plane. |y| ≤ b. Consider the the region R = (x. y.t]. Solution: ◮ The projection of the intersection of the plane 3x+y+z = 1 and the cylinder is the ellipse x2 + 2y 2 = 1. The Maple™ commands to graph this curve and to find its length are: > with(plots): > with(Student[VectorCalculus]): > spacecurve([cos(t). Free to photocopy and distribute 88 x2 + y2 . 2 Thus we may take the parametrisation ¾ x(t) z(t) ¿ ¾ r(t) = y(t) = 1 − 3 cos t − cos t √ 2 sin t 2 √ ¿ ..[x. z =c+1 a2 b2 once around.t=0. z) ∈ R3 : |x| ≤ a.2*Pi.axes=normal). x and thus the length is 2π 0 √ √ 2dt = 2π 2. This ellipse can be parametrised as √ 2 x = cos t. 2 sin t 2 ◭ 131 Example Let a.2*Pi))..0. √ 2 z = 1 − 3x − y = 1 − 3 cos t − sin t. c be strictly positive real numbers. 130 Example Find a parametric representation for the curve resulting by the intersection of the plane 3x + y + z = 1 and the cylinder x2 + 2y 2 = 1 in R3 . first observe that √ ¬¬ →¬¬ ¬¬ ¬¬ ¬¬d− ¬¬ = ¬¬(sin t)2 + (− cos t)2 + 1¬¬dt = 2dt. A point P moves along the ellipse = 1. on the xy-plane. To find the length of the helix as t traverses the interval [0. 0 ≤ t ≤ 2π.y. z = c . 2π].z]=Path(<cos(t).t>. b.sin(t). > PathInt(1. so that the area shadowed is πab(c + 1)2 + 4ab(c + 1)2 + 4abc(c + 1) = c2 ab(π + 12) + 16abc + 4ab.109). Q moves along the circle with equation x2 +y 2 = c2 on the xy-plane (figure 1.4.z Chapter 1 x y Figure 1. and the central square has side 2c + 2.110: Problem 1. ◭ Free to photocopy and distribute 89 x2 + y2 . a(c + 1) (parallel to the x-axis) and b(c + 1) (parallel to the y-axis). For fixed P (u. This creates a region as in figure 1. Resizing to a region R = (x. As P moves along the circle. being the centre of a (2c + 2) × (2c + 2) square. y. and an ellipse = 1. of area πc2 + 4(c + 1)2 + 8c(c + 1). v.109). Figure 1. −cv. Solution: ◮ First consider the same problem as P moves around the circle x2 + y 2 = 1. z) ∈ R3 : |x| ≤ 1. Figure 1. z = c . |y| ≤ 1.110.109: Problem 1. z = c + 1 and the region is R′ = (x. |y| ≤ b. y. the image of R′ (a 2 × 2 square) on the xy plane is a (2c + 2) × (2c + 2) square with centre at the point Q(−cu.16. z = c + 1 a2 b2 we use instead of c + 1. z = c . c + 1) on the circle.108: Problem 1.16.4. where each quarter circle has radius c.16. z) ∈ R3 : |x| ≤ a.4. 0) (figure 1. x = 0. 2. As a varies through R. as vividly as possible these sur.16.16. tersect.6 Let a be a real number paintersect. y = 4t3/2 . P2 : x − ay + az = −1. 4. 1.4 Give a parametrisation for the part of the ellipsoid x2 + y2 z2 + =1 9 4 t ∈ [0. Find a direction vector for l. rameter. and x = 1. and r r r → (t ) are coplanar if and only if − r 4 1 1 1 1 + + + = 0. 0. let → z−x2 −y 2 −1 = 0. Graph all points R on the faces and their intersection.xy-plane. Let l be their intersection line. Prove that → (t1 ). As a point Q moves along C . if they at all Problem 1. 132 Definition Rn is the n-dimensional space. z) be the point of intersection of this surface and the plane z = c. Find a parametric equation for the curve on which they intersect.Multidimensional Vectors Homework Problem 1.the xy-plane. Problem 1. R be the point of intersection of ← and PQ Describe. z+x2 +y 2 −3 = 0. The ideas expounded earlier about the plane and space carry almost without change. . and consider the planes Problem 1. 32. 1 + t2 2 t 1 + t2 Let tk . Let (x. . y. Problem 1. →(t3 ).17 Multidimensional Vectors We briefly describe space in n-dimensions. 3.1 Let C be the curve in R3 defined by x = t2 . 1 ≤ k ≤ 4 be non-zero real num− − − bers. 1) and consider the curve C : z = y 2 on the yz-plane. t1 t2 t3 t4 P1 : ax + y + z = −a.16.3 Consider the space curve ¾ ¿ t4 1 + t2 t3 →:t→ − r . Find an equation relating x and y.16. →(t2 ).16. Find the volume bounded by the two planes.5 Let P be the point (2. xn ã : xk ∈ R . 36). z = 9t. +∞[. l describes a surface S in R3 . 9) to (16. and the surface S as c varies. if they at all in. Problem 1. the collection x1 Rn = x2 . Free to photocopy and distribute 90 . 1.2 Consider the surfaces in R3 implicitly defined by which lies on top of the plane x + y + z = 0. Calculate the distance along C from (1.16. → = − ek 1 . (1. . . their dot product is a →•→ = − − a b n ak bk . } e e e with ¾ ¿ (1. αan 135 Definition The standard ordered basis for Rn is the collection of vectors − − − {→1 .Chapter 1 → − − − − → 133 Definition If → and b are two vectors in Rn their vector sum → + b is defined a a by the coordinatewise addition ¾ a1 + b1 a2 + b2 . k=1 We now establish one of the most useful inequalities in analysis. . 0 (a 1 in the k slot and 0’s everywhere else). . . . . . . →2 . b of Rn . If α ∈ R and → ∈ Rn a we define scalar multiplication of a vector and a scalar by the coordinatewise multiplication ¾ ¿ αa1 − α→ = a αa2 . . →n . αn − − → 136 Definition Given vectors →. an + bn ¿ − →+→= − a b . . . Free to photocopy and distribute 91 . Observe that ¾ n k=1 α1 ¿ − αk →k = e α2 . . .20) − 134 Definition A real number α ∈ R will be called a scalar. . .21) 0 . ¬¬− ¬¬2 ¬¬− ¬¬2 ¬¬− ¬¬¬¬− ¬¬ − − − − (2→•→)2 − 4(¬¬→¬¬ )(¬¬→¬¬ ) ≤ 0 ⇐⇒ |→•→| ≤ ¬¬→¬¬¬¬→¬¬. that is. a from where the desired result follows. x y x y x y x y giving the theorem. u The above proof works not just for Rn but for any vector space (cf. yk . u Again.22) for real numbers xk . Then a we have ¬¬ ¬¬→¬¬ ¬¬→¬¬ →¬¬ − ¬¬ ¬¬→ ¬¬− ¬¬ − − ¬¬ a + b ¬¬ ≤ ¬¬ a ¬¬ + ¬¬ b ¬¬. we have ¬¬→ − ¬¬ ¬¬− + t→¬¬ ≥ 0 x y ⇐⇒ ⇐⇒ ⇐⇒ − − − − (→ + t→)•(→ + t→) ≥ 0 x y x y →•→ + 2t→•→ + t2 →•→ ≥ 0 − − − − − − x x x y y y ¬¬→¬¬2 ¬¬− ¬¬ − − ¬¬− ¬¬ + 2t→•→ + t2 ¬¬→¬¬2 ≥ 0. Then we have ¬¬− ¬¬¬¬− ¬¬ − − |→•→| ≤ ¬¬→¬¬¬¬→¬¬. → − − 138 Corollary (Triangle Inequality) Let → and b be any two vectors in Rn . →) = ¬¬→¬¬¬¬→¬¬ . Proof: → − − ||→ + b ||2 a → − − → − − = (→ + b )•(→ + b ) a a − − − − − − → → → = →•→ + 2→• b + b • b a a a → − → − − − ≤ ||→||2 + 2||→|||| b || + || b ||2 a a → − − = (||→|| + || b ||)2 .Multidimensional Vectors − − 137 Theorem (Cauchy-Bunyakovsky-Schwarz Inequality) Let → and → be any two x y n vectors in R . (1. x y − ¬¬¬¬− ¬¬ ¬¬ x y Free to photocopy and distribute 92 . x y x y This last expression is a quadratic polynomial in t which is always nonnegative. − − 139 Definition Let → and → be two non-zero vectors in a vector space over the x y − − real numbers. Then the angle (→. →) between them is given by the relation x y → •→ − − x y − − cos (→. below) that has an inner product. As such its discriminant must be non-positive. the preceding proof is valid in any vector space that has a norm. The form of the Cauchy-Bunyakovsky-Schwarz most useful to us will be ¬ ¬ ¬ ¬ ¬ n k=1 ¬ ¬ ¬ x k yk ¬ ≤ ¬ n 1/2 n 2 yk k=1 1/2 x2 k k=1 . x y x y Proof: Since the norm of any vector is non-negative. n. qk > 0.Chapter 1 This expression agrees with the geometry in the case of the dot product for R2 and R3 . n 141 Lemma Let ak > 0. which gives the required inequality. are positive real numbers. k = 1. We need some preparatory work. with k=1 n x→0 qk = 1. ck . Shew that n 4 n n n 2 ak bk ck k=1 ≤ n a4 k k=1 k=1 b4 k k=1 c2 k . . bk . . Solution: ◮ Using CBS on k=1 n k=1 (ak bk )ck once we obtain n 1/2 n 1/2 ak bk ck ≤ n a2 b2 k k k=1 1/2 k=1 c2 k . ◭ We now use the CBS inequality to establish another important inequality. Using CBS again on k=1 n a2 b2 k k n we obtain 1/2 n 1/2 ak bk ck k=1 ≤ ≤ a2 b2 k k k=1 n 1/4 c2 k k=1 n b4 k k=1 1/4 n 1/2 a4 k k=1 c2 k k=1 . Proof: Recall that log(1 + x) ∼ x as x → 0. k=1 Free to photocopy and distribute 93 . 140 Example Assume that ak . . Thus n x→0 1/x lim log k=1 qk a x k = = = = x x k=1 qk (ak − 1) lim x→0 x n (ax − 1) lim qk k x→0 x k=1 x→0 lim log Èn Èn k=1 qk a x k ¡ n qk log ak . Then 1/x n lim log k=1 qk a x k = k=1 qk log ak . . 23) yield the monotone decreasing sequence 1 n n k=1 ak ≥ 1 n n k=1 √ 2 ak ≥ 1 n n k=1 √ 4 4 ak ≥ . an > 0. For. u 143 Example For any positive integer n > 1 we have 1 · 3 · 5 · · · · (2n − 1) < nn . 1 · 3 · 5 · · · · (2n − 1) < 1 + 3 + 5 + · · · + (2n − 1) n n 1 n n log ak k=1 = √ n a1 a2 · · · an . = n2 n n = nn .. (1. by AMGM. 144 Definition Let a1 > 0. As a corollary to AMGM we obtain Free to photocopy and distribute 94 . then by CBS 1 n n k=1 bk ≥ 1 n n k=1 Ô 2 bk . a2 > 0. Notice that since the factors are unequal we have strict inequality.Multidimensional Vectors u 142 Theorem (Arithmetic Mean-Geometric Mean Inequality) Let ak ≥ 0. n Proof: If bk ≥ 0.23) Successive applications of (1.. Their harmonic mean is given by n 1 a1 + 1 a2 + ··· + 1 an .. . Then a1 + a2 + · + an √ n a1 a2 · · · an ≤ . n as wanted. . which by Lemma 141 has limit exp giving a1 + a2 + · + an √ n a1 a2 · · · an ≤ . . . . For then 1 1 b1 b2 u ··· 1 bn 1/n 1 1 1 + + ··· + b b2 bn ≤ 1 . bn > 0. 1 1 1 + + ··· + b1 b2 bn 1 bk Proof: This follows by putting ak = in Theorem 142 . . .Chapter 1 145 Corollary (Harmonic Mean-Geometric Mean Inequality) Let b1 > 0. Then n ≤ (b1 b2 · · · bn )1/n . . bn > 0. we deduce 146 Corollary (Harmonic Mean-Arithmetic Mean Inequality) Let b1 > 0. Prove that n k=1 s s − ak ak s − ak ≥ n2 n−1 n n−1 and n k=1 ≥ . n Combining Theorem 142 and Corollary 145. b2 > 0. . Solution: ◮ Put bk = n s s − ak 1 bk = . Free to photocopy and distribute 95 . . Then b1 + b2 + · · · + bn n ≤ . and s = a1 + a2 + · · · + an . b2 > 0. . n from where the first inequality is proved. 1 1 1 n + + ··· + b1 b2 bn 147 Example Let ak > 0. Then n k=1 s − ak s k=1 =n−1 and from Corollary 146. Èn n n−1 k=1 ≤ s s − ak . . 2 Demonstrate that if x1 .3 (USAMO 1978) Let a. . b. ´ 1.17. k A1 + A2 + · · · + An = n.17. −1 −n = k=1 2 ≥ = ◭ n n−1 n n−1 Homework Problem 1. . + xn ) + + . c. . xn . x2 . In this exercise you will follow the steps of a proof by George Polya.17. . n! < n+1 2 n .4 Find all positive real numbers a1 ≤ a2 ≤ . Put nak Ak = . is strictly in2 n ≥n .17. . . . 3. Problem 1. (a1 a2 · · · an )1/n ≤ n Equality occurs if and only if a1 = a2 = . a2 +b2 +c2 +d2 +e2 = 16. k=1 k=1 a2 = 144.1 The Arithmetic Mean Geometric Mean Inequality says that if ak ≥ 0 then a1 + a2 + · · · + an . Problem 1. are strictly positive real numbers then 1 1 1 (x1 + x2 + . . Prove that ∀x ∈ R. we have n n k=1 = k=1 n s s − ak s s − ak −n .17. e be real numbers such that a+b+c+d+e = 8.. 2. Problem 1. = (a1 + a2 + · · · + an )n ak = 96. a1 + a2 + · · · + an and Gn = a1 a2 · · · an . . Prove the AMGM inequality by assembling the results above. Free to photocopy and distribute 96 . ≤ an such that n n n A1 A2 · · · An and that n Gn . n = 1. k k=1 a3 = 216. 4. + x1 x2 xn Problem 1.6 Let f (x) = (a + x)5 (a − x)3 . d.7 Prove that the sequence 1 n xn = 1 + . Deduce that a1 + a2 + · · · + an Gn ≤ n n Problem 1. Prove that n Problem 1. . .17. x ≤ ex−1 .Multidimensional Vectors Since s s − ak −1= ak s − ak ak s − ak . creasing. 2.5 Demonstrate that for integer n > 1 we have.. = an . x ∈ [−a. . a].17. Maximise the value of e. . . Find the maximum value of de f using the AM-GM inequality. Figure 2. b1 [×]a2.2: Open rectangle in R2 .1: Open ball in R2 . where the ak . bk are real numbers.2 Differentiation 2. bn [. An open ball centred at a of radius ε is the set Bε (a) = {x ∈ Rn : ||x − a|| < ε}. (a1 . b2 [× · · · ×]an−1. bn−1 [×]an . 97 b2 − a2 ε .1 Some Topology An open box is a Cartesian product of open intervals 148 Definition Let a ∈ Rn and let ε > 0. a2 ) b1 − a1 Figure 2. ]a1 . 1[ is open in R.1 Determine whether the following subsets of R2 are open. y) ∈ R2 : |x| < 1. |y| ≤ 1} Free to photocopy and distribute 98 . 1] × [0. y) ∈ R2 : x2 ≤ y ≤ x} 7. however. C = {(x. The reader will recognise that open boxes. 1] =] − ∞. 156 Example The region [−1. an open box in R2 is a rectangle without its boundary (see figure 2. 150 Definition A set O ⊆ Rn is said to be open if for every point belonging to it we can surround the point by a sufficiently small open ball so that this balls lies completely within the set. −1[∪]1. y) ∈ R2 : |x| ≤ 1. R \ [−1. as no interval centred at 1 is totally contained in ] − 1. ∀a ∈ O ∃ε > 0 such that Bε (a) ⊆ O.2) and an open box in R3 is a box without its boundary (see figure 2. an open ball in R2 is an open disk (see figure 2. The interval ] − 1. 155 Example The closed interval [−1. 1[×]0. 152 Example The region ] − 1. An open box in R is an open interval. y) ∈ R2 : xy > 1} 6. +∞[ is open in R. 1] is closed in R. y) ∈ R2 : |x| < 1. or neither. a2 . y 149 Example An open ball in R is an open interval.1) and an open ball in R3 is an open sphere (see figure 2. closed. y x Figure 2.4: Open box in R3 . 1] is neither open nor closed in R. +∞[ is open in R2 . That is. B = {(x. |y| < 1} 2. open ellipsoids and their unions and finite intersections are open sets in Rn .4). in R2 . 1.1. Homework Problem 2. y) ∈ R2 : x2 + 4y 2 < 4 is open in R2 . A = {(x. +∞[×[0. x < y 2 } 3. D = {(x. 2] is closed in R3 .3: Open ball in R3 . The interval ] − 1. 154 Definition A set F ⊆ Rn is said to be closed in Rn if its complement Rn \ F is open. 1] is not open. G = {(x. y) ∈ R2 : xy ≤ 1} 4. 1].3). E = {(x. 153 Example The ellipsoidal region (x. F = {(x. |y| ≤ 1} 5. 151 Example The open interval ] − 1. as its complement. a3 ) Some Topology x Figure 2. y) ∈ R2 : |y| ≤ 9.z z ε (a1 . however. j =1 2. and integrability of multivariable functions. (x. we say that f is a vector field.2. if there is such a curve. y. y) → cos(x2 − y 2 ) Problem 2. 1. y) → sin(x2 + y 2 ) 7. (x. Needless to say. What are the possibilities for the image (range) of p(x. y)? Problem 2. 157 Definition Let A ⊆ R2 and let f : A → R be a function. Shew that there is a pairwise disjoint subcollection Dk . y) and the plane z = c. For example.Chapter 2 Problem 2. |y|. k ≥ 1 in F such that n E⊆ 3Dj . (x.1. defined over the entire plane R2 . recall that the graph of a function f : A → Rm . For most of this course. we have an object of more than three-dimensions! In the case n = 2. y. differentiability. We will now briefly examine this case. y) → x3 − x 5. we would like to examine limits. In particular.2. we say that f is a scalar field. y. y) be a polynomial with real coefficients in the real variables x and y. |y|) 4. y) → x + y 2. If m ≥ 2. is the set {(x. the introduction of more variables greatly complicates the analysis. y) → xy 3. our concern will be functions of the form If m = 1. (x. z) → x + y + z 2. Homework Problem 2. (x.2 (Putnam Exam 1969) Let p(x. y) = x2 +y 2 (an elliptic paraboloid) are the concentric circles x2 + y 2 = c. y) → x2 + 4y 2 6.2 Sketch the level surfaces for the following maps. z) → min(|x|. A ⊆ Rn . Let A ⊆ Rn . m = 1. (x. 158 Example The level curves of the surface f (x. z) → xyz 3. Given c ∈ R. If m + n > 3. c > 0.3 (Putnam 1998) Let F be a finite collection of open disks in R2 whose union contains a set E ⊆ R2 . We would like to develop a calculus analogous to the situation in R.1. we have a tri-dimensional surface. continuity. 1. f (x)) : x ∈ A)} ⊆ Rn+m . |z|) Free to photocopy and distribute 99 . (x. the level curve at z = c is the curve resulting from the intersection of the surface z = f (x. y) → min(|x|. (x. (x.2 Multivariable Functions f : A → Rm . (x.1 Sketch the level curves for the following maps. z) → x2 + y 2 + z 2 2.0) lim x2 y x2 + y 2 = 0.Limits 4. > limit(xˆ2*y/(xˆ2+yˆ2). Observe that 0 ≤ x2 ≤ x2 x2 + y 2 . y. The notions of infinite limits. The Maple™ commands to graph this surface and find this limits appear below. z) → x2 + y 2 5.axes=boxed.0) (x.x=0. Free to photocopy and distribute 100 . and hence (x.0) lim |y|.0) lim x5 y 3 x6 + y 4 . 159 Definition A function f : Rn → Rm is said to have a limit L ∈ Rm at a ∈ Rn if ∀ǫ > 0∃δ > 0 such that 0 < ||x − a|| < δ =⇒ ||f (x) − L|| < ǫ.y)→(0.3 Limits We will start with the notion of limit.color=xˆ2+yˆ2). 160 Example Find Solution: lim x2 y . (x. limits at infinity. Thus x + y2 ¬ ¬ ¬ x2 y ¬ ¬ ¬≤ lim 0≤ lim (x.y)→(0. (x. as one must approach the test point from infinitely many directions. In such a case we write. y. Notice that Maple is unable to find the limit and so returns unevaluated.. > with(plots): > plot3d(xˆ2*y/(xˆ2+yˆ2). are analogously defined. y.. (x.10.y=-10. and so 0 ≤ 2 ≤ 1.0) ¬ x2 + y 2 ¬ (x. ◭ 161 Example Find (x.x=-10.y)→(0.10. z) → x2 + 4y 2 6. z) → sin(z − x2 − y 2 ) 7. x→a lim f (x) = L.0) ◮ We use the sandwich theorem.y)→(0. (x.y=0). x2 + y 2 (x. Limits in more than one dimension are perhaps trickier to find. and continuity at a point.y)→(0. y.y)→(0. The limit does not exist. ¬ x6 + y 4 ¬ y4 If |y| ≤ |x|. Thus ¬ ¬ ¬ x5 y 3 ¬ 4 2 4 2 ¬ ¬ ¬ x6 + y 4 ¬ ≤ max(y . Solution: ◮ When y = 0. Y = y 2 . then ¬ ¬ ¬ x5 y 3 ¬ x8 2 ¬ ¬ ¬ x6 + y 4 ¬ ≤ x6 = x . x2 as x → 0.8: Example 163. as (x.y)→(0. 1+x → +∞. 0).5: (x. ◭ Free to photocopy and distribute 101 . Figure 2. y) → (0. ◭ 162 Example Find (x.7: Example 162. y) → x2 y . x2 + y2 Figure 2. ¬ x6 + y 4 ¬ X2 + Y 2 as (x. Aliter: Let X = x3 . Y = ρ sin θ. x ) ≤ y + x −→ 0. Observe that if |x| ≤ |y|. ¬ ¬ ¬ x5 y 3 ¬ y8 ¬ ¬≤ = y4 . When x = 0. 0).6: (x. we obtain ¬ ¬ ¬ x5 y 3 ¬ X 5/3 Y 3/2 ¬ ¬= = ρ5/3+3/2−2| cos θ|5/3| sin θ|3/2 ≤ ρ7/6 → 0. −y 2 as y → 0. y) → (0. 1+y → −∞. ¬ ¬ ¬ x5 y 3 ¬ X 5/3 Y 3/2 ¬ ¬= . y) → x5 y3 . ¬ x6 + y 4 ¬ X2 + Y 2 Passing to polar coordinates X = ρ cos θ. Solution: then ◮ Either |x| ≤ |y| or |x| ≥ |y|. x6 + y4 Figure 2.Chapter 2 Figure 2.0) lim 1+x+y x2 − y 2 . 0). Solution: ◮ When y = 0 we have 2(x − 1)2 ln(|1 − x|) 2x ∼− . y) → (0.y)→(0. |x| |x| and so the function does not have a limit at (0. ◭ 165 Example Find lim sin(x4 ) + sin(y 4 ) Ô (x.0) lim . 0).0) lim ((x − 1)2 + y 2 ) loge ((x − 1)2 + y 2 ) |x| + |y| .0) x4 + y 4 .y)→(0.12: Example 163.0) Free to photocopy and distribute 102 . Figure 2. as (x.y)→(0. 164 Example Find (x.11: Example 166.y)→(0. Figure 2. But when y = 0. Figure 2. as t → 0. Solution: ◮ Putting x = t4 . the function is 0. ◭ 166 Example Find lim sin x − y x − sin y .10: Example 165. we find xy 6 x6 + y8 = 1 2t2 → +∞. ◭ Figure 2. (x.Limits xy 6 x6 + y 8 163 Example Find (x.9: Example 164. Solution: ◮ sin(x4 ) + sin(y 4 ) ≤ x4 + y 4 and so ¬ ¬ ¬ sin(x4 ) + sin(y 4 ) ¬ ¬ ¬≤ Ô ¬ ¬ x4 + y 4 x4 + y 4 → 0. y = t3 . Thus the limit does not exist. The following possibilities might occur. if x2 + 4y 2 > 1 be continuous everywhere on the xy-plane? Problem 2. 1.y)→(x0 . lim f (x.3 Sketch the domain of defi1 . y) → log(x + y).5 Find (x.Chapter 2 Solution: ◮ When y = 0 we obtain sin x x → 1. It may occur that does not.4 Find y 2 ) sin 1 .2 Sketch the domain of definition of (x.6 For what c will the function Ô 1 − x2 − 4y 2 .7 Find (x. If the iterated limits exist and lim then (x. x2 + y 2 Free to photocopy and distribute 103 .1 Sketch the domain of defiÔ nition of (x. xy lim (x2 + Problem 2.0) lim Ô x2 + y 2 sin 1 . y) exists.0) (x. y0 ). y) = c.y)→(x0 .y0 ) y→y0 x→x0 x→x0 y→y0 lim f (x. y) 2.3. y) exists. When y = x the function is identically −1.y0 ) y→y0 x→x0 x→x0 y→y0 lim f (x.3. y) → 2 x + y2 Problem 2. then each of the iterated limits lim y→y0 x→x0 lim f (x.y )→(0. lim f (x. if x2 + 4y 2 ≤ 1.y )→(0. It may occur that lim (x. Problem 2. both exist. y) does not exist.y0 ) Homework Problem 2. y) . (x. f (x. ◭ If f : R2 → R. If (x. but one of the iterated limits 4.3. y) = lim lim f (x. y) . y) = lim lim f (x. x Problem 2.y)→(x0 . as x → 0.y)→(x0 . x→x0 lim y→y0 lim f (x. y) does not exist.y )→(0. y) and lim y→y0 lim f (x.2) lim sin xy . y) → 4 − x2 − y 2 . y) . y) exists. nition of (x.y0 ) x→x0 lim f (x.3. Problem 2. lim f (x.3. These are called the iterated limits of f as (x. y) → (x0 .3. it may be that the limits y→y0 lim x→x0 lim f (x. Thus the limit does not exist.3. but that 3. 3. Problem 2.3.9 Find lim 2x2 sin y 2 + y 4 e−|x| Ô .12 Let f (x. y = 0 otherwise lim f (x. 104 x→a or equivalently.+∞) 1 1 + y sin x y if x = 0.11 Prove that lim lim x−y x+y = 1 = − lim x−y exist?. a function g : R → R is said to be differentiable at x = a if the limit x→a lim g(x) − g(a) = g ′ (a) x−a exists. For example. |y|) Ô . y) exists.z)→(0.y )→(0.y )→(0.y . but x→0 y →0 (x.y )→(+∞.8 Find lim max(|x|. y) lim f (x. Recall that in one variable. We write f (h) = o (h) if f (h) goes faster to 0 than h. x→0 y →0 y →0 x→0 lim x→0 x 2 y 2 lim = 0.Definition of the Derivative Problem 2. x4 + y 4 Problem 2. The limit condition above is equivalent to saying that lim g(x) − g(a) − g ′ (a)(x − a) = 0. Let f : R → R be a function.3. (x. h→0 h→0 We now define the derivative in the multidimensional space Rn .0 that the iterated limits lim and lim y →0 x→0 lim f (x. y) = x sin 0 Problem 2.3.4 Definition of the Derivative Before we begin.y )→(0. h3 + 2h2 = o (h).3. x−a g(a + h) − g(a) − g ′ (a)(h) h = 0.y )→(0.0. since lim h→0 h lim h3 + 2h2 h = lim h2 + 2h = 0. h→0 lim Free to photocopy and distribute .0) lim but still not exist.10 Demonstrate that (x.13 Prove that x→0 lim Problem 2. y →0 y →0 x 2 y 2 lim x2 y 2 + (x − y)2 x2 y 2 + (x − y)2 = 0. that is.3. x2 + y 2 Prove that (x. let us introduce some necessary notation. Does (x.0) (x.0) x 2 lim x2 y 2 z 2 = 0.0) x 2 y 2 lim x2 y 2 does + (x − y)2 2. if f (h) = 0. + y2 + z2 Problem 2. y) do not exist. x+y lim x−y x+y and that . . L(v) = Da (f )(v). i. as x → a. as h → 0 This means that ||L(v) − Da (f )(v)|| → 0. as h → 0. called the derivative of f at a. 168 Theorem If A is an open set in definition 167. completing the proof. as h → 0. Proof: Let L : Rn → Rm be another linear transformation satisfying definition 167. Observe that since we may not divide by vectors. The condition for differentiability at a is equivalent to f (x) − f (a) = Da (f )(x − a) + o (||x − a||) . Now. f is differentiable at a if there is a linear transformation Da (f ) such that f (a + h) − f (a) = Da (f )(h) + o (||h||) . x→a ||x − a|| Equivalently. ||Da (f )(v) − L(v)|| ≤ ||Da (f )(h) − f (a + h) + f (a)|| = o (||h||) + o (||h||) = o (||h||) . observe that Da (f )(v) − L(v) = Da (f )(h) − f (a + h) + f (a) + f (a + h) − f (a) − L(h). By definition. A function f : A → Rm is said to be differentiable at a ∈ A if there is a linear transformation.Chapter 2 We may write this as g(a + h) − g(a) = g ′ (a)(h) + o (h) . Since A is open. and f (a + h) − f (a) = L(h) + o (||h||) .e. we have f (a + h) − f (a) = Da (f )(h) + o (||h||) . L(v) = Da (f )(v). Da (f ) is uniquely determined. Da (f ) : Rn → Rm such that ||f (x) − f (a) − Da (f )(x − a)|| lim = 0. The above analysis provides an analogue definition for the higher-dimensional case. the corresponding definition in higher dimensions involves quotients of norms. a + h ∈ A for sufficiently small ||h||. By the triangle inequality. u Free to photocopy and distribute 105 +||f (a + h) − f (a) − L(h)|| . 167 Definition Let A ⊆ Rn . We must prove that ∀v ∈ Rn . a singleton. | h k | ≤ ¬¬ h ¬¬¬¬ k ¬¬ = o ¬¬ h ¬¬ . x x y Solution: ◮ We have → − − → − − − − f (→ + h . →) x y → → R →•→ − − x y be the usual dot product in R3 . 0 ). ◭ Just like in the one variable case. whence the claim follows. But by linearity ||L(x) − L(a) − L(x − a)|| = ||L(x) − L(a) − L(x) + L(a)|| = ||0|| = 0. identically. Thus if L satisfies definition 167. 106 Free to photocopy and distribute . we¬¬ ¬¬ by the¬ Cauchy-Buniakovskii-Schwarz ¬¬→¬¬ have ¬ →¬¬ →•→ − − − ¬¬− ¬¬¬¬→¬¬ ¬¬− ¬¬ inequality. k ) → ( 0 . we know that Da (L) uniquely determined. and ||f (x) − f (a) − T (x − a)|| = ||0|| = 0.Definition of the Derivative If A = {a}. Any linear transformation T will satisfy the definition. implies continuity at that point. for any a ∈ Rn . Solution: ◮ Since Rn is an open set. then Da (f ) is not uniquely determined. ◭ 170 Example Let f : R3 × R3 − − (→. as T (x − a) = T (0) = 0. 171 Theorem Suppose A ⊆ Rn is open and f : A → Rn is differentiable on A. and so f (x) = f (a). differentiability at a point. → + k ) − f (→. Then f is continuous on A. − x = = → → − − → → − − As ( h . k ) = →• k + h •→. For ||x − a|| < δ holds only for x = a. we must shew that x→a lim f (x) = f (a).→) f ( h . Shew that f is differentiable and that − − → → − − − → →− − − y D(→. Proof: Given a ∈ A. 169 Example If L : Rn → Rm is a linear transformation. then Da (L) = L. which proves the assertion. →) = x y x y → − − → − − − − (→ + h )•(→ + k ) − →•→ x y x y − − − − − − →•→ + →•→ + →•→ + →•→ − →•→ − − − − x y h y h k x y x k → →•→ →•→ − − − − − →• k + h y + h k . then the claim is established. 4.5 The Jacobi Matrix We now establish a way which simplifies the process of finding the derivative of a function at a given point. u Homework Problem 2. . To find partial derivatives with respect to the j-th variable. xn ) Here fi : Rn → R. . . . . f (→ ) = ¬¬→¬¬ be the usual norm in Rn . x2 . v ¬¬− ¬¬ x → − − for → = 0 . xn ) h whenever this limit exists. . xn ) . ¿ f (x) = . . → Shew that F is differentiable and that → − → − → − − − Dx (F )( h ) = → × L( h ) + h × L(→). x x 2. . x2 . x x ¬¬→ ¬¬2 − − ¬¬− ¬¬ = →•→. . and so f (x) − f (a) → 0. xn ) f2 (x1 . xj . x2 . . . xj + h. . . we simply keep the other variables fixed and differentiate with respect to the j-th variable. . x2 . . . 172 Definition Let A ⊆ Rn . . x2 . . and put ¾ f1 (x1 . − − x x Problem 2. . n ≥ ¬¬− ¬ − 1. . but that f is not differentiable x → − at 0 . . h→0 fi (x1 . . f : A → Rm . .Chapter 2 Since f is differentiable at a we have f (x) − f (a) = Da (f )(x − a) + o (||x − a||) .2 ¬Let f : Rn → R. . Free to photocopy and distribute 107 . .4. . . The partial derivative ∂fi ∂xj (x) = lim ∂fi (x) is defined as ∂xj .1 Let L : R3 → R3 be a linear transformation and R3 F : → − x → R3 → × L(→ ) . Prove that with x x x → •→ − − x v − Dx (f )(→ ) = ¬¬→¬¬ . as x → a. xn ) − fi (x1 . . . . fm (x1 . proving the theorem. . > > > > f:=(x. ∂f1 ∂xn ∂f2 ∂xn . we must compute Dx (f )(→j ). x2 . xn ) Suppose A ⊆ Rn is an open set and f : A → Rm is differentiable. ∂fn (x) ∂x1 ∂fn (x) ∂x2 ··· ∂fn (x) ∂xn − Proof: Let →j .y. x2 . xn ) .z).y). . .x). . 