MA6151 - Mathematics - I Notes

www.rejinpaul.com MA2111/ Engineering Mathematics-I UNIT 1 MATRICES CHARACTERISTIC EQUATION: The equation is called the characteristic equation of the matrix A Note: 1. Solving , we get n roots for and these roots are called characteristic roots or eigen values or latent values of the matrix A 2. Corresponding to each value of , the equation AX = has a non-zero solution vector X If be the non-zero vector satisfying AX = , , when is said to be the latent vector or eigen A vector of a matrix A corresponding to Working rule to find characteristic equation: PR A Method 1:The characteristic equation is TH For a 3 x 3 matrix: YU which we call as characteristic polynomial of matrix A when expanded will give a polynomial, SH CHARACTERISTIC POLYNOMIAL: The determinant Method 2:Its characteristic equation can be written as where , , For a 2 x 2 matrix: Method 1:The characteristic equation is Method 2:Its characteristic equation can be written as where , Problems: 1. Find the characteristic equation of the matrix ( ) [Anna University May 2003] Get useful study materials from www.rejinpaul.com www.rejinpaul.com MA2111/ Engineering Mathematics-I Solution: Let A = ( ). Its characteristic equation is where , = 1(2) – 2(0) = 2 Therefore, the characteristic equation is 2. Find the characteristic equation of ( ) Solution: Its characteristic equation is , where = 8 + 7 + 3 = 18, | | | ) SH PR A Solution: Let A = ( ) TH Find the characteristic polynomial of ( YU Therefore, the characteristic equation is The characteristic polynomial of A is + 2 = 5 and | = 8(5)+6(-10)+2(10) = 40 -60 + 20 = 0 , 3. | A | where = 3(2) – 1(-1) = 7 Therefore, the characteristic polynomial is EIGEN VALUES AND EIGEN VECTORS OF A REAL MATRIX: Working rule to find eigen values and eigen vectors: 1. Find the characteristic equation 2. Solve the characteristic equation to get characteristic roots. They are called eigen values 3. To find the eigen vectors, solve [ ] for different values of Note: 2 Get useful study materials from www.rejinpaul.com =3 www.rejinpaul.com MA2111/ Engineering Mathematics-I 1. Corresponding to n distinct eigen values, we get n independent eigen vectors 2. If 2 or more eigen values are equal, it may or may not be possible to get linearly independent eigen vectors corresponding to the repeated eigen values 3. If is a solution for an eigen value , then c is also a solution, where c is an arbitrary constant. Thus, the eigen vector corresponding to an eigen value is not unique but may be any one of the vectors c 4. Algebraic multiplicity of an eigen value polynomial (i.e., if 5. is the order of the eigen value as a root of the characteristic is a double root, then algebraic multiplicity is 2) Geometric multiplicity of Non-symmetric matrix: is the number of linearly independent eigen vectors corresponding to If a square matrix A is non-symmetric, then A ≠ Note: In a non-symmetric matrix, if the eigen values are non-repeated then we get a linearly independent set of A 1. 2. SH eigen vectors In a non-symmetric matrix, if the eigen values are repeated, then it may or may not be possible to get YU linearly independent eigen vectors. If we form a linearly independent set of eigen vectors, then diagonalization is possible through similarity TH transformation Note: 1. PR A Symmetric matrix: If a square matrix A is symmetric, then A = In a symmetric matrix, if the eigen values are non-repeated, then we get a linearly independent and pair wise orthogonal set of eigen vectors 2. In a symmetric matrix, if the eigen values are repeated, then it may or may not be possible to get linearly independent and pair wise orthogonal set of eigen vectors If we form a linearly independent and pair wise orthogonal set of eigen vectors, then diagonalization is possible through orthogonal transformation Problems: 1. Find the eigen values and eigen vectors of the matrix ( ) [Anna University Tvli. May/June 2011] 3 Get useful study materials from www.rejinpaul.com 4 i. the characteristic equation is or A Therefore...e.e. [ ][ ] [ ] [ ] [( ) YU ) ( )] [ ] [ ] TH [( ] [ ] --------------. -2 ( [ )] [ ] ][ ] [ Case 1: If i.com MA2111/ Engineering Mathematics-I Solution: Let A = ( ) which is a non-symmetric matrix To find the characteristic equation: The characteristic equation of A is where .e. i.www.rejinpaul.e.. Therefore. we get only one equation Therefore Case 2: If [ ] [ ][ ] [ ] [From (1)] 4 Get useful study materials from www.rejinpaul. = 1(-1) – 1(3) = .(1) PR A [ SH To find the eigen vectors: ][ ] [ ] [From (1)] i.com . the eigen values are 2.. com MA2111/ Engineering Mathematics-I i. | | | | | | . = 2 (-5)-2 (-6)-7(2) = -10 + 12 – 14 = -12 Therefore... i.com . May/June 2011] ] which is a non-symmetric matrix YU Solution: Let A = [ SH A 2.e.e. we get only one equation Hence.. [ ] TH To find the characteristic equation: where PR A Its characteristic equation can be written as .e. [ ][ ] [ ] i.rejinpaul. the characteristic equation of A is 3 -4 √ 0 √ 5 Get useful study materials from www.rejinpaul. Find the eigen values and eigen vectors of [ ] [Anna University Tvli.www. (1) PR A TH ------. [ [ ] A ][ ] [ Case 1: If YU --------.rejinpaul.(3) [ ] Considering equations (1) and (2) and using method of cross-multiplication.e.[ ] ][ ] [ ] 6 Get useful study materials from www. 2 -7 1 2 0 2 2 0 Therefore. and -4 A is a non-symmetric matrix with non-repeated eigen values To find the eigen vectors: [ ] ][ ] [ ][ ] [ ] SH i.www. [ Case 2: If . we get.rejinpaul..(2) -------. the eigen values are 3. 1.com MA2111/ Engineering Mathematics-I Therefore.com . www.rejinpaul.com MA2111/ Engineering Mathematics-I i.e., [ ][ ] [ ] -------- (1) -------- (2) -------- (3) -7 -1 2 -2 2 2 -2 Case 3: If [ TH [ ] PR A Therefore, YU SH 2 A Considering equations (1) and (2) and using method of cross-multiplication, we get, ][ ] [ ] -------- (1) -------- (2) -------- (3) Considering equations (1) and (2) and using method of cross-multiplication, we get, 2 -7 6 2 5 2 2 5 7 Get useful study materials from www.rejinpaul.com www.rejinpaul.com MA2111/ Engineering Mathematics-I ] [ Therefore, Find the eigen values and eigen vectors of [ 3. ] [Anna University November/December 2010, May/June 2010], [Anna University CBT January 2011] ] which is a non-symmetric matrix A Solution: Let A = [ SH To find the characteristic equation: where , | | | | | | TH YU Its characteristic equation can be written as PR A , = 2(4)-2(1)+1(-1) = 5 Therefore, the characteristic equation of A is 1 1 -6 5 √ 0 √ Therefore, the eigen values are 1, 1, and 5 A is a non-symmetric matrix with repeated eigen values To find the eigen vectors: [ ] 8 Get useful study materials from www.rejinpaul.com www.rejinpaul.com MA2111/ Engineering Mathematics-I ][ ] [ ][ ] [ Case 1: If i.e., [ ][ ] [ ] [ ] [ ] --------- (1) A ------------- (2) SH ------------ (3) -3 2 -2 1 1 -2 TH 1 PR A 2 YU Considering equations (1) and (2) and using method of cross-multiplication, we get, Therefore, [ ] Case 2: If ,[ i.e., [ ][ ] ][ ] [ ] [ ] 9 Get useful study materials from www.rejinpaul.com Therefore.8 – 4 + 4 = .8 Therefore.rejinpaul. the characteristic equation of A is 1 0 -4 0 Therefore. January 2011] SH Find the eigen values and eigen vectors of [ ] which is a non-symmetric matrix TH 4. 2.com . YU Solution: Let A = [ ][Anna University Madurai.rejinpaul. Put Therefore. | | | | | | . and -2 A is a non-symmetric matrix with repeated eigen values 10 Get useful study materials from www.com MA2111/ Engineering Mathematics-I All the three equations are one and the same. the eigen values are 2.www. [ ] [ ] Therefore. A Put To find the characteristic equation: PR A Its characteristic equation can be written as where . = 2(-4)+2(-2)+2(2) = . (3) .(2) -----------. [ PR A TH Considering equations (1) and (2) and using method of cross-multiplication.rejinpaul. [ ][ ] [ ] ][ ] [ [ ] [ ] SH A --------.rejinpaul..(1) YU ------------. Equations (2) and (3) are one and the same. ] ][ ] ][ ] [ ] [ ] 11 Get useful study materials from www.[ i.com .www. we get.com MA2111/ Engineering Mathematics-I To find the eigen vectors: [ ] ][ ] [ Case 1: If i. [ Case 2: If .e..e. -1 1 2 -1 3 1 1 3 Therefore. SH A -2 TH We get one eigen vector corresponding to the repeated root Solution: Let A =[ ] [Anna University Tvli.4 + 2 -42 = .36 Therefore. we get.(1) ---------------.www.(3) Considering equations (1) and (2) and using method of cross-multiplication.com MA2111/ Engineering Mathematics-I ---------. 2 0 -2 -1 1 1 -1 [ ] YU Therefore. ] which is a symmetric matrix To find the characteristic equation: Its characteristic equation can be written as where .com .rejinpaul. January 2010.(2) -----------..rejinpaul.2011] PR A Find the eigen values and eigen vectors of [ 5. the characteristic equation of A is 12 Get useful study materials from www. | | | | | | . = 1(4)-1(-2)+3(-14) = . com MA2111/ Engineering Mathematics-I -9 1 18 0 √ √ Therefore.rejinpaul. the eigen values are -2. and 6 A is a symmetric matrix with non.www.(3) Considering equations (1) and (2) and using method of cross-multiplication.e.(1) ------------.(2) -----------.repeated eigen values To find the eigen vectors: ] ][ ] ][ ] TH ][ ] [ ] [ ] PR A i.rejinpaul. we get. 1 3 3 1 7 1 1 7 13 Get useful study materials from www. [ YU [ Case 1: If [ ] SH [ A [ --------.com .. 3. [ ] TH 3 PR A 1 YU Considering equations (1) and (2) and using method of cross-multiplication.[ i.com MA2111/ Engineering Mathematics-I Therefore. ][ ] ][ ] [ ] [ ] ---------.(3) SH 3 -2 1 2 1 1 2 Therefore.com .www.rejinpaul. [ Case 2: If .(2) 14 Get useful study materials from www.e.(1) ---------------.. [ Case 3: If ..(2) A -----------. [ ] ][ ] ][ ] [ ] [ ] ---------. we get.[ i.e.(1) ---------------.rejinpaul. com . the characteristic equation of A is 1 -1 √ -2 0 √ 15 Get useful study materials from www.www.(3) Considering equations (1) and (2) and using method of cross-multiplication.com MA2111/ Engineering Mathematics-I 3 -----------. = 0 -1(-1)+ 1(1) = 0 + 1 + 1 = 2 Therefore. geometric multiplicity TH ] which is a symmetric matrix PR A Solution: Let A = [ ]. Determine the algebraic and YU Find the eigen values and eigen vectors of the matrix[ 6. 1 3 -5 1 -1 1 1 -1 [ ] SH A Therefore.rejinpaul. To find the characteristic equation: Its characteristic equation can be written as where . | | | | | | .rejinpaul. we get. [ [ ] A ][ ] [ Case 1: If YU --------.e. 1 1 -2 1 -2 1 1 -2 Therefore.com MA2111/ Engineering Mathematics-I Therefore..rejinpaul. [ ] 16 Get useful study materials from www. the eigen values are 2. -1.(1) PR A TH ------------. and -1 A is a symmetric matrix with repeated eigen values.com . The algebraic multiplicity of is 2 To find the eigen vectors: [ ] ][ ] [ ][ ] [ ] SH i.www.(3) [ ] Considering equations (1) and (2) and using method of cross-multiplication.rejinpaul.(2) -----------. we get. All the three equations are one and the same. we get.www.com .(1) ][ ] -------.(3).(2) Solving (1) and (2) by method of cross-multiplication. [ ]. let ] -----------.(1) ---------------..(2) -----------.rejinpaul. SH A Therefore.e. Put [ ] YU Therefore.rejinpaul. l m 1 1 1 1 1 -1 0 1 17 Get useful study materials from www. . [ ][ ] ][ ] [ ] [ ] ---------.[ Case 2: If i. is orthogonal to [ ][ [ PR A TH Since the given matrix is symmetric and the eigen values are repeated.com MA2111/ Engineering Mathematics-I . rejinpaul. i.e. Let the n roots be Already. SH Proof: Let A be a square matrix of order n. So. [ ] Thus.com MA2111/ Engineering Mathematics-I .www. i. i.(1) where i.. we get n roots..e. Solving (1). Sum of the eigen values = Sum of the main diagonal elements Product of the roots = by (1) and (2) 18 Get useful study materials from www. ----------..