174 Theorem Let ¾ ¿ f1 (x1 . and the matrix representation of Dx (f ) with ∂xj respect to the standard bases of Rn and Rm is the Jacobi matrix ¾ ∂f1 ∂x1 ∂f2 (x) (x) ∂f1 ∂x2 ∂f2 ∂x2 .z). z) = 1 + 3y 2 z 3 . .z). (x. y. z) = 3z 2 + 9xy 2 z 2 . f (x) = . . xn ) f2 (x1 . .z). . To obtain the e − Jacobi matrix. . The Maple™ commands to find these follow. .z)->x+yˆ2+zˆ3+3*x*yˆ2*zˆ3. Since the derivative of a function f : Rn → Rm is a linear transformation. . . y.The Jacobi Matrix 173 Example If f : R3 → R. diff(f(x. . which will give us the j-th e Free to photocopy and distribute 108 .y. be the standard basis for Rn . The following theorem will allow us to determine the matrix representation for Da (f ) under the standard bases of Rn and Rm . . f ′ (x) = ∂x1 . and f (x. 1 ≤ j ≤ n. . . z) = x + y 2 + z 3 + 3xy 2 z 3 then ∂f ∂x ∂f ∂y and ∂f ∂z (x. fm (x1 .y. . . x2 . ¿ (x) (x) . z) = 2y + 6xyz 3 . Then each ∂fi partial derivative (x) exists. diff(f(x. (x. y. . . (x) (x) ··· ··· .y. it can be represented by aid of matrices. . diff(f(x. y. . We will abuse language. xj . . . xn ) ∂fi = (x) . . xj . . − and put y = x + ε→j . xn ) − fi (x1 . xj + h. . . Compute the Jacobi matrix of f . . x2 . . Notice that e ||f (y) − f (x) − Dx (f )(y − x)|| = ||y − x|| − ||f (x1 . y) ∂y ¾ ∂f1 (x. . the Jacobi matrix is not the derivative of a function at a point. |fi (x1 . . ◭ Free to photocopy and distribute 109 . 175 Example Let f : R3 → R2 be given by f (x. xn ) − fi (x1 . . however. .Chapter 2 column of the Jacobi matrix. . . . . and so. x2 . y) ∂x ∂f1 (x. . Since the sinistral side → 0 as ε → 0. . xj + h. . . . . y) ∂y ∂f2 (x. It is a matrix representation of the derivative in the standard basis of Rn . xj + ε. . y) x ∂z ¿ x+z 1 y y 0 ¿ . . . Solution: ◮ The Jacobi matrix is the 2 × 3 matrix ¾ f ′ (x. . . . Let f ′ (x) = (Jij ). and refer to f ′ when we mean the Jacobi matrix of f . . ε ∈ R. . . xn ) − εDx (f )(→j )|| e |ε| . . . x2 . y) = (xy + yz. . . . . . . . x2 . loge xy). . . y) ∂x ∂f2 (x. xn ) − εJij | |ε| This entails that Jij = lim fi (x1 . Jnj ¿ − Dx (f )(→j ) = e . xj . ε→0 This finishes the proof. x2 . y) = ∂f1 (x. . . ε ∂xj → 0. xn ) − f (x1 . . . . u Strictly speaking. . . x2 . and observe that ¾ J1j J2j . the so does the i-th component of the numerator. . y) y ∂z = 1 ∂f2 (x. is not true. The converse. y = 0. θ) = ρ cos θ cos φ −ρ sin φ sin θ ρ sin θ cos φ −ρ sin φ ρ cos θ sin φ 0 ¿ . xy = 0. φ. 178 Example Let f : R2 → R be given by y f (x. then f is differentiable on A. θ) = (ρ cos θ sin φ. Free to photocopy and distribute 110 . z) = (ρ cos θ. z) be the function which changes from cylindrical coordinates to Cartesian coordinates. ¾ f1 f2 . 0) = 1. We have ∂f ∂f however. 0) = (0. Thus even if both partial derivatives exist at ∂x ∂y (0. y) = 1 for values arbitrarily close to (0. Put f = . We have ¾ cos θ sin φ sin θ sin φ cos φ f ′ (ρ. 0). φ. 0) is no guarantee that the function will be differentiable at (0. ρ sin θ sin φ. θ. ρ cos φ) be the function which changes from spherical coordinates to Cartesian coordinates. You should also notice that both partial derivatives are not continuous at (0. θ. The Jacobi matrix provides a convenient computational tool to compute the derivative of a function at a point. (0. z) = −ρ sin θ ρ cos θ 0 0 ¿ 0 . and hence. however. ¿ 179 Theorem Let A ⊆ Rn be an open set. 0) = 0 but f (x. fm If each of the partial derivatives Dj fi exists and is continuous on A. Thus differentiability at a point implies that the partial derivatives of the function exist at the point.. 0). and let f : Rn → Rm . 1 177 Example Let f (ρ.. it is not differentiable there. 0)). y) = 1 if if x = 0. We have. We have ¾ cos θ sin θ 0 f ′ (ρ. 0) (f (0. x if Observe that f is not continuous at (0. ρ sin θ. the following. however.The Jacobi Matrix 176 Example Let f (ρ. √ e 2 which is at the desired point. ◭ 2 181 Example The equation z xy + (xy)z + xy 2 z 3 = 3 defines z as an implicit ∂z ∂z function of x and y. ∂x ∂x Hence. 1) we have ∂x ∂z ∂x Similarly. Find and at (1. 111 2xyz 3 + 3xy 2 z 2 Free to photocopy and distribute . 1. 180 Example Let f (u. w) = eu v cos w. 1. 1). w) at (1. ∂2 ∂u∂v f (u. π 4 ). −1. ∂ xy z ∂y = = ∂ ∂y (xy)z = = ∂ ∂y xy 2 z 3 = ∂ xy log z e ∂y x log z + ∂ xy ∂z z ∂y +1+1+3 ∂z ∂x = 0 =⇒ ∂z ∂x 1 =− . The following examples should be self-explanatory. ∂x ∂y Solution: ◮ We have ∂ ∂x z xy = = ∂ (xy)z = ∂ xy log z e ∂x xy ∂z y log z + z xy . ∂x x ∂ ∂z xy 2 z 3 = y 2 z 3 + 3xy 2 z 2 . ez log xy ∂y ∂z z log xy + ∂y y (xy)z . Determine Solution: ◮ We have ∂2 ∂u∂v (eu v cos w) = ∂ ∂u (eu cos w) = eu cos w. ∂z ∂y . v. v. z ∂x ∂ ez log xy ∂x ∂z z = log xy + (xy)z . at (1. Similarly with the concept of implicit partial differentiation.Chapter 2 The concept of repeated partial derivatives is akin to the concept of repeated differentiation. 2 z xy . at (1. Let A ⊆ Rn . Free to photocopy and distribute 112 . and h′ (x. 1) we have ∂z ∂y ◭ Just like in the one-variable case. v) = . • Chain Rule: Dx ((h ◦ f )) = Df (x) (h) ◦ (Dx (f )). v) = u + v . (h ◦ f )′ = (h′ ◦ f )(f ′ ). 1.The Jacobi Matrix Hence. we have the following rules of differentiation. y). h(x. (2. 4 182 Example Let ¾ uev uv ¿ f (u. the Chain Rule takes the alternative form when applied to the Jacobi matrix. ey+z 1 y+z (x2 + y)ey+z 1 x +y 2 ¿ f ′ (h(x.1) +1+2+3 ∂z ∂y = 0 =⇒ ∂z ∂y 3 =− . α ∈ R. Then we have • Addition Rule: Dx ((f + αg)) = Dx (f ) + αDx (g). h : B → Rl be differentiable on B. B ⊆ Rm be open sets f. Since composition of linear mappings expressed as matrices is matrix multiplication. y) = Observe also that ¾ 2x 0 1 0 1 1 . y) = Find (f ◦ h)′ (x. and f (A) ⊆ B. be differentiable on A. y)) = . g : A → Rm . Solution: ◮ We have ¾ x2 + y y+z . ev 1 v uev 1 u ¿ f ′ (u. v) = 2u ev . y) v(x. Observe 0 1 x 1 ç . y) = z 0 ä u(x. y) = f u(x. y. z) . y) ∂z = h′ (x. u. y) = y + z. y) = that h = f ◦ g. Now. g ′ (x. y)))(g ′ (x. v(x. v) = u2 + ev . y) ∂x ∂h (x. f (u. Solution: ◮ Put g : R3 → R2 . x2 + 2y + z x2 + y ◭ 183 Example Let f : R2 → R. Find the partial derivatives of h. y))h′ (x. z) v(x. y) = (f ′ (g(x. Put h(x. y = xz y+z .Chapter 2 Hence (f ◦ h)′ (x. f ′ (u. f ′ (h(x. y)) = 2xz Thus å è ä ey+z . y. y) = xz. y)) å ¾ è z ¿ = å 2xz 2xz 2 ey+z 0 1 x 1 è . y) ∂y ∂h (x. g(x. y) ¾ ¿ ey+z = ¾ (x2 + y)ey+z ¾ 2x 1 0 1 1 1 x2 + y 2 ¿ 1 y+z 2xe y+z 0 ¿ y+z (1 + x + y)e 2 (x + y)e 1 2 y+z = 2x 2xy + 2xz . ç ∂h (x. v : R3 → R u(x. y) = f ′ (h(x. 0 ey+z = Free to photocopy and distribute 2x2 z + ey+z 113 . b]. y) = ey+z . (x. The above implies that (x2 − 1) · F ′ (x) = 2x = = = = = Free to photocopy and distribute π/2 0 π/2 0 dθ sin2 θ + x2 cos2 θ 2 x2 cos2 θ + sin θ − 1 sin2 θ + x2 cos2 θ π/2 dθ (x2 − 1) cos2 θ dθ π 2 π − 0 π/2 − 2 tan2 θ + ¬ 2 x 0 π 1 tan θ ¬π/2 − arctan ¬ 2 x x 0 π π − . b] being a closed interval. y) = 2x2 z + ey+z . 2 2x sin2 θ + x2 cos2 θ sec2 θdθ 114 . f (x.y) dy exists as a continuously differentiable function of x on [a. and Y being a closed and bounded ∂ subset of R.The Jacobi Matrix Equating components. Suppose that both f (x. y) are continuous in the vari∂x ables x and y jointly. 184 Theorem (Differentiation under the integral sign) Let f : [a. y) = 2xz 2 . y) dy. Solution: ◮ Differentiating under the integral. with derivative d dx 185 Example Prove that π/2 Y Y (x. b] × Y → R be a function. with [a. F (x) = 0 log(sin2 θ + x2 cos2 θ)dθ = π log x+1 2 . Then f (x. F ′ (x) = π/2 0 ∂ ∂x log(sin2 θ + x2 cos2 θ)dθ cos2 θ sin2 θ + x2 cos2 θ dθ π/2 0 = 2x . ∂x ∂h ∂y ∂h ∂z ◭ Under certain conditions we may differentiate under the integral sign. we obtain ∂h (x. y) and f (x. y) dy = Y ∂ ∂x f (x. Finally x+1 2 . x+1 Since F (1) = thus 0 log 1dθ = 0. F (x) = π log(x + 1) − π log 2 = π log ◭ Under certain conditions. π =⇒ F (x) = π log(x + 1) + C. we gather that C = −π log 2. cos2 θdθ = π 2 For x = 1 one sees immediately that F ′ (1) = 2 agreeing with the formula. we gather that C = 0. F ′ (x) = π/2 π/2 0 . Differentiating x2 both sides with respect to a. F ′ (x) = 2x x2 − 1 π 2 − π 2x = π x+1 . π Since I(0) = 0. +∞ 186 Example Given that 0 sin x x dx = π 2 +∞ . The desired integral is I(1) = .Chapter 2 which in turn implies that for x > 0. 2 π 2 a + C. the interval of integration in the above theorem need not be compact. Solution: ◮ Put I(a) = 0 sin2 ax I ′ (a) = = +∞ 0 +∞ 0 +∞ 2x sin ax cos ax x2 sin 2ax dx x sin u du u dx = = Integrating each side gives I(a) = 0 π . Now. with a ≥ 0. +∞ dx. 2 ◭ Homework Free to photocopy and distribute 115 . and making the substitution u = 2ax. x = 1. compute 0 sin2 x x2 dx. explain. ∂w v = rs. y) = xy é and ¾ g(x.5.5. +∞[→ 2 R. evaluate a a b 0 Problem 2. 1. if at all defined.5.6 Gradients and Directional Derivatives Rn x → Rm A function f : → f (x) is called a vector field.5. Determine n such that ∂f 1 ∂ = 2 ∂t r ∂r Problem 2. Find ∂z at (1. ∂x Problem 2.2 Let f : R2 → R. (x2 + a2 )2 b Problem 2. f (r. f (x. y 2 ). 0. ∂r Problem 2. 1). find . y) = x−y x2 y 2 x+y ¿ x+y Find (g ◦ f )′ (0.5. Problem 2. y) Find and .8 Assuming that the equation xy 2 + 3z = cos z 2 defines z implicitly as a ∂z function of x and y.5.3 Let f : R2 → R2 and g : R3 → R2 be given by dx = x2 + a2 0 dx . it is called a scalar field. if at all defined. y) = g(t) dt. .6 Given that 1 b arctan . t) = tn e−r /4t . x 2 . y) = xy 2 x2 y é æ arctan ax − arctan x π dx = log π. f : R2 → R. 2.Gradients and Directional Derivatives Problem 2. If undefined. y) = min(x. ∂x ∂y Problem 2.4 Let f (x.7 Prove that +∞ 0 é æ f (x. g(x. 1). y) ∂f (x. where n is a constant.5. a ∂f (x.9 If w = euv and u = r + s. If m = 1. z) = x − y + 2z xy Compute (f ◦ g)′ (1. ∂x æ Problem 2.5. 0).1 Let f : [0. 1). 187 Definition Let f : Free to photocopy and distribute Rn x → R → f (x) 116 .5. +∞[×]0. Problem 2.5. If undefined. Compute (g ◦ f )′ (1. determine . explain. f (x. y.5 Suppose g : R → R is continuous and a ∈ R is a constant. Find the partial derivatives with respect to x and y of x2 y r2 ∂f ∂r .10 Let z be an implicitlydefined function of x and y through the equation (x + z)2 + (y + z)2 = 8. Let K ∈ R be a constant. and using the chain rule f ′ (c(t0 ))c′ (t0 ) = 0. Proof: Let c: R t → Rn → c(t) be a curve lying on this surface. Choose t0 so that c(t0 ) = x. Since c′ (t0 ) is tangent to the surface and its dot product with ∇f (x) is 0. Then (f ◦ c)(t0 ) = f (c(t)) = K. and assume that f is differentiable in A. Free to photocopy and distribute 117 . 189 Example Find a unit vector normal to the surface x3 + y 3 + z = 4 at the point (1. 188 Theorem Let A ⊆ Rn be open and let f : A → R be a scalar field. Since |(∇f (x))•(c′ (t0 ))| = ||∇f (x)||||c′ (t0 )|| cos θ. u Let θ be the angle between ∇f (x) and c′ (t0 ). . . Then ∇f (x) is orthogonal to the surface implicitly defined by f (x) = K. ∇f (x) is the direction in which f is changing the fastest. ∂f ∂xn (x) The gradient operator is the operator ¾ ∂ ∂x1 ∂ ∂x2 . 1. 2). . we conclude that ∇f (x) is normal to the surface. The gradient of f is the vector defined and denoted by ¾ ∂f ∂x1 ∂f ¿ (x) (x) . ∂ ∂xn ¿ ∇= .Chapter 2 be a scalar field. which translates to (∇f (x))•(c′ (t0 )) = 0. . ∇f (x) = ∂x2 . y. Free to photocopy and distribute 118 ¾ ¿ . 1. 19 1 √ 19 ◭ 190 Example Find the direction of the greatest rate of increase of f (x. y. z) = x3 + y 3 + z − 4 has gradient ¾ 3x2 3y 2 1 ¿ ∇f (x. z) = x + y 2 − z 2 . 2) is 3 . 2). 1. Normalising this vector we obtain 1 ¾ 3 ¿ √ 19 3 √ . Find the equation of the tangent plane to f at (1. Solution: ◮ The direction is that of the gradient vector. y. 3).Gradients and Directional Derivatives Solution: ◮ Here f (x. z) = xyez at the point (2. z) = ¾ e2 ¿ which at (2. 1. y. 2) becomes 2e2 . Normalising this vector we obtain 2e2 1 1 √ 2 . Here ¾ yez xez xyez ¿ ∇f (x. 5 2 ◭ 191 Example Let f : R3 → R be given by f (x. 2. z) = ¾ ¿ 3 which at (1. y. 2 2 Free to photocopy and distribute 119 . 3). 193 Example If f (x. z) = (x2 . . The equation of the tangent plane is thus 1(x − 1) + 4(y − 2) − 6(z − 3) = 0. y. z) = which is ¾ 1 4 −6 ¿ at (1. y. Now ¾ 1 2y −2z ¿ ∇f (x.Chapter 2 Solution: ◮ A vector normal to the plane is ∇f (1. 2. 3). or x + 4y − 6z = −9. . 2. y 2 . yez ) then divf (x) = 2x + 2y + 2yzez . fn (x) The divergence of f is defined and denoted by divf (x) = ∇•f (x) = ∂f1 ∂x1 (x) + ∂f2 ∂x2 (x) + · · · + ∂fn ∂xn (x) . ◭ 192 Definition Let f : be a vector field with Rn x → Rn → f (x) ¾ ¿ f1 (x) f (x) = f2 (x) . . . z) = (x2 . . 196 Example If f (x. z. . . Then the directional derivative of f in the direction of → at the point x is defined and denoted by − v − D→ f (x) = lim v t→0 − f (x + t→) − f (x) v . y. gi2 . g2 . . w2 ). . y. g(x. z. . 0) then e1 ∂ curl(f.. . . . Then the curl of (g1 . z)) = ∇ × f (x. 3 Free to photocopy and distribute 120 . 0. z. y. Let → ∈ Rn \ { 0 } be such that → + t→ ∈ A for v x v − sufficiently small t ∈ R. y 2 . gn−2 ) ¾ curl(g1 . z. y. w) = (exyz . en ∂ ∂xn g1n (x) g2n (x) . . z) = (ez )i. gn−2 )(x) = det e1 ∂ ∂x1 g11 (x) g21 (x) .Gradients and Directional Derivatives 194 Definition Let gk : Rn → Rn . . . . z) = x3 + y 3 − z 2 in the ¾ ¿ 1 direction of 2 . . 195 Example If f (x. g(n−2)2 (x) ··· ··· ··· . and − − assume that f is differentiable in A. Then the directional derivative of f in the direction of → v → is given by − at the point x − v ∇f (x)•→. 1 ≤ k ≤ n − 2 be vector fields with gi = (gi1 .. . gin ). . w) = (0. 0. g(n−2)n (x) ¿ . g2 . v 198 Theorem Let A ⊆ Rn be open and let f : A → R be a scalar field. 197 Definition Let A ⊆ Rn be open and let f : A → R be a scalar field.. . . . 199 Example Find the directional derivative of f (x. y.. . 0. and assume → − − − − that f is differentiable in A. t − Some authors require that the vector → in definition 197 be a unit vector. g)(x. Let → ∈ Rn \ {0} be such that x + t→ ∈ A v v for sufficiently small t ∈ R. y. . w) = det ∂x1 exyz 0 ¾ ¿ 2 2 e2 ∂ ∂x2 0 0 e3 ∂ ∂x3 0 z e4 ∂ ∂x4 w2 0 = (xz 2 exyz )e4 . yez ) then curlf ((x. y. g(n−2)1 (x) e2 ∂ ∂x2 g12 (x) g22 (x) . Problem 2. Problem 2. Problem 2. y) = ex cos 2y at (0. y.11 Find the angles made by √ the gradient of f (x.6. If ∇f (1. and let U.6.1 Let f (x. 1. z) = and so − ∇f (x. y.6. z)•→ = 3x2 + 6y 2 − 6z.6. 1).2 Let f (x. ¾ Problem 2.1.3 Find the tangent plane to x2 the surface − y 2 − z 2 = 0 at the point 2 (2. 1. y. 4 ) f (1.2). ∇ × φV = φ∇ × V + (∇φ) × V → − 3. 2). ∇ × (∇φ) = 0 4. −1.8 Prove that ∇ • (u × v) = v • (∇ × u) − u • (∇ × v). y. 0) to estimate f (0. 1). z) = 3xy − 9xz 2 + y increases most rapidly at the point (1. v ◭ Homework Problem 2.6. 1).6. y) = x 3 + y at the point (1.6. 1. 1) with the coordinate axes.4 Find the point on the surface x2 + y 2 − 5xy + xz − yz = −3 for which the tangent plane is x − 7y = −6.6. z) = xey z . Problem 2. find D( 3 .5 Find a vector pointing in the direction in which f (x. 0). 5 5 Free to photocopy and distribute 121 . ∇ •(∇ × V) = 0 5. y) in the di− rection of the unit vector → .9 Find the point on the surface 2x2 + xy + y 2 + 4x + 8y − z + 14 = 0 for which the tangent plane is 4x + y − z = 0.6.6 Let D→ f (x.Chapter 2 Solution: ◮ We have ¾ 3x2 3y 2 −2z ¿ ∇f (x. 2) = u → → − − 2 i − j .6. ∇(U•V) = (U•∇)V + (V•∇)U + U × (∇ × V) + +V × (∇ × U) Problem 2. y. = xz z ¿ exy . Problem 2. Problem 2. ∇ •φV = φ∇ •V + V•∇φ 2.7 Use a linear approximation of the function f (x.6. Find (∇f )(2. Problem 2. V : R3 → R3 be vector fields. z) Find (∇ × f )(2. y) denote the u directional derivative of f at (x.10 Let φ : R3 → R be a scalar field. − Problem 2. Prove that 1. 0. 7 Levi-Civitta and Einstein In this section. bj . the ij-th entry (AB)ij of the product of two matrices A ∈ Mm×n (R) and B ∈ Mn×r (R) can be written as n (AB)ij = k=1 Aik Bkj = Ait Btj . →. dl are the components of vectors in R3 . 200 Definition (Einstein’s Summation Convention) In any expression containing subscripted variables appearing twice (and only twice) in any term. Free to photocopy and distribute 122 . the trace tr (A) of a square n matrix A ∈ Mn×n (R) is tr (A) = Att = t=1 Att . we will enclose any terms under consideration with · . 204 Example Using Einstein’s Summation convention. − − − → − → 202 Example Given that ai . →. that if A. 2. 3}. what is the meaning of ai bi ck dk ? Solution: ◮ We have 3 ai bi ck dk ◭ = i=1 ai bi ck dk − − → = → • b ck dk a − − → = →• b a 3 k=1 − − − → − → ck dk = (→• b )(→• d ).Levi-Civitta and Einstein 2. 205 Example Demonstrate. 201 Example Using Einstein’s Summation convention. the subscripted variables are assumed to be summed over. In order to emphasise that we are using Einstein’s convention. B are two n × n matrices. unless otherwise noted. ck . b . we are dealing in the space R3 and so. the dot product of two − − vectors → ∈ Rn and → ∈ Rn can be written as x y →•→ = − − x y n x i yi = i=1 x t yt . then tr (AB) = tr (BA) . d a c respectively. a c 203 Example Using Einstein’s Summation convention. subscripts are in the set {1. via Einstein’s Summation convention. 321 → 123.Chapter 2 Solution: ◮ We have tr (AB) = tr ((AB)ij ) = tr ( Aik Bkj ) = and tr (BA) = tr ((BA)ij ) = tr ( Bik Akj ) = Btk Akt . Atk Bkt . Since 132 → 123. and 213 are odd. The 3! = 6 permutations of the index set {1. and odd if it takes an odd number of transpositions to regain the identity permutation. 213 → 123. 3 207 Example It is easy to see that δik δkj = k=1 δik δkj = δij . 321. 209 Definition (Levi-Civitta’s Alternating Tensor) The symbol εjkl is defined as follows: 0 if {j. 2. the permutations 123 (identity). l} = {1. k. for any other permutation. a Recall that a permutation of distinct objects is a reordering of them. since the indices are dummy variables and can be exchanged. We start with the identity permutation 123 and say it is even. Now. 312 → 132 → 123. 3} can be classified into even or odd. 231. we will say that it is even if it takes an even number of transpositions (switching only two elements in one move) to regain the identity permutation. 3} εjkl = −1 +1 if if 1 j 1 j 2 2 k 3 l 3 l k is an odd permutation is an even permutation Free to photocopy and distribute 123 . and 312 are even. 208 Example We see that 3 3 δij ai bj = i=1 j=1 δij ai bj = k=1 − − → ak bk = →• b . 2. ◭ 206 Definition (Kroenecker’s Delta) The symbol δij is defined as follows: ´ δij = 0 1 if i = j if i = j. the permutations 132. Since 231 → 132 → 123. from where the assertion follows. This means that we have a function A → L (Rn . Then x y z → • (→ × →) = − − − x y z − − xi (→ × →)i y z = xi εikl (yk zl ) . As in the one-variable case. given x1 ∈ Rn .7. the derivative will provide us with information about the extrema. Hence. Rm ) T : . Thus the function Bx 0 : Rn × Rn → L (Rn . we obtain a linear transformation Dx0 (T ) = 2 Dx0 (Dx0 (f )) = Dx0 (f ) from Rn to L (Rn . let us consider the following. x y z 2. x2 ) → Free to photocopy and distribute 124 . Dx0 (f ) is a linear transformation from Rn to Rm . x → Dx (f ) where L (Rn . the cross product can also be expressed. if one subindex is repeated we have εrrs = εrsr = εsrr = 0. Rm ). we have 2 2 Dx0 (f )(x1 ) ∈ L (Rn . if we differentiate T at x0 again. if i = →. ε132 = ε321 = ε213 = −1. using the Levi-Civitta alternating tensor. ε123 = ε231 = ε312 = 1. − − − 212 Example Let →. x2 ) (x1 . det A = εijk a1i a2j a3k . Let A ⊂ Rn and f : A → Rm be differentiable on A.Extrema Also. Again. → be vectors in R3 . Demonstrate that x i yi z j − − − = (→ •→)→ . To define an analogue for the second derivative.8 Extrema We now turn to the problem of finding maxima and minima for vector functions. Dx0 (f )(x1 ))(x2 ) ∈ m R . Hence. this means that given x2 ∈ Rn . → .1 Let →. In particular. then − e3 → × → = ε (a b )→ . Rm ) denotes the space of linear transformations from Rn to Rm . 210 Example Using the Levi-Civitta alternating tensor and Einstein’s summation → − − − − → − → convention. Homework − − − Problem 2. then. We know that for fixed x0 ∈ A. Rm ) 2 Dx0 (f )(x1 . − − − x y jkl k l ej 211 Example If A = [aij ] is a 3×3 matrix. and the “second derivative” will provide us with information about the nature of these extreme points. → be vectors in x y z R3 . Rm ). →. j = →. k = e1 e2 →. when f maps into R. . ∂2f ∂xn ∂x2 (x) (x) ··· ··· . z) = xy 2 z 3 .Chapter 2 is well defined. ∂2f ∂xn ∂x1 (x) (x) ··· (x) 214 Example Let f : R3 → R be given by f (x.. Then ¾ 0 2yz 3 3y 2 z 2 2yz 3 2xz 3 6xyz 2 3y 2 z 2 6xyz 2 6xy 2 z ¿ H(x. and f : A → R be twice differentiable 2 on A.z) f = From the preceding example. if there is some open ball Bx1 (r) on which Free to photocopy and distribute 125 . x2 ) ∈ n n R × R we have 2 2 Dx0 (f )(x1 . If there is some open ball Bx0 (r) on which ∀x ∈ Bx0 (r). . etc. that is. . y. it is a bilinear function. .y. we have the following analogue representation of the second derivative. that is. First. as guaranteed by the following theorem. as the ∂2 ∂2 mixed partial derivatives f = f . and f : A → R be twice differentiable 2 2 on A. . ∂2f ∂xn ∂xn ¿ (x) (x) Hx f = ∂x2 ∂x1 . we notice that the Hessian is symmetric. 213 Theorem Let A ⊆ Rn be an open set. Then the matrix of Dx (f ) : Rn × Rn → R with respect to the standard basis is given by the Hessian matrix: ¾ ∂2f ∂x1 ∂x1 ∂2f (x) (x) ∂2f ∂x1 ∂x2 ∂2f ∂x2 ∂x2 . 215 Theorem Let A ⊆ Rn be an open set. we make some analogous definitions. ∂2f ∂x1 ∂xn ∂2f ∂x2 ∂xn . This is no coinci∂x∂y ∂y∂x dence. are equal. . and f : A → R. We are now ready to study extrema in several variables. x2 ) = Dx0 (f )(x2 . and linear in each variable x1 and x2 . f (x0 ) ≥ f (x). ∀(x1 . 216 Definition Let A ⊆ Rn be an open set. x1 ). The basic theorems resemble those of one-variable calculus. . If Dx0 (f ) is continuous. then Dx0 (f ) is symmetric. Similarly. Just as the Jacobi matrix was a handy tool for finding a matrix representation of Dx (f ) in the natural bases. . we say that f (x0 ) is a local maximum of f . 218 Example Find the critical points of f : R2 → R (x. then Dx0 (f ) = 0. y = − . if f is twice-differentiable with continuous second derivative and x0 is a critical point such that Hx0 f is negative definite. 126 . y) = x2 + xy + y 2 + 2x + 3y. that is. 3 3 H(x. which is positive definite. the test is inconclusive. If x0 is an extreme point. A point is called an extreme point if it is either a local minimum or local maximum. y) → x2 + xy + y 2 + 2x + 3y and investigate their nature. x0 is a critical point. Solution: ◮ We have (∇f )(x. and f : A → R be differentiable on A. − 3 3 − ◭ Free to photocopy and distribute 7 1 4 = f − .Extrema ∀x ∈ Bx0 (r′ ).− 3 3 3 2 1 1 2 = is a relative minimum and we have ≤ f (x. we say that f (x1 ) is a local maximum of f . 217 Theorem Let A ⊆ Rn be an open set.y) f = 2 1 1 2 . A point t is called a critical point if f is differentiable at t and Dt (f ) = 0. If Hx0 f is semi-definite (positive or negative). and so to find the critical points we solve 2x + y + 2 = 0. then f has a local maximum at x0 . Moreover. If Hx0 f is indefinite. x + 2y + 3 = 0. then f has a local minimum at x0 . Now. A critical point which is neither a maxima nor a minima is called a saddle point. since ∆1 = 2 > 0 and ∆2 = det 1 4 3 > 0. Thus x0 = − . If Hx0 f is positive definite. 1 4 which yields x = − . then f has a saddle point. f (x1 ) ≤ f (x). y) = 2x + y + 2 x + 2y + 3 . 0) is a critical point. y) = x3 − y 3 + axy. 0. ¾ 2 2 Hr f = −1 −1 2 1 2 6 ¿ 1 . If a = 0 then 3 3 y2 a 2 = −ay =⇒ =⇒ =⇒ =⇒ 27y 4 = −a3 y y(27y 3 + a3 ) = 0 y(3y + a)(9y 2 − 3ay + a2 ) = 0 a y = 0 or y = − . ∆2 = det ¾ 2 −1 −1 2 = 2 2 3 > 0. Now. 3 127 Free to photocopy and distribute . y. Determine the nature of the critical points of f . 0) and −1 2 1 2 1 6 ¿ = 4 > 0. Thus f has a relative 0 = f (0. which vanishes when x = y = z = 0. y. 3y 2 = ax. then x = y = 0 and so (0. with a ∈ R a parameter. Solution: ◮ We have ¾ (∇f )(x. z) = 2x − y + 2z 6z + 2x + y 2y − x + z ¿ . y) = 3x2 + ay −3y + ax 2 = 0 0 =⇒ 3x2 = −ay.Chapter 2 219 Example Find the extrema of f : R3 (x. Solution: ◮ We have (∇f )(x. z) → → R x2 + y 2 + 3z 2 − xy + 2xz + yz . 0) ≤ f (x. which is positive definite. 0. and ∆3 = det −1 minimum at (0. y. If a = 0. z) = x2 + y 2 + 3z 2 − xy + 2xz + yz. since ∆1 = 2 > 0. ◭ 220 Example Let f (x. Extrema If y = 0 then x = 0, so again (0, 0) is a critical point. If y = − a Now, x= 3 · − a 3 2 a 3 then = a 3 so a 3 ,− a 3 is a critical point. Hf (x,y) = 6x a a −6y =⇒ ∆1 = 6x, ∆2 = −36xy − a2 . At (0, 0), ∆1 = 0, ∆2 = −a2 . If a = 0 then there is a saddle point. a a a a , − , ∆1 = 2a, ∆2 = 3a2 , whence ,− At will be a local 3 3 3 3 minimum if a > 0 and a local maximum if a < 0. ◭ Homework Problem 2.8.1 Determine the nature of the critical points of f (x, y) = y 2 −2x2 y+4x3 + 20x2 . Problem 2.8.2 Determine the nature of the critical points of f (x, y) = (x − 2)2 + 2y 2 . Problem 2.8.3 Determine the nature of the critical points of f (x, y) = (x − 2)2 − 2y 2 . Problem 2.8.4 Determine the nature of the critical points of f (x, y) = x4 + 4xy − 2y 2 . Problem 2.8.5 Determine the nature of the critical points of f (x, y) = x4 + y 4 − 2x2 + 4xy − 2y 2 . Problem 2.8.6 Determine the nature of the critical points of f (x, y, z) = x2 + y 2 + z 2 − xy + x − 2z. Problem 2.8.7 Determine the nature of the critical points of f (x, y) = x4 + y 4 − 2(x − y)2 . Problem 2.8.8 Determine the nature of the critical points of f (x, y, z) = 4x2 z − 2xy − 4x2 − z 2 + y. Problem 2.8.9 Find the extrema of f (x, y, z) = x2 + y 2 + z 2 + xyz. Problem 2.8.10 Find the extrema f (x, y, z) = x2 y + y 2 z + 2x − z. of Problem 2.8.11 Determine the nature of the critical points of f (x, y, z) = 4xyz − x4 − y 4 − z 4 . Problem 2.8.12 Determine the nature of the critical points of f (x, y, z) = xyz(4 − x − y − z). Problem 2.8.13 Determine the nature of the critical points of g(x, y, z) = xyze−x 2 −y 2 −z 2 . x 2 +y Problem 2.8.14 Let f (x, y) = y 2 −x g(t)dt, where g is a continuously differentiable function defined over all real numbers and g(0) = 0, g ′ (0) = 0. Prove that (0, 0) is a saddle point for f . Problem 2.8.15 Find the minimum of F (x, y) = (x−y)2 + √ Ô 144 − 16x2 − 4 − y2 3 2 , for −3 ≤ x ≤ 3, −2 ≤ y ≤ 2. Free to photocopy and distribute 128 Chapter 2 2.9 Lagrange Multipliers In some situations we wish to optimise a function given a set of constraints. For such cases, we have the following. 