(2) Sum of the roots = by (1) and (2) i. The characteristic equation of A is PR A TH … We know that the roots of the characteristic equation are called eigen values of the given matrix. the geometric multiplicity of eigen value is 2 PROPERTIES OF EIGEN VALUES AND EIGEN VECTORS: Property 1: (i) The sum of the eigen values of a matrix is the sum of the elements of the principal diagonal (or) The sum of the eigen values of a matrix is equal to the trace of the matrix Product of the eigen values is equal to the determinant of the matrix A (ii) YU --------..e. for the repeated eigen value there corresponds two linearly independent eigen vectors .e.com ..rejinpaul. we know that.e. Therefore. are the eigen values of A. the eigen values of A and A Hence. Product of the eigen values = |A| Property 2: A square matrix A and its transpose and its transpose have the same eigen values (or) A square matrix A have the same characteristic values Proof: Let A be a square matrix of order n. | | On expansion..(2) Since the determinant value is unaltered by the interchange of rows and columns.e. we have |A| = | are the same SH Therefore.www. January 2011] Proof: Let us consider the triangular matrix A = [ ]. it gives. i..com MA2111/ Engineering Mathematics-I i.e. YU Note: A determinant remains unchanged when rows are changed into columns and columns into rows TH Property 3: The characteristic roots of a triangular matrix are just the diagonal elements of the matrix (or) the eigen values of a triangular matrix are just the diagonal elements of the matrix PR A [Anna University Madurai. which are the diagonal elements of matrix A 19 Get useful study materials from www.com .rejinpaul. (1) and (2) are identical.e. The characteristic equation of A and (1) and are ------- --------. The characteristic equation of A is i..rejinpaul. we get.com MA2111/ Engineering Mathematics-I Property 4: If is an eigen value of a matrix A. But. what can you say about the eigen value of matrix (or) If is an Prove your statement [Anna University March 1996.com . is also an eigen value of A Property 6: If are the eigen values of a matrix A. is an eigen value of A. 20 Get useful study materials from www. then .rejinpaul. shows that is the eigen value of the inverse matrix A square matrix A is said to be orthogonal if Let A be an orthogonal matrix. is the eigen value of eigen value of a matrix A.rejinpaul. Anna University May 1997] Proof: If X be the eigen vector corresponding to Pre multiplying both sides by --------. Since . then has the eigen values (m being a positive integer) Proof: Let be the eigen value of A and the corresponding eigen vector.. we get.e. . since the determinant .www. then is also its eigen value TH Property 5: If YU This being of the same form as (1). the matrices A and have the are the same Hence.(1) . Given is an eigen value of i. SH A Dividing by then AX = Definition of orthogonal matrix: PR A Proof: is an eigen value of an orthogonal matrix. is an eigen value of same eigen values. We have. The corresponding eigen vector is the same ..(2) (1) and (2) are of the same form. we get. Property 8: The eigen vectors corresponding to distinct eigen values of a real symmetric matrix are orthogonal [Anna University. Hence. Taking the transpose on both sides.e.. is an eigen value of . ̅̅̅ ̅ ̅̅̅ since ̅ ̅ ̅ ̅= ̅ ̅ or ̅ ̅ since ̅ = A if A is real i. we get. November 2002] Proof: For a real symmetric matrix A. the eigen values are real. from (1). ̅̅̅X = ̅ ̅ ..e.com MA2111/ Engineering Mathematics-I Then. ̅ ̅̅ i. In general. AX = . if A is symmetric. But. Similarly. Hence. ------. we get.rejinpaul.com . YU Let A vector be X. Property 7: The eigen values of a real symmetric matrix are real numbers [Anna University November 2001] Proof: Let be an eigen value (may be complex) of the real symmetric matrix A.e.www. ̅ i.(1) We have.(1) PR A conjugate of TH Pre multiplying this equation by 1 x n matrix ̅̅̅where the bar denotes that all elements of ̅̅̅ are the complex ̅̅ Taking the conjugate complex of this. .rejinpaul. ̅̅̅̅AX = ̅ X ------------. Let eigen values [ be the eigen vectors to two distinct are real] 21 Get useful study materials from www. Let the corresponding eigen SH be the transpose of A. ̅̅̅AX = ̅̅̅X. Since̅̅̅̅ is a 1 x 1 matrix whose only element is a positive value. -------. rejinpaul.www. Then there exists a non-singular matrix P such that YU B– TH = |B– |=| = PR A = = |A = |A Therefore. they have the same eigen values.e. MIT] SH Proof: Let A and B be two similar matrices. Property 10: If a real symmetric matrix of order 2 has equal eigen values. B have the same characteristic polynomial and hence characteristic roots. we get.(3) But i. i.(4) [Since ] From (3) and (4). then the matrix is a scalar matrix Proof: Rule 1: A real symmetric matrix of order n can always be diagonalized 22 Get useful study materials from www. . we get.com . A.(1) ------------..com MA2111/ Engineering Mathematics-I ------------.rejinpaul.e. -------. = ------------. Pre multiplying (2) by .. Since are orthogonal A Property 9: Similar matrices have same eigen values [Anna University.(2) Pre multiplying (1) by . Therefore. e. March 1996] Solution: Sum of the eigen values = Sum of the main diagonal elements = -3 23 Get useful study materials from www. then the matrix is a scalar matrix Given: A real symmetric matrix A of order 2 has equal eigen values By rule 1. it may or may not be possible to get linearly PR A independent eigen vectors corresponding to the equal roots Property 14: Two eigen vectors are called orthogonal vectors if Property 15: Eigen vectors of a symmetric matrix corresponding to different eigen values are orthogonal Property 16: If A and B are n x n matrices and B is a non-singular matrix then A and have same eigen values Problems: 1. A can always be diagonalized. Property 11: The eigen vector X of a matrix A is not unique. Proof: Let be the eigen value of A. i.rejinpaul. be their eigen values. Therefore. By rule 2. the given matrix is a scalar matrix. Find the sum and product of the eigen values of the matrix [ ] [AU. then. we get the diagonalized .www.e. Then.com MA2111/ Engineering Mathematics-I Rule 2: If any diagonalized matrix has its diagonal elements equal. Given.com . an eigen vector is determined by a multiplicative scalar.. Let matrix = [ ]. we get [ ]. A K(AX) = K ( If be distinct eigen values of a n x n matrix..rejinpaul. then the corresponding YU Property 12: SH i. there is a corresponding eigen vector X such that AX = Multiply both sides by a non-zero scalar K. eigen vector is not unique eigen vectors form a linearly independent set TH Property 13: If two or more eigen values are equal. www. [AU.(-1)) = 2 + 2 = 4 2.rejinpaul. Tvli. Dec 1999] Solution: Sum of the eigen values = Sum of the main diagonal elements = -1 Product of the eigen values = │A│= -2(0 – 12) – 2(0 – 6) -3 (-4+1) = 24+12+9 = 45 3. April/May 2009] . Find the third eigen value Solution: We know that sum of the eigen values = Sum of the main diagonal elements = 6+3+3 = 12 Given 24 Get useful study materials from www. Find the third eigen value [AU.rejinpaul. April/May 2003].com MA2111/ Engineering Mathematics-I Product of the eigen values = │A│ = -1 (1 – 1) -1(-1 – 1) + 1(1. The product of two eigen values of the matrix A = [ ] is 16. YU | = 6(9-1)+2(-6+2) +2(2-6) = 48-8-8 = 32 PR A | (Since product of the eigen values is equal to the determinant of the matrix) TH We know that SH Given Find the sum and product of the eigen values of the matrix A = ( ) without finding the roots of the characteristic equation Solution: We know that the sum of the eigen values = Trace of A = a + d Product of the eigen values = │A│ = ad – bc 5.com . Find the sum and product of the eigen values of the matrix [ ] [AU. A Solution: Let the eigen values of the matrix be 4. Two of the eigen values of [ ] are 2 and 8. Clearly. If 3 and 15 are the two eigen values of A = [ ]. without expanding the determinant Solution: Given If 2. Then. find the eigen values of Solution: By the property “A square matrix A and its transpose have the same eigen values”. 3 are the eigen values of A = [ PR A 7. the third eigen value = 2 6.rejinpaul. by the property “The characteristic roots of a triangular matrix are just the diagonal elements of the matrix”.www. = 12 Therefore.com MA2111/ Engineering Mathematics-I Therefore. find │A│. the eigen values of A are 2. the eigen values of 8. 2. 2 25 Get useful study materials from www. TH YU SH A We know that sum of the eigen values = Sum of the main diagonal elements ]. 2. Find the eigen values of A = [ Solution: Given A = [ ] without using the characteristic equation idea ] .com . A is an upper triangular matrix.rejinpaul. 3 + 6 + k = 11 values of A Solution: Sum of the eigen values = Sum of the main diagonal elements = 3 +5+3 = 11 Find the eigen values of the matrix[ are ”.com MA2111/ Engineering Mathematics-I 9.www. by the property “the characteristic roots of a triangular matrix are just the diagonal elements of the matrix”. May 2001 P. the matrix whose eigen values are is 26 Get useful study materials from www. 2 are PR A 11. form the matrix whose eigen values are [AU.T] Solution: Let A =[ ]. A is a lower triangular matrix. Two of the eigen values of A = [ ] are 3 and 6. 6. Therefore. the eigen values of A are 3. The characteristic equation of the given matrix is where and √ Therefore.rejinpaul.rejinpaul. the eigen ].Clearly. Let the third eigen value be k.6 are two eigen values of A. the eigen values of A are 2. the eigen values of A are 6. 3. Hence. the characteristic equation is = Therefore. then the eigen values of TH By the property “If the eigen values of A are YU Then. Find the eigen values of A = [ Solution: Given A = [ ] ] . -1 Hence. Hence. . 4 10.com . Find the eigen values of SH Given 3. rejinpaul. the eigen values of A are 2. Find the eigen values of given A = [ ] [AU. Therefore.www.rejinpaul. Nov/Dec 2002] Solution: Given A = [ SH A ] YU Let the eigen values of A be Therefore. 1. the eigen values of A are 1. Find the eigen values of the inverse of the matrix A = [ ] Solution: We know that A is an upper triangular matrix. adj A = [ [ Therefore. 4. PR A Hence. Find the eigen values of [AU. 3. 5 TH We know that. the eigen values of 14. by using the property “If the eigen values of A are . Sum of the eigen values = Sum of the main diagonal elements = 7 are 1. ] ] Two eigen values of the matrix A = [ ] are equal to 1 each. then the eigen values of are ”.com .6.com MA2111/ Engineering Mathematics-I │A│ = 4 -10 = . 1. Oct/Nov 1997] 27 Get useful study materials from www. The eigen values of 13. 12. Hence. where YU The characteristic equation of A is TH .rejinpaul. 125 28 Get useful study materials from www.com MA2111/ Engineering Mathematics-I Solution: Given A = [ ]. Solution: Given A = [ ] ]. 25 The eigen values of are i. find the eigen values of 5A are the eigen values of A.e. 3 Therefore.1. Therefore. the eigen values are -3.5..e. i. 5 The eigen values of are i. the eigen values of A are 2. PR A Therefore. 2 (real) If 1. the characteristic equation of A is √ 17.27 If 1 and 2 are the eigen values of a 2 x 2 matrix A.. the eigen values of 5A are 5(1).www.. then are the eigen values of kA.25 18.e.com . A is an upper triangular matrix.rejinpaul. the eigen values of are 15.5 are the eigen values of A = [ Solution: By the property “If ]. 4. what are the eigen values of ?[AU. A is an upper triangular matrix.. 2.. 5. Hence. 8. i.8. 1. 1 and 4 are the eigen values of and 1 and are the eigen values of ] are real A Show that the eigen values of the real symmetric matrix A = [ 16. the eigen values of A are 1. Hence. SH Solution: Given A is a real symmetric matrix. if A = [ Find the eigen values of A. Therefore. 