221 Theorem Let A ⊆ Rn and let f : A → R, g : A → R be functions whose respective derivatives are continuous. Let g(x0 ) = c0 and let S = g −1 (c0 ) be the level set for g with value c0 , and assume ∇g(x0 ) = 0. If the restriction of f to S has an extreme point at x0 , then ∃λ ∈ R such that ∇f (x0 ) = λ∇g(x0 ). The above theorem only locates extrema, it does not say anything concerning the nature of the critical points found. 222 Example Optimise f : R2 → R, f (x, y) = x2 − y 2 given that x2 + y 2 = 1. Solution: ◮ Let g(x, y) = x2 + y 2 − 1. We solve ∇f for x, y, λ. This requires 2x −2y = 2xλ 2yλ . x y = λ∇g x y From 2x = 2xλ we get either x = 0 or λ = 1. If x = 0 then y = ±1 and λ = −1. If λ = 1, then y = 0, x = ±1. Thus the potential critical points are (±1, 0) and (0, ±1). If x2 + y 2 = 1 then f (x, y) = x2 − (1 − x2 ) = 2x2 − 1 ≥ −1, and f (x, y) = 1 − y 2 − y 2 = 1 − 2y 2 ≤ 1. Thus (±1, 0) are maximum points and (0, ±1) are minimum points. ◭ 223 Example Find the maximum and the minimum points of f (x, y) = 4x + 3y, subject to the constraint x2 + 4y 2 = 4, using Lagrange multipliers. Solution: ◮ Putting g(x, y) = x2 + 4y 2 − 4 we have ∇f (x, y) = λ∇g(x, y) =⇒ Free to photocopy and distribute 4 3 =λ 2x 8y . 129 Lagrange Multipliers Thus 4 = 2λx, 3 = 8λy. Clearly then λ = 0. Upon division we find x 16 = . Hence y 3 x2 + 4y 2 = 4 =⇒ 256 16 3 y 2 + 4y 2 = 4 =⇒ y = ± √ , x = ± √ 9 73 73. The maximum is clearly then 16 3 4 √ +3 √ 73 73 √ and the minimum is − 73. ◭ = √ 73, 224 Example Let a > 0, b > 0, c > 0. Determine the maximum and minimum x y z x2 y2 z2 values of f (x, y, z) = + + on the ellipsoid 2 + 2 + 2 = 1. a b c a b c Solution: ◮ We use Lagrange multipliers. Put g(x, y, z) = z 2 x2 a2 + y2 b2 + c2 − 1. Then ¾ 1/a 1/b 1/c ¿ ¾ 2x/a2 2z/c2 ¿ ∇f (x, y, z) = λ∇g(x, y, z) ⇐⇒ a = λ 2y/b2 . It follows that λ = 0. Hence x = b c x2 y2 ,y = ,z = . Since 2 + 2 + 2λ 2λ a b √ 2λ z2 3 3 = 1, we deduce = 1 or λ = ± . Since a, b, c are positive, f c2 4λ2 2 will have a maximum when all x, y, z are positive and a minimum when all x, y, z are negative. Thus the maximum is when a b c x = √ ,y = √ ,z = √ , 3 3 3 and √ 3 f (x, y, z) ≤ √ = 3 3 and the minimum is when a b c x = −√ , y = −√ , z = −√ , 3 3 3 and √ 3 f (x, y, z) ≥ − √ = − 3. 3 130 Free to photocopy and distribute Chapter 2 Aliter: Using the CBS Inequality, ¬ ¬ ¬x ¬ y z ¬ ·1+ · 1 + · 1¬ ≤ ¬a ¬ b c x2 a2 + y2 b2 + z2 c2 1/2 12 + 12 + 12 1/2 √ √ x y z = (1) 3 =⇒ − 3 ≤ + + ≤ a b c ◭ 225 Example Let a > 0, b > 0, c > 0. Determine the maximum volume of the parallelepiped with sides parallel to the axes that can be enclosed inside the y2 z2 x2 ellipsoid 2 + 2 + 2 = 1. a b c ◮ Let 2x, 2y, 2z, be the dimensions of the box. We must x2 maximise f (x, y, z) = 8xyz subject to the constraint g(x, y, z) = 2 + a y2 z2 + 2 − 1. Using Lagrange multipliers, b2 c Solution: ¾ 8yz 8xz 8xy ¿ ¾ 2x/a2 2z/c 2 ¿ ∇f (x, y, z) = λ∇g(x, y, z) ⇐⇒ = λ 2y/b2 =⇒ 4yz = λ x a2 , 4xz = λ y b2 , 4xy = λ z c2 . Multiplying the first inequality by x, the second by y, the third by z, and adding, 4xyz = λ Hence x2 a , 4xyz = λ 2 y2 z2 , 4xyz = λ 2 , =⇒ 12xyz = λ b2 c =λ x2 =λ y2 =λ z2 x2 a2 + y2 b2 + z2 c2 = λ. . 3 a2 b2 c2 If λ = 0, then 8xyz = 0, which minimises the volume. If λ = 0, then a x= √ , 3 and the maximum value is b y= √ , 3 abc 8xyz ≤ 8 √ . 3 3 Aliter: Using the AM-GM Inequality, (x2 y 2 z 2 )1/3 = (abc)2/3 ◭ x 2 λ c z= √ , 3 a2 · y 2 b2 · z 2 1/3 x2 ≤ (abc)2/3 · a 2 + y2 b2 3 + z2 8 c2 = 1 =⇒ 8xyz ≤ √ (abc). 3 3 3 a2 Homework Free to photocopy and distribute 131 Lagrange Multipliers Problem 2.9.1 A closed box (with six outer faces), has fixed surface area of S square units. Find its maximum volume using Lagrange multipliers. That is, subject to the constraint 2ab + 2bc + 2ca = S, you must maximise abc. Problem 2.9.2 Consider the problem of finding the closest point P ′ on the plane Π : ax + by + cz = d, a, b, c non-zero constants with a + b + c = d to the point P (1, 1, 1). In this problem, you will do this in three essentially different ways. 1. Do this by a geometric argument, arguing the the point P ′ closest to P on Π is on the perpendicular passing through P and P ′ . 2. Do this by means of Lagrange multipliers, by minimising a suitable function f (x, y, z) subject to the constraint g(x, y, z) = ax + by + cz − d. 3. Do this considering the unconstrained extrema of a suitable funcd − ax − by tion h x, y, . c Problem 2.9.3 Given that x, y are positive real numbers such that x4 +81y 4 = 36 find the maximum of x + 3y. Problem 2.9.4 If x, y, z are positive real 1 numbers such that x2 y 3 z = 2 , what is 6 the minimum value of f (x, y, z) = 2x + 3y + z? Problem 2.9.5 Find the maximum and the minimum values of f (x, y) = x2 + y 2 subject to the constraint 5x2 + 6xy + 5y 2 = 8. Problem 2.9.6 Let a > 0, b > 0, p > 1. Maximise f (x, y) = ax + by subject to the constraint xp + y p = 1. Problem 2.9.7 Find the extrema of f (x, y, z) = x2 + y 2 + z 2 subject to the constraint (x − 1)2 + (y − 2)2 + (z − 3)2 = 4. abc3 ≤ 27 Problem 2.9.8 Find the axes of the ellipse 5x2 + 8xy + 5y 2 = 9. Problem 2.9.9 Optimise f (x, y, z) = x + y + z subject to x2 + y 2 = 2, and x + z = 1. Problem 2.9.10 Let x, y be strictly positive real numbers with x + y = 1. What is the √ maximum value of x + xy? Problem 2.9.11 Let a, b be positive real constants. Maximise f (x, y) = xa e−x y b e−y on the triangle in R2 bounded by the lines x ≥ 0, y ≥ 0, x + y ≤ 1. Problem 2.9.12 Determine the extrema of f (x, y) = cos2 x+cos2 y subject to the conπ straint x − y = . 4 Problem 2.9.13 Determine the extrema of f (x, y, z) = x − 2y + 2z subject to the constraint x2 + y 2 + z 2 = 1. Problem 2.9.14 Find the points on the curve determined by the equations x2 + xy + y 2 − z 2 = 1, x2 + y 2 = 1 which are closest to the origin. Problem 2.9.15 Does there exist a polynomial in two variables with real coefficients p(x, y) such that p(x, y) > 0 for all x and y and that for all real numbers c > 0 there exists (x0 , y0 ) ∈ R2 such that p(x0 , y0 ) = c? Problem 2.9.16 Maximise f (x, y, z) = log x + log y + 3 log z on the portion of sphere x2 + y 2 + z 2 = 5r 2 which lies on the first octant. Demonstrate using this that for any positive real numbers a, b and c, there follows the inequality a+b+c 5 5 . Free to photocopy and distribute 132 1 1 scalar homogeneity: if λ ∈ R. linearity: d(a + b) = da + db. a2 . . . ak ) is the xj1 xj2 . . . Moreover.1 Differential Forms We will now consider integration in several variables. j2 . . anj ¿ aj = ∈ Rn . and let ¾ a1j a2j . . By virtue of being a determinant. 2. . x n in n-dimensional space (used as the names of the axes). . . we need to consider the concept of differential forms. . 1 ≤ j ≤ k. . . a2 . 133 . . . then dλa = λda. In order to smooth our discussion.3 Integration 3. . . be k ≤ n vectors in Rn . associativity: (da ∧ db) ∧ dc = da ∧ (db ∧ dc). . jk } ⊆ {1. . ajk 2 dxj1 ∧ dxj2 ∧ · · · ∧ dxjk (a1 . . ak . . . . . . . xjk component of the signed k-volume of a k-parallelotope in Rn spanned by a1 . . . dxj1 ∧ dxj2 ∧ · · · ∧ dxjk (a1 . . ajk k In other words. . x2 . . a2 . Notice that associativity does not hold for the wedge product of vectors. the wedge product ∧ of differential forms has the following properties anti-commutativity: da ∧ db = −db ∧ da. 1 ≤ j ≤ k is defined and denoted by ¾ aj1 1 aj2 1 . ajk 1 aj1 2 aj2 2 . ak ) = det ··· ··· ··· ··· aj1 k ¿ aj2 k . . . let {j1 . . n} be a collection of k sub-indices. 226 Definition Consider n variables x1 . . . An elementary k-differential form (k > 1) acting on the vectors aj . y... 229 Example In R3 we have dx ∧ dz + 5dz ∧ dx + 4dx ∧ dy − dy ∧ dx + 12dx ∧ dx = −4dx ∧ dz + 5dx ∧ dy.jk dxj1 ∧ dxj2 ∧ · · · dxjk . 231 Definition A k-differential form field in Rn is an expression of the form ω= 1≤j1 ≤j2 ≤···≤jk ≤n aj1 j2 . the length) of the projections of a onto the coordinate axes.Differential Forms Anti-commutativity yields da ∧ da = 0. w) = x + y 2 + z 3 + w4 is a 0-form in R4 . dy(a) = det(0) = 0. 232 Example g(x.. 228 Example In R3 we have dx ∧ dy ∧ dx = 0. 230 Definition A 0-differential form in Rn is simply a differentiable function in Rn . Free to photocopy and distribute 134 . are the (signed) 1-volumes (that is.. since we have a repeated variable. we will simplify the summands and express the other differential forms in terms of the one differential form whose indices appear in increasing order. dx(a) = det(1) = 1. In order to avoid redundancy we will make the convention that if a sum of two or more terms have the same differential form up to permutation of the variables.jk are differentiable functions in Rn . 227 Example Consider a= Then ¾ 1 0 −1 ¿ ∈ R3 . where the aj1 j2 . 233 Example An example of a 1-form field in R3 is ω = xdx + y 2 dy + xyz 3 dz. z. dz(a) = det(−1) = −1. . . . x2 . the exterior derivative dω of ω is the (k + 1)-form dω = df (x1 . y = uv. x2 . . is ω2 = −2xdx + 2ydy. then d(x3 y 4 ) = 3x2 y 4 dx + 4x3 y 3 dy. xn ) be a 0-form in Rn . .Chapter 3 234 Example An example of a 2-form field in R3 is ω = x2 dx ∧ dy + y 2 dy ∧ dz + dz ∧ dx. 235 Example An example of a 3-form field in R3 is ω = (x + y + z)dx ∧ dy ∧ dz. . 239 Example If in R2 . . Furthermore. 238 Example If in R2 . xn )dxj1 ∧ dxj2 ∧ · · · ∧ dxjk 240 Example Consider the change of variables x = u + v. ω = x3 y 4 . . x2 . . Free to photocopy and distribute 135 . Then dx = du + dv. dy = vdu + udv. We shew now how to multiply differential forms. ω1 ∧ ω2 = (2x2 + 2y 2 )dx ∧ dy. . xn ) ∧ dxj1 ∧ dxj2 ∧ · · · ∧ dxjk . 237 Definition Let f (x1 . if is a k-form in Rn . ω = x2 ydx + x3 y 4 dy then dω = = = = d(x2 ydx + x3 y 4 dy) (2xydx + x2 dy) ∧ dx + (3x2 y 4 dx + 4x3 y 3 dy) ∧ dy (3x2 y 4 − x2 )dx ∧ dy x2 dy ∧ dx + 3x2 y 4 dx ∧ dy ω = f (x1 . The exterior derivative df of f is n df = i=1 ∂f ∂xi dxi . . 236 Example The product of the 1-form fields in R3 ω1 = ydx + xdy. whence dx ∧ dy = (u − v)du ∧ dv. . 3) − ω(0. −{(1. y+z x+y+z Then du = dx + dy + dz. 0. 0)} with −(1. 0)) + ω(1. 0. (y + z)2 (x + y + z)2 − dx ∧ dy ∧ dz = 3. Then ω = −ω((1. the second means that (−1. Free to photocopy and distribute 136 . and let ω be a 0-form.Zero-Manifolds 241 Example Consider the transformation of coordinates xyz into uvw coordinates given by z y+z u = x + y + z. x (x + y + z)2 dz. The first one means that the point (1. 0)}2 be an oriented 0-manifold. 0). 0. 0. say. Then M ω = ω(b) − ω(a). 2. 0. 3) = −(1) + (14) − (−4) = 17. 0. 243 Definition Let M = +{b} ∪ −{a} be an oriented 0-manifold. 0. 2. with a choice of the + or − sign. 0) is the additive inverse of (1. +x 244 Example Let M = −{(1. dv = − dw = − y+z (x + y + z)2 z (y + z)2 dy + x (x + y + z)2 y (y + z)2 dz. 0. M 2 Do not confuse. dx + dy + Multiplication gives du ∧ dv ∧ dw = zx y(y + z) − (y + z)2 (x + y + z)2 (y + z)2 (x + y + z)2 z(y + z) xy + − 2 (x + y + z)2 2 (x + y + z)2 (y + z) (y + z) z 2 − y 2 − zx − xy dx ∧ dy ∧ dz.2 Zero-Manifolds 242 Definition A 0-dimensional oriented manifold of Rn is simply a point x ∈ Rn . w= . 3)} ∪ −{(0. 0) is given negative orientation. v = . 0). −2. 0)} ∪ +{(1. 0. −x has opposite orientation to +x and −x ω=− ω. and let ω = x + 2y + z 2 . A general oriented 0-manifold is a union of oriented points. 0) = (−1. whence ω Γ 2 = å −1 (3t3 + 2t2 )dt 3 4 è2 = = 2 3 69 . Then xydx + (x + y)dy = t3 dt + (t + t2 )dt2 = (3t3 + 2t2 )dt. 9]. 4 t3 + t4 −1 What would happen if we had given the curve above a different parametrisation? First observe that the curve travels from (−1. Γ æ é dx → − − If f : R2 → R2 and if d→ = r . Then xydx + (x + y)dy = = = √ √ √ √ √ ( t − 1)3 d( t − 1) + (( t − 1) + ( t − 1)2 )d( t − 1)2 √ √ √ (3( t − 1)3 + 2( t − 1)2 )d( t − 1) √ √ 1 √ (3( t − 1)3 + 2( t − 1)2 )dt.√These conditions are met with the parametrisation √ x = t − 1. 4) on the parabola y = x2 . 246 Example Calculate xydx + (x + y)dy Γ where Γ is the parabola y = x2 .3 One-Manifolds 245 Definition A 1-dimensional oriented manifold of Rn is simply an oriented smooth curve Γ ∈ Rn . y = ( t − 1)2 . or with a choice of a − sign if the curve traverses in the direction of decreasing t. with a choice of a + orientation if the curve traverses in the direction of increasing t. Solution: ◮ We parametrise the curve as x = t. r Γ We now turn to the problem of integrating 1-forms. y = t2 . The curve −Γ has opposite orientation to Γ and −Γ ω=− ω. the classical way of writing this is dy → − − f • d→ . 1) to (2. x ∈ [−1. A general oriented 1-manifold is a union of oriented curves. t ∈ [0. 2 t 137 Free to photocopy and distribute .Chapter 3 3. 2] oriented in the positive direction. 4 9 = as before.x+y> ). 1).One-Manifolds whence ω Γ √ √ 1 √ (3( t − 1)3 + 2( t − 1)2 )dt 2 æ0 2 t 3/2 é9 √ 3t 7t 5t = − + − t 4 3 2 0 69 = . ◭ It turns out that if two different parametrisations of the same curve have the same orientation. −1) in the direction ABCDA. Γ where Γ is the line segment from (0. 0) to (1. t=-1. so we choose the parametrisation x = y = t.x*cos(y)> ).1>)). The integral is thus 1 y sin xdx + x cos ydy Γ = 0 (t sin t + t cos t)dt [t(sin x − cos t)]1 − 0 2 sin 1 − 1. (sin t − cos t)dt To solve this problem using Maple you may use the code below. Hence. then their integrals are equal. and D = (1. > with(Student[VectorCalculus]): > LineInt( VectorField( <x*y. C = (−1. Path( <t. > with(Student[VectorCalculus]): > LineInt( VectorField( <y*sin(x).<1. 1) in the positive direction. 1).0>. Solution: ◮ This line has equation y = x. we only need to worry about finding a suitable parametrisation.. B = (−1.tˆ2>. −1). 1 0 = = upon integrating by parts. Line(<0. 247 Example Calculate the line integral y sin xdx + x cos ydy. Free to photocopy and distribute 138 .2)). To solve this problem using Maple you may use the code below. ◭ 248 Example Calculate the path integral x+y Γ x2 + y 2 dy + x2 + y 2 x−y dx around the closed square Γ = ABCD with A = (1. y = 3 sin t. ◭ When the integral is along a Á closed path.xˆ2> ). and on DA. 2π]. y = −1. To solve this problem using Maple you may use the code below. on BC. on CD. x = 1.(x-y)/(xˆ2+yˆ2)> ). t ∈ [0.<1.<1. 249 Example Calculate the path integral Á x2 dy + y 2 dx.-1>. the area enclosed by the curve is to the left of the curve. y = 1. dy = 0. like in the preceding example. To solve this problem using Maple you may use the code below. x = −1.-1>. so this parametrisation traverses the curve in the positive sense. > with(Student[VectorCalculus]): > LineInt( VectorField( <(x+y)/(xˆ2+yˆ2). > with(Student[VectorCalculus]): > LineInt( VectorField( <yˆ2. The positive direction of integration is that sense that when traversing the path.<-1. Solution: ◮ Parametrise the ellipse as x = 2 cos t. Observe that when traversing this closed curve.1>)). dx = 0. We have Á 2π 2 2 ω Γ = 0 2π ((4 cos2 t)(3 cos t) + (9 sin t)(−2 sin t))dt (12 cos3 t − 18 sin3 t)dt = 0 = 0.1>.Chapter 3 Solution: ◮ On AB. dx = 0. rather than Γ Γ it is customary to use the symbol . > LineSegments(<1. the area of the ellipse is on the left hand side of the path.<-1.1>. ◭ Free to photocopy and distribute 139 .Ellipse(9*xˆ2 + 4*yˆ2 -36)). dy = 0. Γ where Γ is the ellipse 9x + 4y = 36 traversed once in the positive sense. The integral is thus ω Γ = = AB −1 1 ω+ x2 1 −1 ω+ BC ω+ CD −1 1 ω DA 1 −1 x−1 = 4 +1 1 dx + dx y2 y−1 +1 dy + x+1 x2 +1 1 dx + −1 y+1 y2 + 1 dy x2 + 1 = 4 arctan x|1 −1 = 2π. -1>. −2). > with(Student[VectorCalculus]): > PathInt( x. [x. and LBC : y = −4x + 6.One-Manifolds 250 Definition Let Γ be a smooth curve.2> ) ).<1. 3 3 and on LBC √ 2 x||dx|| = x (dx)2 + (dy)2 = x( 1 + (−4) )dx = x 17dx. 251 Example Find Γ x||dx|| where Γ is the triangle starting at A : (−1. ◭ Homework Free to photocopy and distribute 140 . On LAB 3 √ x 10dx 1 2 dx = x||dx|| = x (dx)2 + (dy)2 = x 1 + − . The integral f (x)||dx|| Γ is called the path integral of f along Γ. and ending in C : (1. Solution: ◮ The lines passing through the given points have equations −x − 4 LAB : y = .y]=LineSegments( <-1. Hence x||dx|| Γ = LAB 2 x||dx|| + x||dx|| = = LBC √ 1 √ x 10dx + x 17dx 3√ −1 2 √ 10 3 17 − . 2 2 To solve this problem using Maple you may use the code below.-2>. −1) to B : (2. 2). <2. Proof: We will prove this by induction on p. 4 sin π and continuing on an 6 6 ¡ arc of a circle to B 4 cos π . start- where Γ is the intersection of the sphere x2 + y 2 + z 2 = 1 and the plane x + y = 1.3 Find Γ the semicircle that starts at (−1. xy||dx|| using this parametriProblem 3. 0) goes to (0. .3. C x||dx|| where Γ is the path shewn in figure 3. traversed in the positive direction.2: Problems 3. 3.3 2 Chapter 3 1 -3 -2 -1 -1 1 2 3 -2 -3 Figure 3. Calculate also parametrisation. 0) goes to (0.3.3. . 0) going on a straight line to ¡ A 4 cos π .1 Consider C xdx + ydy and xy||dx||. 4 sin π . . xn ) then dω = k=1 n ∂f ∂xk dxk 141 Free to photocopy and distribute . .4 Closed and Exact Forms ´ 252 Lemma (Poincare Lemma) If ω is a p-differential form of continuously differentiable functions in Rn then d(dω) = 0.1: Example 251.3.3. C ing at O(0. 1) and ends at (1. 2. Evaluate C C A O xy||dx|| using this Figure 3. 5 5 B 1.3. 0). 1) and ends at (1. Evaluate C xdx + ydy where C is the straight line path that starts at (−1.2 Find Γ xdx + ydy where Γ is the path shewn in figure 3. For p = 0 if ω = f (x1 . Calculate also sation. xdx + ydy where C is Problem 3. x2 .2 and 3.2. Problem 3. by parametrising this path.3. 0). by parametrising this path.2.4 Find Á zdx + xdy + ydz Γ Problem 3. jp ∧ dxj1 ∧ dxj2 ∧ · · · dxjp n i=1 = 1≤j1 ≤j2 ≤···≤jp ≤n ∂aj1 j2 .jp are continuous differentiable functions in Rn ..Closed and Exact Forms and n d(dω) = k=1 n d ∂f ∂xk n j=1 ∧ dxk ∂2f ∧ dxj − ∧ dxk dxj ∧ dxk = k=1 n ∂xj ∂xk ∂2f ∂xj ∂xk = 1≤j≤k≤n ∂2f ∂xk ∂xj = 0... But d da ∧ dxj1 ∧ dxj2 ∧ · · · dxjp = = dda ∧ dxj1 ∧ dxj2 ∧ · · · dxjp da ∧ d dxj1 ∧ dxj2 ∧ · · · dxjp .. we have dω = 1≤j1 ≤j2 ≤···≤jp ≤n daj1 j2 . Since such a form can be written as ω= 1≤j1 ≤j2 ≤···≤jp ≤n aj1 j2 . But an independent induction argument proves that d dxj1 ∧ dxj2 ∧ · · · dxjp = 0. since ω is continuously differentiable and so the mixed partial derivatives are equal. +da ∧ d dxj1 ∧ dxj2 ∧ · · · dxjp since dda = 0 from the case p = 0. where the aj1 j2 . p > 0. Consider now an arbitrary p-form. u 253 Definition A differential form ω is said to be exact if there is a continuously differentiable function F such that dF = ω.jp ∂xi dxi ∧ dxj1 ∧ dxj2 ∧ · · · dxjp .jp dxj1 ∧ dxj2 ∧ · · · dxjp .. it is enough to prove that for each summand d da ∧ dxj1 ∧ dxj2 ∧ · · · dxjp = 0.. completing the proof... Free to photocopy and distribute 142 . since x x2 + y dx + 2 y x2 + y dy = d 2 1 2 loge (x2 + y 2 ) . if dω = 0 on a star-shaped domain then ω is exact. By the Chain Rule dF = This demands that ∂F ∂x ∂F ∂x = dx + ∂F ∂y dy. Let ω = dF be an exact form. 256 Example The differential form x x2 + y 2 is exact. 257 Example Determine whether the differential form ω= is exact. Solution: ◮ Assume there is a function F such that dF = ω. 2x(1 − ey ) ey dx + dy (1 + x2 )2 1 + x2 2x(1 − ey ) (1 + x2 )2 ey 1 + x2 . ∂F ∂y Free to photocopy and distribute 143 . since ydx + xdy = d (xy) .Chapter 3 254 Example The differential form xdx + ydy is exact. ´ dω = ddF = 0. that is. By the Poincare Lemma Theorem 252. A result of Poincare says that for certain domains (called star´ shaped domains) the converse is also true. since xdx + ydy = d 255 Example The differential form ydx + xdy is exact. = . dx + y x2 + y 2 dy 1 2 (x2 + y 2 ) . . Since integrating the second expression (with respect to y) is easier. 144 Free to photocopy and distribute . +(2yz 3 dx + 2xz 3 dy + 6xyz 2 dz) ∧ dy = 0. we find 1 + c.Closed and Exact Forms We have a choice here of integrating either the first. 1 + x2 ey − 1 where c is a constant. z). = 2xyz 3 =⇒ F = xy 2 z 3 + b(x. We then take F (x. z). F (x. we differentiate the obtained expression for F with respect to x and find ∂F 2xey =− + φ′ (x). y) = 1 + x2 where φ(x) is a function depending only on x. ∂F ∂x ∂F ∂y = y 2 z 3 =⇒ F = xy 2 z 3 + a(y. ∂x (1 + x2 )2 Comparing this with our first expression for φ′ (x) = that is φ(x) = − 2x (1 + x2 )2 ∂F ∂x . or the second expression. +(3y 2 z 2 dx + 6xyz 2 dy + 6xy 2 zdz) ∧ dz so this form is exact in a star-shaped domain. y) = ◭ 258 Example Is there a continuously differentiable function such that dF = ω = y 2 z 3 dx + 2xyz 3 dy + 3xy 2 z 2 dz ? Solution: ◮ We have dω = (2yz 3 dy + 3y 2 z 2 dz) ∧ dx 1 + x2 + c. So put dF = Then ∂F ∂x dx + ∂F ∂y dy + ∂F ∂z dz = y 2 z 3 dx + 2xyz 3 dy + 3xy 2 z 2 dz. To find it. we find ey + φ(x). Chapter 3 ∂F ∂z = 3xy 2 z 2 =⇒ F = xy 2 z 3 + c(x. By virtue of Theorem 259. Assume ω = dF is an exact form. y. Then B ω= Γ A dF = F (B) − F (A). y). Comparing these three expressions for F . ◭ We have the following equivalent of the Fundamental Theorem of Calculus. where Γ is a loop of x3 + y 3 − 2xy = 0 traversed once in the positive sense. the integral is 0. 0). traversed in the order ABCDEA. Solution: ◮ Observe that d 2x x2 + y 2 dx + 2y x2 + y 2 dy =− 4xy (x2 + y 2 )2 dy∧dx− 4xy (x2 + y 2 )2 dx∧dy = 0. Á In particular. 2). if Γ is a simple closed path. and so the form is exact in a start-shaped domain. E = (1. the form is exact. then ω = 0. 2). D = (3. ◭ Free to photocopy and distribute 145 . 1). B = (5. Solution: ◮ Since ∂ ∂y (x2 − y) = −1 = ∂ ∂x (y 2 − x). C = (7. we obtain F (x. z) = xy 2 z 3 . and since this is a closed simple path. Γ 260 Example Evaluate the integral Á Γ 2y 2x dx + 2 dy x2 + y 2 x + y2 where Γ is the closed polygon with vertices at A = (0. 259 Theorem Let U ⊆ Rn be an open set. the integral is 0. ◭ 261 Example Calculate the path integral Á Γ (x2 − y)dx + (y 2 − x)dy. 0). and Γ a path in U with starting point A and endpoint B. we need the following. the integral f dA D is the sum of all the values of f restricted to D. In particular.5 Two-Manifolds 262 Definition A 2-dimensional oriented manifold of R2 is simply an open set (region) D ∈ R2 . y)dx dy Fubini’s Theorem allows us to convert the double integral into iterated (single) integrals. This corresponds to using the area form dxdy. The region −D has opposite orientation to D and ω=− ω. Given a function f : D → R. Then b d d b f dA = D a c f (x. Let D ⊆ R2 . d]. y)dy dx = c a f (x. y)dA D where dA denotes the area element. 263 Theorem (Fubini’s Theorem) Let D = [a. where the + orientation is counter-clockwise and the − orientation is clockwise.Two-Manifolds 3. D −D We will often write f (x. we will choose the positive orientation for the regions considered. In order to evaluate double integrals. and let f : A → R be continuous. A general oriented 2-manifold is a union of open sets. b] × [c. dA D is the area of D. Free to photocopy and distribute 146 . In this section. unless otherwise noticed. [x..1)). 4 Notice that if we had integrated first with respect to x we would have obtained the same result: = 3 2 1 3 xydx dy 0 = 2 3 x2 y 2 y 2 3 2 1 dy 0 = 2 dx = = y2 4 .y]=Region(0.3] 0 1 2 xydy dx xy 2 2 9x 2 1 3 = 0 1 dx 2 = 0 2 − 2x dx 5x 4 0 5 = ..3))..Chapter 3 264 Example 1 3 xydA = [0. 5 4 Also.. [x.y]=Region(3. > with(Student[VectorCalculus]): > int(x*y.4.3] xdx 1 2 0 ydy 2 = = 5 4 5 2 To solve this problem using Maple you may use the code below.1]×[2.1]×[2. 265 Example We have 4 3 1 4 1 0 (x + 2y)(2x + y) dxdy 0 = 3 4 (2x2 + 5xy + 2y 2 ) dxdy dy = = 409 12 3 2 5 + y + 2y 2 3 2 .2.1.0. Free to photocopy and distribute 147 . this integral is “factorable into x and y pieces” meaning that 1 3 xydA = [0. > with(Student[VectorCalculus]): > int((x + 2*y)*(2*x + y). To solve this problem using Maple you may use the code below. <1.2>).4. 266 Example Find D xy dxdy in the triangle with vertices A : (−1. one variable is kept constant. −2). B : (2.<2. 2). C : (9. ◭ 267 Example Consider the region inside the parallelogram P with vertices at A : (6. P Solution: ◮ The lines joining the points have equations LAB : y = x . as in figure 3.-1>. The integral is thus −1 −2 (6−y)/4 2 (6−y)/4 y −4−3y x dx dy + −1 y (2y−1)/3 x dx dy = − 11 8 .5. 2 LBC : y = 2x − 12. we must divide the region as in figure 3. we decompose so that. The lower point of the dashed line is (1. Free to photocopy and distribute 148 . Solution: ◮ The lines passing through the given points have equations −x − 4 3x + 1 LAB : y = . because there are two upper lines which the upper value of y might be. 3). we must divide the region as in figure 3. If we integrate first with respect to x. LBC : y = −4x + 6. Find xy dxdy. [x. D : (7.-2>. The right point of the dashed line is (7/4. we 3 2 draw the region carefully. > with(Student[VectorCalculus]): > int(x*y. Now.3. If we integrate first with respect to y. −1). C : (1. −1). −5/3). 6). 5). To solve this problem using Maple you may use the code below.Two-Manifolds In the cases when the domain of integration is not a rectangle. The integral is thus 1 (3x+1)/2 2 x −1 (−x−4)/3 y dy dx + 1 x −4x+6 (−x−4)/3 y dy dx = − 11 8 . LCA : y = . because there are two left-most lines which the left value of x might be. B : (8.y]=Triangle(<-1. 4). Notice that we have split the parallelogram into two triangles.5>)) > + int(x*y. x2 + y 2 ≤ 1}.<7. ◭ 268 Example Find y D x2 + 1 dxdy where D = {(x. LDA : y = 2x − 9.3>. ◭ Free to photocopy and distribute 149 .4>. y) ∈ R2 |x ≥ 0. > with(Student[VectorCalculus]): > int(x*y.<8.y]=Triangle(<8.5>)).7 3 3 6 5 4 1 1 3 2 -3 -2 -1 -1 1 2 3 -3 -2 -1 -1 1 2 3 4 5 6 7 8 9 10 1 2 3 1 2 2 Chapter 3 -2 -2 -3 -3 Figure 3. The integral is thus 4 3 2y 5 (y+12)/2 6 (y+12)/2 xy dxdy+ (y+9)/2 4 (y+9)/2 xy dxdy+ 5 2y−3 xy dxdy = 409 4 . LCD : y = x 2 + 3 2 .<9. Figure 3. Also. −y) = −f (x.4: Example 266.<7. −y) ∈ D.y]=Triangle(<6. [x. y) ∈ D then (x. f (x. Solution: ◮ The integral is 0.6>. Figure 3.5: Example 267. y).3: Example 266. To solve this problem using Maple you may use the code below. Integration order dxdy.4>. Integration order dydx. [x. Observe that if (x. 269 Example Find 4 0 √ y y/2 ey/x dx dy.8: Example 271.5 4 3 2 2 1 0 1 0 1 2 3 4 5 1 Two-Manifolds 0 0 Figure 3. √ √ ( 1 − x + x − 1)dx 150 . 1 Figure 3. We then have 4 0 √ y y 2 ≤x≤ √ y =⇒ 0 ≤ x ≤ 2. y) ∈ R2 : 2e2 − (2e2 − e2 + 1) e2 − 1 Ô √ √ √ x + y ≥ 1. 