5(5) ie.e. 5(1). Model paper] Solution: Given 1 and 2 are the eigen values of A. 1.T] Solution: Let A = [ [ ][ [ ].www.com MA2111/ Engineering Mathematics-I The eigen values of are i.e.I is an upper triangular matrix. form the matrix whose eigen values are ] and let be the eigen values of A ] ] ][ [ ] [ ] Hence the required matrix is [ ] 29 Get useful study materials from www. 6.com .I are its main diagonal elements i. 15 The eigen values of A–I=[ are ]-[ ]=[ ] Since A . PR A = 375+ 125-30 + 2 = 472 are the eigen values of [ If [AU.6(2) + 2 = 24 + 20 -12 + 2=34 A Eigen values of 19.rejinpaul.e. 3(5) i.6(5) + 2 YU SH = 3(2)3+ 5(2)2 .. 16... May 2001 P.e.rejinpaul.4 where First eigen value = Second eigen value = TH = 3(5)3 + 5(5)2 . 625 The eigen values of 3A are 3(2). the eigen values of A. rejinpaul. April/May 2011] √ The Normalized matrix N = D= ND = N ND √ √ √ √ √ SH √ . [ ] [ TH Also given. the required matrix is A – 5I = [ [ ]=[ ] 21.3.www. May 1998] where are the eigen values of A Solution: The matrix A – KI has the eigen values ] Hence.com . Hence N Therefore. A = ND √ A= [√ √ √ √ √ √ √ √ [ ] ] √ √ √ √ [√ √ √ √ = ] [√ √ √ √ √ √ √ ] √ [√ √ √ √ √ √ ] √ 30 Get useful study materials from www. = ] PR A [√ √ ] YU ].rejinpaul.[ ] and [ ] respectively. Nov 2002. eigen vectors are [ A Solution: Given eigen values of A are 2.3. Form the matrix whose eigen values are =[ ] [AU.com MA2111/ Engineering Mathematics-I 20. Find the matrix A [AU. The eigen vectors of a 3 x 3 real symmetric matrix A corresponding to the eigen values 2.6 √ √ √ √ [√ √ √ ] = (N N is an orthogonal matrix.6 are [ ] . the elements in the main diagonal are the eigen values. A is an upper triangular matrix.com . the eigen values of A.. Here.www. the eigen values of adj A is equal to the eigen values of 12 triangular matrix and C is a diagonal matrix. ]=[ ] Find the eigen values of adj A if A = [ ] ].rejinpaul.e. In all the cases.rejinpaul. 4. Hence. 6 23. B and C are 1. Hence. 12 ].. Find them Solution: Let the third eigen value be We know that Given 2+3+2 = 7 = 31 Get useful study materials from www.e. 1 We know that are SH The eigen values of A Adj A = │A│ YU │A│=Product of the eigen values = 12 ] [ ] [ i. A is an upper triangular matrix. 4. 4. i. the eigen values of A are Solution: Given A =[ 3.com MA2111/ Engineering Mathematics-I =[ 22. 3. B is a lower PR A Note: A =[ TH Therefore. Two eigen values of A = [ ] are equal and they are times the third. www.com .com MA2111/ Engineering Mathematics-I [ ] Therefore. 2 +3 +k = 2+ 2+2 = 6 SH We know that product of the eigen values = │A│ | TH | Prove that the eigen vectors of the real symmetric matrix A = [ PR A 25.rejinpaul. YU 2(3)(k) = │A│ ] are orthogonal in pairs Solution: The characteristic equation of A is where | . 5 If 2. 24. 3 are the eigen values of [ ] ]. | The characteristic equation of A is 3 1 -7 0 36 0 3 -12 -36 1 -4 -12 0 32 Get useful study materials from www. Let the eigen values of A be 2. 3. and hence the eigen values of A are 1.rejinpaul.1. k Solution: Let A =[ We know that the sum of the eigen values = sum of the main diagonal elements A Therefore. Therefore. 6 To find the eigen vectors: [ Case 1: When ][ ] [ ] -------.(2) --------. we get. the eigen values of A are -2.(3) Solving (1) and (2) by rule of cross-multiplication.www. 3.rejinpaul.com .com MA2111/ Engineering Mathematics-I √ Therefore. 33 Get useful study materials from www.(1) --------.(1) --------.(3) 3 3 1 7 1 1 7 [ Case 2: When PR A 1 TH YU Solving (1) and (2) by rule of cross-multiplication.(2) ] [ ][ ] [ ] --------. SH A ---------. we get.rejinpaul. com MA2111/ Engineering Mathematics-I 1 3 -2 1 2 1 1 2 [ ] [ Case 3: When ][ ] [ ] --------.com .rejinpaul.(3) 1 3 -5 1 -1 1 1 -1 [ Therefore. [ ] ]. PR A TH Solving (1) and (2) by rule of cross-multiplication.(2) YU --------.(1) SH A --------.www.rejinpaul. we get. [ ] [ ] To prove that: [ ][ ] 34 Get useful study materials from www. 8 Solution: Product of the eigen values of A = │A│= | ] [AU CBT J/J 2010] | CAYLEY-HAMILTON THEOREM: [AU CBT J/F 2010.com . Hence the matrix is non-singular. TH │A│≠0. YU Product of the eigen values = │A│ = 1(14 .8) -2(14 . A/M 2011] Statement: Every square matrix satisfies its own characteristic equation Uses of Cayley-Hamilton theorem: (1) To calculate the positive integral powers of A (2) To calculate the inverse of a square matrix A Problems: 35 Get useful study materials from www.www. Find the sum and product of all the eigen values of the matrix A = [ ]. the sum of the eigen values = 1+2+7=10 Find the product of the eigen values of A = [ PR A 27.com MA2111/ Engineering Mathematics-I [ ][ ] [ ][ ] Hence.rejinpaul.Is the matrix singular? [AU. the eigen vectors are orthogonal in pairs 26.4) + 3(4 . Oct/Nov 1997] SH A Solution: Sum of the eigen values = Sum of the main diagonal elements =Trace of the matrix Therefore.2) = 6-20+ 6= .rejinpaul. The characteristic equation of A is where Therefore. YU Therefore.com MA2111/ Engineering Mathematics-I 1.rejinpaul. 36 Get useful study materials from www. i. TH 2.T] ].e.. May 2001 P. the given matrix satisfies its own characteristic equation Model paper] PR A Solution: Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. The characteristic equation of A is where The characteristic equation is To prove [ ]=[ ] [ ] ] [ ] [ ] A ][ SH [ If A = [ ] [AU.rejinpaul.com . Show that the matrix [ Solution: Let A = [ ] satisfies its own characteristic equation [AU.www. the characteristic equation is By Cayley-Hamilton theorem. the characteristic equation is ] [ [ ] ] [ ] [ ] ] : ------------.rejinpaul. Jan 2010] [AU Jan 2010] [AU Madurai June 2011] [AU May/June 2010] Solution: The characteristic equation of A is where ] ][ [ [ To find ] ] PR A [ [ YU ][ [ SH ------------. Therefore. when A = [ Verify Cayley-Hamilton theorem. CBT Jan/Feb 2009] [AU Tvli. 37 Get useful study materials from www.com .www.(1) TH To prove that: A Therefore. find ] [AU A/M 2002. Jan 2009.(2) Multiply by A on both sides.rejinpaul.com MA2111/ Engineering Mathematics-I 3. the characteristic equation is [ ] ] satisfies its own characteristic equation and hence find April/May 2003] To prove: [ PR A 4.rejinpaul. [ [ ] ] [ [ ] ] [ ] [ ] To find Multiplying (1) by ] [ ] [ ] ] [ ] ] Verify that A = [ ].rejinpaul. The characteristic equation of A is Solution: Given A =[ Therefore.(1) ][ ] [ ] [ ] 38 Get useful study materials from www.www. ---------. ] TH [ [ YU [ SH A [ [AU where i.e.com MA2111/ Engineering Mathematics-I ] [ Hence..com . (1) we get. we get. we get. using Cayley-Hamilton theorem [AU. Find [ on both sides.Hamilton theorem.rejinpaul. if A = [ ] [ ] ]. ------------.(2) [ ][ [ ] [ ] [ ] ] [ ] [ [ ] ] 39 Get useful study materials from www.rejinpaul. Multiplying by 5. To find Multiplying (1) by -----------.com .com MA2111/ Engineering Mathematics-I [ ] [ ] [ ] [ ] [ ] To find From (1). Nov 2002] where TH YU SH A Solution: The characteristic equation of A is PR A The characteristic equation of A is By Cayley.www. (3) Solving (2) and (3). Since by Cayley-Hamilton theorem.(1) When TH When PR A -----------. let the quotient be and the remainder be SH When A To find YU ---------.e..e. (1) [ ] [ ][ ] 40 Get useful study materials from www. (2) . is divided by . ] [ From (2).(3) (2) – 2 x (3) i.www..com . If A = [ ] Solution: The characteristic equation of A is where The characteristic equation of A is √ i.com MA2111/ Engineering Mathematics-I 6. we get.rejinpaul.rejinpaul.(2) -----------. com MA2111/ Engineering Mathematics-I 7.rejinpaul. -----------. [AU Tvli. TH Let f(A) = PR A Let g(A) = (i) (-) (-) f(A) = ( ) = Now.www.rejinpaul. we get.(2) [ ][ (by (1)) ] [ ] 41 Get useful study materials from www.com . Jan 2010] where A Solution: The characteristic equation of A is SH The characteristic equation is ----------. Use Cayley-Hamilton theorem to find the value of the matrix given by (i) (ii) if the matrix A = [ ] [AU M/J 2009].(1) YU By Cayley-Hamilton theorem. com MA2111/ Engineering Mathematics-I ] [ ] [ ] [ ] [ (ii) (-) (-) SH A (-) g(A) = ( = ] g(A) = [ 8. [ ] ] [ ] TH [ PR A = YU =0+ Use Cayley-Hamilton theorem for the matrix [ (ii) Solution: Given A = [ ] to express as a linear polynomial in A (i) [AU M/J 2009] ].(1) (i) 42 Get useful study materials from www. we get. The characteristic equation of A is where The characteristic equation is By Cayley-Hamilton theorem.www.com .rejinpaul. -----------.rejinpaul. com .www.rejinpaul. find = 0 + A + 2I = A +2I (by (1)) which is a linear when A = [ Solution: The characteristic equation of A is ] [AU May/June 2005] where The characteristic equation is 43 Get useful study materials from www.rejinpaul. PR A = Using Cayley-Hamilton theorem.com MA2111/ Engineering Mathematics-I (-) A (by (1)) which is a linear polynomial in A YU SH (ii) TH (-) polynomial in A 9. rejinpaul.com .rejinpaul. ].www. ] ][ [ A Pre-multiplying by where PR A Sum of the main diagonal elements = 1+2+1 = 4 The characteristic equation is To prove that: [ ][ ] [ ] [ ] 44 Get useful study materials from www. Jan 2009] SH Verify Cayley-Hamilton theorem for the matrix ] [ ] [ YU ] The Characteristic equation of A is TH 10. [ [ ] ] [ ] ([ ] [ ]) [ Solution: Given A = [ ] [ [ ] [AU Tvli.com MA2111/ Engineering Mathematics-I By Cayley-Hamilton theorem. we get. [ A =[ [ SH =[ ] [ The characteristic equation of A is where The characteristic equation is To prove that: [ ][ [ ] ] [ [ ] [ ] ] 45 Get useful study materials from www.com . Cayley-Hamilton theorem is verified.com MA2111/ Engineering Mathematics-I ] ][ [ ] [ ] ] =[ ] [ ] YU PR A ] ] [ [ ] ] ] [ ] [AU CBT Jan 2011] TH Verify Cayley-Hamilton theorem for the matrix (i) A = [ Solution: (i) Given A = [ ] [ ] Therefore.rejinpaul.rejinpaul. 11.www. www. (ii) Given A = [ ] where SH A The characteristic equation of A is The characteristic equation is [ ] ] [ ] [ ]. 12.com MA2111/ Engineering Mathematics-I [ ] [ ] [ ] [ ] [ ] [ ] Hence Cayley-Hamilton theorem is verified.com . ] The characteristic equation of A is where 46 Get useful study materials from www. [ ] [ [ [ ] ] ] TH ][ PR A [ YU To prove that: [ [ ] ] [ ] Hence Cayley-Hamilton theorem is verified.rejinpaul. Find using Cayley-Hamilton theorem when A =[ Solution: Given A = [ ] .rejinpaul. e. Therefore. PR A 4b = .com MA2111/ Engineering Mathematics-I The characteristic equation is √ i. 13.. let the quotient be and the remainder be ---------.rejinpaul.(1) SH A ------. √ 6. To find When is divided by .rejinpaul. ) TH Substituting YU (1) – (2) Find ] [ ( ) [ ] [ [ ] [ ( )[ ] ] [ ( ] )[ [ ] [ ] ] ] ] using Cayley-Hamilton theorem when A = [ Solution: Given A = [ ].com . Also find .www. 2 By Cayley-Hamilton theorem. ] 47 Get useful study materials from www.b= Since [ [ ] Therefore.(2) ( in (1). we get.(1) When When ------. rejinpaul.(1) When When YU ------.com MA2111/ Engineering Mathematics-I The characteristic equation of A is where The characteristic equation is √ i.rejinpaul.com . √ By Cayley-Hamilton theorem. we get. To find When is divided by . let the quotient be and the remainder be SH A ---------. ( ) 5b = 9 Therefore.(1) TH ------.e.www..(2) PR A (1) – (2) Substituting in (1). [ Since ( )[ ] [ ( ] )[ [ ] ( )[ ] ] [ [ ] ] ] 48 Get useful study materials from www. rejinpaul. Show that the matrix P = [ ] is orthogonal [AU CBT D/J 2011] 49 Get useful study materials from www. [ ][ ] [ ] [ ] Therefore. Justify. B is an orthogonal matrix 2.com MA2111/ Engineering Mathematics-I =[ ] [ ] [ ] ORTHOGONAL TRANSFORMATION OF A SYMMETRIC MATRIX TO DIAGONAL FORM: Orthogonal matrices: A square matrix A (with real elements) is said to be orthogonal if or Problems: 1.com .rejinpaul. Check whether the matrix B is orthogonal. ][ [ ] PR A [ TH B=[ SH To prove that: ] [ ] Similarly.www. B = [ ] A Solution: Condition for orthogonality is [ ] YU ]. com .rejinpaul.rejinpaul.com MA2111/ Engineering Mathematics-I Solution: To prove that: [ ].www. then form a normalized matrix N Step 5: Find Step 6: Calculate AN Step 7: Calculate D = Problems: 1. Diagonalize the matrix [ Solution: Let A = [ ] [AU Jan 2013] ] 50 Get useful study materials from www. [ ] [ ] [ ][ [ ]= I ] [ ] [ ] Therefore. P is an orthogonal matrix SH A WORKING RULE FOR DIAGONALIZATION Step 1: To find the characteristic equation TH Step 2: To solve the characteristic equation YU [ORTHOGONAL TRANSFORMATION]: PR A Step 3:To find the eigen vectors Step 4: If the eigen vectors are orthogonal. com MA2111/ Engineering Mathematics-I The characteristic equation is where Therefore. the characteristic equation is i.rejinpaul.(2) --------.www. √ SH A √ YU Hence. 51 Get useful study materials from www.(1) -------.(3) Solving (1) and (2) by rule of cross-multiplication. the eigen values of A are 0. 15 ][ ] Case 1: When [ [ ] PR A [ TH To find the eigen vectors: ][ ] [ ] ---------.com .. 3.e.rejinpaul. (1) ---------. ] Case 3: When [ ][ ] [ ] ---------.rejinpaul.(3) 2 5 -6 4 -4 -6 4 [ PR A -6 TH Solving (1) and (2) by rule of cross-multiplication.(1) SH ------.com .com MA2111/ Engineering Mathematics-I -6 2 8 -6 7 -4 -6 7 [ ] [ Case 2: When ][ ] [ ] A -------.(2) 52 Get useful study materials from www.www.(2) YU --------.rejinpaul. the eigen vectors are orthogonal to each other [ The Normalized matrix N = [ [ [ ] ] ] ] 53 Get useful study materials from www.com .rejinpaul. -6 2 -7 -6 -8 -4 -6 -8 [ [ ][ [ ][ ] ] YU ] 4-2-2 = 0 TH ][ PR A [ SH A ] Hence.rejinpaul.www.(3) Solving (1) and (2) by rule of cross-multiplication.com MA2111/ Engineering Mathematics-I ---------. e.www. M/J 2011] TH Diagonalize the matrix [ ] PR A Solution: Let A = [ ] [ The diagonal elements are the eigen values of A 2..rejinpaul.rejinpaul. the characteristic equation is 2 1 0 1 -11 2 -9 36 -18 18 -36 36 0 54 Get useful study materials from www. ] ][ [ ][ The characteristic equation is where Therefore.com ] . [ A [ ] SH i.com MA2111/ Engineering Mathematics-I AN = ] [ [ ] [ [ ] ] [ ] [ ] ] [ YU ] [AU Tvli. www. 6 To find the eigen vectors: [ Case 1: When [ ] ][ ] [ ] A ][ ] SH [ YU ---------. ] Case 2: When [ ][ ] [ ] 55 Get useful study materials from www. 3.rejinpaul. the eigen values of A are 2.(3) TH -------.(2) -1 1 1 -1 3 -1 -1 3 [ PR A Solving (1) and (2) by rule of cross-multiplication.com MA2111/ Engineering Mathematics-I √ √ = Hence.(1) --------.com .rejinpaul. -1 1 -3 -1 -1 -1 -1 -1 56 Get useful study materials from www.(3) Solving (1) and (2) by rule of cross-multiplication.www.rejinpaul.(2) --------.(1) -------.(3) Solving (1) and (2) by rule of cross-multiplication. 1 0 -1 2 -1 -1 2 SH A -1 ][ ] PR A ---------.(2) --------.rejinpaul.com .(1) [ ] TH [ Case 3: When YU [ ] -------.com MA2111/ Engineering Mathematics-I ---------. √ √ √ [√ √ SH √ √ √ √ YU √ √ √ [ √ √ [√ i.e. the eigen vectors are orthogonal to each other ] √ √ √ √ √ ] [√ √ √ √ √ ] [√ ] √ [ √ ] ] √ ] The diagonal elements are the Eigen values of A QUADRATIC FORM.com .REDUCTION OF QUADRATIC FORM TO CANONICAL FORM BY ORTHOGONAL TRANSFORMATION: 57 Get useful study materials from www.www. √ ] √ √ √ √ √ √ [√ √ √ PR A AN = [ √ √ ] TH The Normalized matrix N = √ A Hence.com MA2111/ Engineering Mathematics-I ] [ [ ][ ] [ ][ [ ] =0 ][ ] ] [√ √ √ √ √ √ √ √ √ √ √ √ √ √ √ [√ √ √ .rejinpaul..rejinpaul. A Problems: YU Solution: Q = . .rejinpaul.com MA2111/ Engineering Mathematics-I Quadratic form: A homogeneous polynomial of second degree in any number of variables is called a quadratic form is a quadratic form in three variables Example: Note: The matrix corresponding to the quadratic form is [ ] Write the matrix of the quadratic form SH 1. ] ] Write down the quadratic form corresponding to the following symmetric matrix [ Solution: Let [ ] [ ] 58 Get useful study materials from www. Nov 2001] [ Solution: Q = [ 3.com ] . ] PR A Here ] TH [ Write the matrix of the quadratic form [AU.rejinpaul. [ 2.www. Determine the nature of the following quadratic form f( [AU April/May 2003] Solution: The matrix of the quadratic form is Q = [ ] The eigen values of the matrix are 1. 0 Therefore.www. is called the signature of the quadratic form A Signature of the quadratic form: The difference between the number of positive and negative square terms YU The quadratic form is said to be Positive definite if all the eigen values are positive numbers (2) Negative definite if all the eigen values are negative numbers (3) Positive Semi-definite if all the eigen values are greater than or equal to zero and at least one eigen (4) Negative Semi-definite if all the eigen values are less than or equal to zero and at least one eigen value is zero (5) PR A value is zero TH (1) Indefinite if A has both positive and negative eigen values Problems: 1.rejinpaul. the quadratic form is Positive Semi-definite 59 Get useful study materials from www.com .com MA2111/ Engineering Mathematics-I The required quadratic form is = NATURE OF THE QUADRATIC FORM: Rank of the quadratic form: The number of square terms in the canonical form is the rank (r) of the quadratic form Index of the quadratic form: The number of positive square terms in the canonical form is called the index (s) of the quadratic form SH = s – (r-s) = 2s-r. 2.rejinpaul. com .rejinpaul. the quadratic form is indefinite YU form without reducing it to canonical Solution: The matrix of the quadratic form is Q = [ | | | | PR A TH ] Therefore. M/J 2011] | | | Discuss the nature of the quadratic form SH 3. the quadratic form is positive definite REDUCTION OF QUADRATIC FORM TO CANONICAL FORM THROUGH ORTHOGONAL TRANSFORMATION [OR SUM OF SQUARES FORM OR PRINCIPAL AXES FORM] Working rule: Step 1: Write the matrix of the given quadratic form Step 2: To find the characteristic equation 60 Get useful study materials from www.www.F is indefinite ] Solution: The matrix of the quadratic form is Q = [ | [AU Tvli. Prove that the Q.com MA2111/ Engineering Mathematics-I 2.rejinpaul. A Therefore. rejinpaul. SH Problems: Reduce the quadratic form Q = into canonical form by an Solution: The matrix of the Q.e.www.[AU Trichy.com .rejinpaul. A [ ] ] PR A [ TH YU orthogonal transformation [AU March 1996].com MA2111/ Engineering Mathematics-I Step 3: To solve the characteristic equation Step 4: To find the eigen vectors orthogonal to each other Step 5: Form the Normalized matrix N Step 6: Find Step 7: Find AN Step 8: Find D = ][ ] [ ] A Step 9: The canonical form is [ 1..F is A = i. Jan 2010] where The characteristic equation of A is Sum of the main diagonal elements = 6+3+3=12 The characteristic equation of A is 61 Get useful study materials from www. com .(2) [ ] YU ][ ] SH A [ Solving (1) and (2) by rule of cross-multiplication.com MA2111/ Engineering Mathematics-I 2 1 -12 36 -32 0 2 -20 32 1 -10 16 0 √ √ Therefore the eigen values are 2. 8 To find the eigen vectors: ][ ] Case 1: When [ [ ] PR A ---------.www. -2 2 -2 -2 -5 -1 -2 -5 [ ] 62 Get useful study materials from www. 2.rejinpaul.rejinpaul.(1) ---------.(3) TH --------. (1) -------. SH [ ]. l m n -1 1 2 -1 1 1 0 1 [ ] 63 Get useful study materials from www.(3) All three equations are one and the same. Let = 1. ------.(2) ---------.com MA2111/ Engineering Mathematics-I ][ ] [ Case 2: When [ ] --------. Since Let A [ ] Solving (1) and (2) by rule of cross-multiplication.(1) ---------.com .www.(2) ][ ] YU [ TH ][ ] PR A [ . Then .rejinpaul.rejinpaul. Put . rejinpaul. ] YU The canonical form is [ ] A ] AN = [ ] SH The Normalized matrix N = PR A .F is A = [ i.com . [AU Madurai.com MA2111/ Engineering Mathematics-I √ √ √ √ √ √ √ √ √ √ √ √ [√ √ √ ] [√ √ √ √ √ √ √ √ √ √ √ √ √ √ [√ √ √ [√ √ √ √ √ √ √ √ √ √ √ √ √ [√ √ √ ] [√ √ √ ] ][ ] ][ √ [ ] Reduce the quadratic form to a canonical form by an orthogonal reduction TH 2. [AU A/M 2011] Solution: The matrix of the Q.www.e. A [ ] ] The characteristic equation of A is where 64 Get useful study materials from www.rejinpaul. June 2011]..Also find its nature. (3) 1 1 2 1 2 -1 1 2 [ ] 65 Get useful study materials from www.com MA2111/ Engineering Mathematics-I The characteristic equation of A is 1 1 0 0 1 -3 1 1 2 1 -2 -2 0 √ √ The eigen values are 1.www. 1.com .rejinpaul. -2 To find the eigen vectors: ][ ] ------.rejinpaul.(2) SH [ Case 1: When A [ ] YU ][ ] TH [ ------.(1) [ ] PR A ------. (3) All three equations are one and the same. Put .(1) -----. TH [ is orthogonal to PR A [ ]. .rejinpaul. Since Let SH A [ .com MA2111/ Engineering Mathematics-I ][ ] [ Case 2: When [ ] ------.(2) l m ][ ] YU ][ ] . Let n 1 1 -1 1 -1 1 0 -1 [ ] 66 Get useful study materials from www.(1) --------.com .www.rejinpaul. Then ] [ -------.(2) -----. .. √ YU ] √ TH AN = [ √ A √ Nature: The eigen values are -2.www. Therefore. 1. A = [ ] ] The characteristic equation of A is where 67 Get useful study materials from www. 1.F is A = [ i.rejinpaul.e.com MA2111/ Engineering Mathematics-I √ √ √ √ √ [√ √ √ The Normalized matrix N = √ √ √ [√ √ √ .rejinpaul. it is indefinite in nature. Reduce the given quadratic form Q to its canonical form using orthogonal transformation Q = [AU Jan 2009] Solution: The matrix of the Q. 3.com . ] √ √ √ √ √ √ √ √ √ √ √ √ [√ √ √ [√ √ √ ] √ √ √ √ √ √ √ √ √ √ [√ √ √ ] [√ √ √ ] √ √ [√ √ [ √ √ ] ] SH √ ] ][ ][ ] PR A The canonical form is [ ] ] √ [ i.e. 2. 2 -1 0 2 -1 2 0 -1 68 Get useful study materials from www.com .(1) -------. we get.(3) Solving (2) and (3) by rule of cross multiplication. 4 ][ ] Case 1: When [ [ ] PR A [ TH YU To find the eigen vectors: ][ ] [ ] = 0 -------.www.rejinpaul.rejinpaul.com MA2111/ Engineering Mathematics-I The characteristic equation of A is 1 1 -7 14 -8 0 1 -6 8 1 -6 8 0 √ A √ SH The eigen values are 1.(2) -------. (1) -------.rejinpaul. 69 Get useful study materials from www.(1) -------.(3) Solving (1) and (2) by rule of cross multiplication.(2) -------. we get.(2) -1 0 1 -1 0 1 TH 0 PR A 0 YU Solving (1) and (2) by rule of cross multiplication.com . [ ] Case 3: When SH A -------.(3) [ ][ ] [ ] = 0 -------.rejinpaul.www.com MA2111/ Engineering Mathematics-I [ ] ][ ] [ Case 2: When [ ] = 0 -------. we get. [AU M/J 2009].www.com MA2111/ Engineering Mathematics-I 0 0 -3 0 -1 -1 0 -1 ] AN = [ √ √ √ [√ √ √ √ √ √ √ √ √ [√ √ √ [ ] [ ] √ √ √ √ √ √ √ √ ] A √ ] [ √ √ √ √ ][ [ i. [AU Tvli. PR A TH ] √ SH The Normalized matrix N = √ YU [ √ √ √ √ ] ] ] The canonical form is [ 4. Jan 2009] 70 Get useful study materials from www.rejinpaul.e. [ ][ ][ ] Reduce the quadratic form to the canonical form through an orthogonal transformation and hence show that it is positive semi-definite.com .rejinpaul.. Also give a non-zero set of values ( which makes the quadratic form zero. e. A = [ ] ] The characteristic equation of A is SH A where YU The characteristic equation of A is To find the eigen vectors: √ PR A The eigen values are 0. 1.(2) 71 Get useful study materials from www.F is A = [ i.com MA2111/ Engineering Mathematics-I Solution: The matrix of the Q. 3 TH √ [ ][ ] Case 1: When [ [ ] ][ ] [ ] ------.rejinpaul.(1) -----.com ..rejinpaul.www. rejinpaul.(3) Solving (1) and (2) by rule of cross multiplication.rejinpaul. we get.(3) TH ------.www.com MA2111/ Engineering Mathematics-I -----.(2) [ ] YU [ Case 2: When SH A [ Solving (1) and (2) by rule of cross multiplication. we get.(1) -----.com . -1 0 1 -1 2 1 -1 2 ] ][ ] PR A -----. -1 0 0 -1 1 1 -1 1 [ ] 72 Get useful study materials from www. √ The Normalized matrix N = √ √ [√ √ AN = [ ] √ √ [√ √ √ √ √ √ ] √ [√ √ √ ] √ √ √ √ √ ] √ = √ √ √ √ [ √ √ √ √ √ [√ ] √ √ √ √ √ √ ] [ √ √ [ ] √ ] 73 Get useful study materials from www.rejinpaul.(3) 0 -2 -1 -1 1 -1 -1 TH ] PR A [ YU SH -1 A Solving (1) and (2) by rule of cross multiplication.(2) ---------.com .www.