0 0 1 2 Figure 3. ey/x dx dy 2 2x x2 = 0 2 ey/x dy dx ¬ y/2 = 0 2 xey/x ¬2x dx x2 (xe2 − xex ) dx = 0 = = ◭ 270 Example Find the area of the region R = {(x. 1 − x + 1 − y ≥ 1}.7: Example 270.6: Example 269. Solution: ◮ The area is given by 1 1−(1− (1− √ 1−x)2 dA = D 0 1 0 √ 2 x) dy dx = = ◭ Free to photocopy and distribute 2 2 3 . Solution: ◮ We have 0 ≤ y ≤ 4. x2 ≤ y ≤ 2x. 1). y ≥ 0.4 Find xydxdy D where D = {(x.x≥0.1 Evaluate the iterated inte3 x 1 dydx. ◭ Homework Problem 3. y) ∈ R2 : x ≥ 0. we have 0 ≤ x2 +y 2 ≤ ( 2)2 +( 2)2 = 4. Problem 3.5. x2 + y 2 dA 1dA + 1≤x2 +y2 <2. For (x.x≥0. Free to photocopy and distribute 151 . x2 + y 2 dA = R 1≤k≤3 Rk k ∈ {1. where R is the rectangle [0. gral 1 0 x Problem 3. Now the integrals can be computed by realising that they are areas of quarter annuli. y) ∈ R. y) → x2 + y 2 jumps √ every time x2 + y 2 √ is an integer.5. y ≥ 0. √ 2] Solution: ◮ The function (x. and so.5. x ≥ y 2 }.2 Let S be the interior and boundary of the triangle with vertices (0. 1 1 Problem 3. 0). Thus we decompose R as the union of the Rk = {(x. (2. R √ 2] × [0.5 Find (x + y)(sin x)(sin y)dA D ydA. 0). 2.Chapter 3 271 Example Evaluate . 1 ≤ x2 +y 2 ≤ 4}. kdA = k · 1 πk · π(k + 1 − k) = . 3}.y≥0 = 2dA + 3≤x2 +y2 <4.3 Let S = {(x. y) ∈ R2 |y ≥ x2 . x2 + y 2 dA. y) ∈ R2 : x ≥ 0.x≥0. Find S Problem 3.y≥0 3dA. 4 4 k≤x2 +y2 <k+1. and (2.6 Find 0 0 min(x2 .x≥0. where D = [0.5. π]2 . k ≤ x2 +y 2 < k+1}.y≥0 2≤x2 +y2 <3.5. y 2 )dxdy.y≥0 Hence x2 + y 2 dA = R π 4 (1 + 2 + 3) = 3π 2 . Problem 3.5. Find S x2 dA. 9. 2].10 with O(0.7 Find D xydxdy where D = {(x. y) ∈ R2 |x ≥ 0.9 Find 0 2ex dxdy 2 Problem 3. where R is the rectangle [0 . y > 0.10: Problem 3. y ≥ 0.5.5. + y)dA log (1 + x D Problem 3.13 Evaluate [0. B Problem 3. y 2 )dA.11: Problem 3.15 Evaluate R xdA where R Figure 3.5.9: Problem 3.2]2 is the (unoriented) circular segment in figure 3.Two-Manifolds Problem 3.10 Evaluate [0. and B(4. A(3. B Problem 3. 3 x + y 2 dA.5. A O Figure 3. Free to photocopy and distribute 152 .5.5.11 Find R xydA.1]2 min(x.11.5. y) ∈ R2 : x2 + y 2 ≤ 16} and (x. which formed by the the area between the circles x2 + y 2 = 1 and x2 + y 2 = 4 in the first quadrant. y) ∈ R2 : y ≥ − √ 3 x+4 . 4).8 Evaluate R xdA where R where D = {(x.5. which is created by the intersection of regions {(x. where R Figure 3.5. A Problem 3.15. 1] × [0 .12 Find Problem 3. 0).5.5.14 Evaluate R x + y dA.5. 9 < x2 +y 2 < 16. Problem 3.8. 1 < x2 −y 2 e< 16}. x + y ≤ 1}.11. y) ∈ R2 : x > 0. 1 1 y Problem 3. 1).5.16 Evaluate R xdA where R is the E-shaped figure in figure 3.12. Problem 3. is the quarter annulus in figure 3. is the (unoriented) △OAB in figure 3. 5.23 Evaluate Problem 3. y) ∈ R2 : x ≥ 0.17 Evaluate 0 x cos y dydx.12: Problem 3.21 Find xydA D D = {(x. √ 2y x 2 Problem 3. 3).26 Find D (2x + 3y + 1) dA. where D = {(x. y) ∈ R2 : x ≥ 0. D where Problem 3.16.5. as shewn in figure 3.5.5. Problem 3.14.5. 4 4 as in figure 3.5.5.5.22 Find π /2 π /2 Problem 3.18 Find 2 1 πx sin dy √ 2y x x 4 dx+ 2 D = {(x. y ≥ 0. −4). and C(1.5. y where loge (1 + x2 + y)dA D Problem 3.Chapter 3 Figure 3. −1).5. |y| ≤ 1}. y) ∈ R2 : |x| ≤ 1.19 Find 2x(x2 + y 2 )dA D R xdA.13: Problem 3. 1 1 √ x Problem 3. 1}.23.5.14: Problem 3. y) ∈ R2 : x4 + y 4 + x2 − y 2 ≤ 1}.5.13. xy+y+x ≤ A(−1.20. where R is the region between the circles x2 +y 2 = 4 and x2 +y 2 = 2y. B(2. Free to photocopy and distribute 153 . Figure 3. x2 +y ≤ 1}.25 Find |x − y|dA Figure 3.20 Find the area bounded by y2 x2 the ellipses x2 + = 1 and + y 2 = 1. πx sin dy dx. where where D is the triangle with vertices at D = {(x. Problem 3.5.24 Evaluate 0 ex/y dydx. Problem 3. y ≥ 0. y Problem 3. 5. with f increasing.32 Find 2 xy dA. y ≥ 1. Problem 3.5. Problem 3. x−1 dx log x = Then demonstrate that log 2. y ≥ 0.5. 0 (f ◦g)(x)dx ≤ f (x)dx+ 0 0 g(x)dx. 0 x Ô y x2 + y 2 dxdy = log(1 + √ 2).38 Evaluate √ 4 4−y Ô 12x − x3 dxdy. 2] × [1.5. . y) ∈ R2 |x ≥ 0.40 Evaluate dydx.5.29 Let f.35 Evaluate 1 0 0 1 1 xf 2 (x)dx 0 1 f 2 (x)dx ≤ 0 1 ··· 0 (x1 + x2 + · · · + xn ) dx1 dx2 . y) ∈ R : y ≥ 0. 1] be continuous. 1] →]0. 2] and let f. +∞] be a decreasing function. where 2 D = {(x. . Problem 3. Prove that 1 1 Problem 3. It is known that at least one side of each of the rectangles Rk is an integer. y2 16 Problem 3.5.36 Let I be the rectangle [1. g : [0.5. .33 A rectangle R on the plane is the disjoint union R = ∪N Rk of rectank=1 gles Rk . 2] such that f (x) ≤ g(x).5. 2] → [1. y) ∈ R2 |x ≥ 1. 1 + x4 where a and b are positive. y) ∈ R : |x| 2 1/2 + |y| 1/2 ≤ 1}. x + y ≤ 4}.5. Problem 3. Free to photocopy and distribute 154 . Problem 3.5. Prove that 1 0 1 1 Problem 3.5. Shew that at least one side of R is an integer. √ D Problem 3. f (x)dx Problem 3. . g be continuous functions f. Problem 3.39 Evaluate 0 y y Ô 1 + x3 dxdy.43 Prove that 1 y y2 ··· 0 (x1 x2 · · · xn )dx1 dx2 .28 Find (xy(x + y))dA D (g(y) − f (x)) dxdy ≥ 0. .42 Prove that +∞ 0 +∞ 2x xe−y sin y 1 dydx = .5.27 Let f : [0. (x + y) ≤ 2x}.34 Evaluate 1 0 0 1 1 where D = {(x.5.Two-Manifolds Problem 3. Problem 3. Problem 3.5.5. 1] → [0. x + y ≤ 1}. dxn . Demonstrate that I xf (x)dx 0 0 Problem 3. 0 0 2 2 where S = {(x. dxn .30 Compute S (xy + y 2 )dA Problem 3.37 Find 0 1 0 xy dxdy. 1 1 where D = {(x. g : [1.a y ) 2 2 Problem 3.41 Find 1 dA (x + y)4 D 0 √ xy dxdy.31 Evaluate a 0 0 b 1 2 1 y emax(b 2 x .5.5. bounded sets in Rn with volume and let g : ∆ → D be a continuously differentiable bijective mapping such that 1 det g ′ (u) = 0.48 Let a ∈ R. un ))| det g ′ (u)|du1 ∧ du2 ∧ . . . . . dxn . f ◦ g| det g ′ (u)| is integrable on ∆ and ··· that is ··· D f = D ··· ∆ (f ◦ g)| det g ′ (u)|. 155 Free to photocopy and distribute . π Prove that (x1 + x2 + · · · + xn ) dx1 dx2 . Prove that æ é2 b b f (x) g(x) 1 det dx dy 2 a f (y) g(y) a 0 0 ··· 0 0 (f (x1 )f (x2 ) · · · f (xn ))dxn dxn−1 .6 Change of Variables We now perform a multidimensional analogue of the change of variables theorem in one variable. ∧ dun . . . This is an integral analogue of Lagrange’s Identity. . g be continuous functions in the interval [a. a] → R be continuous. b]. are bounded on ∆. . and both | det g ′ (u)|.5.5.Chapter 3 Problem 3. u2 . . 272 Theorem Let (D. a > 0. dx2 dx equals 1 n! a n f (x)dx 0 .5. x2 . . n ∈ N. . .5. . 2n a x x x 1 n−2 n−1 Problem 3.5.47 Let f.46 Evaluate 1 n→+∞ 1 0 1 lim 0 ··· cos2 0 Problem 3. x−y+1 ≥ 0. n > 0. b (x + y)3 0 0 0 0 (f (x))2 dx a Is this a contradiction to Fubini’s Theorem? Problem 3. b 2 b b where 2 a (f (x)g(x))dx ≤ (f (x))2 dx a a (g(x))2 dx . f (x1 . Problem 3. . . Deduce Cauchy’s Inequality for integrals.45 Find xdA D x−y 1 dydx = =− (x + y)3 2 (g(x))2 dx − (f (x)g(x))dx a . xn )dx1 ∧ dx2 ∧ . D = {(x. ∆) ∈ (Rn )2 be open. Let f : [0. For f : D → R | det g ′ (u)| bounded and integrable.44 Prove that 1 1 equals 1 1 b a b 2 x−y dxdy. . ∧ dxn = ··· ∆ f (g(u1 . x+2y−4 ≤ 0}. 3. y) ∈ R |y ≥ 0. and (0. The corners of the rectangle in the area of integration in the xyplane are (0. and (8.Change of Variables One normally chooses changes of variables that map into rectangular regions. giving du ∧ dv = −3dx ∧ dy. 6). 3).15: Example 273. Now. (9. 5). (7. Put u = x + 2y =⇒ du = dx + 2dy.16: Example 273. 3). or that simplify the integrand. The form dx ∧ dy has opposite orientation to du ∧ dv so we use dv ∧ du = 3dx ∧ dy Free to photocopy and distribute 156 . 0 Solution: ◮ Observe that we have already computed this integral in example 265. (u. (traversed counter-clockwise) and they map into (6. and hence it maps quadrilaterals into quadrilaterals. v = 2x + y =⇒ dv = 2dx + dy. 7 6 5 4 3 2 1 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 Figure 3. in the uv-plane (see figure 3. (1. 4). Figure 3. 4). uv-plane. v) = 1 2 2 1 x y is a linear transformation. xy-plane. (1. 273 Example Evaluate the integral 4 3 1 (x + 2y)(2x + y)dxdy. 3). respectively.16). Let us start with a rather trivial example. 4). 0 ≤ v ≤ 2. Hence BC is . Evaluate it using the change of variables u = x2 − y 2 . ◭ 274 Example The integral (x4 − y 4 )dA = 1 0 P uv dvdu = 409 12 . We now determine the region ∆ into which the square D = [0. dv = 2ydx + 2xdy. 4 Free to photocopy and distribute 157 . v = 0. Since 0 ≤ x ≤ 1. DA is mapped into the line segment −1 ≤ u ≤ 0. We use the fact that boundaries will be mapped into boundaries. Put AB = {(x. 0 ≤ v ≤ 2. v = 2xy. v = 2(1 − x). 1 5 [0. 4 Finally. The region ∆ is thus v2 the area in the uv plane enclosed by the parabolas u ≤ − 1. v = 0.1]2 − y 4 dy = 0. Thus u = 1 − mapped to the portion of the parabola u = 1 − v 2 v2 4 . y) : 0 ≤ y ≤ 1}. On AB we have u = x. The integral sought is 1 3 from example 267. on DA. CD = {(1 − x. Since 0 ≤ y ≤ 1. On BC we have u = 1 − y 2 . AB is thus mapped into the line segment 0 ≤ u ≤ 1. v = 0. 1) : 0 ≤ x ≤ 1}. v = 0. 0 ≤ v ≤ 2. 1 − y) : 0 ≤ y ≤ 1}. and so du ∧ dv = (4x2 + 4y 2 )dx ∧ dy. u ≤ 4 2 v 1− with −1 ≤ u ≤ 1. 0) : 0 ≤ x ≤ 1}.Chapter 3 instead. DA = {(0. 1]2 is mapped. This means that v2 u= − 1. BC = {(1. Solution: ◮ First we find du = 2xdx − 2ydy. we have u = −(1 − y)2 . v = 2y. 4 On CD we have u = (1 − x)2 − 1. 0 ≤ v ≤ (2p)1/3 }.Change of Variables We deduce that (x4 − y 4 )dA = = = = 1 ∆ [0. x2 − 2py ≤ 0. using the change of variables x = u2 v.1]2 (x4 − y 4 ) ∆ 1 4(x2 + y2 ) dudv 4 1 4 1 (x2 − y 2 )dudv ududv 1−v 2 /4 ∆ 2 0 4 = 0. y = uv 2 . v) ∈ R2 |0 ≤ u ≤ (2p)1/3 . The region transforms into ∆ = {(u. y)dxdy D = ∆ exp 3 u6 v 3 + u3 v 6 u3 v 3 3 (3u2 v 2 ) dudv = 3 ∆ eu ev u2 v 2 dudv (2p)1/3 2 2 u3 = = 3u e 3 0 1 2p (e − 1)2 . you may try the (more natural) substitution x3 = u2 v. 3 1 du As an exercise. p ∈]0. dx ∧ dy = 3u2 v 2 du ∧ dv. y 3 = v 2 u and verify that the same result is obtained. ◭ 275 Example Find e(x D 3 udu dv v 2 /4−1 +y3 )/xy dA where D = {(x. dy = v 2 du + 2uvdv. ◭ Free to photocopy and distribute 158 . +∞[ fixed}. The integral becomes f (x. y) ∈ R2 |y 2 − 2px ≤ 0. as before. Solution: ◮ We have dx = 2uvdu + u2 dv. 276 Example In this problem we will follow an argument of Calabi. xy-plane. Prove that n=1 1 (2n − 1)2 = 0 dxdy 1 − x2 y 2 sin u cos v 3. Observe that the sum of the even terms is +∞ n=1 1 (2n)2 = 1 4 +∞ n=1 1 n2 = 1 4 S. Prove that if S = n=1 +∞ 3 .17: Example 276. uv-plane. then S = 2 n 4 1 1 0 1 +∞ n=1 1 (2n − 1)2 . 1 − x2 y 2 Solution: ◮ 1. a quarter of the sum. hence the sum of the odd terms must be three 3 quarters of the sum. Beukers. y = sin v cos u in order to evaluate dxdy .Chapter 3 π 1 0 0 1 2 π 2 Figure 3. Observe that 1 2n − 1 1 = 0 x2n−2dx =⇒ 1 2n − 1 2 1 = 0 x2n−2 dx 1 0 y 2n−2 dy 159 1 1 0 = 0 (xy)2n−2 dx Free to photocopy and distribute . 4 2. Figure 3. 2. and +∞ π2 1 = . S.18: Example 276. Kock to prove that n2 6 n=1 +∞ 1. . Use the change of variables x = 1 0 1 0 . y= forms in the uv-plane. We deduce. 1 − x2 y 2 We now have to determine the region that the transformation x = sin u sin v . This exchange of integral and sum needs justification. 1 − x2 y 2 as claimed. This means that the square in the xy-plane in figure 3. 1 0 1 0 dxdy 1 − x2 y 2 π/2 π/2−v π/2 = 0 0 dudv = 0 (π/2 − v) dv = π 2 v− v2 2 ¬π/2 π2 π2 π ¬ = − = ¬ 0 4 8 Finally. 1 − x2 y 2 = 1 − This gives dxdy = dudv. Also. Observe that cos v cos u u = arctan x 1 − y2 × dy = (sec u)(tan u)(sin v)du+(sec u)(cos v)d sin2 v cos2 v · sin2 v cos2 u = 1 − (tan2 u)(tan2 v).18. 3 π2 π2 S= =⇒ S = . v) ∈ R2 : u ≤ 1.Change of Variables Thus +∞ n=1 1 = (2n − 1)2 +∞ n=1 1 0 1 0 (xy)2n−2 dxdy = 1 0 1 +∞ 0 n=1 (xy)2n−2 dxdy = 1 0 1 0 dxdy .6. We will accept it for our purposes. Consider R2 Φ: 3 → → (u. v) R2 u+v u−v .17 is transformed into the triangle in the uv-plane in figure 3. 2 2 . then cos v cos u dx = (cos u)(sec v)du+(sin u)(sec v)(tan v)dv.1 Let D ′ = {(u. 2 v = arctan y 1 − x2 1 − y2 . Free to photocopy and distribute 160 . 1−x . −u ≤ v ≤ u}. 4 8 6 ◭ Homework Problem 3. from where dx∧dy = du∧dv−(tan2 u)(tan2 v)du∧dv = 1 − (tan2 u)(tan2 v) du∧dv. If x = .y= .3 sin v sin u 3. 3 Using the change of variables u = x − y and v = x + y.6. 6 2 Free to photocopy and distribute 161 . v = x − y to shew that 1 0 0 1 dxdy =2 1 − xy 1 0 u −u dv 4 − u2 + v 2 2 2−u u−2 du + 2 1 dv 4 − u2 + v 2 du. that is.2 Using the change of variables x = u2 − v 2 . 2). find D = Φ(D ′ ). (1. prove that the above integral is π2 u by using the substitution θ = arcsin . where R is the square with vertices at (0. y) ∈ R2 : −1 ≤ x ≤ 1. y = 2uv. y ≥ x ≥ 0. n2 Use the change of variables u = x + y. Problem 3. (2. 2 4−u 4 − u2 Finally. y)dA where D = {(x. y) ∈ R2 |a ≤ xy ≤ b. Ô evaluate x2 + y 2 dA. Shew that the above integral reduces to 1 2 0 √ 2 u arctan √ du + 2 2 4−u 4 − u2 2 1 √ 2 2−u arctan √ du.6. v ≥ 0. 2 n 6 Use the series expansion 1 = 1 + t + t2 + t3 + · · · 1−t in order to prove (formally) that 1 0 0 1 ∞ n=1 |t| < 1. dxdy = 1 − xy 1 . 0 ≤ y ≤ 2 1 − |x|} Problem 3. evaluate x−y dA. 2). Find (x + y)2 ex D 2 −y 2 dA. (1. (a. 0 < a < b} and f (x. y 2 − x2 ≤ 1. where R R = {(x. b) ∈ R2 . R x + y Problem 3. y) = y 4 − x4 by using the change of variables u = xy. Problem 3.4 Find D f (x. u ≥ 0.6.5 Use the following steps (due to Tom Apostol) in order to prove that ∞ n=1 1 π2 = . 1). 3).Chapter 3 Find the image of Φ on D ′ .6. v = y 2 − x2 . 5. One of the most common changes of variable is the passage to polar coordinates where whence dx ∧ dy = (ρ cos2 θ + ρ sin2 θ)dρ ∧ dθ = ρdρ ∧ dθ. where R is the region bounded by the circles x2 + y 2 = 4 and x2 + y 2 = 2y. from 2 sin θ to 2. y = ρ sin θ =⇒ dy = sin θdρ + ρ cos θdθ. Since x2 + y 2 = r 2 . y ≥ 0. that is. . 4 Therefore the integral becomes ρ4 cos θ sin θ dρdθ ∆ π/4 1 = 0 cos θ sin θ dθ 0 ρ4 dρ = ◭ 1 20 .23. x2 + y 2 ≤ 1}. 278 Example Evaluate R xdA. The region D transforms into the region π ∆ = [0. Free to photocopy and distribute 162 . 277 Example Find xy D x2 + y 2 dA where D = {(x. y ≤ x. the radius sweeps from r 2 = 2r sin θ to r 2 = 4. Solution: ◮ We use polar coordinates. 1] × 0.7 Change to Polar Coordinates x = ρ cos θ =⇒ dx = cos θdρ − ρ sin θdθ. y) ∈ R2 |x ≥ 0. Solution: ◮ Observe that this is problem 3.Change to Polar Coordinates 3. The integral becomes dxdy = 2 √ 3 e−(U {U 2 +V 2 ≤1} 2 −xy−y2 +V 2 ) dU dV. Put U = x + . Figure 3.22: Example 280.20: Example 278. Solution: ◮ Completing squares x + xy + y = x + y 2 √ 3y 2 2 y 2 2 + √ 3y 2 2 . Figure 3.21: Example 279.V = e−x 2 2 . Thus the integral becomes r 2 cos θdrdθ 2 sin θ xdA = R 3 = 2. {x2 +xy+y2 ≤1} Passing to polar coordinates. 3 . the above equals 2 √ 3 ◭ Free to photocopy and distribute 163 2π 0 1 0 2π 2 ρe−ρ dρdθ = √ (1 − e−1 ). ◭ 279 Example Find D = 1 0 2 π/2 0 (8 cos θ − 8 cos θ sin3 θ)dθ e−x 2 −xy−y2 dA.Chapter 3 Figure 3. where D = {(x. The angle clearly sweeps from 0 to π/2 π 2 .19: Example 277. Figure 3. y) ∈ R2 : x2 + xy + y 2 ≤ 1}. whence dx ∧ dy = (abr cos2 θ + abr sin2 θ)dr ∧ dθ = abrdr ∧ dθ. a2 b Solution: ◮ Put x = ar cos θ. R Solution: ◮ The radius sweeps from r = integral is dA = (x2 + y 2 )3/2 = = ◭ 281 Example Evaluate R 1 sin θ 2 csc θ to r = 2. Then x = ar cos θ =⇒ dx = a cos θdr − ar sin θdθ. Observe that on the ellipse x2 a2 + y2 b2 = 1 =⇒ a2 r 2 cos2 θ a2 + b2 r 2 sin2 θ b2 = 1 =⇒ r = 1. 3 sin θ − (x3 + y 3 )dA where R is the region bounded by the ellipse x2 y2 + 2 = 1 and the first quadrant.Change to Polar Coordinates dA over the region (x. The desired 1 5π/6 π/6 5π/6 1 r2 drdθ 1 2 dθ R √ π/6 π 3− .22). y ≥ 1 (x2 + y 2 )3/2 1 280 Example Evaluate (figure 3. y = br sin θ =⇒ dy = b sin θdr + br cos θdθ. 15 Free to photocopy and distribute 164 . y) ∈ R2 : x2 + y 2 ≤ 4. Thus the required integral is (x3 + y 3 )dA = R π/2 0 1 0 1 0 abr 4 (cos3 θ + sin3 θ)drdθ π/2 0 = = = ◭ ab ab 1 r 4 dr (a3 cos3 θ + b3 sin3 θ)dθ 2a3 + 2b3 5 3 2ab(a3 + b3 ) . y = br sin θ. a > 0 and b > 0. Problem 3.4 Find and f (x.Chapter 3 Homework Problem 3. Prove that Ö a 2 b2 − a 2 y 2 − b2 x 2 πab(π − 2) dA = . 2 a b Free to photocopy and distribute 165 . +∞[ fixed} Problem 3.7. D = {(x. y) ∈ R2 |(x2 + y 2 )2 ≤ 2xy}.7. y) ∈ R2 |(x − 1)2 + y 2 ≤ 1}. as in the figure 3. Set up the integral in both Cartesian and polar coordinates.7. y) ∈ R2 : b2 x2 + a2 y 2 = a2 b2 . b) ∈]0. xydA where R is the region Figure 3. (a.1 Evaluate R ¨ © R = (x.1. Problem 3.23: Problem 3. y ≥ 1 . y) ∈ R2 : x2 + y 2 ≤ 16. y) = x3 + y 3 .7. x ≥ 1.2 Find (x2 − y 2 )dA D where D = {(x.3 Find D √ xydA where D = {(x.5 Let a > 0 and b > 0.23.7. a 2 b2 + a 2 y 2 + b2 x 2 8 R where R is the region bounded by the ellipse x2 y2 + 2 = 1 and the first quadrant. f dA where D Problem 3.7. y) ∈ R2 |x ≥ 1. x2 + y 2 − 2y ≥ 0}.6 Prove that Ô R x2 √ y dxdy = 2 − 1.7.7. y) ∈ R2 |y ≤ x2 + y 2 ≤ 1}. o < y < x2 }. Problem 3. D Problem 3.7. (x2 + y 2 )2 Problem 3. 2 +y where R = {(x. x2 + y 2 − 2x ≤ 0} and f (x. x2 + y 2 ≤ 1. y) ∈ R2 |y ≥ 0. D Problem 3. x2 + y 2 − x ≤ 0}.11 Find D f dA where D = {(x. y) = x2 y. y) ∈ R2 : x2 + y 2 − y ≤ 0. y) ∈ R2 |x ≥ 0. y) ∈ R2 : x ≥ 1. y) = 1 .10 Let D = {(x.7.7.8 Find D Ô x2 + y 2 dA where D = {(x. 5 Problem 3. Find the integral (x + y)2 dA.12 Let D = {(x. y) ∈ R2 : 0 < x < 1. y ≥ 0. x2 + y 2 − 2x ≤ 0} and f (x. Find xdA. f dA where D Problem 3.7.7 Prove that the ellipse (x − 2y + 3)2 + (3x + 4y − 1)2 = 4 bounds an area of 2π . Problem 3.9 Find D = {(x.13 Let D = {(x.7. Compute D dA .Change to Polar Coordinates Problem 3.7. x2 + y 2 ≤ 4}. (1 + x2 + y 2 )2 Free to photocopy and distribute 166 . 2 1 . if R is the set of points inside and on a convex polygon. y)dA. let D(x.” Prove that +∞ 0 e−x dx = by following these steps.18 In the xy-plane. y) = Find D f (x.y ) dxdy = 2π + L + A. y) to the nearest point R. y) ∈ R2 : 4 ≤ x2 + y 2 ≤ 16} and f (x.7. Let Ja = ∆a e−(x 2 +y 2 ) dxdy. Problem 3.7. x2 + xy + y 2 Problem 3.Chapter 3 Problem 3. Free to photocopy and distribute 167 . Liouville was a mathematician. 2 √ π 2 Let a > 0 be a real number and put Da = {(x.y ≥0. Problem 3. −∞ where L is the perimeter of R and A is the area of R.7. y) be the distance from (x. Find Ia = Da e−(x 2 +y 2 ) dxdy.17 Prove that every closed convex region in the plane of area ≥ π has two points which are two units apart.y )∈R2 :x≥0. Show that +∞ −∞ +∞ e−D (x. y 2 = ρ sin θ.7. |y| ≤ a}. y) ∈ R2 |x2 + y 2 ≤ a2 }.x 4 +y 4 ≤1} Ô 1 − x4 − y 4 dA using x2 = ρ cos θ.7. y) ∈ R2 ||x| ≤ a. Deduce that 0 +∞ e−x dx = 2 √ π . Problem 3.16 Let D = {(x.15 William Thompson (Lord Kelvin) is credited to have said: “A mathematician is someone to whom √ +∞ 2 π e−x dx = 2 0 is as obvious as twice two is four to you. Let a > 0 be a real number and put ∆a = {(x. Prove that Ia ≤ Ja ≤ Ia √ 2 .14 Evaluate x3 y 3 {(x. Given a function f : V → R. where the + orientation is in the direction of the outward pointing normal to the body. dV V is the oriented volume of V . This corresponds to using the volume form dx ∧ dy ∧ dz. unless otherwise noticed. 2 = e− ◭ Free to photocopy and distribute 168 . 283 Example Find x2 yexyz dV. In particular. the integral f dV V is the sum of all the values of f restricted to V . The region −M has opposite orientation to M and −M ω=− ω.1]3 Solution: ◮ The integral is 1 0 1 0 1 0 x2 yexyz dz dy dx 1 1 0 = 0 1 x(exy − 1) dy dx = 0 (ex − x − 1)dx 5 . M We will often write f dV M where dV denotes the volume element. [0. and the − orientation is in the direction of the inward pointing normal to the body. we will choose the positive orientation for the regions considered. A general oriented 3-manifold is a union of open sets. In this section.Three-Manifolds 3.8 Three-Manifolds 282 Definition A 3-dimensional oriented manifold of R3 is simply an open set (body) V ∈ R3 . Let V ⊆ R3 . z ≥ 0. 3. 0. y ≥ 0. xdxdydz V = 0 0 xdxdydz 1 2 1 ay az − b c 0 3 a2 (−z + c) b dx c3 a− b−bz/c 2 = = = ◭ dydz 6 0 a2 bc 24 286 Example Evaluate the integral S xdV where S is the (unoriented) tetrahe- dron with vertices (0. xdV = V a2 bc . Solution: ◮ The integral is 1 (1− √ 2 z) (1− 0 √ √ z− x)2 zdxdydz R = 0 1 z 0 √ (1− z)2 dy (1 − √ z)4 dz √ √ z − x)2 dx dx dz dz = 0 z 1 6 1 1 0 0 = = ◭ 285 Example Prove that z(1 − 840 .Chapter 3 284 Example Find R z dV if √ √ √ x + y + z ≤ 1}. Solution: ◮ We have c b−bz/c 0 c 0 c a−ay/b−az/c Ò x a + y b + z c ≤1 .24. (3. 2. y. y. y ≥ 0. 0). 24 Ó where V is the tetrahedron V = (x. 0). R = {(x. and (0. z ≥ 0. (0. See figure 3. 0). Free to photocopy and distribute 169 . 2). z) ∈ R3 : x ≥ 0. 0. z) ∈ R3 |x ≥ 0. The line 3 2 passing through AC has equation y = x.25 we draw the projection of the tetrahedron on the xy x plane.0>. its Free to photocopy and distribute 170 . dzdxdy 2.<3. > with(Student[VectorCalculus]): > int(x. ◭ 287 Example Evaluate R xyzdV .0>. on x and y.2>)).0. With the order dzdxdy.24: Problem 286. 0). where R is the solid formed by the intersection of the parabolic cylinder z = 4 − x2 .<0. 3 We find. 2. 0).3. dxdydz Solution: ◮ We must find the projections of the solid on the the coordinate planes. the limits of integration of z can only depend. Hence.z C B Three-Manifolds O x A Figure 3. if at all. y = x. Solution: ◮ A short computation shews that the plane passing through (3.y. A C B y Figure 3. (0.0>. 1. The line passing through AB has equation y = − + 3.25: xy-projection. the planes z = 0. finally. In figure 3. 0. Use the following orders of integration: 1. and (0. We must now figure out the xy limits of integra9 tion.[x.2.0. 2) has equation 2x + 6y + 9z = 18. 3 3−x/3 2x/3 3−x/3 2x/3 (18−2x−6y)/9 xdV S = 0 3 0 xdzdydx dydx 9 18xy − 2x2 y − 3y 2 x ¬3−x/3 ¬ dx ¬ 2x/3 9 3 x − 2x2 + 3x dx 3 18x − 2x2 − 6yx = 0 3 = 0 3 = = 9 4 0 To solve this problem using Maple you may use the code below.<0. and y = 0. Given an arbitrary point in the solid.z]=Tetrahedron(<0. 3. 18 − 2x − 6y 0≤z≤ . Chapter 3 lowest z coordinate is 0 and its highest one is on the cylinder. The projection of the solid on the xy-plane is the area bounded by the lines y = x. 2 0 y 0 4−x2 xyzdzdxdy 0 = = = 1 2 1 2 2 2 0 2 0 y 0 y 0 xy(4 − x2 )2 dxdy y(16x − 8x3 + x5 )dxdy y7 12 dy 0 4y 3 − y 5 + = 8. With the order dxdydz. Prove that xyz = first by 2 3 S integrating in the order dzdydx. and the z and y axes. and then by integrating in the order dydxdz. where R is the tetrahedron with vertices at (0. 1. 1). The projection of the solid on the yz-plane is the area bounded by z = 4 − y 2 .8. ◭ Homework Problem 3. so the limits for x are from x = y to x = 4 − z. 2. and (0. 1). 0.3 Evaluate the integrals R 1dV and R xdV . 4 0 √ 0 4−z 2 xyzdxdydz y = = 1 2 4 4 0 √ 0 4−z 0 2z − z3 8 (4y − y 3 )zdydz dz = 8. the limits of integration of x can only depend. x = 2. (1. 0. Given an arbitrary point in the solid. and y = 0. Problem 3.1 Compute E zdV where E is the region in the first octant bounded by the planes y + z = 1 and x + z = 1. 0).8. if at all. x sweeps √ from the plane to x = 2. and the x and y axes. 0).8. Free to photocopy and distribute 171 .8. Problem 3. on y and z. 0. so the limits for z are from z = 0 to z = 4 − x2 . Problem 3.2 Consider the solid S in the first octant. y = x. bounded by the parabolic cylinder x2 2 z = 2− and the planes z = 0.4 Compute E xdV where E is the region in the first octant bounded by the plane y = 3z and the cylinder x2 + y 2 = 9. (1. y. du ∧ dv ∧ dw = −4dx ∧ dy ∧ dz. and 2x − z = 1. u + v + 2w = 1 in the uvw-coordinate frame. 0 ≤ y ≤ 1. The tetrahedron in the xyz-coordinate frame is mapped into a tetrahedron bounded by u = 0. say. z) ∈ R3 : 0 ≤ x ≤ 1. v = x + y − z =⇒ dv = dx + dy − dz.8.9 Change of Variables (x + y + z)(x + y − z)(x − y − z)dV. y. From what we know about polar coordinates dx ∧ dy = ρdρ ∧ dθ. x + y − z = 0. x − y − z = 0. so we choose. Consider a transformation to cylindrical coordinates (x. R 288 Example Find where R is the tetrahedron bounded by the planes x + y + z = 0. 3. ◭ Free to photocopy and distribute 172 . z ≥ 0}. 2x − z = 1 =⇒ u + v + 2w = 2. These forms have opposite orientations.Change of Variables dV where (1 + x2 z 2 )(1 + y 2 z 2 ) D Problem 3. ρ sin θ. Solution: ◮ We make the change of variables u = x + y + z =⇒ du = dx + dy + dz. This gives w = x − y − z =⇒ dw = dx − dy − dz. Also.5 Find D = {(x. The integral becomes 1 4 2 0 1−v/2 0 2−v−2w uvw dudwdv = 0 1 180 . Since the wedge product of forms is associative. du ∧ dw ∧ dv = 4dx ∧ dy ∧ dz which have the same orientation. v = 0. z). z) = (ρ cos θ. dx ∧ dy ∧ dz = ρdρ ∧ dθ ∧ dz. Solution: ◮ The region of integration is mapped into ∆ = [0. 16 ◭ 291 Example Three long cylinders of radius R intersect at right angles. In this case it is easier to integrate with respect to z first. 1] through a cylindrical coordinate change. +∞[. z) ∈ 0. 0 ≤ z ≤ 1}. R] × [0. Find the volume of their intersection. y. By symmetry. y 2 +z 2 ≤ R2 . z) ∈ R3 : 0 ≤ y ≤ x. x2 +y 2 ≤ R2 .Chapter 3 289 Example Find R z 2 dxdydz if R = {(x. x = y. The integral is therefore 2π 1 1 f (x. Solution: ◮ The integral is 1 0 π/2 π/4 2 1 ρ3 dρdθdz = 15π . 2π] × [0. Using cylindrical coordinates ∆′ = (θ. Ò π 4 × [0. 173 Ó Free to photocopy and distribute . y. 0 ≤ z ≤ R2 − ρ2 cos2 θ . where V′ = D′ dxdydz. Solution: ◮ Let V be the desired volume. z) ∈ R3 |x2 + y 2 ≤ 1. z)dxdydz R = = π 3 0 dθ 0 ρ dρ 0 z 2 dz . y. 0 ≤ z. V = 24 V ′ . D ′ = {(x. D 2 (x2 + y 2 )dxdydz over the first octant region bounded = 1 and x2 + y 2 = 4 and the planes z = 0. z = 1. z 2 +x2 ≤ R2 }. ρ. 1] × [0. ◭ 290 Example Evaluate by the cylinders x2 + y x = 0. and so we choose dx ∧ dy ∧ dz = ρ2 sin φdρ ∧ dφ ∧ dθ instead. z) ∈ R3 : x2 a2 + y2 b2 + z2 c2 xyz dV if R = cos θ sin φdρ − ρ sin θ sin φdθ + ρ cos θ cos φdφ. ≤ 1. y. y = ρ sin θ sin φ. 292 Example Let (a. V′ π/4 R 0 R 0 √ 0 = 0 π/4 R2 −ρ2 cos2 θ dz ρdρ dθ = 0 π/4 = = u = cos θ = = Finally = R3 3 R3 0 3 √ ä ç 22 R3 1 − u−1 + u 1 3 √ 2−1 3 R . z = ρ cos φ. y ≥ 0. From this derivation. c) ∈]0.Change of Variables Now. +∞[3 be fixed. 174 Free to photocopy and distribute . √ 2 V = 16V ′ = 8(2 − ä çR 1 (R2 − ρ2 cos2 θ)3/2 dθ 2θ 0 3 cos π/4 1 − sin3 θ dθ cos2 θ √ 0 2 2 1 − u2 π/4 [tan θ]0 + du u2 1 ρ R2 − ρ2 cos2 θdρ dθ − √ 2)R3 . z ≥ 0 . = sin θ sin φdρ + ρ cos θ sin φdθ + ρ sin θ cos φdφ. = cos φdρ − ρ sin φdφ. Find R= (x. the form dρ ∧ dθ ∧ dφ is negatively oriented. ◭ Consider now a change to spherical coordinates x = ρ cos θ sin φ. b. We have dx dy dz This gives dx ∧ dy ∧ dz = −ρ2 sin φdρ ∧ dθ ∧ dφ. x ≥ 0. find this volume. We have dx ∧ dy ∧ dz = abcρ2 sin φdρ ∧ dφ ∧ dρ. bρ sin θ sin φ. Set up integrals for the volume of this region in Cartesian. y. Problem 3. y. b > 0. The integration region is mapped into ∆ = [0. Set up integrals for the volume of this region in Cartesian. z) = (aρ cos θ sin φ. cρ cos φ).3 Consider the region R bounded by the sphere x2 + y 2 + z 2 = 4 and the plane z = 1. Set up integrals for the volume of this region in Cartesian. 48 dxdydz V = 0 ρ2 sin φ dρdφdθ = 63π 4 . z) ∈ R3 : x2 + y 2 + z 2 ≤ 9. 1 ≤ z ≤ 2}. Problem 3. Homework Ô Problem 3. Problem 3. Also. Here a > 0.Chapter 3 Solution: ◮ We use spherical coordinates.9. cylindrical and spherical coordinates. where R is the region above the paraboloid z = x2 + y 2 and under the sphere x2 + y 2 + z 2 = 4. find this volume. 1] × [0. 3 Free to photocopy and distribute 175 . The integral becomes (abc)2 ◭ 293 Example Let V = {(x.4 Prove that the volume enclosed by the ellipsoid x2 y2 z2 + 2 + 2 =1 2 a b c is 4πabc . where (x. Then 2π π/2−arcsin 1/3 π/2−arcsin 2/3 2/ cos φ 1/ cos φ π/2 1 π π ] × [0. Also.9.9.1 Consider the region R below the cone z = x2 + y 2 and above the 2 2 paraboloid z = x + y for 0 ≤ z ≤ 1.2 Consider the integral R xdV . cylindrical and spherical coordinates. find this volume. 2 2 cos θ sin θ dθ 0 0 ρ5 dρ π/2 0 cos3 φ sin φ dφ = (abc)2 .9. Also. c > 0. cylindrical and spherical coordinates. ]. This corresponds to using the ordered basis {dy ∧ dz.9. y. dx ∧ dy}.9.9.9 Compute the 4-dimensional integral ex x 2 +y 2 +u 2 +v 2 ≤1 2 +y 2 +u 2 +v 2 dxdydudv. Free to photocopy and distribute 176 .9. below the plane x − z = −2 and above the xy-plane. Problem 3. Σ In this section.8 Compute E z Ô x2 + y 2 + z 2 dV where E is is the upper solid hemisphere bounded by the xy-plane and the sphere of radius 1 about the origin.6 Prove that √ 2 2 2 e− x +y +z dV = π 2 − 2e−R − 2Re−R − R2 e−R . The surface −Σ has opposite orientation to Σ and −Σ ω=− ω.9. dz ∧ dx. unless otherwise noticed. Problem 3.5 Compute and x2 + y 2 = 4. z) ∈ R3 : x ≥ 0. we will choose the positive orientation for the regions considered. x + y + z ≤ 1}. z ≥ 0. A general oriented 2-manifold in R3 is a union of surfaces.9.y≥0 x 2 +y 2 +z 2 ≤R 2 Problem 3. where the + orientation is in the direction of the outward normal pointing away from the origin and the − orientation is in the direction of the inward normal pointing towards the origin. x≥0. E y 2 z 2 dV where E is bounded by the paraboloid x = 1 − y 2 − z 2 Problem 3. y ≥ 0.7 Compute and the plane x = 0. 3. R where R = {(x.Surface Integrals ydV where E is the region between the cylinders x2 +y 2 = 1 E Problem 3.10 (Putnam Exam 1984) Find x1 y 9 z 8 (1 − x − y − z)4 dxdydz. Problem 3.10 Surface Integrals 294 Definition A 2-dimensional oriented manifold of R3 is simply a smooth surface D ∈ R3 . and z = v. dz = dv. We parametrise the cylinder by putting x = cos u. We see that dx ∧ dy = du ∧ dv. Hence dx = − sin udu. z ¬¬d2 x¬¬ = D ¬¬ ¬¬ (u2 + v 2 ) 1 + 4u2 + 4v 2 dudv. Solution: ◮ We parametrise the paraboloid as follows. and so (u2 + v 2 ) 1 + 4u2 + 4v 2 dudv D Ô 2π 1 0 = 0 ρ3 1 + 4ρ2 dρdθ = ◭ √ 1 (5 5 + ). Observe that the domain D of Σ is the unit disk u2 +v 2 ≤ 1. The integral of f over the smooth surface Σ (oriented in the positive sense) is given by the expression f ¬¬d2 x¬¬.Chapter 3 295 Definition Let f : R3 → R. and so Now. y = v. 12 5 π 297 Example Find the area of that part of the cylinder x2 + y 2 = 2y lying inside the sphere x2 + y 2 + z 2 = 4. dz ∧ dx = −2vdu ∧ dv. Solution: ◮ We have x2 + y 2 = 2y ⇐⇒ x2 + (y − 1)2 = 1. Let x = u. dy ∧ dz = −2udu ∧ dv. 296 Example Evaluate Σ z ¬¬d2 x¬¬ where Σ is the outer surface of the section of the paraboloid z = x2 + y 2 . Σ Ô ¬¬ 2 ¬¬ ¬¬d x¬¬ = 1 + 4u2 + 4v 2 du ∧ dv. Σ ¬¬ ¬¬ Here ¬¬ 2 ¬¬ ¬¬d x¬¬ = (dx ∧ dy)2 + (dz ∧ dx)2 + (dy ∧ dz)2 ¬¬ ¬¬ is the surface area element. 0 ≤ z ≤ 1. dy = cos udu. y − 1 = sin u. Free to photocopy and distribute 177 . Ô To evaluate this last integral we use polar coordinates. z = u2 +v 2 . 0. (0. that is. and so ¬¬ 2 ¬¬ ¬¬d x¬¬ = = (dx ∧ dy)2 + (dz ∧ dx)2 + (dy ∧ dz)2 cos2 u + sin2 u du ∧ dv = du ∧ dv. 2. 0. dy ∧ dz = cos udu ∧ dv. 0). 2 xydxdy = − . Σ √ √ − 2−2 sin u πÔ dvdu = 0 2 2 − 2 sin udu Ô where Σ is the top side of the triangle with vertices at (2. Σ ◭ Free to photocopy and distribute 178 . v 2 = 4−2(1+sin u) = 2−2 sin u. The cylinder and the sphere intersect when x2 + y 2 = 2y and x2 + y 2 + z 2 = 4. Also 0 ≤ u ≤ π.e. The integral is thus ¬¬ 2 ¬¬ ¬¬d x¬¬ Σ π √ 2−2 sin u π = 0 = 2 2 1 − sin u du √ 0√ = 2 2 4 2−4 . ◭ 298 Example Evaluate xdydz + (z 2 − zx)dzdx − xydxdy.Surface Integrals whence dx ∧ dy = 0. 4). We project this plane onto the coordinate axes obtaining 4 2−z/2 0 2 0 2 xdydz = Σ 0 (2 − y − z/2)dydz = 4−2x 0 8 3 . 3 2−y 0 − Σ xydxdy = − 0 and hence xdydz + (z 2 − zx)dzdx − xydxdy = 10. when z 2 = 4−2y. Solution: ◮ Observe that the plane passing through the three given points has equation 2x + 2y + z = 4. 0). (0. i. dz ∧ dx = sin udu ∧ dv. (z 2 − zx)dzdx = Σ (z 2 − zx)dzdx = 8. Stokes’. 0 ≤ y ≤ 2.11 Green’s. Problem 3. Á 300 Example Evaluate C (x − y 3 )dx + x3 dy where C is the circle x2 + y 2 = 1.10. Find the surface area of the part of the cone which lies between the planes z = 1 and z = 2. ¬¬ ¬¬ x2 ¬¬d2 x¬¬ where Σ is the surface of the unit sphere x2 + y 2 + Σ Problem 3. this takes the name of Green’s Theorem.3 Evaluate z 2 = 1. if ω is a 1-form.10. In R2 . Σ where Σ is the top of the triangular region of the plane 2x + 2y + z = 6 bounded by the first octant. having boundary ∂M . 3.10. then ω= ∂M M dω.5 You put a perfectly spherical egg through an egg slicer. Ô Problem 3.2 Consider the cone z = x2 + y 2 . Your argument must use surface integrals.10. 0 ≤ x ≤ 1. If ω is a differential form. 299 Theorem (General Stoke’s Theorem) Let M be a smooth oriented manifold. Problem 3.Chapter 3 Homework Problem 3.6 Evaluate xydydz − x2 dzdx + (x + z)dxdy. S ¬¬ ¬¬ Ô z ¬¬d2 x¬¬ over the conical surface z = x2 + y 2 between Problem 3.10. and Gauss’ Theorems We are now in position to state the general Stoke’s Theorem. resulting in n slices of identical height. but you forgot to peel it first! Shew that the amount of egg shell in any of the slices is the same.4 Evaluate z = 0 and z = 1.1 Evaluate Σ ¬¬ ¬¬ y ¬¬d2 x¬¬ where Σ is the surface z = x + y 2 . 179 Free to photocopy and distribute .10. y) dx ∧ dy ∂x ∂y ∂ M 2 ∂ ∂ g(x. Free to photocopy and distribute 180 . Á C (x − y 3 )dx + x3 dy 2π = 0 2π (cos θ − sin3 θ)(− sin θ)dθ + (cos3 θ)(cos θ)dθ (sin4 θ + cos4 θ − sin θ cos θ)dθ. = 0 To evaluate the last integral. Then dω = df (x. y)dy = ∂M 3 ∂x g(x. Thus by Green’s Theorem. y)dy ∧ dx + ∂x ∂y ∂ ∂ = g(x. let ω = f (x. 2 0 (1 − 2 sin2 θ cos2 θ − sin θ cos θ)dθ ◭ In general. y)dx + g(x. observe that 1 = (sin2 θ + cos2 θ)2 = sin4 θ + 2 sin2 θ cos2 θ + cos4 θ. and using polar coordinates. We have dω = (dx − 3y 2 dy) ∧ dx + (3x2 dx) ∧ dy = d(x − y 3 ) ∧ dx + d(x3 ) ∧ dy = (3y 2 + 3x2 )dx ∧ dy. y) − ∂ ∂y f (x. y) ∧ dy ∂ ∂ = f (x. y) dxdy.Green’s. y)dx + g(x. y)dx + g(x. the above theorem takes the name of Gauß’ or the Divergence Theorem. In R . y)dy be a 1-form in R . Á The region M is the area enclosed by the circle x2 + y 2 = 1. Stokes’. and Gauss’ Theorems Solution: ◮ We will first use Green’s Theorem and then evaluate the integral directly. y)dx + f (x. if ω is a 2-form. y) ∧ dx + dg(x. y) − f (x. again resorting to polar coordinates. 2 Aliter: We can evaluate this integral directly. y)dy ∧ dy ∂x ∂y which gives the classical Green’s Theorem f (x. whence the integral equals 2π 0 (sin4 θ + cos4 θ − sin θ cos θ)dθ 2π = = 3π . C (x − y 3 )dx + x3 dy = = = 3π 0 (3y 2 + 3x2 )dxdy M 2π 1 0 3ρ2 ρdρdθ . y. z) → −y is an odd function of y and the domain of integration is symmetric with respect to y. dω = = (dx − dy) ∧ dy ∧ dz + dz ∧ dz ∧ dx − dy ∧ dx ∧ dy dx ∧ dy ∧ dz. y. z)dy ∧ dz + g(x. Σ Σ since (x. Similarly −ydxdy = 0. The integral becomes dxdydz M = 4π 3 (27) = 36π. y. Σ since (x. y. ◭ In general. z)dz ∧ dx + h(x. let ω = f (x. y. −3 |ρ| 9 − ρ2 dρdθ Also zdzdx = 0. 3 2π 0 xdydz Σ = = 36π. y. z)dx ∧ dy Free to photocopy and distribute 181 . z) → −y is an odd function of y and the domain of integration is symmetric with respect to y. Σ since (x. Solution: ◮ The region M is the interior of the sphere x2 +y 2 +z 2 = 9. Aliter: We could evaluate this integral directly.Chapter 3 301 Example Evaluate of the sphere S (x − y)dydz + zdzdx − ydxdy where S is the surface x2 + y 2 + z 2 = 9 and the positive direction is the outward normal. z) → z is an odd function of z and the domain of integration is symmetric with respect to z. Now. Now. We have (x − y)dydz = xdydz. z)dx ∧ dy ∂ ∂ ∂ = f (x. z)dx + f (x. x2 + y 2 = 3. z)dz ∧ dx ∧ dy ∂x ∂y ∂z ∂ ∂ ∂ = f (x. if ¾ f (x.e. Also. Stokes’. y. y. the integral evaluates to dxdy = 3π.. d S = dzdx . z = 1 implies dz = 0 and so dω = dxdy. z)dydz+g(x. z)dz ∧ dx + dh(x. i. z)dx + g(x. y. y. z) + ∂ ∂y g(x. y. the surface Σ on which we are integrating is the inside of the circle x2 + y 2 + 1 = 4. z) + g(x. z)dy + f (x. y. z) + ∂ ∂z h(x. y. y. z) dx ∧ dy ∧ dz. y. z) + h(x. z)dxdy = ∂M ∂x ¿ f (x. y. ∂x ∂y ∂z ∂ M which gives the classical Gauss’s Theorem f (x. z)dy + g(x. and Gauss’ Theorems be a 2-form in R3 . z)dx + h(x. z)dy ∧ dz + dg(x. Σ Free to photocopy and distribute 182 .Green’s. y. z)dzdx+h(x. Á 302 Example Evaluate ydx+(2x−z)dy+(z−x)dz where C is the intersection C 2 of the sphere x2 + y 2 + z = 4 and the plane z = 1. Σ Σ Since this is just the area of the circular region x2 + y 2 ≤ 3. z) . z)dy + h(x. z) ¿ ¾ →= − a then M → − g(x. y. Solution: ◮ We have dω = = = (dy) ∧ dx + (2dx − dz) ∧ dy + (dz − dx) ∧ dz dx ∧ dy + dy ∧ dz + dz ∧ dx. − a S ∂M dydz − (∇ • →)dV = a The classical Stokes’ Theorem occurs when ω is a 1-form in R3 . y. y. z = 1. −dx ∧ dy + 2dx ∧ dy + dy ∧ dz + dz ∧ dx Since on C. y. z)dz ∧ dy ∧ dz ∂x ∂y ∂z ∂ ∂ ∂ + g(x. y. z) h(x. y. y. Then dω = df (x. dxdy − → • d→. y. y. y. y. y. z)dz ∧ dz ∧ dx ∂x ∂y ∂z ∂ ∂ ∂ + h(x. y. z) dx Using classical notation. if ¾ ¿ ¾ ¿ ¾ ¿ ∂ g(x. y. y. r → − dS = dydz dxdy dzdx . then M → − − (∇ × →) • d S = a ∂M → • d→. z) − g(x. z)dy + f (x. y. z)dz ∂M = M ∂ ∂y + h(x. z) ∧ dz ∂ ∂ ∂ f (x. z)dz ∧ dy ∂x ∂y ∂z ∂ ∂ ∂ + h(x.1 Evaluate (2. z) − h(x. y. C x3 ydx + xydy where C is the square with vertices at (0. z) h(x. z) ∧ dx + dg(x. y. − − a r Homework Á Problem 3. y. z)dx + h(x. y. z) − ∂ f (x. y. y. z) dxdy. let ω = f (x. y. y.11. z)dx + g(x. z)dz ∧ dx ∂x ∂y ∂z ∂ ∂ ∂ + g(x. Free to photocopy and distribute 183 . z) ∧ dy + dh(x. y. z)dy + +h(x. y. z)dy + h(x. z) dz ∧ dx + ∂z ∂x ∂ ∂ g(x. − d→ = dy . y. z) . 0). y. y.Chapter 3 ◭ In general. z)dy + h(x. z) dxdy ∂z ∂x ∂ ∂ + h(x. y. z) dx ∧ dy ∂x ∂y = which gives the classical Stokes’ Theorem f (x. z) dydz f (x. z) dy ∧ dz ∂y ∂z ∂ ∂ f (x. z)dx + g(x. z)dx + g(x. z) − ∂ ∂z g(x. y. 2). z) − f (x. 0). y. 2) and (0. y. ∂x ∂y Using classical notation. y. z)dz be a 1-form in R3 . z) − f (x. z) dx dz →= − a g(x. y. z)dy + g(x. y. y. y. y. y. y. y. z)dz ∧ dz ∂x ∂y ∂z ∂ ∂ h(x. y. y. (2. Then dω = = df (x. y. y. z)dx + f (x. the upper face of the box U = {(x. 1 0 0 1−r 2 3. 3. z) ∈ R3 : 0 ≤ x ≤ 1. 0 ≤ y ≤ 1. 1.3 Problems 1 through 4 refer to the differential form ω = xdy ∧ dz + ydz ∧ dx + 2zdx ∧ dy. 1).11. z = 2}. y. z) ∈ R3 : 0 ≤ x ≤ 1. ∂M ω= −1 2π 1−x 2 −y 2 √ 4dzdydx. Prove that the integral on the upper face of the box is 1 1 0 U (arctan y − x2 )dydz + (cos x sin z − y 3 )dzdx + (2zx + 6zy 2 )dxdy = 4x + 12y 2 dxdy = 6. 0 ≤ y ≤ 1. Problem 3. Stokes’. and Gauss’ Theorems Problem 3. B : (1.11. the boundary of the box without the upper top S = ∂M \ U . 0 Free to photocopy and distribute 184 . and LB C . Prove that in Cartesian coordinates. z = 0.Green’s. 2). 0 ≤ z ≤ 2}. and the solid M whose boundaries are the paraboloid z = 1 − x2 − y 2 . 4. 0 ≤ z ≤ 1 and the disc x2 + y 2 ≤ 1. 1. Prove that dω = 3y 2 dx ∧ dy ∧ dz. M dω = 0 4rdzdrdθ.11. y. Prove that ∂M xdydz + ydzdx + 2zdxdy = 2π.2 Consider the triangle △ with vertices A : (0. Evaluate If LP Q denotes the equation of the line joining P and Q find LAB . C : (−2. 1 √ − 1−x 2 1−x 2 0 2. 2. LAC . Prove that in cylindrical coordinates. Á y 2 dx + xdy. and the differential form ω = (arctan y − x2 )dy ∧ dz + (cos x sin z − y 3 )dz ∧ dx + (2zx + 6zy 2 )dx ∧ dy. △ Find (1 − 2y)dx ∧ dy D where D is the interior of △. Prove that 2 0 0 1 0 1 ∂M (arctan y − x2 )dydz + (cos x sin z − y 3 )dzdx + (2zx + 6zy 2 )dxdy = 3y 2 dxdydz = 2. The surface ∂M of the solid is positively oriented upon considering outward normals. 0). Problem 3.4 Problems 1 through 4 refer to the box M = {(x. Here the boundary of the box is positively oriented considering outward normals. Prove that dω = 4dx ∧ dy ∧ dz. Problem 3. and C(0.5 Problems 1 through 3 refer to a triangular surface T in R3 and a differential form ω. 3). 12. Problem 3. The vertices of T are at A(6. as viewed from a point far above the triangle. Γ where Γ is the circle (x − 2)2 + y 2 = 4. Prove this directly by parametrising the boundary of the surface and evaluating the path integral. 0. 0).8 Use Green’s Theorem to evaluate Á (x3 − y 3 )dx + (x3 + y 3 )dy. 2.11. Also.6 Use Green’s Theorem to prove that (x2 + 2y 3 )dy = 16π. oriented clockwise when viewed from the origin.11. Free to photocopy and distribute 185 . Use Stoke’s Theorem to prove that ydx + zdy + xdz = −2π Γ √ 2. Prove that dω = ydy ∧ dz + (2x − y) dz ∧ dx + zdx ∧ dy. prove this directly by using a path integral. 0). ∂ M \U Problem 3. The boundary of of the triangle ∂T is oriented positively by starting at A.11. 3. Prove that the integral on the open box is y 3 )dzdx + (2zx + 6zy 2 )dxdy = −4.Chapter 3 (arctan y − x2 )dydz +(cos x sin z − 4.11. 2 1. Prove that 3 0 0 12−4z ∂T (2xz + arctan ex ) dx+(xz + (y + 1)y ) dy+ xy + 6 0 0 3−x/2 y2 + log(1 + z 2 ) dz = 2 ydydz + 2xdzdx=108.7 Let Γ denote the curve of intersection of the plane x + y = 2 and the sphere x2 − 2x + y 2 − 2y + z 2 = 0. 0. The surface T is oriented positively by considering the top of the triangle. continuing to B. The differential form is ω = (2xz + arctan ex ) dx + (xz + (y + 1)y ) dy + xy + y2 + log(1 + z 2 ) dz. following to C. and ending again at A. B(0. C where C is the positively oriented boundary of the region between the circles x2 + y 2 = 2 and x2 + y 2 = 4. Problem 3. Prove that the equation of the plane that contains the triangle T is 2x + y + 4z = 12. [B]. − → − → − → − → − → − → − → − → − → 1.1. Hence − → − → − → − → − → AC + BD = AD + BC = 2BC.2 We have 2BC = BE + EC. 0 .6 α = 4 3 4 2 1 . The zero vector 0 .1. has magnitude but no direction. 4 4 Now −→ − −→ − MA + 3MB = = = and −→ −→ − − 3MA + MB = = = 1.1.m = . ⇐⇒ − → − → 4IA = −3AB − → − → 4AI = 3AB − → → 3− AI = AB. 0 . 0 . − → − → − → − → − → − → − → 1. AD = BC.1. 0 . → − → − → − → − → − − 1.4 We have IA = −3IB ⇐⇒ IA = −3(IA + AB) = −3IA − 3AB. By Chasles’ Rule AC = AE + EC.A Answers and Hints → − 1. Thus we deduce − → − → − → IA + 3IA = −3AB ⇐⇒ ⇐⇒ Similarly → − → 1− JA = − JB 3 ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ . 2→ (= 2 d ) c 186 . [D].β = . 7 7 9 9 3 − → − → − → MI + IA + 3IB − → − → − → 4MI + IA + 3IB − → 4MI. We deduce that − → − → − → − → − → − → − → − → AC + BD = AE + EC + BE + ED = AD + BC.n = .1.5 x + 1 = t = 2 − y =⇒ y = −x + 1.1 No. l = . 4 − → − → 3JA = −JB − → − → − → 3JA = −JA − AB − → − → 4JA = −AB − → → 1− AJ = AB 4 − → − → − → − → 3MJ + 3JA + MJ + JB − → − → − → 4MJ + 3JA + JB − → 4MJ.8 [A]. and BD = BE + ED. 1. [E].1. − → → − → → 3− 1− Thus we take I such that AI = AB and J such that AJ = AB. [C]. − → − → But since ABCD is a parallelogram. 1.4. Horizontal Figure A.9 a c b −a é . we find (a1 . æ é a b æ é æ é b − a −1 a+b 1 = + .2 The desired transformations are in figures A.10.3: Horizontal and Vertical Stretch.8 through A.1 through A.2: Vertical Stretch.1: Stretch. 1. 2 2 1 1 1 1.4. b = − .1 Upon solving the equations −3a1 + 2a2 = 0.Appendix A 1.8 The transformations are shewn in figures A. bc = −a2 1. 2 1. Figure A.4.3.4. æ 1.1 a = −3.3 The desired transformations are shewn in figures through A. 3) or (a1 .7.4 A.3. a2 ) = (−2. −3).5. a2 + a2 = 13 1 2 Free to photocopy and distribute 187 . 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 Figure A.4. a2 ) = (2.2 Plainly. Figure A.5. 2 Figure A.5 4 3 2 1 0 -1 -2 -3 5 -4 4 3 2 1 0 -1 -2 -3 -4 -4 5 -3 4 -2 -1 0 1 2 3 4 5 -4 -3 -2 -1 0 1 2 3 4 5 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 Answers and Hints Figure A. i. we have a → − − ||→ + b ||2 a = = = = from where the desired result follows.4: Levogyrate rotaπ tion radians. (a2 · 1 + b2 · 1) ≤ (a4 + b4 )1/2 (12 + 12 )1/2 . 1.3 By the CBS Inequality. In particular.e. Figure A. − − → 1. a Free to photocopy and distribute 188 . a a a → − → − − But the norm of a vector is 0 if and only if the vector is the 0 vector. whence the assertion follows. → = b .4 We have ∀→ ∈ R2 .5. choosing → = → − b .7: Dextrogyrate ro3 π tation radians.5. 2 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 5 4 3 2 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 Figure A. we v v a v a gather → − − → − → − − − (→ − b )•(→ − b ) = ||→ − b ||2 = 0. Therefore → − b = a → − → − − 0 . a → − − → − − (→ + b )•(→ + b ) a a − − →•→ + 2→ •→ + → •→ − − − − a b b b a a → •→ − − → •→ + 0 + b b − − a a → − − ||→||2 + || b ||2 . → • (→ − b ) = 0.6: Dextrogyrate roπ tation radians. 4 Figure A.5: Levogyrate rotaπ tion radians. 2 4 1 0 -1 -2 -3 -4 -4 -3 -2 -1 0 1 2 3 4 5 Figure A. → − → − − − − − − 1.10: Reflexion about the origin.2 Since →• b = 0.8: Reflexion about the x-axis.9: Reflexion about the y-axis .. 7 The line L has a parametric equation x y Let L′ have parametric equation x y = 0 b +t = 0 1 +t æ 1 é . The lines are perpendicular if and only if. −1 æ 1 é . Now. 6 and −1 2 æ 1 é This gives two possible values for the slope of L′ . respectively.8 We must prove that → •→ = 0. → → − − v w − → − ¬¬ • ¬¬ → − v w − ¬¬→ ¬¬2 w → → − − v w − − →•→ − ¬¬ • ¬¬ → •→ − − v w w w − ¬¬→¬¬2 w → → − − →•→ − v • w ¬¬→¬¬2 − − v w w ¬¬→¬¬2 ¬¬− ¬¬ ¬¬− ¬¬ w → •→ − → •→ − − − − v w v w 0.5. − − 4u v giving the result. y = (−2 ± √ √ √ − − 1. m æ We need the angle between 1−m = 1 é m Ô π to be and so by Theorem 21.5.5. 1.5 We have − − − − ||→ + → ||2 − ||→ − → ||2 u v u v = = = − − − − − − − − (→ + → )•(→ + →) − (→ − →)•(→ − → ) u v u v u v u v → •→ + 2→ •→ + → •→ − (→ •→ − 2→ •→ + →•→ ) − − − − − − − − − − − − u u u v v v u u u v v v → •→ . since L′ must pass through 3)x + b =⇒ 2 = (−2 ± 3)(−1) + b =⇒ b = ± 3 √ √ √ √ and the lines are y = (−2 + 3)x + 3 and y = (−2 − 3)x − 3.6 A parametric equation for L1 is x y A parametric equation for L2 is x y = 0 b2 +t = 0 b1 +t æ 1 m1 é . a w − •→ w = = = = Free to photocopy and distribute 189 . according to Corollary 22. æ é æ é 1 1 • = 0 ⇐⇒ 1 + m 1 m 2 = 0 ⇐⇒ m 1 m 2 = −1.Appendix A 1. æ 1 m2 é . Using the distributive law for the dot product. m1 m2 1. 6 −1 √ √ π 1 + m 2 2 cos =⇒ m = −2 ± 3.5. this is a straight line with positive slope. y = log cos t. and finishes in the 2 2 middle point (0. = tan θ =⇒ dt = 1 sec2 θdθ. This gives (dx)2 + (dy)2 = To integrate 1 2 4 t− √ 2 2 0 we now use the trigonometric substitution 2 t− 1 2 1 2 Ô 2 4t2 + (2 − 2t)2 dt = 2 2t2 − 2t + 1 dt = √ 2 4 t− 1 2 2 + 1 dt. This is one branch of a hyperbola. this is the line segment on the plane joining (−1. 1]. 0 ≤ t ≤ Then (dx)2 + (dy)2 = (1 + tan2 t)(dt)2 = sec2 t(dt)2 .Answers and Hints 1.6. we want the portion of the line y = 2x + 1 with −1 ≤ x ≤ 1 (and. 3 Hence the arc length is sec tdt = log(2 + 0 √ 3). π /3 π . Since in the interval [0. −1 ≤ x ≤ 1. 2 − 2 = 1. + y5 y−x and so 4 y−x 4 3 −2 y−x 4 5 x5 If t = 0.2 Observe that for t = {0. 3) again at t = . −1}. 1. 0). 1) at t = . y = (1 − t)2 . ay − cx = ad − bc.6. 1) when t = 4π. it goes to its low point (−1.6. This is a hyperbola. so the trace is part of this line.6. + 1 dt. t ∈ [0. x > 0. thus −1 ≤ y ≤ 3). 2 Free to photocopy and distribute 190 . Again. reaches its high 2 2 5π 7π point (1. 1] and [0. 0) to (1. y 2 y x = t =⇒ x = y x 1+ x =⇒ x = y 2 x3 =⇒ x5 + y 5 = x2 y 2 .4 We may simply give the trivial parametrisation: x = t. 1].6. √ √ 1.6. x2 y2 − 2 = 1. 2 a b y2 x2 4.3 1. −1 ≤ sin t ≤ 1. One is x = t2 . reaches its low point (−1. y = 0. What happens as t → −1? 1.5 Observe that y = 2x + 1. 3) at t = .6 First observe that x + y = 1 demands x ∈ [0. then x = 0. 1. The curve starts at the middle point (0. reaches π 3π the high point (1. y = 0 and our Cartesian equation agrees. 2. 1. 4π]. 1) at t = . 1) (at t = 0).1 We have y − x = 4t =⇒ t = x= is the Cartesian equation sought. a b 3. one can give many parametrisations. 1.Appendix A The integral thus becomes √ π /4 2 0 sec3 θdθ.7 First notice that x = 1 =⇒ t3 + 1 = 1 =⇒ t = 0 and x = 2 =⇒ t3 + 1 = 2 =⇒ t = 1. and it appears in most Calculus texts: √ π /4 2 0 sec3 θdθ = = = √ 1 1 sec θ tan θ − log | sec θ + tan θ| 2 2 √ √ 1 √ 1 2 · 2 + log( 2 + 1) 2 2 √ 1 √ log( 2 + 1) + 1. The area is given by æ é Á Á x dx 1 1 det = xdy − ydx 2 Γ 2 y dy 0 4 = (sin3 t(− sin t(1 + sin2 t) + 2 sin t cos2 t) 2 π /2 − cos t(1 + sin2 t)(3 sin2 t cos t))dt 0 = = = 2 π /2 0 (−3 sin2 t + sin4 t)dt − 9 1 + cos 2t + cos 4t dt 8 8 2 9π .9 Observe that dx = tdt. the arc length is 1/2 (1 + t)dt = −1/2 t+ t2 2 1/2 = 1.8 Observe that dx = 6tdt.10 Observe that the parametrisation traverses the curve once clockwise if t ∈ [0. (You write sec θd tan θ = tan θ sec θ − tan θd sec θ. dy = √ 2t + 1dt =⇒ (dx)2 + (dy)2 = Ô √ t2 + ( 2t + 1)2 dt = t2 + 2t + 1dt = (t+1)dt. (dx)2 + (dy)2 = t=0 t=0 6t Ô √ 1 + t2 dt = 4 2 − 2. and so the arc length is t=1 t=1 dy = 6t2 dt =⇒ (dx)2 + (dy)2 = 6t Ô 1 + t2 dt. 8 π /2 Free to photocopy and distribute 191 .6. as I assume most of you have seen it.6.6. etc. 2π]. The area under the graph is t=1 t=1 t=1 ydx = t=0 t=0 (1 − t2 )d(t3 + 1) = t=0 3t2 (1 − t2 )dt = 2 . which is a standard example of integration by parts where you “solve” for the integral. 2 2 π /4 0 1.6. the famous secant cube integral. Hence. −1/2 1. 5 1.) I will simply quote it. If the gun were fired at t = 0. and the airplane is on a point with coordinates (u. x (1 + t3 )2 1. and P = (0. Then the centre of the circle has displaced rθ units horizontally. This is the desired parametrisation. h) with u ≥ 0. and so is now located at (ρθ.Answers and Hints 1. ρ). dx = and dy = Hence xdy−ydx = 9t2 (1 + t3 ) 9t2 − 18t5 9t2 + 9t5 9t2 18t2 − 9t5 ·dt− ·dt = ·dt = ·dt = ·dt. d cos θ).12 See figure 1.14 We have dx = et (cos t−sin t)dt. and let C be the centre of the circle. then x → 0 and y → 0. As t → +∞.6. 1. The polar coordinates of the point P are (d sin θ. in relation to the centre of the circle (notice that the circle moves clockwise). Let θ be the angle (in radians) of rotation of the circle.11 Using the quotient rule. 1. we must have u = V t cos a. 2 where a is the angle of elevation. At θ = 0 the centre of the circle is at (0.6. the area is given by 1 2 +∞ 0 9t2 3 · dt = (1 + t3 )2 2 +∞ 0 3t2 3 dt = (1 + t3 )2 2 +∞ 1 du 3 = .15 Choose coordinates so that the origin is at the position of the gun. Suppose the circle is displaced towards the right.58. 2 Free to photocopy and distribute 192 . t is the time and g is the acceleration due to gravity. ρ). h = V t sin a − gt2 . Since we know that the shell strikes the plane. then x = V t cos a.6.6. making the point P to rotate an angle of θ radians. u2 2 A shorter way of obtaining xdy − ydx would have been to argue that y 9t2 xdy − ydx = x2 d = dt. The point P has moved x = ρθ − d sin θ horizontal units and y = ρ − d cos θ units. (1 + t3 )3 (1 + t3 )3 (1 + t3 )3 (1 + t3 )3 (1 + t3 )2 6t(1 + t3 ) − 3t2 (3t2 ) 6t − 3t4 18t2 − 9t5 · dt = · dt =⇒ xdy = · dt 3 )2 3 )2 (1 + t (1 + t (1 + t3 )3 3(1 + t3 ) − 3t2 (3t) 3 − 6t3 9t2 − 18t5 · dt = · dt =⇒ ydx = · dt (1 + t3 )2 (1 + t3 )2 (1 + t3 )3 Observe that when t = 0 then x = y = 0. y = V t sin a − gt2 . ρ − d). Hence to obtain the loop Using integration by substitution (u = 1 + t3 and du = 3t2 dt) . the y-axis is vertical. dy = et (sin t+cos t)dt =⇒ The arc length is thus π (dx)2 + (dy)2 = et (cos t − sin t)2 + (sin t + cos t)2 dt = (dx)2 + (dy)2 = 0 √ π 2 0 et dt = √ 2(eπ − 1). The slope of the tangent at P is . a ¬¬ ¬¬ − ¬¬→ →¬¬ − ¬¬ a − b ¬¬ = x 1.6. y= 2x0 The line normal to this line and passing through the origin is hence y = −2xx0 .Appendix A whence u2 + h + and thus 1 2 gt 2 2 = V 2 t2 . Free to photocopy and distribute 193 .16 Suppose the parabolas have a point of contact P = (4px2 . 1. g 2 t4 + (gh − V 2 )2 t2 + h2 + u2 = 0. 0 the vertex V of the rolling parabola is the reflexion of the origin about the line tangent to 1 their point of contact.1 Observe that. eliminating t yields x=− or 2p y ¡2 y ¡2 x x 1+ (x2 + y 2 )x + 2py 2 = 0. 4px0 ).7. and so the lines intersect at − from where V = − 4px2 8px3 0 0 . By symmetry. from where the equation of 2x0 the tangent is x + 2px0 . Ǒ = 2 3 + t −1 . ¬¬ ¬¬ ¬¬ ¬¬ − − ¬¬→ →¬¬2 ¬¬→ →¬¬2 − − = ¬¬ a − b ¬¬ + ¬¬ a + b ¬¬ = whence ¬¬ ¬¬2 ¬¬ ¬¬ ¬¬ ¬¬2 ¬¬→¬¬2 − − − − − − − ¬¬− ¬¬ − 2→ •→ + ¬¬→ ¬¬ + ¬¬→ ¬¬2 + 2→•→ + ¬¬→ ¬¬ a a b a a b ¬¬ b ¬¬ ¬¬ b ¬¬ ¬¬ ¬¬2 ¬¬− ¬¬2 − ¬¬→¬¬ 2¬¬→¬¬ + 2¬¬ b ¬¬ .2 y z ¬¬ ¬¬2 ¬¬ ¬¬ ¬¬− ¬¬2 − → ¬¬ − ¬¬→¬¬ ¬¬− 2¬¬→¬¬ + 2¬¬ b ¬¬ − ¬¬→ + b ¬¬ = a a Ǒ 1 ¾ ¿ −2 0 2(13)2 + 2(19)2 − (24)2 = 22.7. in general. As −2x0 x(t) = y(t). y(t)). 1 + 4x2 1 + 4x2 0 0 . from where the assertion follows. 2 1 + 4x0 1 + 4x2 0 := (x(t). giving the equation of the locus. 1. 4 The quadratic equation in t2 has a real root if (gh − V 2 )2 ≥ g 2 (h2 + u2 ) =⇒ g 2 u2 ≤ V 2 (V 2 − 2gh). 8px2 16px3 0 0 . two b × c × 1 bricks. four quarter-cylinders of length b and radius 1. the vectorial form of the equation of the line is ¾ ¿ ¾ ¿ ¾ ¿ 0 1 1 →= − r 0 + t 1/2 = t 1/2 .7.7. z = t/3. Four quarter-cylinders of length a and radius 1. 6 3 2 1. and thus ¾ 1 ¿ × ¾ 1 ¿ = ¾ 1/6 3/2 ¿ −1 −1 1/2 1/3 −4/3 is normal to the plane. 