(1) -----------. we get.rejinpaul.com MA2111/ Engineering Mathematics-I ][ ] [ Case 3: When [ ] --------. www.rejinpaul.com MA2111/ Engineering Mathematics-I The canonical form is [ ][ ][ ] Nature: The eigen values are 0, 1 and 3. Therefore, it is positive semi-definite. To find the non-zero set of values which makes the quadratic form zero: X = NY √ √ √ √ ------- (1) √ √ Put √ SH √ √ ] √ √ ------ (2) ------- (3) and YU [√ [ ] A √ TH [ ] √ √ PR A √ √ in (1), (2) and (3), we get, These values make the quadratic form zero. Verification: Q.F = = = 1 + 2 +1 -2 -2 = 0 WORKING RULE FOR DIAGONALIZATION [SIMILARITY TRANSFORMATION]: 74 Get useful study materials from www.rejinpaul.com www.rejinpaul.com MA2111/ Engineering Mathematics-I Let A be any square matrix of order n. Step 1: To find the characteristic equation Step 2: To solve the characteristic equation Step 3: To find the Eigen vectors Step 4: Form the modal matrix P, its columns the Eigen vector of A Step 5: Find Step 6: Calculate AP A Step 7: Calculate D = ] to the diagonal form [AU April/May 2002] YU Reduce the matrix [ TH 1. SH Problems based on similarity transformation: ]. The given matrix is a non-symmetric matrix. Hence similarity PR A Solution: Let A = [ transformation is only possible. The characteristic equation of the given matrix A is where | | | | | The characteristic equation of A is 1 1 0 1 -1 1 0 -5 0 -5 5 -5 0 75 Get useful study materials from www.rejinpaul.com | www.rejinpaul.com MA2111/ Engineering Mathematics-I i.e., √ Hence the Eigen values are 1, √ √ The given matrix is non-symmetric and the Eigen values are distinct. Hence we can find . To find the Eigen vectors: (A – [ ] A ][ ] PR A ---------------- (2) [ ] TH ------------ (1) ][ ] YU [ Case 1: When SH [ -------------- (3) Solving (1) and (2) by rule of cross multiplication, we get, 2 -2 -2 2 1 1 1 1 Hence [ ] 76 Get useful study materials from www.rejinpaul.com (5) √ -------------------------.(6) -1 -1 -1 -√ √ SH 2-√ √ ( YU 1 √ √ ) ( PR A √ √ TH 2-√ 1 A Solving (5) and (6).(4) √ ) --------------------.com . we get.com MA2111/ Engineering Mathematics-I Case 2: When √ [ √ ][ ] √ [ ] √ ( √ ) ( ---------------. √ √ ) [ √ Case 3: When √ ( √ ) √ √ √ √ Hence √ √ ] √ . [ √ ][ ] √ [ ] √ ( √ ) ---------------.www. we get.rejinpaul.rejinpaul.(7) 77 Get useful study materials from www. (9) Solving (7) and (8).com MA2111/ Engineering Mathematics-I ( √ ) --------------------. 2 -2 2+√ -1+√ 2 1 1 2+√ √ ( √ )( √ ) √ √ ) ( √ √ ) ( √ √ YU √ √ √ ( √ ) √ √ TH √ SH A √ √ √ Hence PR A √ √ [ √ ] √ The modal matrix P = [ √ ] To find |P| = Co-factor of P = [ (√ √ √ ) ( √ √ √ ) √ √ √ ] 78 Get useful study materials from www.rejinpaul.www.com .rejinpaul.(8) √ -------------------------. we get. √ √ √ √ √ √ √ ] √ ] √ Diagonalize the matrix A = [ ] Solution: The characteristic equation of A is where | | | | | 79 Get useful study materials from www.rejinpaul.e.com MA2111/ Engineering Mathematics-I [ √ √ √ √ √ √ √ D= 2..rejinpaul. √ √ √ √ [ √ √ √ √ ][ √ √ √ √ √ √ √ √ [ √ √ √ ]= ] √ [ [ ] ] √ √ [ √ √ TH = √ √ PR A = √ ][ √ √ √ A AP = [ [ ] ] SH √ √ YU i.com | .www. www.com .(1) -------------. SH To find the Eigen vectors: YU (A – [ [ ] ][ ] [ ] PR A Case 1: When ][ ] TH [ ---------------.rejinpaul. Hence the Eigen values are -1.rejinpaul. 2.. Hence A transformation is possible only if the Eigen vectors are distinct.com MA2111/ Engineering Mathematics-I The characteristic equation of A is -1 i.e. The given matrix is a non-symmetric matrix and the Eigen values are not distinct. we get.(3) Solving (2) and (3) by rule of cross-multiplication. 2. 2 2 -1 2 -1 3 -1 -1 80 Get useful study materials from www.e.(2) --------------.. 1 -3 0 4 0 -1 4 -4 4 0 1 -4 √ i. com MA2111/ Engineering Mathematics-I Hence [ ] [ Case 2: When ][ ] [ ] -----------. diagonalization is not possible.rejinpaul.(6) We get only one Eigen vector corresponding to the repeated root .rejinpaul.www. Hence similarity transformation is not possible. SH A -------------.(5) -1 -1 -1 0 -1 -1 Hence [ ] TH 2 PR A -1 YU Solving (5) and (6) by rule of cross-multiplication.(4) --------------. 81 Get useful study materials from www. Therefore. we get.com . 4. .. 7. .rejinpaul.com . PR A Solution: Let { } be the sequence of real numbers. 1. .A 1. is a sequence of numbers 2. -2. . . is a sequence of numbers. 5. 6. 8.Define limit of a sequence. i. } YU = TH 3. Solution: If in accordance with a definite rule.e. -4. . The numbers }={ represented as { { will be referred mo as the terms of the sequence.www. . -1. . The sequence may be } or }={ } Examples: A 1. Then approaches the limit L as n approaches infinity. every number ‘ n ‘ of a series of natural numbers 1. { } SH 2. { = 2. 3. . n is put into correspondence with a certain real number . Define a sequence.2. Example: { } is a convergent sequence because { } 82 Get useful study materials from www. 4. . .Define convergence. -3. then 3. Solution: A sequence { } is said to be convergent if it has a finite limit.com MA2111/ Engineering Mathematics-I UNIT 2 – SEQUENCES AND SERIES PART . .3. . is called a number sequence or simply a sequence. then the set of the ordered real quantities .rejinpaul. . there exists a positive integer N such that If (n ) approaches the limit L. if for every > 0. . 9. rejinpaul. { } oscillates finitely because ={ } oscillates infinitely because . 5. then { } is divergent.Examine the convergence of the sequence { SH 2. Example: { n } is divergent because . A { = Now √ PR A √ ( ( } TH Solution: Here sn = √ YU 6.com } is . √ √ ) ) Hence the given sequence is convergent since the limit is 3. is not unique ( oscillates finitely ) or Solution: If called an oscillatory sequence.Examine the convergence of the sequence { } Solution: Here sn = ( So ) { 83 Get useful study materials from www.When is a sequence said to be oscillatory? Give example. 7. Solution: If .Define divergence. ( oscillates infinitely) then { Examples: 1.rejinpaul.www.com MA2111/ Engineering Mathematics-I 4. Hence the given sequence is oscillatory. SH The expression 11. 12. 10. a sequence which is monotonic and bounded is convergent. for all values of n .Define a bounded sequence. A monotonic sequence always tends to a limit.Examine the convergence of the series 1+ + Solution: Here = 1+ + = = 84 Get useful study materials from www.com . Solution: A sequence { sn } is said to increase steadily or to decrease steadily according as sn+1 sn or sn+1 sn . Thus.Examine the convergence of the series and = + + + + TH = + YU It is denoted by ∑ Solution: Here A Solution: Suppose we have an infinite sequence of numbers PR A = = = Hence the given series diverges to . finite or infinite. 8. if there exists a number k such that sn < k for every n.www. 9.rejinpaul. is called a series .com MA2111/ Engineering Mathematics-I This shows that sn does not tend to an unique limit.Define a monotonic sequence.Define a series. Solution: A sequence { sn } is said to be bounded.rejinpaul. Both increasing and decreasing sequences are called monotonic sequences. 85 Get useful study materials from www.com . = { This implies that sn des not converge to a unique limit. .com MA2111/ Engineering Mathematics-I ( = Since the )= . upto n terms. .rejinpaul.www. partial sum sn converges to the given series converges to 13. Hence we have. sn =2 -3 + 1 +2 -3 + 1 +2 -3 + 1+ . Hence the series is oscillatory.Examine the convergence of the series 2 – 3 + 1 + 2 -3 + 1+ 2 -3 + 1+ . partial sum of the series as follows: s3 = 2 – 3 – 1 =0. . Solution : Construct the S1 = 2.rejinpaul. s2 = 2 – 3= -1. .Examine the convergence of the series and this can be rewritten as TH YU SH A Solution: Here un = PR A = =1–0=1 =1- Since the partial sum sn converges to the given series converges to 14. Show that ∑ are both finite ( in this case ∑ is converges using the Integral test. converges . convergent) or both infinite ( in this case ∑ 18.com MA2111/ Engineering Mathematics-I then is the series of positive terms ∑ 15.www.com . State the Integral test. . Solution: If then the series of positive terms ∑ shall give a counter example to prove this. . for all n Then the given series ∑ is convergent. . Solution: Let f(x)= ∑ =∑ .rejinpaul. 86 Get useful study materials from www. PR A 17. f(x) > 0 and f(x) is decreasing in [1. .negative decreasing function in [1.State the Comparison test.rejinpaul. f(2)=u2. f(n) = un Then the improper integral ∫ and the series ∑ is divergent). Let ∑ YU ( ii ) be such that TH (i)∑ and ∑ SH Solution: If two series of positive terms ∑ A √ be a series with positive and decreasing terms Let f be a non. f(3)=u3. Consider the series 1+ Sn = 1+ √ √ √ √ √ > 1+ need not necessarily be convergent. We √ √ √ √ = √ √ √ Thus the series is divergent even though 16. f(1)=u1. If always convergent. 19. 20. ∫ ∫ =[ ] =[log log 1] = PR A TH ∫ A =∑ YU ∑ 1+ SH Solution: The given series is ∑ using the Integral test. 21. Solution: A series in which the terms are alternatively positive or negative is termed as an alternating series.com MA2111/ Engineering Mathematics-I ∫ ∫ =[ ] =[ ] = ∫ converges to By the integral test ∑ also converges.Test the convergence of the Harmonic series 1+ .Define alternating series. f(x) > 0 and f(x) is decreasing in [1.com . Hence by the integral test ∑ also diverges. and in this case we can use the comparison test.State D’Alembert’s ratio test. 87 Get useful study materials from www.www. diverges in [1.rejinpaul.rejinpaul. Solution: In a series with positive terms ∑ If ( i ) ( ii ) ( iii ) . 23.Discuss the convergence of the series 1- √ √ A . 25.Discuss the convergence of the series 1- using Leibnitz test.Test the convergence of the series ∑ Solution: The given series ∑ using Leibnitz test. Solution: If in the alternating series the terms are such that: (i) (ii) and then the series is convergent. PR A 24. using Leibnitz test.www.rejinpaul.Define Leibnitz test. √ YU Solution: The terms are alternatively positive and negative SH If (i) 1> TH (ii) √ √ √ √ Hence by Leibnitz test the series is convergent. we use a test known as the Leibnitz test.com MA2111/ Engineering Mathematics-I Example: 2To test the convergence of the alternating series . = 1- = 1-1+ =1–1+ The terms are alternatively positive and negative (i) 1> (ii) Hence by Leibnitz test the series is convergent.rejinpaul. 22. 88 Get useful study materials from www. then the series is oscillatory.com . Solution: The given series is an alternating series with term is >… and = 0.Discuss the convergence of the series = term is PR A with (i) (ii) using Leibnitz test. TH Solution: The given series SH (i) with = Hence by Leibnitz test the series is convergent. 89 Get useful study materials from www.Discuss the convergence of the series 0< x < 1. We Know that if 0 < x <1 . 28.Discuss the convergence of the series 2- using Leibnitz test. 26.com MA2111/ Engineering Mathematics-I Solution: The given series 1- 1– =1– The terms are alternatively positive and negative (i) 1> (ii) Hence by Leibnitz test the series is convergent. then = This implies that (i) (ii) Hence by Leibnitz test the given series is convergent.com . The second condition of Leibnitz test is not YU satisfied.rejinpaul. hence the series is not convergent (oscillatory).rejinpaul.www. Solution: The given series 2- = A 2> term is (ii) . 27. com . The second condition of Leibnitz rule is not satisfied.B 1.com MA2111/ Engineering Mathematics-I 29. PART .rejinpaul.Define conditional convergence. since the series 1+ SH .When is a series said to be absolutely convergent? Solution: If the series of arbitrary terms is convergent. is divergent but ∑ For example. consider the series 1- is is said to be conditionally convergent. but the series of finite values 1+ is divergent.Test the convergence of the series ∑ Solution: The given series ∑ .Prove that the series ∑ converges for k >1 and diverges for k Proof: The given series is ∑ Case 1: If k >1. then ∑ PR A Solution: If ∑ TH It is absolutely convergent. = 1- The terms are alternatively positive and negative (i) 1> (ii) . is convergent. hence the original series is conditionally convergent. YU For example. the series 1- is known to be convergent.www. then 90 Get useful study materials from www. 31. is convergent. hence the series is not convergent (oscillatory). then the series ∑ A be such that the series ∑ said to be absolutely convergent. 