0 1/3 1/3 ¾ This means that 1 1/2 1/3 ¿ also lies on the plane. of volume abc.4 Observe that Thus the vector 0 0 (x − 1) − 2(y − 1) − (z − 1) = 0. Two a × b × 1 bricks. ¾ −1 − 0 −1 − 0 1−0 ¿ = ¾ 1 ¿ −1 −1 lies on the plane. Free to photocopy and distribute 194 . y = t/2. The desired equation is thus 1 4 3 x − y + z = 0. then x = t. and hence on the plane. and two c × a × 1 bricks. this means that −2 −1 Ǒ 0 1. 1 ¾ Since the line follows the direction of 1 ¿ −1 ¾ 1 ¿ is normal to the plane. if x = 2y = 3z = t.Answers and Hints 1.7.5 The set B can be decomposed into the following subsets: The set A itself.3 The vectorial form of the equation of the line is ¾ ¿ ¾ ¿ 1 1 →= − r 0 + t −2 . −1 and thus the equation of the desired plane is −2 . Eight eighth-of-spheres of radius 1. and four quarter-cylinders of length c and radius 1. Now. Hence. (as 0 = 2(0) = 3(0)) is on the line. so x = t/a. from where we deduce that 2 2 2 − − →•→ + →•→ + → •→ = −3 − 4 − 5 = −25. ¾ 1. − − − − a b b c c a 2 ¾ ¿ a 1. the shortest distance from Free to photocopy and distribute 195 . 1. this perpendicular line to the plane has equation ¾ ¿ Ǒ Ǒ x 1 1 y z = 2 3 + t −1 1 =⇒ x = 1 + t. Hence.7.9 The vector 1 ¿ 1 (1.7. Therefore. The line sought has the same direction as a2 this vector.7. ¾ Thus the vector is 1/a 1/b 1/c ¿ is perpendicular to the plane. z t ∈ R.7.8 Put ax = by = cz = t. the equation of the plane ¾ 1/a 1/b 1/c ¿ ¾ • x−1 y−1 z−1 ¿ = ¾ ¿ 0 0 . y = t/b. 3 1. a b c a b c bcx + cay + abz = ab + bc + ca. y = 2 − t.6 We have. −1 is perpendicular to the plane. + t a2 . 3) is obtained by the perpendicular line to the plane that pierces the plane. z = t/c. thus the equation of the line is ¾ ¿ ¾ ¿ x 0 y z = 0 1 ¾ a a 2 ¿ t ∈ R. The parametric equation of the line is ¾ ¿ ¾ ¿ x 1/a y =t 1/b 1/c . z = 3 + t. ¬¬ ¬¬ ¬¬→ ¬¬2 ¬¬→¬¬2 ¬¬→¬¬2 − ¬¬→ → →¬¬2 ¬¬− ¬¬ → → → − − − → → → − − − → → → → → → − − − − − − − − − − ¬¬ a + b + c ¬¬ = ( a + b + c )•( a + b + c ) = ¬¬ a ¬¬ +¬¬ b ¬¬ +¬¬ c ¬¬ +2( a • b + b • c + c • a ).Appendix A Thus the required formula for the volume is abc + 2(ab + bc + ca) + π(a + b + c) + 4π .7 A vector normal to the plane is a2 . 0 or We may also write this as x y z 1 1 1 + + = + + . 2. From problem (a2 + b2 + c2 )2 and thus maximising f is equivalent to minimising 2 1. y3 ) let θ be the angle between them. 1− which in turn gives √ 2 1 3 = √ = . This gives (since squares are positive. 3) is therefore is 2 7 8 . 3 The closest point on the plane to (1. c) 1 = a 2 + b2 + c 2 4 1 4 1 4 s(s − a)(s − b)(s − c) (a + b + c)(b + c − a)(c + a − b)(a + b − c) (a2 + b2 + c2 )2 − 2(a4 + b4 + c4 ) Ö 1−2 a 4 + b4 + c 4 .7. and so the lines are skew.Answers and Hints The intersection of the line and the plane occurs when 1 1 + t − (2 − t) + (3 + t) = 1 =⇒ t = − . b.7. b. which proves the claim at once.11 Observe the CBS Inequality in R3 given the vectors → = (x1 .12 First observe that S(a.11. (1 − )2 + (2 − )2 + (3 − )2 = 3 3 3 3 1. 3 − − 1. → = x y (y1 . 3 12 4 3 Free to photocopy and distribute 196 . x3 = c2 and y1 = y2 = y3 = 1. c) = = = Hence S(a. a 4 + b4 + c 4 1 ≥ .7. (a2 + b2 + c2 )2 3 S(a. 2. Then → •→ = ¬¬→ ¬¬¬¬→ ¬¬ cos θ =⇒ |x y + x y + x y | ≤ − − ¬¬− ¬¬¬¬− ¬¬ x y x y 1 1 2 2 3 3 x2 + x2 + x2 1 2 3 2 2 2 y1 + y2 + y3 . Let θ be the angle between them. 1. and the distance sought 3 3 3 √ 7 8 2 3 . which is clearly impossible. x2 = b2 . c) 1 ≤ a 2 + b2 + c 2 4 the desired maximum. x2 .7. y2 . . Now take x1 = a2 . Then cos θ = Ô 2 · 2 + 1 · (−1) + 1 · 1 4 Ô = √ √ =⇒ θ = arccos (2)2 + (1)2 + (1)2 (2)2 + (−1)2 + (1)2 6 6 2 . . we don’t need the absolute values) |x1 y1 +x2 y2 +x3 y3 | ≤ x2 + x2 + x2 1 2 3 2 2 2 y1 + y2 + y3 =⇒ (a2 +b2 +c2 )2 ≤ (a4 +b4 +c4 )(3). b.10 If the lines intersected. there would be a value t′ for which ¾ ¿ ¾ ¿ ¾ ¿ ¾ ¿ Ǒ Ǒ 1 2 0 2 1−0 2−2 1 1 + t′ 1 1 = 0 1 + t′ −1 1 =⇒ 1−0 1−1 = t′ −1 − 1 1−1 ¾ ¿ 1 =⇒ 1 0 = t′ ¾ 0 0 ¿ −2 . x3 ). (a2 + b2 + c2 )2 a 4 + b4 + c 4 . 8).11: Problem 1. 1 . V2 V6 .7. V2 = (0. 0.8. V1 V6 . 0).2 We have → × → = −→ × → by letting → = → in ??.7. 0. 1+1+1−1−1−1 which is impossible. → are three unit vectors in R3 such that the a c − − 2π 1 → − 1 − → − − angle between any two of them is > . → •→ = →•→ .Appendix A 1. 6 − − → − 1. Thus 2→ × → = 0 and x x x x y x x x → − − − hence → × → = 0 . z x Figure A. 2). and so r0 n a n r ||projn0 − b || = → − → − → − − → − − → − − − − − − − | → •→ − b •→ | r0 n n |(→ − b )•→ | a n |→•→ − b •→ | a n n ¬¬→ ¬¬ ¬¬→¬¬ ¬¬→¬¬ = . b . and V4 V6 ). 1.18 There are 7 vertices (V0 = (0. V1 = (11. V0 V2 . and → •→ < a c c a 3 2 2 1 − . Then bx0 = x0 b distance we seek is ¬¬ ¬¬ → − − − ¬¬ → → •→ ¬¬ − → −→ − − |(→ − b )•→| r0 n r0 n − ¬¬ (− − b ) − → ¬¬ r0 b ¬¬→¬¬ n ¬¬ = . 3. 9. 0). b •→ < − . and the − on the plane that is nearest to b.16 Assume contrariwise that → . Then →• b < − . V1 V5 . y → − − − − − − − − − 1. 0. 8). x x Free to photocopy and distribute 197 . V5 = (9. 1. 7. Thus 2 ¬¬ ¬¬ ¬¬− ¬¬2 ¬¬→¬¬2 ¬¬− ¬¬2 − ¬¬→ → → ¬¬2 ¬¬− ¬¬ − − = ¬¬→ ¬¬ + ¬¬ b ¬¬ + ¬¬→¬¬ a c ¬¬ a + b + c ¬¬ − − → •→ + 2→ •→ + 2→ •→ − − − − +2 a b b c c a < = 0. V4 = (0. 0)) and 11 edges (V0 V1 . since a norm of vectors is always ≥ 0.17 Let projt = s → •→ − − t s → − − − be the projection of → over → = →.7. 0). = − ¬¬| − ¬¬ − ¬¬ ¬¬ n ¬¬ n ¬¬ n as we wanted to shew. V3 V5 .7. V3 V4 . Let x be the point − s t s 0 0 (||s||)2 − → − → − → is orthogonal to the plane. V6 = (4. ||projn || = ¬¬ ¬¬→ ¬¬2 ¬¬− ¬¬ ¬¬− ¬¬ ¬¬ ¬¬ n n − − − − Since R0 is on the plane. V0 V3 . V4 V5 . V3 = (0.15 x + y + z = 1. V2 V4 .7. 0.18. as obtained before.4 The vector is normal to the plane. The equation of the plane is thus given by ¾ ¿ ¾ ¿ −2a x−a −3a2 a • z−a y−0 = 0. Now − − − − || v w v w − − − − → × → = (−a→ + a→) × (→ + a→) − − v w j k i j → → − − → → − − → → − − → → − − 2 = −a( j × i ) − a ( j × j ) + a( k × i ) + a2 ( k × j ) → − → − → − = a k + a j − a2 i Ǒ −a2 = a a . 1.8. c a c a ¾ ¿ ¾ ¿ ¾ ¿ 1 1 −1 1 × 1 1 = 1 0 0 ¿ ¾ ¿ −1 • 1. A vector normal to the plane is ¾ ¿ ¾ ¿ ¾ ¿ 2a a −2a −1 a × 0 = 2a a −3a2 .5 The vectors ¾ a − (−a) a−0 0−1 ¿ = ¾ 2a a ¿ −1 and ¾ 0 − (−a) 2a − 0 1−1 ¿ = ¾ a 0 2a ¿ lie on the plane. →×→ − − →×→ − − v w v w → × →|| or − ||→ × → || will do.Answers and Hints 1.6 Either of Free to photocopy and distribute 198 . The plane has thus equation ¾ x z−2 y+1 1 0 = 0 =⇒ −x + y + 1 = 0 =⇒ x − y = 1.8. that is. 1.3 One has → − − → → − − − → − → − − − − − − ( b − → ) × (→ − →) = 0 =⇒ → × b + b × → + → × → = 0 a c a a c c a This gives → → − → − − → → → → − − − − → → − − − − b × − = −(→ × b + → × →) = −( j + k + i − j ) = − i − k .8. 2ax + 3a2 y − az = a2 .8. putting x a c z → − → − − − (→ × b ) • (→ × d ) a c = → − → − − − ((→ × b ) × → ) • d . x x x x ¬¬− ¬¬2 − − − − → − → − → x and since (→• i )2 +(→• j )2 +(→• k )2 = ¬¬→ ¬¬ = 1.12 → − → → → − − − → → − − → = ( a • b ) a + 6 ¬b +¬2 a × c − x ¬→¬ 2 12 + 2¬¬− ¬¬ a − → → → − − − → − → → − → = (a•c)a +6c +3a × b − y ¬¬→¬¬2 18 + 3¬¬− ¬¬ a 1.8 By Lagrange’s Identity. x x x x ¬¬− ¬¬2 ¬¬→¬¬2 → − − − ¬¬− ¬¬ − − → − → ||→ × j ||2 = ¬¬→¬¬ ¬¬ k ¬¬ − (→ • k )2 = 1 − (→• k )2 . 1.8. x x x x ¬¬− ¬¬2 ¬¬→¬¬2 → − − − ¬¬− ¬¬ − − → − → ||→ × k ||2 = ¬¬→¬¬ ¬¬ j ¬¬ − (→ • j )2 = 1 − (→• j )2 . ¬¬− ¬¬2 ¬¬→¬¬2 → − − − ¬¬− ¬¬ − − → − → ||→ × i ||2 = ¬¬→¬¬ ¬¬ i ¬¬ − (→ • i )2 = 1 − (→• i )2 .8. →. y = → and → = d we gather that − − − − Now. x x x 1. − − − − − − c a c c a b and adding yields the result. → .8. Hence we may take either v w Ǒ −a2 1 √ 2 a 2a + a4 a or −√ 1 2a2 + a4 −a2 a a Ǒ .9 − − − − − − − − − → ×(→ ×→) = →×(→ ×→ ) ⇐⇒ (→•→ )→−(→ •→)→ = (→ •→)→ −(→•→ )→ ⇐⇒ →•→ = → •→ = 0. − − − − − − x y z x y z − − − This is so because both sides give the volume of the parallelogram spanned by →. x y z → → − − → − → = → × b .13 First observe that → • (→ × →) = (→ × →) • →.8. a c Free to photocopy and distribute 199 .8. x x a 1. − − − − − − − − a b c a c b a b c → − → − − − → − − − − − b × (→ × → ) = ( b •→ )→ − ( b •→ )→ .7 From Theorem ?? we have − → × (→ × → ) = (→ •→ )→ − (→ •→ )→ . c a a c c a → − → → − − → × (→ × b ) = (→ • b ) a − (→ •→ )→ . the desired sum equals 3− 1 = 2. − − − − − − − − − − − − − a x b b x a a b x a x b b a x b x a a x b x → − − − − The answer is thus {→ : → ∈ R→ × b }. 1.Appendix A Ô Ô − − and ||→ × →|| = a4 + a2 + a2 = 2a2 + a4 . Answers and Hints Now, again, → − − − − − − − − − − → − → − − − → − − → − → − (→ × b ) × → = −→ × (→ × b ) = −((→ • b )→ − (→•→ ) b ) = (→ •→) b − (→ • b )→ . a c c a c a c a c a c a This gives → − → − − − → − − − − − − − − − → − → − − − → → − → − → ((→ × b ) × →) • d = ((→ •→) b − (→ • b )→ ) • d = (→•→ )( b • d ) − (→ • b )(→ • d ), a c c a c a c a c a proving the identity. 1.8.14 By problem 1.8.13, − − − − − − − − − − − − y u (→ × →) • (→ × →) = (→•→ ) • (→•→ ) − (→ •→) • (→ •→). x y u v x u y v x v Using this three times: → − − → − − ( b × →) • (→ × d ) c a → → − − − − (→ × →) • ( b × d ) c a → − → − − − (→ × b ) • (→ × d ) a c = = = → − − − → → − − − − → − → ( b •→) • (→ • d ) − ( b • d ) • (→ • d ) a c c − − − − − → − → − → − → (→ • b ) • (→ • d ) − (→ • d ) • (→ • b ) c a c a → •→ − − → − → •→ − − − − − (→ •→) • ( b d ) − (→ • d ) • ( b c ) a c a Adding these three equalities, and using the fact that the dot product is commutative, we see that all the terms on the dextral side cancel out and we obtain 0, as required. 1.8.15 1. We have ¾ − → CA = 6 0 ¿ , − → CB = ¾ 0 4 ¿ − → − → → − → − → − → − → − → − → − =⇒ CA×CB = (6 i −3 k )×(4 j −3 k ) = 24 k +18 j +12 i = ¾ 12 24 ¿ 18 . −3 2. We have −3 ¬¬ ¬¬ Ô √ √ → − ¬¬ → ¬¬− ¬¬CA × CB¬¬ = 122 + 182 + 242 = 1044 = 6 29. 3. The desired line has equation ¾ ¿ Ǒ Ǒ x 0 6 y z = 0 3 +t 0 −3 =⇒ x = 6t, y = 0, z = 3 − 3t. 4. The desired line has equation Ǒ Ǒ x 3 y z = 0 0 +s ¾ 0 0 ¿ =⇒ x = 3, y = 0, z = −3s. −3 5. From the preceding items, the line LC A is x = 6t, y = 0, z = 3 − 3t and the line LD E is x = 3, y = 0, z = −3s. If the line intersect then 6t = 3, 0 = 0, 3 − 3t = 1 1 3 −3s gives t = and s = − . The point of intersection is thus 3, 0, . item The 2 2 2 area is ¬¬ ¬¬ √ √ → − ¬¬ → 1 ¬¬− 1 ¬¬CA × CB¬¬ = · 6 29 = 3 29. 2 2 Free to photocopy and distribute 200 Appendix A 6. Observe that 3 P = 0 z Ǒ , Q= 3 y 0 Ǒ , R= x 3 0 Ǒ , S = 0 3 z Ǒ . Since all this points lie on the plane 2x + 3y + 4z = 12, we find 3 2(3) + 3(0) + 4z = 12 =⇒ P = 0 3 2 Ǒ , Ǒ 3 2 0 3Ǒ 2 , 3 0 , 2(3) + 3y + 4(0) = 12 =⇒ Q = 2x + 3(3) + 4(0) = 12 =⇒ R = 0 2(0) + 3(3) + 4z = 12 =⇒ S = Ǒ 3 . 3 4 7. A possible way is to decompose the pentagon into three triangles, say △CP Q, △CQR and △CRS and find their areas. Another way would be to subtract from the area of △ABC the areas of △AP Q and △RSB. I will follow the second approach. Let [△AP Q], [△RSB] denote the areas of △AP Q and △RSB respectively. Then ¬¬¾ ¿ ¾ ¬¬¾ ¿¬¬ ¿¬¬ ¬¬ 3 ¬¬ −3 ¬¬ 0 ¬¬ ¬¬ ¬¬ ¬¬ ¬¬ ¬¬ ¬¬ → − ¬¬ → 1 ¬¬− 1 ¬¬ 1 2 ¬¬ = ¬¬ 9 ¬¬ = [△AP Q] = ¬¬PA × PQ¬¬ = ¬¬ 0 × ¬¬ ¬¬ ¬¬ 2 2 ¬¬ 3 2 ¬¬ 2 ¬¬ 3 ¬¬ ¬¬ ¬¬ ¬¬ 6 ¬¬ − 2 2 ¬¬¾ ¿¬¬ ¬¬¾ ¿ ¾ ¬¬ 3 ¬¬ ¿¬¬ ¬¬ 3 ¬¬ ¬¬ ¬¬ 0 ¬¬ ¬¬ ¬¬ ¬¬ ¬¬ 4 ¬¬ 2 ¬¬ → − ¬¬ → 1 ¬¬ 1 ¬¬ 9 ¬¬ 1 ¬¬− 1 ¬¬ = ¬¬ [△RSB] = ¬¬SR × SB¬¬ = ¬¬ 0 × ¬¬ = ¬¬ 2 2 ¬¬ 2 ¬¬ 8 ¬¬ 3 ¬¬ ¬¬ 3 ¬¬ 3 ¬¬ − ¬¬ − ¬¬ ¬¬ ¬¬ 4 4 2 Hence the area of the pentagon is √ 3√ 3 √ 33 √ 3 29 − 29 − 29 = 29. 4 16 16 1 2 9+ 81 3√ + 36 = 29, 4 4 1 2 9 81 9 3 √ + + = 29. 16 64 4 16 1.9.2 The assertion is trivial for n = 1. Assume its truth for n − 1, that is, assume An−1 = 3n−2 A. Observe that ¾ ¿¾ ¿ ¾ ¿ 1 1 1 1 1 1 3 3 3 A2 = 1 1 1 1 1 1 1 1 1 1 1 1 = 3 3 3 3 3 3 = 3A. Free to photocopy and distribute 201 Answers and Hints Now An = AAn−1 = A(3n−2 A) = 3n−2 A2 = 3n−2 3A = 3n−1 A, and so the assertion is proved by induction. 1.9.3 First, we will prove that ¾ 1 0 0 2 1 0 ... 0 3 2 1 . . . 0 4 3 2 . . . 0 ... ... ... . . . ... n−1 n−2 n−3 ... 0 n ¿ A = 2 n−1 ... 1 n−2 . ... 0 Observe that A = [aij ], where aij = 1 for i ≤ j and aij = 0 for i > j. Put A2 = [bij ]. Assume first that i ≤ j. Then n j bij = k=1 aik akj = k=i 1 = j − i + 1. Assume now that i > j. Then n n bij = k=1 aik akj = k=1 0 = 0, proving the first statement. Now, we will prove that ¾ (n − 1)n 1 3 6 10 . . . 2 (n − 2)(n − 1) 0 1 3 6 ... 2 (n − 3)(n − 2) A3 = 0 0 1 3 ... 2 . . . . . . ... ... . . . ... 0 0 0 0 ... 0 n(n + 1) 2 (n − 1)n 2 (n − 2)(n − 1) 2 ... 1 ¿ . For the second part, you need to know how to sum arithmetic progressions. In our case, we need to know how to sum (assume i ≤ j), j j S1 = k=i a, S2 = k=i k. The first sum is trivial: there are j − i + 1 integers in the interval [i; j], and hence j j S1 = k=i a = S1 = a k=i 1 = a(j − i + 1). For the second sum, we use Gauß trick: summing the sum forwards is the same as summing the sum backwards, and so, adding the first two rows below, S2 S2 2S2 2S2 = = = = i j (i + j) (i + j)(j − i + 1) + + + i+1 (i + j) j−1 + + + i+2 (i + j) j−2 + + + ··· ··· ··· + + + j−1 + + + j i (i + j) , ··· (i + j) i−1 Free to photocopy and distribute 202 Appendix A (i + j)(j − i + 1) . 2 which gives S2 = Put now A3 = [cij ]. Assume first that i ≤ j. Since A3 = A2 A, n cij = k=1 j bik akj (k − i + 1) j j = k=i j = k=i k− i+ k=i k=i 1 = = (j + i)(j − i + 1) − i(j − i + 1) + (j − i + 1) 2 (j − i + 1)(j − i + 2) . 2 Assume now that i > j. Then n n cij = k=1 bik akj = k=1 0 = 0. This finishes the proof. 1.9.4 For the first part, observe that ¾ m(a)m(b) = 1 −a 0 1 0 1 ¾ = 0 a a2 − 2 1 0 1 0 0 1 0 ¿¾ 1 −b 0 0 1 0 −a − b 0 a+b a2 b2 − − + ab 2 2 1 a+b (a + b)2 − 2 1 b b2 − 2 1 ¿ ¿ ¾ = 1 −(a + b) 0 ¿ = m(a + b) For the second part, observe that using the preceding part of the problem, ¾ ¿ 1 0 0 2 m(a)m(−a) = m(a − a) = m(0) = −0 1 − 0 = I3 , 2 0 0 1 giving the result. a π 1.11.1 √ , . 2 4 Free to photocopy and distribute 203 Answers and Hints a 1.11.3 √ 3 1.12.1 Consider a right triangle △ABC rectangle at A with legs of length CA = h and AB = r, as in figure A.12. The cone is generated when the triangle rotates about CA. The gyrating curve is theÔ hypotenuse, whose centroid is its centre. The length of the generating curve is thus r 2 + h2 and the length of curve described by the centre of Ô r gravity is 2π = πr. The lateral area is thus πr r 2 + h2 . 2 C r/2 ′ h/3 G G B A Figure A.12: Generating a cone. rh . We need 2 to find the centroid of the triangle. But from example 17, we know that the centroid G of the triangle is is two thirds of the way from A to the midpoint of BC. If G′ is the perpendicular projection of G onto [CA], then this means that G′ is at a vertical height h 2 h GG′ h/3 r of · = . By similar triangles = =⇒ GG′ = . Hence, the length of the 2 3 3 r h 3 2 curve described by the centre of gravity of the triangle is πr. The volume of the cone is 3 2 rh π thus πr · = r 2 h. 3 2 3 −− −→ −− −→ − 1.14.1 Find a vector → mutually perpendicular to V1 V2 and V1 V3 and another vector a → − −− −→ −− −→ 1 and a vector b mutually perpendicular to V1 V3 and V1 V4 . Then shew that cos θ = , 3 → − − where θ is the angle between → and b . a Ǒ x To find the volume, we gyrate the whole right triangle, whose area is 1.15.1 Let y z ellipse, and its distance to the axis of rotation would be |x| = be a point on S. If this point were on the xz plane, it would be on the 1Ô 1 − z 2 . Anywhere else, 2 Free to photocopy and distribute 204 1. The directrix has direction → → − − i − k. the equation takes the form f (A.15. The direction vector is thus 1 Free to photocopy and distribute 205 .15. x2 + z 2 = 1 |1 − 4y|. B) = eA +B − A = 0.3 A spiral staircase.15. Σ : x2 + y 2 + z 2 = 0. Ǒ x 1.6 The planes A : x + z = 0 and B : y = 0 are secant.15. 1. 2 or 4x2 + 4y 2 + z 2 = 1.8 Considering the planes A : x − y = 0. we find the intersection of find ¾ ¿ x 1 the planes x = y and y = z. Anywhere else. be a point on S. shewing that the surface is → → → − − − of revolution. and it is thus a cylinder. A B A+B thus the equation represents a cylinder. y z = t 1 . (x2 + y 2 + z 2 )2 − 1 ((x + y + z)2 − (x2 + y 2 + z 2 )) − 1 = 0. This gives → → → − − − i + j + k. We must have to the axis of rotation is the same as the distance of y z to y 0 . To¾ ¿ its directrix. 1. B : y − z = 0.7 Rearranging.Appendix A x the distance from y z Ǒ to the z-axis is the distance of this point to the point 0 0 z Ǒ : x2 + y 2 .2 Let y z 1 line. 2 and so we may take A : x + y + z = 0. it would be on the Ô Ô Ô which is to say 1. The surface has equation of 2 2 the form f (A.15. This distance is the same as the length of the segment on the xz-plane going from the z-axis. 1. If this point were on the xy plane. 3 9x2 + 9z 2 − 16y 2 + 8y − 1 = 0. B) = 1 1 1 + − − 1 = 0. We thus have Ô 1Ô x2 + y 2 = 1 − z 2.4 A spiral staircase. and its distance to the axis of rotation would be |x| = |1 − 4y|.15. the 3 Ǒ Ǒ Ǒ x x 0 distance of that is y z x2 + z2. Its axis is the line in the direction i + j + k . The → → → − − − axis of revolution is then in the direction of i + j + k . 4 1 y2 1 1 3 = 4x2 + y 2 − 1 =⇒ 2 = x2 + − = 1− = . Parallel cross sections of the ellipsoid are similar ellipses.10 After rearranging. one may verify that the two planes given by a2 (b2 − c2 )z 2 = c2 (a2 − b2 )x2 give circular cross sections of radius b. (x + y + z)2 − (x2 + y 2 + z 2 ) + 2(x + y + z) + 2 = 0. b = 1. x2 + y2 = 1. We need t = 1 to t = 4. 0. the radius of the circle is at most b. The desired length is ¬4 4 ¬ (2t + 9) dt = (t2 + 9t)¬ = (16 + 36) − (1 + 9) = 42. hence we may increase the size of these by moving towards the centre of the ellipse. z−1z−1 A : x = 0.15. The equation is thus 4 3z 2 = 4x2 + y 2 − 1. 2 c c 4 4 4 4 since at z = ±2. Every plane through the origin which makes a circular cross section must intersect the yz-plane. 1 1 Free to photocopy and distribute 206 . Therefore. 1).15. we obtain (z − 1)2 − xy = 0. with apex at (0. and the diameter of any such cross section must be a diameter of z2 y2 the ellipse x = 0. 1.1 The arc length element is (dx)2 + (dy)2 + (dz)2 = 1 . 2 z2 = 4x2 + y 2 − 1. Σ : x2 + y 2 + z 2 = 0 as our plane and sphere. To shew that circular cross section of radius b actually exist. 1.11 The largest circle has radius b. 4 1.15.16.Answers and Hints 1.12 Any hyperboloid oriented like the one on the figure has an equation of the form z2 x2 y2 = 2 + 2 − 1. c2 Ô 4t2 + 36t + 81 dt = (2t + 9) dt.9 Rearranging. c2 a b When z = 0 we must have 4x2 + y 2 = 1 =⇒ a = Thus Hence. Arguing b c similarly on the xy-plane shews that the radius of the circle is at least b. C : z = 1. 2 + 2 = 1. B : y = 0. or − Considering the planes x y + 1 = 0. so we may take A : x + y + z = 0.15. we see that our surface is a cone. letting z = ±2. 1. 16. with vertex (lowest point) at (0. Hence. 1 + t2 1 + t2 1 + t2 − which means that if →(t) is on the plane ax + by + cz = d. that is.6 ¾ 1. x2 + y 2 = c2 + 1 3. the angle¾ ¿measured from the positive x-axis needs to be in the θ. t1 t2 t3 +t1 t2 t4 +t1 t3 t4 +t2 t3 t4 = 0 =⇒ as required. t1 t2 t3 t4 t1 t2 t3 t4 The projection of the plane x + y + z = 0 onto the xy-plane is the line y = −x.3 Let →(t) lie on the plane ax + by + cz = d. y = 3 sin θ sin φ. so they intersect at the circle x2 +y 2 = 1.16. π 0 (c2 + 1)dc = 4π 3 1.1 Free to photocopy and distribute 207 . then t must satisfy the quartic r polynomial p(t) = at4 + bt3 + (c − d)t2 − d = 0.Appendix A 1. Since the coefficient of t in this polynomial is 0. z = 2.2 Observe that z = 1 + x2 + y 2 is a paraboloid opening up. the plane makes 4 4 1 π an angle of with the z-axis. 2π]. Hence to be above the plane we need 0 ≤ φ ≤ .16. (1 + x2 + y 2 ) − (3 − x2 − y 2 ) = z − z = 0 =⇒ 2x2 + 2y 2 − 2 = 0 =⇒ x2 + y 2 = 1. 0. with highest point at (0.4 Observe that in this problem you are only parametrising the ellipsoid! The tricky part is to figure out the bounds in your parameters so that only the part above the plane x + y + z = 0 is described. On the other hand. Then we must have r a t4 t3 t2 +b +c = d =⇒ (at4 +bt3 +ct2 ) = d(1+t2 ) =⇒ at4 +bt3 +(c−d)t2 −d = 0. A common parametrisation found was: x = cos θ sin φ. 4 1. so they intersect at the plane z = 2. Recall that φ is measured from φ = 0 (positive z-axis) to 4 3π φ = π (negative z-axis). t1 t2 t3 + t1 t2 t4 + t1 t3 t4 + t2 t3 t4 1 1 1 1 = 0 =⇒ + + + = 0. Since 1 is normal to the plane x + y + z = 0. 1. z = cos φ. 1). we may parametrise this circle as x = cos t. t ∈ [0. − 1. y = sin t.16. a2 − 1 a2 + 1 1 −2a ¿ 2. then the sum of the roots of p(t) taken three at a time is 0. 3). 1 π 3π interval − ≤ θ ≤ . Subtracting the equations. Adding the equations we obtain 2z = z + z = (1 + x2 + y 2 ) + (3 − x2 − y 2 ) = 4 =⇒ 2z = 4 =⇒ z = 2. 0. To be “above” this line. Since they meet at the circle x2 +y 2 = 1. the tk are coplanar if and only if they are roots of p(t). z = 2. z = 3 − x2 − y 2 is another paraboloid opening down.17. Thus for all real numbers x 0 = f (1) ≤ f (x) = ex−1 − x. f ′′ (x) = ex−1 .Answers and Hints 1. for equality to occur. ¡ 16 (8 − e)2 ≤ 4 16 − e2 ⇐⇒ e (5e − 16) ≤ 0 ⇐⇒ 0 ≤ e ≤ . . which gives the desired result.. = an . . .. + x1 x2 xn n ≥ = √ i=1 1 xi √ xi 2 n2 . we need each of the inequalities Ak ≤ exp(Ak − 1) to hold. If f ′ (x) = 0 then ex−1 = 1 implying that x = 1. . Now. A1 A2 · · · An ≤ exp(A1 + A2 + · · · + An − n). to obtain. we may multiply all these inequalities together. By the preceding results. A2 ≤ exp(A2 − 1).2 By CBS. = ¡ 4 a 2 + b2 + c 2 + d 2 . Now. This occurs. (x1 + x2 + . we have A1 ≤ exp(A1 − 1). Easy Algebra! 3. 1. f (x) = ex−1 − x. f ′ (x) = ex−1 − 1. Thus f has a single minimum point at x = 1. the preceding inequality is equivalent to nn Gn ≤ exp(n − n) = e0 = 1. . Since all the quantities involved are non-negative. Put f : R → R. . In view of the observations above. which translates into a1 = a2 = . This completes the proof. ¡ (a + b + c + d)2 ≤ (1 + 1 + 1 + 1) a2 + b2 + c2 + d2 Hence.17. 5 5 Free to photocopy and distribute 208 . in view of the preceding lemma.17. Easy Algebra! 4. Clearly f (1) = e0 − 1 = 0. . (a1 + a2 + · · · + an )n We deduce that Gn ≤ which is equivalent to (a1 a2 · · · an )1/n ≤ a1 + a2 + · · · + an . + xn ) 1 1 1 + +. n a1 + a2 + · · · + an n n An ≤ exp(An − 1). 5 16 6 The maximum value e = is reached when a = b = c = d = . 2. . if and only if Ak = 1. 1.3 By CBS. ∀k. We deduce that n = 8.Appendix A 1. a]. 1 n n/(n+1) . then a + x ≥ 0 and a − x ≥ 0. (a1 .. 1 n n > 1+ . . and thus we may use AM-GM with a−x a+x and a6 = a7 = a8 = .7 Applying AM-GM to the set of n + 1 numbers 1. 2 1. .17.1 + . . for 1. from where na = 96.4 Observe that 96 · 216 = 1442 and by CBS. a3 ) 1 2 n for some real number t.. 1 n+1 1 > n+1 1+ n+1 1 1 1 . Free to photocopy and distribute 209 .. . . = an = a.. . n n n 1 n+1 n/(n+1) 1 n .6 If x ∈ [−a. n 2 a−x 3 3 5 ≤ a+x 5 +3 8 a−x 3 8 = a 4 8 . 48 1. an ) = t(a3 . . n n n k=1 a2 ≤ k a3 k k=1 k=1 ak . . 5 3 5 1 +2 +··· + n n+1 = .17. .1 + . Hence a1 = a2 = .17. a3 . . .17. As there is equality. a2 . 2. n: n!1/n = (1 · 2 · · · n)1/n < with strict inequality for n > 1. na2 = 144 3 gives a = y n = 32. . a1 = a2 = · · · = a5 = 5 3 a+x 5 from where f (x) ≤ with equality if and only if a+x a−x = . giving the assertion. . 1 + has arithmetic mean 1+ and geometric mean 1+ Therefore. 1+ that is 1+ which means xn+1 > xn . 55 33 a8 . . 1.5 Applying the AM-GM inequality. and by the Cauchy-Bunyakovsky-Schwarz inequality. then f (x. a semi-infinite interval with an endpoint (take for example p(x.1 We have → − − − F (→ + h ) − F (→ ) x x = = = → − → − − − − − (→ + h ) × L(→ + h ) − → × L(→ ) x x x x − − → + → ) × (L(→ ) + L(→ )) − → × L(→ ) − − − − (x h x h x x − − − − → × L(→ ) + → × L(→) + → × L(→ ) − − x h h x h h ¬¬ ¬¬ − → − → − → − → − ¬¬→ ¬¬ Now. 2.3.3. +∞[). e Free to photocopy and distribute 210 .3. 2. a single point (take for example. z) → (0.5 2 2.2 Since polynomials are continuous functions and the image of a connected set is connected for a continuous function.1. we will prove that || h × L( h )|| = o ¬¬ h ¬¬ as h → 0 . 3.10 By AM-GM. 2. For let → − h = n − hk → k . +∞[). 2. x2 (x2 + y 2 + z 2 )3 (x2 + y 2 + z 2 )2 x2 y 2 z 2 ≤ = →0 2 + z2 2 + y2 + z 2) +y 27(x 27 as (x.3. y) = x2 whose image is [0. a semi-infinite interval with no endpoint (take for example p(x.4. 4. e k=1 − where the → k are the standard basis for Rn . y. e k=1 and hence by the triangle inequality. The possibilities are thus 1. y) = (xy − 1)2 + x2 whose image is ]0. kx) would be bounded for every constant k.6 c = 0. n → − − ||L( h )|| ≤ |hk |||L(→k )|| e n k=1 1/2 n 1/2 ≤ = k=1 |hk |2 ¬¬ ¬¬ − ¬¬→ ¬¬ ¬¬ h ¬¬( n k=1 − ||L(→k )||2 e k=1 − ||L(→k )||2 )1/2 .7 0 2. y) = x). 0).4 0 2. the image must be an interval of some sort. and so the image would just be the point f (0. 0. Then e → − L( h ) = n − hk L(→ k ). y) = 0).Answers and Hints 2. If the image were a finite interval.3. all real numbers (take for example p(x. p(x. 0). and so by example 169. we have ¬¬ ¬¬2 ¬¬ ¬¬ ¬¬→¬¬2 ¬¬− ¬¬ − − − − ¬¬− ¬¬ + 2→ •→ + ¬¬→¬¬ + ¬¬→ ¬¬ → 2¬¬→ ¬¬. x x h x x ¬¬ h ¬¬ ¬¬ ¬¬2 − − − → ¬¬→¬¬ ¬¬ ¬¬ 2→ • h + ¬¬ h ¬¬ x → •→ − − − x h ¬¬→ ¬¬ → ¬¬→ ¬¬ + o ¬¬ h ¬¬ . We x 2 have ¬¬− ¬¬ → − → − − − − x f (→ + h ) − f (→ ) = ||→ + h || − ¬¬→ ¬¬ x x x ¬¬− ¬¬ → • → → − − − = (→ + h ) (− + h ) − ¬¬→¬¬ x x x ¬¬ ¬¬2 ¬¬ ¬¬ ¬¬→ ¬¬2 − − − − ¬¬− ¬¬ + 2→•→ + ¬¬→ ¬¬ − ¬¬→¬¬ x h = x x ¬¬ h ¬¬ ¬¬ ¬¬2 − − − → ¬¬→ ¬¬ 2→ • h + ¬¬ h ¬¬ x = ¬¬ ¬¬2 ¬¬ ¬¬ . This is a → − contradiction. ¬¬¬¬ ¬¬ ¬¬ ¬¬ ¬¬ → ¬¬ − − − ¬¬¬¬→¬¬ ¬¬→ ¬¬ ¬¬¬¬ h ¬¬ − L( h )¬¬ = o ¬¬ h ¬¬ . We use the fact that (1 + t)1/2 = 1 + + o (t) as t → 0. − ¬¬→ ¬¬ ¬¬ ¬¬ h ¬¬ ¬¬ or → − → − → − → − But the sinistral side → 1 as h → 0 . again by the Cauchy-Bunyakovsky-Schwarz Inequality. This implies that ¬¬→¬¬ → D0 (L)( 0 ) = 0 . ¬¬ ¬¬ ¬¬ ¬¬2 − − → − → − ¬¬→ ¬¬ ¬¬→ ¬¬ Since ¬¬ h ¬¬ = o ¬¬ h ¬¬ as h → 0 . L : Rn → R such that ¬¬ ¬¬ − → − − → → − → − ¬¬→¬¬ ||f ( 0 + h ) − f ( 0 ) − L( h )|| = o ¬¬ h ¬¬ . → − → − → − → − → − → − − || h × L( h )|| ≤ || h ||||L( h )| ≤ || h ||2 |||L(→k )||2 )1/2 = o || h || . → − t − 2. as ¬¬ h ¬¬ → 0. ¬¬→ ¬¬2 − − − − ¬¬− ¬¬ + 2→•→ + ¬¬→¬¬ + ¬¬→¬¬ x x h x ¬¬ h ¬¬ → − → − As h → 0 . L( 0 ) = 0 . assume that there is a linear transformation D0 (f ) = L. ¬¬ ¬¬ ¬¬ ¬¬1 − ¬¬ ¬¬ ¬¬ → ¬¬ − L( h ) ¬¬ ¬¬ ¬¬ ¬¬ = o (1) . ¬¬− ¬¬ ¬¬ h ¬¬ ¬¬ ¬¬ → − →¬¬ − → − ¬¬ Since f ( 0 ) = ||0|| = 0.Appendix A whence. → − ¬¬ ¬¬ − L( h ) → − → − → − → − → − ¬¬→ ¬¬ D0 (L)( 0 ) = L( 0 ) = 0 .4. Free to photocopy and distribute 211 . such linear transformation L does not exist at the point 0 . f ( h ) = ¬¬ h ¬¬ this would imply that ¬¬ ¬¬ − ¬¬→¬¬ as ¬¬ h ¬¬ → 0. and the dextral side → 0 as h → 0 . e as it was to be shewn. → − → − Recall that by theorem ??. ¬¬ ¬¬ ¬¬− ¬¬ ¬¬→¬¬2 ¬¬→ ¬¬ → ¬¬→ ¬¬2 − − − ¬¬ h ¬¬ ¬¬− ¬¬ + 2→ • h + ¬¬ h ¬¬ + ¬¬− ¬¬ x x x proving the first assertion.2 Assume that → = 0 . To prove the second assertion. and so. 0.1 We have 2 ∂f = ntn−1 e−r /4t + tn ∂t r 2 −r 2 /4t e 4t2 = ntn−1 + r 2 tn−2 4 e−r 2 /4t . 1) = ′ r 2 tn−2 3 1 3 = − tn−1 + r 2 tn−2 =⇒ n = − .Answers and Hints 2. 0.3 Observe that æ é 3 g(1. y) = ′ y2 2xy 2xy x2 é . 0 and hence g (1. 0) = ′ ′ æ 0 0 é 0 9 . g (x. 4 2 4 2 x y 2 if x ≤ y 2 if x > y 2 1 0 if x ≤ y 2 if x > y 2 0 2y if x > y 2 if x ≤ y 2 æ f (x. y) = Hence ∂ f (x. 1) = (g ′ (f (0. via the Chain-Rule. 1))g ′ (1.2 Observe that f (x. The composition g ◦ f is undefined. 1)) = (g ′ (0. This gives. 1) = æ é 0 1 . 1) = . 0. the output of f is R2 . y) = ∂y 2.5. æ 1 0 −1 1 é 2 0 . the Chain Rule gives ¾ (g◦f )′ (0. 1)) = f (3. + r t 2 4 We conclude that ntn−1 + 2. (f ◦ g)′ (1. 0. 1) = f ′ (g(1. f (g(1. 1)))(f ′ (0. 1)) = 1 0 1 ¿ −1 æ 1 0 1 1 é 0 1 ¾ = 0 0 2 ¿ −1 0 1 Free to photocopy and distribute 212 . For. 0.5.4 Since f (0. y) = ′ æ 1 y −1 x 2 0 é .5.5. 1) = æ 0 0 éæ 0 1 9 0 −1 1 2 0 é = æ 0 0 0 9 é 0 0 . 0. 1))(f ′ (0. y) = ∂x and ∂ f (x. but the input of g is in R3 . r2 1 ∂ r 2 ∂r 2 2 ∂f 2r 1 = r 2 − rtn e−r /4t = − r 3 tn−1 e−r /4t . ∂r 4t 2 2 ∂f 1 ∂ 1 r2 = − r 3 tn−1 e−r /4t ∂r r 2 ∂r 2 2 1 4 n−1 −r 2 /4t 1 3 r t e = − r 2 tn−1 e−r /4t + r2 2 4t 3 n−1 1 2 n−2 −r 2 /4t = − t e . 2. 1. y.6. 1) = ¾ 0 e2 ¿ 2 . y. 2e 0 x yexy ¾ 2.5. Free to photocopy and distribute 213 .4 The vector 1 0 ¿ is perpendicular to the plane. y. Then (∇f )(x. z) = 2x − 5y + z x−y ¿ 2y − 5x − z .6. 1.5. ∂x ∂f (x. ∂y and 1 b 2. y) = x2 g(x2 y). Observe that ∇f (x. 1) the last equation becomes 4(1 + ∂z ∂z 1 ∂z )+4 = 0 =⇒ =− .5.6.10 We have ∂ ∂ ∂ ∂z ∂z (x + z)2 + (y + z)2 = 8 =⇒ 2(1 + )(x + z) + 2 (y + z) = 0. z) = ey z xzey z yz xye ¾ 2. z = 4a. the integral is arctan + 2a3 a b 2a2 (a2 + b2 ) 2. Put f (x. ∂x ∂x ∂x ∂x ∂x At (1.1 ∇f (x. ¾ 2.5 We have ∂f (x. z) = ¿ =⇒ (∇ × f )(2. 1) = 2e . y.6 Differentiating both sides with respect to the parameter a. = a −7 Since the point is on the plane x − 7y = −6 =⇒ a − 7a = −6 =⇒ a = 1. y.2 (∇ × f )(x. and hence there exists a constant a such that ¾ 2x − 5y + z 2y − 5x − z x−y ¿ ¾ 1 0 ¿ =⇒ x = a. ∂x ∂x ∂x 2 ¿ ¾ ¿ e =⇒ (∇f )(2.Appendix A 2. y = a. z) is parallel ¾ to the vector 1 0 ¿ −7 . y) = 2xyg(x2 y). z) = x2 + y 2 − 5xy + −7 ¾ xz − yz + 3. 1. Thus x = y = 1 and z = 4. 11 An angle of 2. y) = (cos 2y)ex cos 2y =⇒ fx (0. v u v u 2.7 Observe that f (0. 0) = 1.2) ≈ 1 + 0. 0) = 1. Recall that when we defined the volume of a parallelepiped − − → − spanned by the vectors →. we saw that a c − − → • (→ × →) = (→ × →) • →. fx (x. −0. − . 2. − − − − a b c a b c − − − − Treating ∇ = ∇→ + ∇→ as a vector. This gives f (0. where ∇ acts the differential operator and × is the product. fy (x.1. y) = −2x sin 2yex cos 2y =⇒ fy (0. z) = ¿ 2x − y + 1 2y − x 2z − 2 ¾ ¿ 0 = 0 0 1 2 =⇒ x = − . ∆2 = det 2 −1 é −1 2 ¾ = 3. z) = −1 0 2 0 . first keeping → constant and then keeping → v u u v constant we then see that → − − − − − ∇→ • (→ × → ) = ( ∇ × →) • → . → . y) ≈ f (0. 1 is the only critical point. from where we gather that the critical point is a local minimum. 6 3 → − − − − − − − ∇→ • (→ × →) = −∇ • (→ × → ) = −( ∇ × →) • → .6.8. y. Hence f (x.6 We have π π with the x-axis and with the y-axis. 3 3 2 1 whence − . 0)(x − 0) + fy (0. y) ≈ 1 + x. b . 0) + fx (0.Answers and Hints 2.6. u − v u v u Thus → − − − → − − − − − − − − − − − u v u v v u u v ∇•(u×v) = (∇→ +∇→ )•(u×v) = ∇→ •(→ ×→ )+∇→ •(→ ×→) = ( ∇ ×→)•→ −( ∇ ×→)•→ . u − v v u v u v ¾ ∇f (x. The hessian is 3 3 ¾ ¿ 2 −1 0 Hf (x.8 This is essentially the product rule: duv = udv +vdu. whence the hessian is positive definite at the critical point. 2 0 The principal minors are æ ∆1 = 2. 0)(y − 0) =⇒ f (x.1 = 1. Free to photocopy and distribute 214 . z = 1. 0) = 0. ∆3 = 2 −1 0 −1 2 0 ¿ 0 0 2 = 2 det æ 2 −1 é −1 2 = 6.1. y = − .6. y. y) = (0. now. − 2). so the matrix is positive definite and we have a local minimum at each of these points. −4. æ é 12x2 − 4 4 Hf (x. f (x. If (x. y) = (− 2. y) = ( 2. 4x3 − 4(x − y) = 0 =⇒ 4x3 − 8x = 0 =⇒ 4x(x2 − 2) = 0 =⇒ x ∈ {− Since x = −y. √ 2}. = 2x2 (x2 −1) < 0. 2.8. y) = Hence 4x3 − 4(x − y) 4y 3 + 4(x − y) é = æ é 0 0 =⇒ 4x3 = 4(x − y) = −4y 3 =⇒ x = −y. y.8. What happens in a neighbourhood of (0. 2) or (x. −x) = 2x4 − 4x2 = 2x2 (x2 − 1). √ √ 2). √ √ √ √ If (x. ( 2. 0) then ∆1 = −4 < 0 and ∆2 = 0.8 We have ¾ ∇f (x. If x is close enough to 0.7 We have æ (∇f )(x. z) = The hessian is 8xz − 2y − 8x 4x − 2z −2x + 1 2 ¿ = ¾ ¿ 0 0 0 =⇒ x = 1/2. then ∆1 = 20 > 0 and ∆2 = 384 > 0. y) = . the matrix is negative definite and the critical point is thus a saddle point. 0). 0)? We have f (x. ¾ H = 8z − 8 −2 8x −2 0 0 8x 0 ¿ . y. z = 1/2. which means that the function both increases and decreases to 0 in a neighbourhood of (0. 4 12y 2 − 4 √ √ √ 2. 2. so the matrix is negative semidefinite and further testing is needed. (0. and Hr f = ¾ 2 z y z 2 x y 2 ¿ x . At z = 1/2.Appendix A 2. Free to photocopy and distribute 215 . the critical points are thus (− 2. z) = ¾ 2x + yz 2z + xy ¿ 2y + xz . −2 The principal minors are 8z − 8. 0).8. 0. x) = 2x4 > 0. meaning that there is a saddle point there. − 2).9 We have (∇f )(x. The Hessian is and its principal minors are ∆1 = 12x2 − 4 and ∆2 = (12x2 − 4)(12y 2 − 4) − 16. and 8. y = −1. −2). and substituting this into 2x + yz = 0 we gather (yz)2 = 16. in which case x = −2. 2. y.10 We have (∇f )(x. or all three are negative. −2. 2). xyz = −8. ∆2 (x. z) = and Hr f = ¾ 2xy + 2 x2 + 2yz y −1 2 ¿ . then we must have 2 z z 2 é = 4 − z 2 and ∆3 (x. z) = det det Hr f = 8 − 2x2 − ¾ ¿ − 2z 2 + 2xyz. At (0.8. (2. y. y. The other critical points are therefore (−2. z) = 8xyz = −x2 y 2 z 2 . then we must have at least one of the variables equalling 0. 0) = 2 > 0. 0. 0. −2). and thus (0. It is easy to see then that either exactly one of the variables is negative. in which case. z) = 2y. then ∆2 (x. 2. if xyz = 0. and solving for x. If xyz = −8. y = ±1. y. −2. ∆1 (0. 0) is a minimum point. 2z = −xy. ∆3 = −32. by virtue of the original three equations. We see that ∆1 (x. ∆1 (0. y. 0). Clearly. or zy = −4. y. ∆2 (0. 0) is a critical point. so these points are saddle points. say. 2y 2 0 If (∇f )(x. ¾ 2y 2x 0 2x 2z 2y æ 0 0 ¿ 2y . xyz(xyz + 8) = 0. and (−2. y. 0.Answers and Hints æ We see that ∆1 (x. z) = 2x ¾ ¿ 0 0 0 xy = −1. in which case x = 2. y. z) = 0. we must have 8 x = − . that is. If (∇f )(x. z) = det det Hr f = −8y 3 . meaning that yz either yz = 4. y. 0. 2y = −xz. all equal 0. 2. 0) = 8 > 0. z) = 0 0 2x = −yz. 0. x2 = −2yz. 0. z) = 2y 2x 2z é = 4yz − 4x2 and ∆3 (x. then we must have Free to photocopy and distribute 216 . and upon multiplication of the three equations. Thus (0. If x2 = y 2 = z 2 = 4. 0) = 4 > 0. ∆2 (x. y. z) = 2. 2). (2. then none of the variables is 0. 4xy = 4z 3 =⇒ xyz = x4 = y 4 = z 4 .11 We find ∇f (x. Since xyz = 1. ) is a saddle point. 0. ¾ Hf (x. −1.12 Rewrite: f (x. z) = 4xy − x2 y − xy 2 − 2xyz z(4 − 2x − 2y − z) −2yz z(4 − 2x − 2y − z) x(4 − x − 2y − 2z) −2xz y(4 − 2x − y − 2z) −2xy 4xz − x2 z − 2xyz − xz 2 . If 1 = xyz = x2 = y 2 = z 2 .8. Similarly. Thus we find the following critical points: (0. 1) = −7. 1. ∆2 (−1. 2 2 2 1 1 1 ∆2 (1. This means that for xyz = 0 the Hessian is negative definite and the function has a local maximum at each of the four points (1. Thus xyz ≥ 0. and ∆3 (−1. 1. (−1. 1. shewing that 2 2 2 1 (1. −1). (−1. −1. (−1. and ∆3 (1. exactly two of the variables can be negative. z) = xyz(4 − x − y − z) = 4xyz − x2 yz − xy 2 z − xyz 2 . −1. (1. Then 4yz = 4x3 . 1. ¾ ¿ 4yz − 2xyz − y 2 z − yz 2 (∇f )(x. y. −1. Thus (0. 1. Similarly.8. 0. ∆1 (1. ). ) = −2. Assume now that xyz = 0. (−1. and ∆3 = = = < −1728x2 y 2 z 2 + 192x4 + 192z 4 + 128zyx + 192y 4 −1728 + 192 + 192 + 128 + 192 −1024 0. y. 1. −1. −1. z) = 0. −1. −1). Now. Observe that at these critical points f = 1. 2 2. − ) = −8. 1. 0. ) = 8. (1. y. y. 1. −1). − ) are the critical points. Thus (1. z) = Assume ∇f (x. (1. 1). 1). −1). ) is also a saddle point. 0) is the only critical point with at least one of the variables 0. z = ±1. −1. 4xz = 4y 3 . 0) = 0 and f (−1. 2. y = ±1. 0). Now f (0. Then (xyz)3 = x4 y 4 z 4 = (xyz)4 =⇒ xyz = 1 =⇒ yz = 1 =⇒ x4 = 1 =⇒ x = ±1. −1.Appendix A 1 1 1 and hence (1. we have ∆1 = −12x2 = −12 < 0. − ) = 2. y. Then. The Hessian is Hx f = ¾ −12x2 4z 4y 4z −12y 2 4x 4y 4x −12z 2 ¿ . − ) = −6. ∆2 = 144x2 y 2 − 16z 2 = 144 − 16 = 128 > 0. x ¾ 4yz − 4x3 4xy − 4z 3 ¿ 4xz − 4y 3 . ) = −6. −1. 1. 2 2 2 1 1 1 ∆1 (−1. 1). 1). and if one of the variables is 0 so are the other two. and (−1. z) = ¿ y(4 − 2x − y − 2z) x(4 − x − 2y − 2z) Free to photocopy and distribute 217 . z) = (1 − 2y 2 )(xz)(e−x )(e−y )(e−z ) . 1) = 4 − x − 2y − z = 0. Since at the critical points we have 1 − 2x2 = 1 − 2y 2 = 1 − 2z 2 = 0. −x. so further testing is needed. Now. we can discard any point with a 0. Since the function clearly assumes positive and negative values. z) = (e−x )(e−y )(e−z ). yz(4 − 2x − y − z) = 0. If ∇(x. If either of x. 4 − x − y − 2z = 0 =⇒ x = y = z = 1. (4z − 6z)(xy) 3 ∆3 = (4x3 − 6x)(4y 3 − 6y)(4z 3 − 6z)(x2 y 2 z 2 ). y. xy(4 − x − y − 2z) = 0. 0) is a saddle point. x) = x3 (−4 + x). we will get ∆3 = 0. 1).13 To facilitate differentiation observe that g(x. y. Now ¾ ¿ 2 2 2 (1 − 2x2 )(yz)(e−x )(e−y )(e−z ) ∇g(x. which means that in some neighbourhood of (0. (4z 3 − 6z)(xy) with t(x. y. f (x. If xyz = 0 then we must have 4 − 2x − y − z = 0. ∆2 = 3 > 0. so the matrix is negative definite and we have a local maximum at (1. which means that (0. z) = 0. −2 −1 −1 −1 −2 −1 xz(4 − x − 2y − z) = 0. Thus as x → 0+ then f (x. 0. x) < 0. 2. 1. In this case Hf (1. x) = x3 (4 − 3x). the Hessian reduces to ¾ ¿ (4x3 − 6x)(yz) 0 0 Hx g = (e−3/2 ) 0 0 We have ∆2 = (4x3 − 6x)(4y 3 − 6y)(xyz 2 ) ∆1 = (4x3 − 6x)(yz) (4y 3 − 6y)(xz) 0 0 . 0) the function is both decreasing towards 0 and increasing towards 0. Free to photocopy and distribute 218 . 1. z) = (xe−x )(ye−y )(ze−z ). or z is 0. x. y.Answers and Hints Equating the gradient to zero. we obtain. x) > 0 and f (x. y. y = ± √ z = ± √ . (1 − 2z 2 )(xy)(e−x )(e−y )(e−z ) 2 2 2 2 2 2 2 2 2 The function is 0 if any of the variables is 0. f (x. We find 2 2 2 ¾ ¿ (4x3 − 6x)(yz) (1 − 2x2 )(1 − 2y 2 )z (1 − 2x2 )(1 − 2z 2 )y Hx g = t(x. ¾ ¿ −1 −1 −2 and the principal minors are ∆1 = −2 < 0. and ∆3 = −4 < 0. −x. then x = 1 1 1 ± √ . z) (1 − 2y 2 )(1 − 2x2 )z 2 2 2 2 2 (1 − 2z )(1 − 2x )y (1 − 2z )(1 − 2y )x (4y 3 − 6y)(xz) 2 2 (1 − 2y 2 )(1 − 2z 2 )x . y. 0. x.8. This means that if an even number of the variables is negative (0 or 2).8. √ Free to photocopy and distribute 219 . 0) = g(0) = (0. √ ).15 Since the coordinates (x. We conclude that we have local maxima at 1 1 1 1 1 1 1 1 1 1 1 1 ( √ . the determinant of of this last matrix is −(g ′ (0))2 < 0. ∂x ∂y so (0. 144 − 16x2 ). Regardless of the sign of g ′ (0). √ . √ ). Hf (0. Hence ∂f (x. y) = G′ (x2 + y) − 2yG′ (y 2 − x) = g(x2 + y) − 2yg(y 2 − x). Now. −2 ≤ y ≤ 2 describe a circle centred at the origin with radius 2. 0) is a critical point. ∂x ∂f (x. − √ . y) = y 2 −x g(t)dt = G(x2 + y) − G(y 2 − x). 2 2 2 2 2 2 2 2 2 2 2 2 2. √ . Geometrically this is easily seen to be 1. ∂y This gives ∂f ∂f (0. (− √ . − √ ). the Hessian is positive definite. y) = 2xG′ (x2 + y) + G′ (y 2 − x) = 2xg(x2 + y) + g(y 2 − x). 4 − y 2 ). − √ ). 0) = æ ′ −g (0) 0 0 g ′ (0) é . √ ). the Hessian of f is æ Hf (x. and if an odd numbers of the variables is positive (1 or 3). 0) = 0. and the coordinates (y.14 By the Fundamental Theorem of Calculus. 1 4 −√ 2 3 1 − 6 −√ 2 √ = 2 2 > 0. √ . then we the Hessian is negative definite. − √ . the problem reduces to finding the minimum between the boundaries of the circle and the ellipse.Appendix A Also.8. − √ . 4 1 √ 2 3 −6 1 √ 2 √ = −2 2 < 0. 2. √ ). ( √ . (− √ . − √ ) 2 2 2 2 2 2 2 2 2 2 2 2 and local minima at 1 1 1 1 1 1 1 1 1 1 1 1 (− √ . and so (0. √ . ( √ . ( √ . 0) is a saddle point. (− √ . y) = and so 2g(x2 + y) + 4x2 g ′ (x2 + y) − g ′ (y 2 − x) 2xg ′ (x2 − y) + 2yg ′ (y 2 − x) g ′ (x2 + y) − 2g(y 2 − x) − 4y 2 g ′ (y 2 − x) 2xg ′ (x2 + y) + 2yg ′ (y 2 − x) é . there exists a continuously differentiable function G such that x 2 +y f (x. −3 ≤ x ≤ 3 describe an ellipse centred 3 Ô at the origin and semi-axes 3 and 4. − √ ). − √ . Answers and Hints 2. Hence. b b+a a b+a = =⇒ a = c. The intersection of the line and the plane happens when a(1 + at) + b(1 + tb) + c(1 + tc) = d =⇒ t = Hence P′ = d − a − b − cá a 2 + b2 + c 2 d−a−b−c 1+b· 2 a + b2 + c 2 d−a−b−c 1+c· 2 a + b2 + c 2 d−a−b−c . The vector √ ¾ ¿ a b is perpendicular to Π. 6 and the maximum volume is √ 3 ( S) abc = √ .1 We have ¾ =⇒ bc ca ab ¿ ¾ 2b + 2c 2b + 2a ¿ ∇(abc) = λ∇(2ab + 2bc + 2ca − S) = λ 2a + 2c = = = 2λ(b + c) 2λ(a + c) 2λ(b + a) bc =⇒ ca ab By physical considerations. c b+c √ S 2a2 + 2a2 + 2a2 = S =⇒ a = √ .9. the equation of the perpendicular c passing through P is Ǒ Ǒ x 1 y z = 1 1 +t ¾ ¿ a b c =⇒ x = 1 + ta. z = 1 + tc. a c+a Therefore c a+c = =⇒ b = c. abc = 0 and so λ = 0. 3 3 6 Equality happens if S 2ab = 2bc = 2ca =⇒ a = b = c = √ . b+c b = =⇒ a = b. Hence. a 2 + b2 + c 2 1+a· Free to photocopy and distribute 220 . y = 1 + tb.2 1. ( 6)3 The above result can be simply obtained by using the AM-GM inequality: S 2ab + 2bc + 2ca S 3/2 = ≥ ((2ab)(2bc)(2ca))1/3 = 2(abc)2/3 =⇒ abc ≤ 3/2 .9. by successively dividing the equations. 6 2. 9. x =1+ λa . Now. 6 Free to photocopy and distribute 221 . a 2 + b2 + c 2 a + b2 + c 2 y= c2 + a2 − ab + bd − bc d−a−b−c = 1+b· 2 a 2 + b2 + c 2 a + b2 + c 2 as before. ¾ ¿ ¾ ¿ 2(x − 1) a ∇f (x. 2 2 2 2. we gather that λ = 0. y. y) = (x − 1)2 + (y − 1)2 + d − ax − by −1 c 2 . 2. y.4 Using AM-GM. z) = ax + by + cz − d. z) = (x − 1)2 + (y − 1)2 + (z − 1)2 be the square of the distance from P to a point on the plane and let g(x. 2(z−1) = λc. Since (1. 1. 2 y =1+ λb . a 1+ λa 2 +b 1+ λb 2 +c 1+ λc 2 = d =⇒ λ = 2 · d−a−b−c . 1) is not on the plane and abc = 0. y. Consider the function t(x. y. 2 Putting these into the equation of the plane. √ √ x + 3y x4 + 81y 4 1/4 361/4 ≤( ) = 1/4 =⇒ x + 3y ≤ 23/4 6 = 25/4 3. Let f (x. Now. 1. 1 61/3 = Ô 6 x2 y 3 z ≤ 2x + 3y + z =⇒ 2x + 3y + z ≥ 62/3 . z) = λ∇g(x. a 2 + b2 + c 2 Then the coordinates of P ′ are x = 1+ λa d−a−b−c = 1+a· 2 . which is the square of the distance from a point (x. as before. Using Lagrange multipliers.3 Using CBS.Appendix A The distance is then Ö d−a−b−c a· 2 a + b2 + c 2 2 + b· d−a−b−c a 2 + b2 + c 2 2 + c· d−a−b−c a 2 + b2 + c 2 2 2. 3. 2(y−1) = λb. Substituting this in the equation of the plane gives the same coordinate of z. 2 z =1+ λc . z) on the plane to the point (1. ¾ ¿ æ é a d − ax − by 2(x − 1) − 2 −1 0 c c ∇t(x. 2 a + b2 + c 2 z = 1+ λc d−a−b−c = 1+c· 2 . 2 a + b2 + c 2 as before.9. z) =⇒ 2(y − 1) 2(z − 1) =λ b c =⇒ 2(x−1) = λa. y. 1). 2 a + b2 + c 2 y = 1+ λb d−a−b−c = 1+b· 2 . y) = = b d − ax − by 0 2(y − 1) − 2 −1 c c which implies x= d−a−b−c −b2 − c2 + ab − ad + ac = 1+a· 2 . 1/(p−1) )1/p 1/(p−1) + b1/(p−1) )1/p +b (a a p p/(p−1) a λp p/(p−1) + b λp p/(p−1) . −1/ 2) The desired maximum is thus √ √ √ √ f (− 2. y. 2) = f ( 2. 1/ 2). Clearly then . 0 = 3x + 5(λ − 1)y.7 Let g(x. y) =⇒ =λ . Substituting this into the third equation we gather that √ 1 10x2 ± 6x2 = 8. these points define a maximum for f and so ax + by ≤ ap/(p−1) bp/(p−1) + . 5 5 5 5 2 8 5 2. Aliter: Observe that. but this solution does not lie on the third equation. We need a = pλxp−1 and b = pλy p−1 . λp λp Thus 1 = xp + y p = which gives λ= This gives x= (a1/(p−1) a1/(p−1) b1/(p−1) . (1/ 2. the feasible values are ( 2. (a1/(p−1) + b1/(p−1) )1/p (a1/(p−1) + b1/(p−1) )1/p 2. 5x2 +6xy +5y 2 = 8 =⇒ x2 +y 2 = 8 6 8 6 x2 + y 2 5 8 − xy ≥ − · =⇒ x2 +y 2 ≥ · = 1. then we deduce that x = ±y. − 2) = 4 and the minimum is f (1/ √ 2. −1/ √ 2) = 1. 2y 6x + 10y This gives the three equations 0 = 5(λ − 1)x + 3y. − 2). If 25(λ−1)2 −9 = 0. resulting in x = ± 2 or x = ± √ . λ = 0. (p−1)/p + b p p/(p−1) .9. y) = 5x2 + 6xy + 5y 2 − 8 and argue using Lagrange multipliers. 2). (− 2. We have æ é æ é 2x 10x + 6y ∇f (x. We then have 1/(p−1) 1/(p−1) a b x= . (−1/ 2.9. 1/ √ 2) = f (−1/ √ 2. 0) as long as 25(λ−1)2 −9 = 0. We solve ¾ ¿ ¾ ¿ x x ∇f y z = λ∇g y z Free to photocopy and distribute 222 . y= . Taking into account the third equa2 √ √ √ √ √ √ √ √ tion. y) = xp + y p − 1. using AM-GM.5 We put g(x. y) = λ∇g(x. Since f is non-negative. 5x2 + 6xy + 5y 2 = 8.Answers and Hints 2.9.6 Put g(x. y= . z) = (x − 1)2 + (y − 2)2 + (z − 3)2 − 4. The linear system (the first two equations) will have the unique solution (0. √ x + z = 1.9. y. If a and b are the semi-axes.9. Now maximise and minimise the distance between a point on the ellipse and the origin. We must find λ. We may shew that (0. 2 This gives the two points (x. 14 14 14 √ 12 14 The first point gives an absolute maximum of 18 + and the second an absolute 7 √ 12 14 . we gather that −λ −1 1−λ from where λ =1± 2 + −2λ −2 1−λ 2 + −3λ −3 1−λ 2 = 4.2 − √ . z). 1 = 2λy. y.9 Put g(x. z) = 2. y. and z = . y = . λ. y. z) = and 2 4 6 1− √ .3 − √ . This gives x = −λ −2λ −3λ . y. This requires ¾ 2x 2y 2z ¿ = ¾ 2(x − 1)λ 2(z − 3)λ ¿ 2(y − 2)λ . y.8 Observe that the ellipse is symmetric about the origin.Appendix A for x.2 + √ . you will find that 2a = 2 and 2b = 6 2. y. z) = λ∇g(x. y.9. and x2 + y 2 = 1.3 + √ 14 14 14 We deduce that x = 0. z = 1. √ 14 . 1 = δ. 2.10 One can use Lagrange multipliers here. minimum of 18 − 7 (x. 1) yields a maximum 2. h(x. 1) yields a minimum. Clearly. y = ± 2. λ = 1. z) = x2 + y 2 − 2. Free to photocopy and distribute 223 . δ such that ∇f (x. z) = x + z − 1. Substituting into 1−λ 1−λ 1−λ 2 2 2 (x − 1) + (y − 2) + (z − 3) = 4. − 2. z) + δ∇h(x. √ 2 4 6 1 + √ . which translates into 1 = 2λx + δ. √ and that (0. But perhaps the easiest approach is to put y = 1 − x and maximise f (x) = x + x(1 − x). so the problem reduces to maximising h(x. we only need to look on the line x + y = 1 for the maxima.9. Hence we maximise f subject to the constraint x + y = 1. y) = λ(∇g)(x. y) =⇒ =λ . y) = = =⇒ x = 0 or y = 0. a b−1 bx y 1 which in turn =⇒ axa−1 y b = λ = bxa y b−1 =⇒ ay = bx =⇒ ay = b(1−y) =⇒ y = Finally. y) = xa y b e−(x+y ) ≤ xa y b e−1 ≤ 2. To prove this claim we have to prove that there are no critical points strictly inside the triangle. In 4 Free to photocopy and distribute 224 . y). 2 4 2 x(1 − x) ′ Since f ′′ (x) = − the value sought is a maximum. If cos 2x = 0 then x = this case f π π 1 . Since the function is identically 0 for x = 0 or y = 0. Then ∇f (x. y) =⇒ =⇒ =⇒ −2 sin x cos x = λ. For this we compute the gradient and set it equal to the zero vector: æ é æ é −axa−1 y b e−(x+y ) 0 (∇f )(x. a+b x= a . This means that we may restrict ä ç ä ç ourselves to x ∈ 0. 4(x(1 − x))3/2 x(1 − x) 2. y + π) = cos2 (x + π) + cos2 (y + π) = cos2 x + cos2 y = f (x. Since x + y = 1. 0 = cos2 = . y) = xa y b subject to the constraint x + y = 1. so f is doubly-periodic of period π in each argument. π and y ∈ 0. π . y) = λ∇g(x. Using Lagrange multipliers. æ a−1 b é æ é ax y 1 (∇h)(x. a+b b a+b b e−1 .12 Observe that f (x + π. we can see that f (x. This maximum is thus √ √ 1 2 1 2 + .11 Claim: the function achieves its maximum on the boundary of the triangle. y) = xa y b e−(x+y ) = xa y b e−1 on the line. Put g(x. −2 sin y cos y = −λ − sin 2x = sin 2y = sin 2 x − π 4 = (sin 2x)(cos a a+b a b . y) = x − y. 0 −bxa y b−1 e−(x+y ) which means that the critical points occur on the boundary. f = + 2 4 2 2 (1 − 2x)2 1 − Ô < 0. 2 sin x cos x = −2 sin y cos y π π ) − (sin )(cos 2x) = − cos 2 2 2 A similar analysis gives sin 2y = − cos 2y. f (x.Answers and Hints For this we have √ 1 1 − 2x 2 f (x) = 0 =⇒ 1 + Ô = 0 =⇒ x = + .9. 4 4 2 π and so y = 0. 707106781. −π /2 (sin t)(cos t)dt − (cos t)(sin t)dt Also. . 8 8 π ≈ 1. 8 2. however.3.− 8 8 If. Let L1 : y = x + 1.9. 2 π π =⇒ y = − . We put x = sin t. L1 xy||dx|| + 1 −1 L2 2 x(x + 1)dx − √ 1 2 0 x(−x + 1)dx ä π πç 2.15 Try p(x. 8 8 = 2 cos2 3π ≈ . −1 xdx(x + 1)dx + xdx − (−x + 1)dx Also. cos 2y = 0 then f = 2 cos2 π π . t ∈ − . ||dx|| = (cos t)2 + (− sin t)2 dt = dt. 8 8 sin 2y = − cos 2y =⇒ tan 2y = −1 =⇒ y = In this case we have f 5π 3π . Then 2 2 π /2 xdx + ydy C = = 0. y) = (y 2 + 1)x2 + 2xy + 1. both on L1 and on L2 we have ||dx|| = xy||dx|| C √ 2dx. −π /2 Free to photocopy and distribute 225 . Then xdx + ydy C = L1 1 xdx + ydy + L2 xdx + ydy 1 0 = = 0. In this case 4 2 f If. and thus π /2 xy||dx|| C = = = (sin t)(cos t)dt ¬ (sin t)2 ¬π /2 ¬ 2 −π /2 0. 8 3π 5π =⇒ x = .Appendix A If cos 2y = 0 then y = π π and so x = . 3. − cos 2x = 0 then − sin 2x = − cos 2x =⇒ tan 2x = 1 =⇒ x = In this case we have π π . thus xy||dx|| = = = √ 0. y = cos t.1 1.2928932190. however. 2 4 = 1 . L2 : −x + 1. Observe that there 5 xdx + ydy = (cos θ)d cos θ + (sin θ)d sin θ = − sin θ cos θdθ + sin θ cos θdθ = 0.0>. Line( <0. <2*sqrt(3).2> )) > +LineInt( VectorField( <x. Assembling these two pieces. > with(Student[VectorCalculus]): > LineInt( VectorField( <x.3 Using the parametrisations from the solution of problem 3. 5 6 5 Assembling these we gather that x||dx|| = Γ Γ1 x||dx|| + Γ2 √ π x||dx|| = 4 3 − 8 + 16 sin . 0) and radius 4 going counterclockwise from θ = to 6 π θ= . Pi/5) ). the integral will be zero. 3 3 3 x||dx|| = Γ1 0 √ 2 √ xdx = 4 3. Arc(Circle( <0. 4). 5 Free to photocopy and distribute 226 . and since the integrand is 0. xdx + ydy = Γ Γ1 xdx + ydy + Γ2 xdx + ydy = 8 + 0 = 8. 5 ← → x Observe that the Cartesian equation of the line OA is y = √ .Answers and Hints √ 3. x||dx|| = Γ2 π /6 16 cos θdθ = 16 sin π π π − 16 sin = 4 sin − 8. 3 π On the arc of the circle we may put x = 4 cos θ. To solve this problem using Maple you may use the code below.3. 2) and Γ2 denote π the arc of the circle centred at (0.y> ). we find on Γ1 that x||dx|| = x whence 2 (dx)2 + (dy)2 = x √ 1+ 1 2 dx = √ xdx. Then on Γ1 3 x x 4 xdx + ydy = xdx + √ d √ = xdx. 3 3 3 Hence 2 √ 3 xdx + ydy = Γ1 0 4 xdx = 8.y> ).2 Let Γ1 denote the straight line segment path from O to A = (2 3. 3 On Γ2 that x||dx|| = x whence π /5 (dx)2 + (dy)2 = 16 cos θ sin2 θ + cos2 θdθ = 16 cos θdθ. Pi/6. y = 4 sin θ and integrate from θ = 6 π to θ = .0>.3. 3.3.3. and to parametrise this curve. 2 2 2x2 − 2x + z 2 = 0 1 2 1 2 x− + z2 = 2 2 1 2 + 2z 2 = 1.5. y say. 1. <2*sqrt(3). To check that these two are indeed 10 5 the same.0>. which returns true.y]=Arc(Circle( <0. use the code > > > is(16*cos(3*Pi/10)=16*sin(Pi/5)). Thus Á zdx + xdy + ydz Γ = = = 3.2> )) +PathInt(x.y]= Line( <0. 4 x − +2z 2 = 2 0 1 is an ellipse. Pi/5) ). [x. from where y = 1 − x and x2 + y 2 + z 2 = 1 give x2 + (1 − x)2 + z 2 = 1 =⇒ =⇒ =⇒ So we now put x= 1 cos t + .3 Free to photocopy and distribute 227 .3. and not your right).4 The curve lies on the sphere. 3π π Maple gives 16 cos rather than our 16 sin . 1.2 1 3 15π 16 sin t 1 cos t + √ d 2 2 2 2π 0 cos t cos t 1 1 + − + d 2 2 2 2 2π 0 1 cos t sin t + − d √ 2 2 2 2π 0 sin t cos t cos t sin t 1 + √ + − √ 4 4 2 2 2 2 2π π √ .Appendix A To solve this problem using Maple you may use the code below. 4 x− 2 We must integrate on the ¿ ¾ side of the plane that can be viewed from the point (1. 4).5. we dispose of one of the variables. To obtain a positive parametrisation we must integrate from t = 2π to t = 0 (this is because when you look at the ellipse from the point (1. Pi/6. 2 y =1−x = 1 cos t − . On the zx-plane.5.0>.1 2 3. with(Student[VectorCalculus]): PathInt(x. > 3. 2 0 dt 3. 2 2 sin t z= √ . 0) the positive x-axis is to your left. [x. 0) 1 1 2 (observe that the vector 1 is normal to the plane). 8 Observe that x2 + y 2 = 16.4 The integral equals 1 √ x xydxdy D = 0 1 x x2 y dy dx = = 3.Answers and Hints 3. √ 16−x 2 − 33 x+4 √ xdydx = = = √ Ô 3 2 + x 16 − x x − 4 dx 3 0 √ ¬2√ 3 1 3 3 ¬ − (16 − x2 )3/2 + x − 2x2 ¬ 3 9 0 8 . 12 0 1 x(x − x4 ) dx 2 π π x sin x sin ydxdy + D D y sin x sin ydxdy = = 2 0 y sin y dy 0 sin x dx 4π.7 3.5. 6 3. 2 3.5.5.5 The integral equals 1 .5. 3 2 √ 3 3.9 e − 1 Free to photocopy and distribute 228 . y = − The integral is 2 0 √ 3 √ 3 x + 4 =⇒ 16 − x2 = 3 − √ 3 x+4 3 2 √ =⇒ x = 0. 3.6 The integral is 1 y 1 1 x2 dxdy + x≤y y ≤x y 2 dxdy = 0 1 0 x2 dxdy + 0 y y 2 dxdy = = = = 21 8 1 2 ¡ y3 dy + y − y 3 dy 3 0 ¬ 0 ¬ y 4 ¬1 y 4 ¬1 y3 − ¬ + ¬ 12 0 3 4 0 1 1 1 + − 12 3 4 1 .5.5. <4. Upon examining the region.5.4>)). one notices that it does not make much of a difference. ’inert’).y]=Triangle(<0.<4. In such a case notice that for 0 ≤ x ≤ 3.1>.1]2 xdA + [0.<3. Free to photocopy and distribute 229 . [x.0>. but this command is limited. We have a choice of whether integrating with respect to x or y first. y 2 )dA [0. It also evaluates expressions that it deems below its dignity to return unevaluated. I will integrate with ← → respect to y first. y goes from the line AB to the line OB 3 x 4 x xydA R = 0 x/3 3 xydydx + xydydx 3 3x−8 ¬x 4 2¬ = = = = = 1 1 dx + xy ¬ dx 2 0 2 3 x/3 3x−8 3 4 2 ¡ 1 x 1 x x2 − dx + x x2 − (3x − 8)2 dx 2 0 9 2 3 3 4 ¡ 4 1 x3 dx + −8x3 + 48x2 − 64x dx 9 0 2 3 9+9 18. ¬x ¬ xy 2 ¬ To solve this problem using Maple you may use the code below. [x.11 Begin by finding the Cartesian equations of the various lines: for OA is y = 3 ← → ← → (0 ≤ x ≤ 1). > > with(Student[VectorCalculus]): int(x*y.10 We have min(x.1>. and for BO is y = x (0 ≤ x ≤ 4). and for 3 ≤ x ≤ 4. y goes from the line OA to the ← → ← → ← → line OB.1]2 = [0.4>).5. > int(x*y. for AB is y = 3x − 8 (3 ≤ x ≤ 4). 30 xdxdy + 0 ¬y 2 1 ¬ x2 ¬ dy + 1 0 y 2 dxdy ¬1 ¬ y 2 x¬ dy y2 4 y2 (y − y )dy ← → x 3. Maple can also provide the limits of integration.Appendix A 3. since Maple is quite whimsical about which order of integration to choose.1]2 y 2 dA y 2 <x x≤y 2 1 y2 1 0 1 0 2 1 = 0 = = = = 1 2 0 0 1 1 4 y dy + 2 0 2 1 + 10 15 7 .