30.rejinpaul. 2. 3.com . 91 Get useful study materials from www. then ∑ PR A TH YU SH A Case 2: If k ( ) ∑ >1+ = 1+ ( ) ( )+… which diverges to Similarly. . 3. = a+(n-1)d = 1+(n-1)1 = n which gives the first factor. 2. . .www. .Examine the convergence of the series Solution: To find the term: In denominator: 1.rejinpaul.com MA2111/ Engineering Mathematics-I ( ∑ ( = ) ( < 1+ Which is a geometric series with common ratio )+… ) So. = a+(n-1)d = 2+(n-1)1 = (n+1) which gives the second factor.rejinpaul. . . 4. for k < 1 it diverges to Hence the given series converges for k >1 and diverges for k 2. . the series is convergent for k > 1. Where a denotes the first term and d denotes the common ratio. ∑ ∑ ∑ A also convergent. Here = . PR A Solution: To find the SH 3. . 3. 4. . = a+(n-1)d = 1+(n-1)1 = n which gives the first factor.com MA2111/ Engineering Mathematics-I Here = Let us take Then = = Since the series ∑ . . . 3. . . 3.www. and this can be written as = Let us take 92 Get useful study materials from www. = a+(n-1)d = 1+(n-1)2 = (2n-1) which gives the first factor. TH In numerator: 1. Where a denotes the first term and d denotes the common ratio. . . In denominator: 1. .rejinpaul. YU term: = a+(n-1)d = 2+(n-1)1 = (n+1) which gives the second factor. .rejinpaul.Examine the convergence of the series = a+(n-1)d = 3+(n-1)1 = (n+2) which gives the third factor. 2. 3. .com is . 2. 5. 5. 4. . Since the series ∑ ∑ ∑ ∑ is also convergent. 10. .rejinpaul. 13. and this can be written as ( )( ) ( )( )( ) Let us take 93 Get useful study materials from www. 4. . . = which gives the first factor. TH 7.rejinpaul. . . .www.com MA2111/ Engineering Mathematics-I Then = ) ( = ( )( ) . 16. 10. . . . . = A term: SH Solution: To find the = a+(n-1)d = 7+(n-1)3 = (3n+4) which gives the second factor. 9. = a+(n-1)d = 4+(n-1)3 = (3n+1) which gives the first factor. 4. 7. . 13. . . = a+(n-1)d = 10+(n-1)3 = (3n+7) which gives the third factor. YU In denominator: 4. 9.com . Examine the convergence of the series In numerator: 1. Here = = PR A Where a denotes the first term and d denotes the common ratio. 94 Get useful study materials from www.rejinpaul. Since the series ∑ ∑ ∑ ∑ ( also convergent.com ) is . and this can be written as √ √ ( ( ) YU = √ A Solution: The given series is ) PR A TH Let us take Then ( ) ( ) = = .rejinpaul.Test the convergence of the series ∑ √ SH = √ √ .www. √ 5. Since the series ∑ ∑ ∑ ∑ also divergent.com MA2111/ Engineering Mathematics-I Then ( = = )( ( )( )( ) )( ) . com MA2111/ Engineering Mathematics-I 6. √ √ (√ (( √ ) √ ) (√ ) √ √ (( A .. . 4.√ =√ =√ =√ . where a denotes the first term and d denotes the common ratio. 95 Get useful study materials from www.. = a+(n-1)d = 3+(n-1)1 = (n+2) which gives the first factor. .com . 5.rejinpaul. In denominator: 3.√ .www. which gives the first factor.Test the convergence of the series Solution: The given series is To find the √ √ √ √ √ √ √ term: In numerator: √ . and this can be written as ) ) SH = √ ) YU = PR A TH Let us take (√ √ (( Then √ ) ) ) = √ = ( √ √ ) . . Since the series Since the series ∑ ∑ √ ∑ ∑ is also convergent.rejinpaul. Here = . Solution: The given series is To find the term: In numerator: 2.com .rejinpaul. and this can be written as = = PR A Then TH YU SH A Let us take = ( ) .www.Test the convergence of the series where a denotes the first term and d denotes the common ratio. 4. = a+(n-1)d = 1+(n-1)1 = n which gives the first factor. = a+(n-1)d = 2+(n-1)1 = (n+1) which gives the first factor. 3. .Discuss the convergence of the series ∑ Solution: The given series is + Case 1: For x < 1 96 Get useful study materials from www. In denominator: 1.rejinpaul. . . . 2. . Since the series ∑ ∑ ∑ ∑ is convergent if p > 2 and divergent if p 8. .com MA2111/ Engineering Mathematics-I 7. 3. Test the convergence of the series ∑ ∑ √ –√ √ Test the convergence of the series √ Solution: The given series can be written as ∑ √ –√ ∑ =∑ √ (√ –√ ) √ (√ √ √ ) √ ) 97 Get useful study materials from www. the series becomes SH Case 3: For x > 1 YU = TH Let us take = We know that the series ∑ ∑ PR A = .com is .com MA2111/ Engineering Mathematics-I = Let us take = = . Hence the given series converges for x < 1 and x > 1 and diverges for x = 1. We know that the series ∑ ∑ ∑ ∑ is also convergent if x < 1.www. ∑ ∑ also convergent if x > 1. 9. which is a divergent series .rejinpaul. A Case 2: For x = 1.rejinpaul. and this can be written as √ √ √ √ (√ ) Let us take √ (√ = ) A Then (√ ) YU = SH √ TH .www.rejinpaul. and this can be written as √ ( ) ( ) ( ) √ ( ) ( ) Let us take 98 Get useful study materials from www. ∑ ∑ PR A Since the series √ divergent.Test the convergence of the series ∑ Solution: The given series is √ =√ =√ – √ ∑ ∑ √ –√ ) is also √ √ √ .rejinpaul.com . 10.com MA2111/ Engineering Mathematics-I ∑ = = √ √ √ . Since the series ∑ ∑( ) is a geometric series with ratio and this can be written as =( =[ ( = n[( – )] .com .Test the convergence of the series ∑ ∑√ SH ∑ A Then √ ) = n[ ( ) ] ] = [ ] = [ ] 99 Get useful study materials from www.rejinpaul.rejinpaul.n ) ] ( =n[ PR A Solution: Here un =√ √ √ ( or ) TH Test the convergence of the series∑ √ is also divergent.www.com MA2111/ Engineering Mathematics-I ( ) ( ) = ( ) ( ) ( ) √ = ( ) ( ) √ . YU 11. com MA2111/ Engineering Mathematics-I Let us take Then [ = = ] [ ] . TH Solution: Let f(x)= ∫ Put log x = t is also SH convergent. ∑√ When x=2.rejinpaul.rejinpaul. PR A ∑ using the Integral test.Test the convergence of the series ∑ =∑ YU . 100 Get useful study materials from www.com .www. ∫ ∫ =[ ] =[log log(log 2)] = ∫ diverges in [2. for x . f(x) > 0 and f(x) is decreasing in [2. Since the series ∑ ∑ A ∑ 12. when x= ∫ . t = log 2. Examine the convergence of the series ∑ Solution: The given series is ∑ ∑ =∑ [ ] also converges.rejinpaul.Discuss the convergence of the series ∑ using the Integral test.com . ∫ ∫ ( ∫ = =-( ∫ = SH ∫ ∫ ∫ A ∑ ∫ = = -( ) ) ( ) ) 101 Get useful study materials from www. 13.rejinpaul. f(x) > 0 and f(x) is decreasing in [1. ∫ = ∫ = [ ] [ = ] ] YU = [ = converges and by the integral test ∑ TH ∫ PR A 14.com MA2111/ Engineering Mathematics-I Hence by the integral test ∑ also diverges. Solution: Let f(x)= =∑ .www. 1 sin 1+ . f(x) > 0 and f(x) is decreasing in [1. using the Integral test. = = = = ( ) ) <1 ( since e = 2.www.Discuss the convergence of the series ∑ Solution: The given series is ∑ Here = and also converges. = < 1. 16. .com .rejinpaul. 102 Get useful study materials from www.718…) TH = ) SH ( = ( YU = A = PR A Hence by ratio test the given series is convergent. = = = = = ( = ) = .com MA2111/ Engineering Mathematics-I = converges and by the integral test ∑ ∫ 15.Discuss the convergence of the series ∑ Solution: The given series is ∑ Here = and .rejinpaul. 17. = = = = = = ( ) 103 Get useful study materials from www. ∑ Clearly this series is convergent.Examine the convergence of the series Solution: The and term is SH A = is convergent.www. is convergent and ∑ is convergent. ∑ = 1+ TH Now. Solution: Consider the series ∑ Let = and = = = = ( ) = .com MA2111/ Engineering Mathematics-I Hence by ratio test the given series is convergent. consider the series ∑ is convergent.rejinpaul. x>0.rejinpaul. YU Hence by ratio test the given series ∑ 18.Test the convergence or divergence of the series ∑ ( ).com . = < 1. by property of series ∑ PR A Since. . x >0. the given series is divergent. A Put x=1 in un. = = and = = ∑ is convergent. 104 Get useful study materials from www. we make use of the comparison test. When x < 1 .Discuss the convergence of the series YU Hence the given series converges for x ≤ 1 and diverges for x > = × = = Hence by D’Alembert’s ratio test. When x = 1. When x < 1 . the given series is convergent When x > 1 . Hence for x = 1 .com MA2111/ Engineering Mathematics-I = Hence by D’Alembert’s ratio test. the given series is convergent.www. When x > 1 . SH But since ∑ . the given series is divergent. by the comparison test ∑ is also convergent.com . TH 19.rejinpaul.the ratio test fails.rejinpaul. then and = = = term is = PR A Solution: The +…. Hence for Put = 1 . When < 1 . Put x = 1 in un .Discuss the convergence of the series = = √ √ √ √ = √ = √ √ √ = ( ) ( ) √( √ = ) TH √ .www. – Now =1 By the necessary condition for convergence the series diverges. PR A = SH √ YU and term is √ A Solution: The √ . When = 1.com . =1 in un. When > 1 . = Hence by D’Alembert’s ratio test. we make use of the comparison test.the ratio test fails. then = = √ ( ) and = 105 Get useful study materials from www. Hence the given series converges for x < 1 and diverges for x 20. the given series is convergent. the given series is divergent. then = .the ratio test fails.com MA2111/ Engineering Mathematics-I When x = 1.rejinpaul.rejinpaul. rejinpaul.Discuss the convergence of the series Solution: The = = = SH = A and term is YU = = PR A Hence by D’Alembert’s ratio test. TH = When x < 1 .com MA2111/ Engineering Mathematics-I ( = But since ∑ ∑ ) = . When x > 1 . Hence the given series converges for x ≤ 1 and diverges for x > 21. Put x = 1. the given series is divergent. by the comparison test ∑ is also convergent.the ratio test fails. we try to examine the series at x =1. Now.com . Hence the given series converges for x < 1 and diverges for x 22.rejinpaul.www. is convergent. When x = 1. the series becomes =( ) ( ) which is a divergent series .Check for convergence & absolute convergence of the series Solution: The given series is 106 Get useful study materials from www. the given series is convergent. www.com .rejinpaul. Since the given series is an alternating series ∑ with (i) (ii) By Leibnitz test the series is convergent. Next we construct the series of absolute terms as | | | | | YU | SH A Solution: The given series ∑ ∑ PR A TH The series of absolute terms is Which is the harmonic series with k=1. Hence the given series is convergent. So the given series is absolutely convergent. This series is known to be divergent. and hence the series is convergent. whereas the series of absolute terms is divergent.Examine whether the series ∑ is absolutely convergent or conditionally convergent.com MA2111/ Engineering Mathematics-I Next we construct the series of absolute terms as | | | | | | | | | | The series of absolute terms is ∑ Which is the harmonic series with k=2.Test for convergence & absolute convergence of the series ∑ Solution: The terms of the given series are alternately positive and negative with 107 Get useful study materials from www. 24. Therefore the series is conditionally convergent.rejinpaul. 23. so the given series is absolutely convergent.com MA2111/ Engineering Mathematics-I The series is Clearly. t = log 2. ∫ = [ ∫ ∫ ] = (finite) converges in [2.rejinpaul. TH =∑ ∫ PR A ∫ Put log x = t When x=2. when x= . f(x) > 0 and f(x) is decreasing in [2.www. for x YU Let f(x)= | SH | A Next we construct the series of absolute terms as .rejinpaul.com . (i) (ii) By Leibnitz test the series is convergent. Hence by the integral test ∑ 25. | ∑ | | | . √ √ for 108 Get useful study materials from www.Test the series 1- √ also converges. com MA2111/ Engineering Mathematics-I (i) absolute convergence (ii)conditional convergence Solution: The terms are alternatively positive and negative (i) 1> √ √ √ (ii) √ Hence by Leibnitz test the series is convergent. ∑ ∑ √ ∑ Is convergent since in k-series.rejinpaul. 109 Get useful study materials from www.rejinpaul. for k = > 1 the series is convergent. PR A TH YU SH A Hence the given series is absolutely convergent. Now.www.com . rejinpaul.com .www. Radius of curvature The radius of curvature of a curve at any point on it is defined as the reciprocal of the curvature ⁄ { ( ) } Cartesian form of radius of curvature ⁄ ( ) ⁄ SH = YU Implicit form of radius of curvature A Parametric equation of radius of curvature Centre of curvature TH The circle which touches the curve at P and whose radius is equal to the radius of curvature and its centre is PR A known as centre of curvature. Properties of envelope and evolute 110 Get useful study materials from www.com MA2111/ Engineering Mathematics-I Unit 3 Differential Calculus Curvature The rate of bending of a curve in any interval is called the curvature of the curve in that interval Curvature of a circle The curvature of a circle at any point on it equals the reciprocal of its radius.rejinpaul. ̅ Equation of circle of curvature Centre of curvature ̅ ̅ & ̅ Evolute The locus of the centre of curvature is called an evolute Involute If a curve C1 is the evolute of C2 . then C2 is said to be an involute of a curve C1 Envelope A curve which touches each member of a family of curves is called envelope of that family curves. Envelope of a family of curves The locus of the ultimate points of intersection of consecutive members of a family of curve is called the envelope of the family of curves. A =2√ Find the radius of curvature of at x = on the curve y = 4 sin x – sin 2x ⁄ Ans: y1=4 cosx – 2 cos 2x at x= y1=2 y2= at x = y2=-4 ⁄ ⁄ = = √ 111 Get useful study materials from www.rejinpaul. at x=0 TH 1.com .we know that the envelope of the family of these normals is the locus of the ultimate points of intersection of consecutive A normals.but an infinite number of involutes Property:4 The envelope of a family of curves touches at each of its point.the evolute of a curve is the envelope of the normals of the curve Find the radius of curvature of y= ⁄ PR A Ans: y= y1= at x= 0 y1=1 y2= at x= 0 y2=1 ⁄ ⁄ = 2. Property:2 The difference between the radii of curvature at two points of a curve is equal to the length of the arc of the evolute between the two corresponding points. The corresponding member of that family.hence the envelope of SH the normals and the locus of the centres of curvature are the same that is . Property:3 There is one evolute .com MA2111/ Engineering Mathematics-I Property:1 The normal at any point of a curve is a tangent to its evolute touching at the corresponding centre of curvature. Evolute as the envelope of normals The normals to a curve form a family of straight lines.www.but the centre of curvature of a curve is also the point of consecutive normals.rejinpaul. YU Part . t.1 -----------. where m is the parameter and A.B and C are functions of x and y.2 2 4( ̅ -2a)3=27a ̅ 2 The locus of the centre of curvature (evolute) is 4(x-2a)3=27a 4.com MA2111/ Engineering Mathematics-I 3.(2) YU 2Am+B=0 PR A AB2/4A2-B2/2A+C=0 TH Substitute (2) in (1) we get A(-B/2A)2+B(-B/2A)+C=0 SH Differentiate (1) partially w.rejinpaul. 2 Write the envelope of Am2+Bm+C=0.AB2+4A2C=0 Therefore B2-4AC=0 which is the required envelope. 5.rejinpaul. (NOV-07) Solution: Radius of curvature Given y=x2 y1= Y2 = ( ) ⁄ =2x =2 112 Get useful study materials from www.com . Given the coordinates of the centre of curvature of the curve is given as ̅ 2a +3at2 ̅ -2at3 Determine the evolute of the curve Ans: ̅ ( ̅ ̅ t2=( ̅ 2a +3at2 t3 = ̅ -2at3 ⁄ ⁄ )3 = ( ̅ )-----------.(NOV-08) A Solution: Given Am2+Bm+C=0……………………(1) m=-B/2A…………. ‘m’ AB2-2AB2+4A2C=0 . Find the radius of curvature at any point of the curve y=x2.r.www. r.4x-6y+10=0 at any point on it . √ √ + y.y sin Sin = cos √ Substitute in (1) x.com .rejinpaul. ‘ ’ X cos between (1) and (2) SH Eliminate A = o…………………. ⁄ ) = ) ⁄ Find the envelope of the family of straight lines x sin + y cos = p.com MA2111/ Engineering Mathematics-I ( ( 6. NOV- 07) Solution: Given x sin + y cos = p……………. (JAN-06) Solution: Given x2 +y2 . √ PR A Tan TH YU = y sin √ =p =p Squaring on both sides. x2 +y2=p2 which is the required envelope 7. (1) Differentiate (1) partially w.www. being the parameter.t.(2) X cos .4x-6y+10=0 The given equation is of the form x2 +y2 +2gx+2fy+c =0 113 Get useful study materials from www.rejinpaul. What is the curvature of x2 +y2 . A=x2-1 B=-2xy C=y2+1 The condition is B2-4AC=0 4 x2y2-4(x2-1)(y2+1)=0 4 x2y2-4 x2y2-4x2+4y2+4=0 X2-y2=4 which is the required envelope 9. Find the envelope of the family of straight lines y= mx √ A There fore Curvature of x2 +y2 . where m is the parameter (JAN-06) SH 8.com MA2111/ Engineering Mathematics-I Here 2g =-4 g=-2 2f =-6 f=-3 Centre C(2.NOV-07) Solution: Given 2x2 +2y2 +5x-2y+1=0 ÷2 x2 +y2 +5/2x-y+1/2=0 114 Get useful study materials from www. radius r =√ r=√ =√ Curvature of the circle = √ YU Solution: Given y= mx √ TH (y-mx)2=m2-1 Y2+m2x2 – 2mxy-m2+1=0 .3).rejinpaul.4x-6y+10=0 is PR A m2 (x2-1)-2mxy+y2+1=0 which is quadratic in ‘m’ Here.rejinpaul.www.com . Find the curvature of the curve 2x2 +2y2 +5x-2y+1=0 (MAY-05. com .-2).B3.then C1 is said to be an involute of a curve C2. 11.B2. 13.6x+4y+6=0 The given equation is of the form x2 +y2 +2gx+2fy+c =0 Here 2g =-6 g=-3 2f =4 f=2 Centre C(3.(JAN-05) Solution: The locus of centre of curvature of a curve (B1. State any two properties of evolute . Define the curvature of a plane curve and what the curvature of a straight line. The difference between the radii of curvature at two points of a curve is equal to the length YU of the arc of the evolute between the two corresponding points. (JAN-05) TH Solution: The rate at which the plane curve has turned at a point (rate of bending of a curve is called the PR A curvature of a curve.…) is called evolute of the given curve. Find the radius of curvature of the curve x2 +y2 -6x+4y+6=0 (NOV-08) Solution: Given X2 +y2 .rejinpaul. Define evolute and involute .rejinpaul. radius r =√ 115 Get useful study materials from www. (MAY-05) A Solution: The normal at any point of a curve tangent to its evolute touching at the corresponding contre SH of curvature. If a curve C2 is the evolute of a curve C1 . 12. The curvature of a straight line is zero.1/2) radius r= √ =√ =√ = √ Therefore Curvature of the circle 2x2 +2y2 +5x-2y+1=0 is √ 10.com MA2111/ Engineering Mathematics-I Here 2g =5/2 g=5/4 2f=-1 f=-1/2 centre C (-5/4.www. Define evolute .rejinpaul. Solution: The locus of centre of curvature ( ̅ .where is the parameter.rejinpaul. (NOV-07) Solution: Given y=mx+ m2x-my+a=0 which is quadratic in ‘m’ The condition is B2-4AC=0 Here A=x B=-y C=a Y2-4ax=0 116 Get useful study materials from www.www.(MAY-07) Solution: Given (x.(MAY-07). ̅) is called an evolute .)2+y2=4 X2-2 x+ 2-4 +y2=0 2 -2 (x+2)+x2+y2=0 which is quadratic in SH A The condition is B2-4AC=0 Here A=1 B=-2 (x+2) C= x2+y2 YU 4(x+2)2-4(x2+y2)=0 TH x2-4x+4.x2-y2=0 PR A y2+4x=4 which is the required envelope.com MA2111/ Engineering Mathematics-I r=√ =√ Radius of Curvature of the circle = radius of the circle=√ 14. Find the envelope of the family of circles (x.)2+y2=4 . 15.com . 16. Find the envelope of the family of straight lines y=mx+ for different values of ‘m’. 05).c)=cos 0=1 =cos h(x/c)(1/c) (0. Therefore the point of intersection is (0. Find the radius of curvature of the curve xy=c2at (c. (i. Solution: Given +yt=2c Yt2-2ct+x=0 which is quadratic in ‘t’ The condition is B2-4AC=0 Here A=y B=-2c C=x C2-xy=0 SH A There fore xy=c2 which is the required envelope. 18. 17.where ‘t’ is the parameter.com MA2111/ Engineering Mathematics-I There fore y2=4ax which is the required envelope.e.com .c) =c cos h(x/c)(1/c)=cos h (x/c) (0.www. ) ⁄ TH Solution: Radius of curvature ( YU (NOV-05) PR A Given y=c cosh(x/c) and the curve crosses the y-axis.rejinpaul.rejinpaul. Find the radius of curvature of the curve y=c cosh(x/c)at the point where it crosses the y-axis. Find the envelope of the line +yt=2c.c).c)=1/c ⁄ =c2√ 19.)x=0 implies y=c.(NOV-02.(NOV-02) 117 Get useful study materials from www. JUNE 2009) Solution: Given x= Y= ……………………(1) ………………….(2) Differentiate (1) and (2) w.rejinpaul.c)= =-[ 1. .MAY-08.com .implies (c. PR A PART-B Find the radius of curvature at the point on the curve ⁄ ⁄ ⁄ (NOV-07.www.r.rejinpaul..com MA2111/ Engineering Mathematics-I Solution: Radius of curvature ( ) ⁄ Given xy=c2 =x +y=0 =.t 118 Get useful study materials from www.c)=-1 ] =-[ ⁄ TH YU √ √ SH ⁄ A ] (c. t. SH A Radius of curvature { ( at the point (-a.com . ‘x’ 2y ( ) ( ) 119 Get useful study materials from www.rejinpaul.com MA2111/ Engineering Mathematics-I ⁄ ⁄ = ( ) =-sec2 = ⁄ { ( ) } ) ⁄ ⁄ YU = ( TH =3a sin Find the radius of curvature of the curve PR A 2.r.www.rejinpaul. ⁄ ) } Solution: Radius of curvature Given Differentiate w.0).(NOV-08). com MA2111/ Engineering Mathematics-I ⁄ { [ ( ( ) } ) } [ ] ⁄ 6. JAN 2010.1=-3x2 ………………(1) ⁄ { ( ) } Therefore 120 Get useful study materials from www.0)on the curve YU NOV 2010) .(MAY-07.rejinpaul.’x’ ) } PR A Solution: Radius of curvature ( TH { 2xy +y2.rejinpaul.r. SH A { ] Find the radius of curvature at the point (a.t.com .www. ⁄ Given Differentiate w. (NOV-07) ⁄ { ( ) } SH Solution: Radius of curvature YU Given.rejinpaul.r.rejinpaul. =2/y ⁄ { ( ) } Therefore Differentiate (1) w.t.www.’y’. )( [( ) [( ) ( )] ] ⁄ { ( ) } { Therefore radius of curvature } ⁄ ⁄ (since the radius of curvature is non-negative) A Find the curvature of the parabola y2=4x at the vertex.r.r.com .com MA2111/ Engineering Mathematics-I Differentiate (2) w.t. 121 Get useful study materials from www.t.’x’ =4 PR A 7.’y’. y2=4x 2y TH Differentiate w. www.r.2a)= 1 Y2= [ ] 122 Get useful study materials from www.(1) SH Given 27ay2= 4x3 54ay =12x2 TH = -----------------------------------(2) (x1.y1) =Tan 45o=1-------------------------------.y1) =Tan 45o=1= PR A (x1.y1) be the point on the curve at which the tangent makes an angle 450 with the X.rejinpaul.com MA2111/ Engineering Mathematics-I } ⁄ { Therefore Curvature K=1/ 8. Solution.t.axis.y1) = = Gives YU Differentiate w.’x’ -----------------------------(3) As ( x1.com .y1) lies on the curve 27ay21= 4x31 ---------------------------------(4) Using gives x1= 3a And using (3) gives y1= 2a Y1 at (3a. A (x1. Let (x1. =2 =1/2 Find the radius of curvature of the curve 27ay2= 4x3 at the point where the tangent of the curve makes an angle 450 with the X.rejinpaul.axis. www.(JAN-06. =c ( )=- PR A = ( ) ( ) TH ( ) Y2= YU Y1= The co-ordinates of the center of curvature Is ̅ ̅ Where ̅ ( ⁄ ) ( ⁄ ̅ ( ) ( ) ( ) ( ) )……………………….rejinpaul..(1) The parametic form of (1) is A X=ct. Find the evolute of the rectangular hyperbola xy=c2.NOV-08) Solution: The equation of the given curve is xy=c2……….com MA2111/ Engineering Mathematics-I Y2= [ ]=1/6a ⁄ ⁄ Therefore radius of curvature ⁄ √ 7.(2) ̅ 123 Get useful study materials from www. y= SH =c.com .rejinpaul. rejinpaul.www.(5) ) ( ) TH ̅ ̅ YU ̅ SH A (2)-(3)gives ̅ ) PR A (4)2/3-(5)2/3gives ̅ ⁄ ̅ ̅ ⁄ ̅ ( ) ⁄ [( ) ( ) ] ⁄ = Therefore ̅ ̅ ⁄ ̅ ̅ ⁄ ⁄ ⁄ The locus of centre of curvature ̅ ̅ is ⁄ 9. Find the radius of curvature for the curve r=a(1+cos )at is a constant.08) 124 Get useful study materials from www..com MA2111/ Engineering Mathematics-I = ( ⁄ ̅ ) ( ( ( ) ) ( ) )………………………………(3) Eliminating ‘t’ between (2)and(3).com . (2)+(3)gives ̅ ̅ ̅ ( ̅ ( ) ( ) ( ) ……………………………(4) ( ( ) ( ) ………………….rejinpaul. ⁄ ⁄ which is the required evolute of the rectangular hyperbola xy=c2. (NOV-07. com . Considering the evolute as the envelope of normal’s.rejinpaul. . y=at2 =2a =2at 125 Get useful study materials from www.rejinpaul.www. find the evolute of the parabola x2=4ay. = PR A Also.com MA2111/ Engineering Mathematics-I Solution: Given r=a(1+cos ) R’=-a sin R’’ =cos The radius of curvature in polarform is = ( ) ⁄ { =[ } ⁄ ] ⁄ ⁄ SH √ is TH at ⁄ √ YU = ⁄ A = Therefore. 8.(NOV08) Solution: Given x2=4ay The parametric equations are x=2at. r.rejinpaul. JAN 10.(1) Differentiate (1) partially w.’t’we get ( t2 ) ⁄ A Y=3at2+2a +x=a( ) ⁄ [ ( ) ⁄ +2a( ) ) ] ⁄ ( YU ⁄ PR A x =( ) ) ⁄ TH y( SH Substitute the value of ‘t’ in (1) =( √ √ ) [( ⁄ ) [ ] ] ⁄ Squaring on both sides. we get which is the required evolute.www.com .JAN 2012. JUNE 2012) Solution: Given y2=4ax………………………(1)ss The parametric equations are x= at2. 