<3.0>.y]=Triangle(<0. Answers and Hints 3.5.12 Integrating by parts, 1 1−x loge (1 + x + y)dxdy D = 0 1 0 loge (1 + x + y) dy dx = 0 1 [(1 + x + y) loge (1 + x + y) − (1 + x + y)]1−x dx 0 (2 loge (2) − 1 − loge (1 + x) − x loge (1 + x) + x) dx = = 1 . 4 0 3.5.13 First observe that on [0; 2]2 , 0 ≤ x + y 2 ≤ 6, so we decompose the region of integration according to where x + y 2 jumps across integer values. We have x + y 2 dA [0;2]2 [0;2]2 [0;2]2 [0;2]2 [0;2]2 [0;2]2 = x+y 2 =1 1dA + x+y 2 =2 2dA + x+y 2 =3 3dA + x+y 2 =4 4dA + x+y 2 =5 5dA = [0;2]2 1≤x+y 2 <2 dA + 2 [0;2]2 2≤x+y 2 <3 dA + 3 [0;2]2 3≤x+y 2 <4 dA + 4 [0;2]2 4≤x+y 2 <5 dA + 5 [0;2]2 5≤x+y 2 <6 dA By looking at the regions (as in figures A.13 through A.17 below) (I am omitting the details of the integrations, relying on Maple for the evaluations), we obtain 1 √ √ 2−x 2 √ 0 2−x dA = 0 [0;2]2 1≤x+y 2 <2 dydx + 1−x 1 dydx = − 4 4√ 2 2 4√ + 2+ =− + 2. 3 3 3 3 3 2 √ √ 3−x 2 [0;2]2 dA = 2 0 2−x √ 8√ 4 dy dx = 4 3 − 2− . 3 3 2≤x+y 2 <3 2 √ √ 4−x 3 [0;2]2 3≤x+y 2 <4 dA = 3 0 3−x √ √ dy dx = 18 − 6 3 − 4 2. 1 2 2 √ √ 5 5−x 4 [0;2]2 dA = 4 0 √ dy dx+4 4−x 1 4−x dy dx = − √ √ 40 64 16 √ +8 3+ −16 3+ 2. 3 3 3 4≤x+y 2 <5 2 5 [0;2]2 dA = 5 1 √ 5−x dy dx = − √ 50 + 10 3. 3 4≤x+y 2 <5 Adding all the above, we obtain x + y 2 dA = [0;2]2 22 4√ 4√ + 3− 2 ≈ 7.7571. 3 3 3 Free to photocopy and distribute 230 2 1 0 2 0 1 0 0 1 2 2 1 0 1 2 0 1 2 2 1 0 0 1 2 2 1Appendix A 0 0 1 2 Figure A.13: 1 ≤ x + y2 < 2. Figure A.14: 2 ≤ x + y2 < 3. Figure A.15: 3 ≤ x + y2 < 4. Figure A.16: 4 ≤ x + y2 < 5. Figure A.17: 5 ≤ x + y2 < 6. 3.5.14 Observe that in the rectangle [0 ; 1] × [0 ; 2] we have 0 ≤ x + y ≤ 3. Hence x + y dA R = = 1 R 1dA + 1dydx + R 2dA 2dydx 1≤x+y <2 2 2−x 1−x 2≤x+y <3 2 2 1 2−x = 1 4. √ 0 2 1 4−x 2 √ √ 4−x 2 2 3.5.15 0 1 xdydx + 1−x 2 2 1 −1 −2 xdydx = 7 . 3 2 3 3 3.5.16 0 −2 xdxdy + 1 xdxdy + 1 0 xdxdy + 1 2 xdxdy = 7. 3.5.17 Exchanging the order of integration, π /2 0 0 y cos y dxdy = y π /2 cos ydy = 1. 0 3.5.18 Upon splitting the domain of integration, we find that the integral equals 2 1 y y2 sin πx dx 2y 2 å − 2 1 dy = 1 2y πx cos π 2y èy 2 dy y = = upon integrating by parts. − 2y πy cos dy π 2 4(π + 2) , π3 − 3.5.19 The integral is 0. Observe that if (x, y) ∈ D then (−x, y) ∈ D. Also, f (−x, y) = −f (x, y). 2/ √ 5 5 1 2 √ 4−y 2 3.5.20 −2/ √ 1 −2 √ dxdy = 4−y 2 8 + 4 arcsin 5 √ 5 5 . Free to photocopy and distribute 231 Answers and Hints 3.5.21 The integral equals 1 xydA D = 0 1 0 1−x 1 + x xy dy dx = = = 3.5.22 Using integration by parts, 1 1−x 2 x dx 2 1+x 0 2 2 (t − 1)(t − 2) dt t2 1 11 4 loge 2 − . 4 1 1−x 2 loge (1 + x2 + y)dA D = 0 1 0 loge (1 + x2 + y) dy dx = 0 (2 loge (2) − 1 − loge (1 + x2 ))dx 1 + 0 (−x2 loge (1 + x2 ) + x2 ) dx = 2 2 8 π loge 2 + − . 3 9 3 √ √ 4−y 2 3.5.23 0 xdxdy = 2. 2y −y 2 3.5.25 Let D1 = {(x, y) ∈ R2 | − 1 ≤ x ≤ 1, x ≤ y}, D2 = {(x, y) ∈ R2 | − 1 ≤ x ≤ 1, x > y}. Then D = D1 ∪ D2 , D1 ∩ D2 = ∅ and so f (x, y)dxdy = D D1 f (x, y)dxdy + D2 f (x, y)dxdy. By symmetry, f (x, y)dxdy = D1 D2 f (x, y)dxdy, and so f (x, y)dxdy D = = = = 2 D1 1 f (x, y)dxdy 1 x 2 −1 1 −1 (y − x) dy dx (1 − 2x + x2 ) dx 8 . 3 3.5.26 The line joining A, and B has equation y = −x − 2, line joining B, and C has equation y = −7x + 10, and line joining A, and C has equation y = 2x + 1. We split Free to photocopy and distribute 232 Appendix A the triangle along the vertical line x = 1, and integrate first with respect to y. The desired integral is then 1 2x+1 (2x + 3y + 1) dxdy D = −1 (2x + 3y + 1)dy + 1 1 −x−2 2 −7x+10 −x−2 dx dx (2x + 3y + 1)dy = 21 2 3 x + 9x − dx 2 2 −1 2 ¡ 60x2 − 198x + 156 dx + 1 = = 4−1 3. 3.5.27 Since f is positive and decreasing, 1 0 0 1 f (x)f (y)(y − x)(f (x) − f (y))dxdy ≥ 0, from where the desired inequality follows. 3.5.28 The domain of integration is a triangle. The integral equals 1 1−x xy(x + y)dxdy D = 0 0 1 xy(x + y) dy dx = = = å 2 è1−x y y3 x x + dx 2 3 0 0 1 2 x(1 − x) (1 − x)3 x + 2 3 0 1 . 30 1 dx 3.5.29 For t ∈ [0; 1], first argue that 1 0 1 1 0 1 0 f (x)dx ≥ (1 − t)f (t) ≥ f (t) − t. Hence 1 1 (f (x)dx) dy ≥ 1 0 (f ◦ g)(y)dy − g(y)dy. 0 Since 0 0 f (x)dxdy = 0 f (x)dx, the desired inequality is established. 3.5.30 Put f (x, y) = xy + y 2 . If I, II, III, IV stand for the intersection of the region with each quadrant, then f (x, y)dxdy = II II f (−x, y)d(−x)dy = − I f (x, y)dxdy, f (x, y)dxdy = IV IV f (x, −y)dxd(−y) = − I f (x, y)dxdy, Free to photocopy and distribute 233 Answers and Hints and f (x, y)dxdy = III III f (−x, −y)d(−x)d(−y) = + I f (x, y)dxdy. Thus (xy + y 2 )dxdy S = I f (x, y)dxdy + II f (x, y)dxdy + III f (x, y)dxdy + IV f (x, y)dxdy f (x, y)dxdy I = I f (x, y)dxdy − I f (x, y)dxdy + I f (x, y)dxdy − = 0. 3.5.31 We split the rectangle [0; a] × [0; b] into two triangles, depending on whether bx < ay or bx ≥ ay. Hence a 0 0 b emax(b 2 x 2 ,a 2 y 2 ) dydx = bx<ay emax(b e a2 y 2 2 x 2 ,a 2 y 2 ) dydx + bx≥ay emax(b dydx ea 2 2 x 2 ,a 2 y 2 ) dydx = dydx + bx≥ay 2 e a b2 x2 bx<ay b ay /b = 0 b 0 ea a2y 2 y2 bx/a dxdy + a 0 0 a2y 2 y2 dydx = = = aye bxe dy + b a 0 0 2 2 2 2 ea b − 1 ea b − 1 + 2ab 2ab a2b2 e −1 . ab dx 1 3.5.32 Observe that x ≥ (x + y)2 ≥ 0. Hence we may take the positive square root 2 √ √ giving y ≤ 2x − x. Since y ≥ 0, we must have 2x − x ≥ 0 which means that x ≤ 2. The integral equals 2 0 0 √ 2x−x √ xydy dx = = = = = 2 3 2 0 √ 2 √ 4 u2 (u 2 − u2 )3/2 du 3 0 1 1 (1 − v 2 )3/2 (1 + v)2 dv 6 −1 π /2 1 cos4 θ(1 + sin2 θ)dθ 6 −π /2 7π . 96 √ √ x( 2x − x)3/2 dx 3.5.33 Observe that a ´ sin 2πx dx = 0 0 1 (1 − cos 2πa) 2π if a is an integer if a is not an integer Free to photocopy and distribute 234 .34 We have 1 0 0 1 1 n 1 n ··· 0 (x1 x2 · · · xn )dx1 dx2 .35 This is 1 0 0 1 1 n n 1 0 0 1 1 ··· xk 0 k=1 dx1 dx2 .41 The integral equals 1 dxdy (x + y)4 D 3 n . . we deduce that at least one of the sides of R is an integer. . . 3. . dxn = k=1 n ··· xk dx1 dx2 . 2 2 3. dxn 2n Free to photocopy and distribute 235 . . dxn = xk dxk k=1 0 = k=1 1 1 = n. dxn 0 = k=1 1 2 = 3. 2 4−x 1 = 1 3 dy dy (x + y)4 dx = = = 3.5. Since N sin 2πx sin 2πy dxdy = R k=1 Rk sin 2πx sin 2πy dxdy.Appendix A Thus 0 a sin 2πx dx = 0 ⇐⇒ a is an integer. Then 1 1 0 1 I = 0 ··· cos2 0 π (x1 + x2 + · · · + xn ) dx1 dx2 . . N Now sin 2πx sin 2πy dxdy = 0 k=1 Rk since at least one of the sides of each Rk is an integer.5.45 The integral equals 2/3 x+1 4−x 1 − (x + y)−3 dx 3 1 1 3 1 1 1 − dx 3 1 (1 + x)3 64 1 . 48 4 xdxdy D = −1 2/3 0 dy x dx + 2/3 4 0 2− x 2 dy x dx = = 275 .5. finishing the proof.5.46 Make the change of variables xk = 1 − yk .5. 54 −1 x(x + 1) dx + 2/3 x 2− x 2 dx 3. . Then x + y = u and x − y. The integral becomes f (x. u = 1. 4 b 1 = = v dv 0 Free to photocopy and distribute 236 . 2 Since sin t + cos t = 1. Clearly also 1 dx ∧ dy = du ∧ dv.Answers and Hints equals 1 0 2 0 2 1 1 ··· sin2 0 π (y1 + y2 + · · · + yn ) dy1 dy2 .6.4 Here we argue that du = ydx + xdy. x + y = 1. y)dx ∧ dy = = = The region transforms into ∆ = [a.1 u+v u−v and y = . 1]. Its image under Φ is the triangle bounded by the equations x = 0. 2n 1 . . v = −u.6. b] × [0. y)dx ∧ dy D (y 4 − x4 ) 1 2 (y − x2 )du ∧ dv 2 v du ∧ dv 2 1 du ∧ dv 2(y 2 + x2 ) = ∆ v du ∧ dv 1 du 2 a b−a . du ∧ dv = 2(y 2 + x2 )dx ∧ dy. v = u. we have 2I = 1. dyn . 4 3. y = 0. and so I = 3. Observe that D ′ 2 2 is the triangle in the uv plane bounded by the lines u = 0. Taking the wedge product of differential forms. 2 Put x = From the above (x + y)2 ex D 2 −y 2 dA = = = = 1 u2 euv dudv 2 D′ 1 u 1 u2 euv dudv 2 0 −u 1 2 2 1 u(eu − e−u )du 2 0 1 (e + e−1 − 2). dv = −2xdx + 2ydy. . Hence f (x. and (1. 1 + Ω2 This is straightforward but tedious! 3.6. (1. 1). −1). u − v = 2y and so 4xy = u2 − v 2 . We will split this region of integration into two disjoint triangles: T1 with vertices at (0. −1). 2 and that u + v = 2x. as desired. (1. This follows by using the identity t 0 dω = arctan t. (1. 1).5 Formally. and T2 with vertices at (1. (1. (2. 1). 2 √ 1 15 15y − y 3 dy Free to photocopy and distribute 237 . The integral becomes 1 0 0 1 dxdy 1 − xy = = 1 2 T 1 ∪T 2 1 1− u −u du ∧ dv u 2 −v 2 4 2 2−u u−2 2 0 dv 4 − u2 + v 2 du + 2 1 dv 4 − u2 + v 2 du. 1] in the xy plane into the square with vertices at (0.1 The integral in Cartesian coordinates is √ 1 15 1 √ 16−y 2 xydxdy = = 1 2 49 . −1) in the uv plane. 1 0 0 1 1 1 0 dxdy 1 − xy = 0 1 (1 + xy + x2 y 2 + x3 y 3 + · · · )dxdy 1 0 = = = x2 y 3 x3 y 4 xy 2 + + +··· 2 3 4 0 1 2 3 x x x (1 + + + + · · · )dx 2 3 4 0 1 1 1 1 + 2 + 2 + 2 + ··· 2 3 4 y+ dx This change of variables transforms the square [0. 0).7. 0). 0).Appendix A 3. (2. 0). Observe that dx ∧ dy = 1 du ∧ dv. 1] × [0. 15 3.7.7. √ D cos4 θ(cos2 θ − sin2 θ) dθ π. 18 9 = = Free to photocopy and distribute 238 .7. y = bρ sin θ.7. y)dA D = 0 π /6 2 sin θ ρ2 dρ dθ 1 (1 − 8 sin3 θ) dθ 3 0 √ π 16 − + 3. 12 sin2 2θ dθ 0 3.8 Using polar coordinates.2 Using polar coordinates. π /4 √ 0 sin 2θ xydxdy = = = = 4 0 ρ π /4 0 Ô ρ2 cos θ sin θ dρ dθ 4 3 √ √ ( sin 2θ)3 cos θ sin θ dθ π /4 4 √ 3 √2 π 2 . π /2 2 cos θ 1 4 arccos 1 4 arccos 1 4 arcsin π 4 π 4 1 cos4 θ sin θ sin θ cos θdθ (cot θ)(csc2 θ)dθ (tan θ)(sec2 θ)dθ 1 4 1 4 π 4 arcsin 1 4 arccos 1 4 π 4 7 7 28 − − 4 4 49 2 x2 − y 2 dxdy D = −π /2 π /2 0 ρ3 dρ 8 0 (cos2 θ − sin2 θ)dθ = = 3. π /6 1 f (x. the integral becomes 2π 1 (ab) 0 a3 cos3 θ + b3 sin3 θdθ 0 ρ4 dρ = 2 (ab)(a3 + b3 ).Answers and Hints The integral in polar coordinates is π 4 4 r 3 sin θ cos θdrdθ + 1/ sin θ arccos 1 4 π 4 4 r 3 sin θ cos θdrdθ 1/ cos θ = arcsin 1 4 1 4 π 4 arcsin 1 4 44 − 1 sin4 θ 44 − sin θ cos θ + = 44 4 − − = = 3.4 Using x = aρ cos θ.3 Using polar coordinates. 14 We have 2xdx = cos θdρ − ρ sin θdθ.7.7. the integral equals √ √ π π π 2 π 2 − − = .13 Observe that D = D2 \D1 where D2 is the disk limited by the equation x2 +y 2 = 1 and D1 is the disk limited by the equation x2 + y 2 = y. x2 + y 2 − y ≤ 0. 2 2 4 4 3. Hence dxdy = (1 + x2 + y 2 )2 dxdy − (1 + x2 + y 2 )2 dxdy . Free to photocopy and distribute 239 .Appendix A 3. Then the integral equals 2 D′ (x + y)2 dxdy.12 Put D ′ = {(x.7. x2 + y 2 − x ≤ 0}. 64 48 π /2 3. 5 3.7. 8 3. whence 4xydx ∧ dy = ρdρ ∧ dθ. y) ∈ R2 : y ≥ x.9 Using polar coordinates the integral becomes π /2 0 0 2 cos θ ρ4 dρ cos2 θ sin θdθ = 4 . (1 + x2 + y 2 )2 D D2 D1 Using polar coordinates we have dxdy = (1 + x2 + y 2 )2 2π 0 0 1 D2 ρ π dρdθ = (1 + ρ2 )2 2 and dxdy (1 + x2 + y 2 )2 π /2 = = 2 0 +∞ 0 D1 ρ sin2 θdθ dρdθ = 2 )2 (1 + ρ 1 + sin2 θ 0 0 √ dt dt π π 2 − 2 = − . Using polar coordinates the integral equals π /2 cos θ 2 π /4 (cos θ + sin θ)2 0 ρ3 dρ dθ = = 1 cos4 θ(1 + 2 sin θ cos θ)dθ 2 π /4 3π 5 − . t2 + 1 2t + 1 2 4 sin θ π /2 (We evaluated this last integral using t = tan θ) Finally.11 Using polar coordinates the integral becomes π /4 −π /4 2 cos θ 1/ cos θ 1 dρ ρ3 π /4 dθ = 0 cos2 θ − sec2 θ 4 dθ = π .7. 2ydy = sin θdρ + ρ cos θdθ. we deduce that Ja → π. Hence the integral becomes π /2 0 0 1 Ô 1 3 (ρ cos θ sin θ 1 − ρ2 )dρdθ 4 = = = 1 cos θ sin θdθ 4 0 1 1 2 · · 4 2 15 1 . The respective domains of integration of Ia and Ia √ 2 are the inscribed and the exscribed circles to the square.7. 0 2 π dθ . 60 π /2 1 ρ3 0 Ô 1 − ρ2 dρ 3. This gives the result. whose Calculus book I recommend greatly. 3. 1 by Michael Spivak.15 Using polar coordinates 2π a 0 Ia = 0 ρe−ρ dρ dθ = π(1 − e−a ). we have 0 ≤ θ ≤ π/2. Free to photocopy and distribute 240 .7. First observe that Ja = −a a e−x dx 2 2 . Since both Ia and Ia √ 2 tend to π as a → +∞. To find this integral. we now 2 + sin 2θ 0 use what has been dubbed as “the world’s sneakiest substitution”1 : we put tan θ = t. 2 2 The domain of integration of Ja is a square of side 2a centred at the origin.16 x2 4≤x 2 +y 2 ≤16 1 dA + xy + y 2 2π 4 2 = 0 2π 4 r drdθ r 2 + r 2 sin θ cos θ = = = = = = so the problem reduces to evaluate I = 1 drdθ r(1 + sin θ cos θ) 2π 4 dθ dr 1 + sin θ cos θ r 0 2 2π dθ log 2 1 + sin θ cos θ 0 2π dθ 2 log 2 2 + sin 2θ 0π dθ 4 log 2 2 + sin 2θ 0 4I log 2. Since the integration takes place on the first quadrant.Answers and Hints It follows that Ô x3 y 3 1 − x4 − y 4 dx ∧ dy = = 1 2 2 Ô (x y )( 1 − x4 − y 4 )(4xy dx ∧ dy) 4 Ô 1 3 (ρ cos θ sin θ 1 − ρ2 )dρ ∧ dθ 4 Observe that x4 + y 4 ≤ 1 =⇒ ρ2 cos2 θ + ρ2 sin2 θ ≤ 1 =⇒ ρ ≤ 1. 2 0 a contradiction. 3.7. Hence 1 + t2 1 + t2 1 + t2 π 0 dθ 2 + sin 2θ π /2 = 0 +∞ dθ + 2 + sin 2θ 2 dt 1+t 2 2t + 1+t 2 π π /2 0 dθ 2 + sin 2θ = 0 +∞ + = = = = = We conclude that dt dt + 2 2(t2 + t + 1) −∞ 2(t + t + 1) +∞ 0 2 dt 2 dt + 2t 1 2 2t 1 3 0 3 −∞ ( √ + √ )2 + 1 ( √3 + √3 ) + 1 3 √ ¬+∞ √ √ √ ¬3 √ √ 0 3¬ 2t 3 3 3¬ 2t 3 3 arctan + + arctan + ¬ ¬ 3 0 3 3 3 −∞ 3 3 √ √ 3 π π 3 π π − + − − 3 6 3 6 2 √ 2 π 3 . 0 By the Pythagorean Theorem. ]∪] . π] = [0. 2 Γ Using polar coordinates xdy − ydx = = (ρ cos θ)(sin θdρ + ρ cos θdθ) − (ρ sin θ)(cos θdρ − ρ sin θdθ) ρ2 dθ.18 Let I(S) denote the integral sought over a region S. Since D(x.Appendix A In so doing we have to pay attention to the fact that θ → tan θ is not continuous on π π [0. so we split the interval of integration into two pieces. dθ = . their squares are no farther than 4 units.7. cos 2θ = . 3 0 dt 1+t 2 2t −∞ 2 + 1+t 2 0 4π 1 dA = x2 + xy + y 2 4≤x 2 +y 2 ≤16 √ 3 log 2 . [0. Parametrise the curve enclosing the region by polar coordinates so that the region is tangent to the polar axis at the origin. y) = 0 inside R. π]. and so the area π /2 1 < 4dθ = π. The area of the region is then given by 1 2 π ρ2 dθ = 0 1 2 π (f (θ))2 dθ = 0 1 2 π /2 ((f (θ))2 + (f (θ + π/2))2 )dθ.14 that the area enclosed by a simple closed curve Γ is given by 1 xdy − ydx. Let the equation of the curve be ρ = f (θ). the integral above is the integral of the square of the chord in question. Then 2 2 2t 1 − t2 dt sin 2θ = . I(R) = A. 3 3. If no two points are farther than 2 units. Let L be a side of R with length l and let S(L ) be the half strip consisting of Free to photocopy and distribute 241 . π].17 Recall from formula 1. 1). C.8. If α is the measure of that angle. 6 Free to photocopy and distribute 242 . C = (0. z sweeps from P2 to P1 . and O = (0. Thus 1 0 0 x y 1−x+y 1 x 0 dzdydx = 0 1 (1 − x)dydx = = = ¡ x − x2 dx 0 ¬ x2 x 3 ¬1 − ¬ 2 3 0 1 . O x−y+z =1 z=y x=y y = 0. The sum of these integrals over all the vertices of R is 2π. 3. 0. Then l +∞ I(S(L )) = 0 0 e−v dudv = l. x = 1 and x = y on the first quadrant of the xy-plane.Answers and Hints the points of the plane having a point on L as nearest point of R. 0). so the limits are z = y to z = 1 − x + y. the points that have V as nearest from R lie inside an angle S(V ) bounded by the rays from V perpendicular to the edges meeting at V . then using polar coordinates α +∞ I(S(V )) = 0 0 ρe−ρ dρdθ = α.8. Set up coordinates uv so that u is measured parallel to L and v is measured perpendicular to L. 0. A. C. B = (1. The projection of the solid on the xy plane produces the region bounded by the lines x = 0. 1. We have four planes passing through each triplet of points: P1 : P2 : P3 : P4 : A. Assembling all these integrals we deduce the result. 12 3. O A. The sum of these integrals over all the sides of R is L. 0.1 We have 1 1−y 0 0 1−z zdV E = 0 1 1−y zdxdzdy z − z 2 dzdy = 0 1 0 = = = (1 − y)2 (1 − y)3 − dy 2 3 0 4 3 ¬1 (1 − y) (1 − y) ¬ − ¬ 12 6 0 1 = . C. B.3 Let A = (1. O B. 0). Using the order of integration dzdxdy. 1). If V is a vertex of R. B. 4 We have 3 ¡ x2 − x3 dx 0 ¬ x3 x4 ¬1 − ¬ 3 4 0 1 . x 2 +y 2 r Cylindrical: rdzdθdr. 0 r2 Spherical: π /2 π /4 0 2π 0 (cos φ)/(sin φ)2 r 2 sin φdrdθdφ. Since x2 + y 2 ≤ z ≤ 4 − x2 − y 2 . 8 3. 2 Free to photocopy and distribute 243 . y. We have 1 0 0 x y 1−x+y 1 x 0 xdzdydx = 0 1 (x − x2 )dydx = = = 3.Appendix A We use the same limits of integration as in the previous integral.8. √ √ − 1−y 2 1−y 2 2π 0 x 2 +y 2 √ 1 dzdxdy.1 Cartesian: 1 −1 π 2 0 1 0 π(x − y) dxdy 2(x2 − y 2 ) π dxdy 2(x + y) ¬1 ¬ π ¬ · ((y + 1) log(y + 1) − (y + 1) − y log y + y)¬ 2 ¬ 0 log(y + 1) − log ydy π log 2. Observe that if (x.5 The desired integral is 1 0 0 1 0 ∞ dxdydz (1 + x2 z 2 )(1 + y 2 z 2 ) 1 1 0 0 = 0 1 1 0 1 1 0 = 0 dxdydz ¬z=∞ ¬ 1 ¬ (x arctan(xz) − y arctan(yz)) ¬ dxdy x2 − y 2 ¬ z=0 ∞ 1 x2 − y 2 x2 y2 − 1 + x2 z 2 1 + y2z2 = 0 1 1 = 0 = = = 3. z) is on the intersection of the surfaces then √ −1 ± 17 2 z + z = 4 =⇒ z = . 3 3.9.8. we start our integration with the z-variable.9. 12 y /3 √ 0 9−x 2 0 xdV = E 0 xdzdydx = 27 .2 Ô 1. π The volume is . Observe that z = x2 +y 2 =⇒ r cos φ = r 2 (cos θ)2 (sin φ)2 +r 2 (sin θ)2 (sin φ)2 =⇒ r ∈ {0. (csc φ)(cot φ) Perhaps it is easiest to evaluate the integral using cylindrical coordinates. Now.3 Cartesian: √ − 3 3 √ − 3−y 2 3−y 2 1 √ 4−x 2 −y 2 √ √ dzdxdy. √ 17 − 1 √ 17 − 1 2 z = r cos φ =⇒ cos φ = =⇒ φ = arccos 2 4 The desired integral √ 2π 0 0 arccos 17 − 1 4 2 r 3 cos θ sin2 φdrdφdθ. The desired integral is thus 2 Since x + y + z 2 2 2 = 4 =⇒ −2 ≤ z ≤ 2. 3. We obtain √ 2π 0 0 17 − 1 2 √ r2 4−r 2 r 2 cos θdzdrdθ = 0.9. The desired integral 2 2π 0 0 √ 17−1 2 √ r2 4−r 2 r 2 cos θdzdrdθ. 2 The projection of the circle of intersection of the paraboloid and the sphere onto the xy-plane satisfies the equation √ 17 − 1 z 2 + z = 4 =⇒ x2 + y 2 + (x2 + y 2 )2 = 4 =⇒ x2 + y 2 = . and so r 2 ≤ z ≤ tion circle onto the xy-plane is again a circle with centre at the origin and radius Ö√ 17 − 1 . The angle φ starts at φ = 0. a conclusion that is easily reached.Answers and Hints 17 − 1 only. x 2 +y 2 2. It is clear that the limits of the angle θ are from θ = 0 to θ = 2π. (csc φ)(cot φ)}. 2 Ö√ 17 − 1 a circle of radius . since the integrand is an odd function of x and the domain of integration is symmetric about the origin in x. Free to photocopy and distribute 244 . The z-limits remain the same as in the Cartesian coordinates. but translated into Ô 4 − r 2 . The projection of the interseccylindrical coordinates. 3. we must have z = √ √ 17−1 2 √ 17−1 −x 2 2 √ 4−x 2 −y 2 − √ 17−1 2 − √ 17−1 −x 2 2 xdzdydx. r 2 cos2 θ sin2 φ sin2 t + r 2 sin2 θ sin2 φ sin2 t + r 2 cos2 φ sin2 t + r 2 cos2 t r 2 cos2 θ sin2 φ + r 2 sin2 θ sin2 φ + r 2 cos2 φ r 2 cos2 θ + r 2 sin2 θ r2.9.5 We have 5π . 0 dθ 0 sin φdφ 0 sin2 tdt = = Free to photocopy and distribute 245 .9. Spherical: π /3 0 r 2 sin φdrdθdφ.8 π 96 π 14 3. 3 2π 2 1 0 2+r cos θ ydV = E 0 r 2 sin θdzdrdθ = 0. y = ρ sin θ sin φ sin t.7 3. u = ρ cos φ sin t.9.9 We put x = ρ cos θ sin φ sin t. sin θ sin φ sin tdr + ρ cos θ sin φ sin tdθ + ρ sin θ cos φ sin tdφ + ρ sin θ sin φ cos tdt cos φ sin tdr − ρ sin φ sin tdφ + ρ cos φ cos tdt cos tdr − ρ sin tdt After some calculation.9. x2 + y 2 + u2 + v 2 = = = = Now. Upon using sin2 a + cos2 a = 1 three times.Appendix A Cylindrical: √ 0 3 0 2π 0 2π 1 2 √ 4−r 2 rdzdθdr. 3. dx dy du dv = = = = cos θ sin φ sin tdr − ρ sin θ sin φ sin tdθ + ρ cos θ cos φ sin tdφ + ρ cos θ sin φ cos tdt . 1/ cos φ The volume is 3. dx ∧ dy ∧ du ∧ dv = r 3 sin φ sin2 tdr ∧ dφ ∧ dθ ∧ dt. Therefore ex x 2 +y 2 +u 2 +v 2 ≤1 2 +y 2 +u 2 +v 2 π 2π 0 1 0 2 π 0 1 dxdydudv = 0 r 3 er sin φ sin2 tdrdφdθdt 2π π π 2 = 0 r 3 er dr 1 π (2π)(2) 2 2 π 2. v = ρ cos t. z = u + v 2 . 2].9. 0 ≤ v ≤ 1. dz ∧ dx = −2vdu ∧ dv. n one has 1 0 2 xm (1 − x)n dx = m!n! . uvw = z =⇒ uvdw + uwdv + vwdu = dz. using the fact that for positive integers m. y = v. and so Ô ¬¬ 2 ¬¬ ¬¬d x¬¬ = 2 + 4v 2 du ∧ dv. z = uvw. which finally give 0 ≤ u ≤ 1. y = uv(1 − w). Free to photocopy and distribute 246 . u v du ∧ dv ∧ dw = dx ∧ dy ∧ dz. 0 ≤ uv ≤ 1. Observe that dx ∧ dy = du ∧ dv. 0 ≤ w ≤ 1. 44172388260000 3.10 We make the change of variables u = x + y + z =⇒ du = dx + dy + dz. 0 ≤ u − uv ≤ 1 − uv + uvw − uvw. 1] × [0. This gives x = u(1 − v). (m + n + 1)! we deduce.1 We parametrise the surface by letting x = u. To find the limits of integration we observe that the limits of integration using dx ∧ dy ∧ dz are 0 ≤ z ≤ 1. which in turn is 1 0 1 1 u20 (1 − u)4 du 0 v 18 (1 − v)dv 0 w 8 (1 − w)9 dw = 1 1 1 · · 265650 380 437580 which is = 1 . Observe that the domain D of Σ is the square [0. dy ∧ dz = −du ∧ dv. The integral sought is then. 0 ≤ y ≤ 1 − z. This translates into 0 ≤ uvw ≤ 1. 0 ≤ u ≤ 1. 0 1 0 1 0 1 u20 v 18 w 8 (1 − u)4 (1 − v)(1 − w)9 dudvdw. 0 ≤ x ≤ 1 − y − z. Thus 0 ≤ uvw ≤ 1. uv = y + z =⇒ udv + vdu = dy + dz.10.Answers and Hints 3. 0 ≤ uv − uvw ≤ 1 − uvw. y = v.10. dy ∧ dz = cos θ sin2 φdφ ∧ dθ. (x. 3. π] is the longitude.Appendix A The integral becomes ¬¬ ¬¬ y ¬¬d2 x¬¬ Σ 2 1 = 0 1 0 v du 13 2 . dz ∧ dx = − sin θ sin2 φdφ ∧ dθ. dz ∧ dx = − du ∧ dv. Observe that dx ∧ dy = sin φ cos φdφ ∧ dθ. and so The integral becomes ¬¬ 2 ¬¬ ¬¬d x¬¬ = sin φdφ ∧ dθ. 3 0 √ Ô 2 + 4v 2 dudv 2 = = y 0 Ô 2 + 4v 2 dv 3. dy = dv. ¬¬ ¬¬ x2 ¬¬d2 x¬¬ Σ 2π π = 0 0 cos2 θ sin3 φdφdθ 4π . the surface area is √ 2π 2 2 0 1 rdrdθ = 3π √ 2. whence u v dx ∧ dy = du ∧ dv. y = r sin θ. Here θ ∈ [0.10. 1+ (dx ∧ dy)2 + (dz ∧ dx)2 + (dy ∧ dz)2 Hence ¬¬ ¬¬ z ¬¬d2 x¬¬ = Σ u 2 +v 2 ≤1 Ô u2 + v 2 √ 2 dudv = √ 2π 1 2 0 0 ρ2 dρdθ = √ 2π 2 . z z and so ¬¬ 2 ¬¬ ¬¬d x¬¬ = = = u2 + v 2 du ∧ dv z2 √ 2 du ∧ dv. Then dx = du. cos φ). z) = (cos θ sin φ. 0 ≤ θ ≤ 2π.4 Put x = u. 3 Free to photocopy and distribute 247 . y. zdz = udu + vdv. 3 = 3. 2π] is the latitude and φ ∈ [0. dy ∧ dz = − du ∧ dv. z 2 = u2 + v 2 . 1 ≤ r ≤ 2.10.3 We use spherical coordinates. sin θ sin φ.2 Using x = r cos θ. 4 − Σ x2 dzdx = − 3 0 x2 dzdx = − 27 . We have dω = = = (3x2 ydx + x3 dy) ∧ dx + (ydx + xdy) ∧ dy d(x3 y) ∧ dx + d(xy) ∧ dy (y − x3 )dx ∧ dy.5 If the egg has radius R.10. with R cos φ1 − R cos φ2 = 2R/n. The area of the part of the surface of the sphere in slice is 2π 0 φ2 φ1 R2 sin φdφdθ = 2πR2 (cos φ1 − cos φ2 ) = 4πR2 /n.11. each slice will have height 2R/n. The integral equals Á 2 2 x3 ydx + xydy = (y − x3 )dxdy C 0 0 = 3. We will use Green’s Theorem. one for each side of the square. 2 27 .1 Evaluating this directly would result in evaluating four path integrals. LAC is y = −x. This means that each of the n slices has identical area 4πR2 /n. Free to photocopy and distribute 248 .2 We have −4.Answers and Hints 3. and LB C is clearly y = − x + . 2 (x + z)dxdy = Σ 0 (6 − x − 2y)dxdy = 27 . 3 3 We have 1 y 2 dx + xdy AB = 0 (x2 + x)dx −2 0 = 2 5 6 − 15 2 14 3 y 2 dx + xdy BC = 1 1 4 − x+ 3 3 − 1 x dx 3 = = y 2 dx + xdy CA = −2 (x2 − x)dx Adding these integrals we find Á △ y 2 dx + xdy = −2.11.6 We project this plane onto the coordinate axes obtaining 6 3−z/2 0 3 xydydz = Σ 0 (3 − y − z/2)ydydz = 6−2x 0 3−y 0 27 .10. 3. A slice can be parametrised by 0 ≤ θ ≤ 2π. The region M is the area enclosed by the square. φ1 ≤ φ ≤ φ2 . 4 and hence xydydz − x2 dzdx + (x + z)dxdy = Σ 3. 1 4 LAB is y = x. 2 a b2 If we were to evaluate this integral directly.6 Observe that 10 44 − 27 27 −2. 0 ≤ t ≤ 2π. by Stokes’ Theorem the integral sought is − 2(y −1)2 +z 2 ≤2 0 = x2 + y 2 + z 2 − 2(x + y) = (2 − y)2 + y 2 + z 2 − 4 = 2y 2 − 4y + z 2 = 2(y − 1)2 + z 2 − 2. To do it directly. − 0 x −x/3+4/3 (1 − 2y)dy Hence by the generalised Stokes’ Theorem the integral equals π /2 4 cos θ 0 d(x2 + 2y 3 ) ∧ dy = 2xdx ∧ dy.11. y = 2 sin t. (To evaluate the integrals you may resort to the fact that the area of the elliptical region (x − x0 )2 (y − y0 )2 + ≤ 1 is πab). we would set √ y = 1 + cos θ.7 At the intersection path 16π. 2xdx ∧ dy = {(x−2)2 +y 2 ≤4} −π /2 2ρ2 cos θdρ ∧ dθ = 16π. Since dx ∧ dy = 0. x = 2 − y = 1 − cos θ. Then the integral becomes 2π 2π ((2 + 2 cos t)2 + 16 sin3 t)d2 sin t 0 = 0 (8 cos t + 16 cos2 t +8 cos3 t + 32 cos t sin3 t)dt = 3. Free to photocopy and distribute 249 . We have d (ydx + zdy + xdz) = dy ∧ dx + dz ∧ dy + dx ∧ dz = −dx ∧ dy − dy ∧ dz − dz ∧ dx. z = 2 sin θ. Similarly we get 2(x − 1)2 + z 2 = 2 on the xz-plane. which describes an ellipse on the yz-plane. dydz − 2(x−1)2 +z 2 ≤2 √ dzdx = −2π( 2). The integral becomes 2π 0 (1 + cos θ)d(1 − cos θ) + 2π √ √ 2 sin θd(1 + cos θ) + (1 − cos θ)d( 2 sin θ) √ √ √ which in turn = 0 sin θ + sin θ cos θ − 2+ 2 cos θdθ = −2π 2.11. put x − 2 = 2 cos t.Appendix A We have 0 (1 − 2y)dx ∧ dy D = −2 1 −x/3+4/3 −x (1 − 2y)dy dx dx + = = 3. A section “Entitled XYZ” means a named subunit of the Document whose title either is precisely XYZ or contains XYZ in parentheses following text that translates XYZ in another language. or “History”. philosophical. and standard-conforming simple HTML. PostScript or PDF designed for human modification. while not being considered responsible for modifications made by others. The “Title Page” means. in any medium. modify or distribute the work in a way requiring permission under copyright law. SGML or XML using a publicly available DTD. The Document may include Warranty Disclaimers next to the notice which states that this License applies to the Document. refers to any such manual or work. 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Que ´ languidezca de dolor gritando por piedad. se le convierta en una serpiente en las manos y lo venza. “Dedications”. the Document’s Cover Texts may be placed on covers that bracket the Document within the aggregate. See http://www. and will automatically terminate your rights under this License. and distribute it individually under this License. or distribute the Document except as expressly provided for under this License. and replace the individual copies of this License in the various documents with a single copy that is included in the collection. HUBBARD and Barbara Burke HUBBARD. Linear Algebra. BUCK. A Unified Approach. Edmund ROBERTSON. Analysis on Manifolds. Calculus on Manifolds: A Modern Approach to Classical Theorems of Advanced Calculus. R. Spherical Trigonometry. Sean DINEEN Multivariate Calculus and Geometry ´ [MarTro] Jerrold E. James R. Analysis on Manifolds. MUNKRES. [Lan] [Spik] [Edw] [Mun] [Tod] Serge LANG. Michael SPIVAK. Basic Linear Algebra.Bibliography [BlRo] [Mun] [Buck] [Din] Thomas BLYTH. Advanced Calculus. Isaac TODHUNTER. Vector Calculus. Vector Calculus [HubHub] John H. MUNKRES. Harold M. Introduction to Linear Algebra. and Differential Forms. 253 . James R. Advanced Calculus: A Differential Forms Approach.C. EDWARDS. MARSDEN and Anthony TROMBA.