9. y-y1= (x-x1) y-at2= (x-2at) yt-at3=-x+2at x+yt=at3+2at……………….rejinpaul.JUNE 2010. JAN 2011. (NOV-07.t. y=2at =2at =2a 126 Get useful study materials from www. Obtain the evolute of the parabola y2=4ax.com MA2111/ Engineering Mathematics-I M= We know that the equation of normal to the curve is. rejinpaul.com MA2111/ Engineering Mathematics-I =y1 ( ) Y2= = ( ) ( ) = The co-ordinates of the center of curvature Is ̅ ̅ Where ̅ ( ⁄) ) ⁄ ̅ ( ) ) A …………………….com .. 127 Get useful study materials from www. (2) gives (3) gives ̅ ̅ ( ̅ ) ̅ ( ̅ ) ̅ The locus of centre of curvature ̅ ̅ is which is the required evolute.www.(2) SH ( ( ⁄ ) ( ) ………………………………(3) PR A ̅ ( TH ̅ YU ̅ Eliminating ‘t’ between (2)and(3).rejinpaul. www.rejinpaul.com MA2111/ Engineering Mathematics-I 10. Find the equation of the envelope of .(NOV-02,07, NOV 2010, JAN 2013) Solution: Given that …………………(1) And ……………..(2) Differentiate (1)and(2) w.r.t ‘b’ ………………..(3) (4)gives ……………………(6) SH ……………..(5) YU (3)gives A 2a +2b=0…………………………(4) ⁄ PR A ⁄ TH From (5)and (6) ⁄ Substitute in (2) we get, Therefore ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ which is the required envelope. 11. Find the equation of circle of curvature of the parabola y2=12x at the point (3,6).(NOV-07,08) ̅ Solution: The equation of circle of curvature is ̅ Where, ̅ 128 Get useful study materials from www.rejinpaul.com www.rejinpaul.com MA2111/ Engineering Mathematics-I ̅ ⁄ y2=12x Given Differentiate w.r.t.’x’ we get 2y =12 implies (3,6)=-1/6 ⁄ SH Y2= A Y1= (3,6)=1 ⁄ ( can not be negative) TH √ √ YU ⁄ =3 ⁄ PR A ̅ ̅ ⁄ Therefore,the equation of circle of curvature is 12. Find the radius of curvature at ‘t’ on x=etcost,y=etsint.(JAN-06) ⁄ Solution: Radius of curvature Given 129 Get useful study materials from www.rejinpaul.com www.rejinpaul.com MA2111/ Engineering Mathematics-I X’= Y’= X’’= Y’’= ⁄ The radius of curvature is [ ] ⁄ [ √ ( ] . ) ⁄ ⁄ √ SH [ ] YU = ] A [ .(MAY-05,07) TH 13. Find the evolute of the ellipse PR A Solution: The given curve is The parametric equations are x=acos ,y=bsin Y1= Y2= ( ) ( ) ( ) Y2 The Co-ordinate of centre of curvature is ̅ ̅ 130 Get useful study materials from www.rejinpaul.com ( ̅ ⁄ ) ( ̅ ⁄ ) ̅ ⁄ ̅ ⁄ ⁄ ⁄ 131 Get useful study materials from www.com MA2111/ Engineering Mathematics-I ̅ Where ( = acos ) )( ( - ) = acos ( = ( ( ) ) ………………….rejinpaul.com .(1) SH A ̅ ) ( ) ( + ̅ ] [ ] PR A [ [ ) TH = YU ̅ [ ] ]………………………(2) Eliminating ‘ ’ between (1) and (2).www.rejinpaul.we get ̅ ( ⁄ ) ̅ ( ⁄ ) we know that. t.(1) Solution: Given that …………………(2) ) ( ) SH ( A Differentiating (1) w. 132 Get useful study materials from www.r.www.com ..rejinpaul.r.rejinpaul.’m’ YU …….’m’ …………..com MA2111/ Engineering Mathematics-I ⁄ ̅ ⁄ The locus of ̅ ̅ is 14. (MAY – 05.JAN 2012) ………………. NOV-05.(4) From (3) and (4) √ √ substitute in equation (2) .b are constants.(3) PR A TH Differentiating (2) w. Find the envelope of ̅ ⁄ ⁄ ⁄ ⁄ which is the evolute of the ellipse where l and m are connected by and a.t. The points are (1.www. P (a.com MA2111/ Engineering Mathematics-I √ √ √ √ which is the required envelope.2). Find the points on the parabola …………….b) be the point on the curve √ at where A ⁄ SH Differentiate (1) w.r.com .t. (MAY – 05) 15. 8 ⁄ √ √ a+1=2 a=1.(1) Solution: Given Let.(1.rejinpaul.-2) 133 Get useful study materials from www. at which the radius of curvature is 4 √ . ‘x’ TH YU Y1=2y PR A Y2= ⁄ ⁄ √ ⁄ But.rejinpaul. com . Considering the evolute of a curve as the envelope of its normals find the evolute of (NOV-02.t.05.(1) Differentiate (1) partially w.’ ’.com MA2111/ Engineering Mathematics-I 16.y=bsin PR A TH YU SH We know that the equation of the normal is y-y1= (x-x1) A m= . .rejinpaul. we get ⁄ ⁄ 134 Get useful study materials from www.we get ………….rejinpaul..MAY-05) Solution: The given curve is The parametric equations are x=acos .r.www. 4)on xy=12.rejinpaul.www.rejinpaul.com MA2111/ Engineering Mathematics-I ⁄ √ ⁄ ⁄ √ ⁄ ⁄ ⁄ Substitute in equation (1). Find the circle of curvature at (3.6) ( ) ) 135 Get useful study materials from www.’x’ we get x implies ( Y1= (3.(JAN-05. SH A ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ [ √ 17.t. JAN 2010) ̅ TH Where.r.4)=-4/3 Y2= (3.we get ⁄ √ ⁄ ⁄ ⁄ √ ⁄ ⁄ ⁄ ⁄ ⁄ ] ⁄ which is the required evolute of the ellipse.com . ̅ PR A ̅ Given ̅ YU Solution: The equation of circle of curvature is ⁄ xy=12 Differentiate w. (JAN-05.rejinpaul.sec2 / ̅ = a cos 3 - = a cos 3 + ̅ =( (sec4 θ cosecθ))/3a ( 1 + tan 2 ------------------------------------------(2) a sin 3 ( 1 + tan 2 136 Get useful study materials from www. y = a sin 3 Y1= / = -tan Y2= .rejinpaul.com .(1) PR A The parametric equations are x = a cos 3 .com MA2111/ Engineering Mathematics-I ) ( ⁄ ⁄ ⁄ ̅ ( ) ̅ ⁄ 18.. NOV-07) ………………. the equation of circle of curvature is ( ( A ⁄ SH ⁄ =3 ) ⁄ ( ) .www. . Find the evolute of the four cusped hyper cycloid ⁄ ) ( ⁄ ⁄ ⁄ ⁄ TH Solution: The equation of the given curve is ) YU Therefore. JUNE’12.sin + ̅ = ------------------------------------(5) ⁄ ̅ ⁄ = ⁄ ⁄ (( ⁄ ) +( (2) ) A Eliminate + ⁄ ⁄ + = ⁄ (2) YU The locus of centre of curvature is SH a sin 3 19.www.cos ). JUNE 10.cos ). x’ =a( 1 + cos ) y’ = a( sin ) x” = -a sin  y” = a cos  ⁄ The radius of curvature is =      ⁄  = 4a cos At = 0 20.com . TH (MAY’07.a sin 3 + 3 .JAN 13) PR A Given: x = a ( + sin ) . Nov ’08. α is the parameter (Nov’ 07) Solution: Given x cos α + y sin α = a sec α 137 Get useful study materials from www. Find the radius of curvature at the origin of the cycloid x = a ( + sin ) and y = a( 1. y = a( 1.̅ = a cos 3 + = a( cos ̅ ̅ ⁄ .com MA2111/ Engineering Mathematics-I -----------------------------------------(3) from 2 & 3 ̅ + ̅ = a cos 3 + = a( cos + a sin 3 + 3 + sin ------------------------------------(4) ̅ . Find the envelope of the straight lines represented by the equation x cos α + y sin α = a sec α .rejinpaul.rejinpaul. y=a sin  is the catenary y = a cosh YU Nov ‘05 = a cot  cos  y=a sin  ) PR A x’ = a(-sin  + TH Solution : x = a (cos  + log tan ) y’ = a cos  y1 = y2 = = tan  =¼ (sec 4  sin ) ̅ 138 Get useful study materials from www.com MA2111/ Engineering Mathematics-I Divided by cos α x + y tan α = a sec2α x + y tan α = a ( 1 + tan2α) a tan2α – y tan α + a –x = 0 which is quadratic in tan α A = a.rejinpaul. Prove that the evolute of the curve x = a (cos  + log tan ). B = -y C = a-x The envelope is given by B2 – 4AC = 0 A y2 = 4a(a-x) which is the required envelope SH 21.www.rejinpaul.com . com MA2111/ Engineering Mathematics-I = a (cos  + log tan )- ( 1 + tan 2 ̅ =a log tan ……………….(3) ̅̅̅̅̅⁄ ) SH ̅ ( ̅⁄ ̅ PR A TH y = a cosh which is the required evolute YU tan = A Eliminate ‘’from (1) and (2) 139 Get useful study materials from www.(2)  = ̅⁄ ………………….(1) ̅ = a sin + ̅ ( 1 + tan 2 ……………….www..rejinpaul.com .rejinpaul. Problems based on Euler`s Theorem III.to x and it is denoted PR A TH by the symbols Notation: Successive Partial Differentitation: Let . Problems based on Lagrangian Multiplier Partial Derivatives be a function of two Variables x and y.com MA2111/ Engineering Mathematics-I Unit 4 Functions of several Variables I. Problems based on Maxima and Minima for Functions of Two Variables VII. Problems based on Partial Derivatives II.com . Problems based on Total Derivatives-Differentiation of Implicit Function IV. then z SH Let A Partial Derivatives: YU is a function of x only . Problems based on Jacobian V. If we keep y as a constant and Vary x alone .www. ( ) ( ) ( ) ( ) Problems: 140 Get useful study materials from www.to x.rejinpaul.r. treating y as a constant is called the partial derivatives w. Problems based on Taylor`s and Laurent Series VI.rejinpaul.r. then being the function of x and y can further be differentiation partially with respect to x and y. The derivative of z w. (2). (1) (2) (3) (1)+ (2)+ (3) 141 Get useful study materials from www. If show that 4.www.and (3) we get TH YU SH A Solution: Given show that .com . Find if where and Solution: 3.rejinpaul.rejinpaul. Solution: Let PR A Adding (1). If u = find Solution: 2.com MA2111/ Engineering Mathematics-I 1. com .www. it satisfies the Eulers equation ( ) ( ) ( ) Euler`s Theorem for Homogeneous Function Euler`s Theorem: If u be a homogeneous function of degree n an x and y then ( ) ( ) 142 Get useful study materials from www.rejinpaul. If u = . Show that Solution: Given z is a homogeneous function of degree n=2 6. ( If ). find Sol: given u = PR A TH YU SH A As z is homogeneous function of order n=2.com MA2111/ Engineering Mathematics-I 5.rejinpaul. it satisfies the Eulers theorem (1) (1) (2) By Eulers theorem of second order 143 Get useful study materials from www. Show that √ PR A 2. TH Hence Eulers theorem is verified. Verify Eulers theorem for the function Solution: Given SH A This is a homogeneous function of degree 2.com .rejinpaul. As z is a homogeneous function of order n = 1. YU Adding (1) and (2) we get If Solution: Given √ .www.rejinpaul.com MA2111/ Engineering Mathematics-I Problems: 1. rejinpaul. prove that √ √ √ √ ) ) ( ) = Hence by Euler’s theorem. 3. put in (3) If u = √ √ √ √ √ √ √ ( √ √ √ ( Sol: given .rejinpaul.com MA2111/ Engineering Mathematics-I (3) . Total Derivative: If . then we can express u as a function of Thus we can find the ordinary derivatives TH t alone by substituting the value of x and y in YU Total Derivatives-Differentiation of Implicit Function Now to find the PR A is called the total derivative of u distinguish it from partial derivatives without actually substituting the values of x and y in following formula we establish the Problems: 1.com which . Find if where and Solution: 2. we have SH xu=0 A is a homogeneous function of degree 0. If where and find and Solution: 144 Get useful study materials from www.www. SH A ( TH Ans : 5.com . Find if Solution: Let ) ( ) ( ) Find YU 4. such that ( show that ) Solution: z may be represented as the function of u.v ( )( ) (1) Similarly 145 Get useful study materials from www.rejinpaul.rejinpaul. ( )( ) PR A ( ) ( )( ) If z be a function of x and y and u and v are other two variables.com MA2111/ Engineering Mathematics-I 3.www. rejinpaul.rejinpaul.(4) )( ) …………………………….(3) )…………….www.com . ) Ans: Here Z is a composite function of u and v ……………………………………(1) A ……………………………………….(2) SH Now YU Sub these values in (1) & (2) We get ( Which implies (3)x(4) We get ( ) ( PR A ) ( Now TH ……………………………….(A) Similarly we get ………………………(B) (A)+(B) Gives 146 Get useful study materials from www..com MA2111/ Engineering Mathematics-I (2) ( (1)+(2) ) ( 6.... If u is a function x and y and x and y are functions of r and  given by (a) x  e r cos  ..com .  x and y r   r u u x u y u u Now  .  y  x  x  y  x y From (1).  x   y   r  r    r[    Solution (a) Here u is a function of x. we get .www..  .. y  e r sin  shown that 2  2u  2u  2u  2 r   u e     2  x 2 y 2  2   r (b) x  r cos  .   y. x y r x r y r x y SH A   2 z    2 z    2 z  1   2 z  1  z  and  2    2    2   2  2     .(1) . r  y y  e r sin   y..(2) ..  e r cos   x r  x x y y  x.  .. y  r sin  prove that 2 1  z   z   z   z         2    x   y   r  r  r  2 2 2 PR A YU TH x r x  e cos   x.rejinpaul.(3) 147 Get useful study materials from www. u u x u y u u  .rejinpaul.  e r sin    y.com MA2111/ Engineering Mathematics-I ( ) 14.    x y r x y . y which is given as a function of r and  . .   cos  .rejinpaul.(6)     2 x 2 xy y 2 Adding (5) and (6) we get .     y  x  x y Now..com ..  sin  .e.  r cos  r  r  z z x z y z z  ...com MA2111/ Engineering Mathematics-I From (2) ..  e  2  2 x 2 y 2    r SH   2u  2u    2u  2u   ( x 2  y 2 )  2  2   e2 r  2  2  y  y   x  x A 2 2  2u  2u 2 2  u 2 2  u   (x  y ) 2  (x  y ) 2 r 2  2 x y 2 1  z   z z   z z   z   sin     sin   cos      2    cos  x y   x y   r  r     2  z   z       x   y  2 2 2 148 Get useful study materials from www.  sin  .(5) r 2 x 2 xy y 2  2u   u      u u     y  x     y  x 2       x y  x y  2 2  2u  2u 2  u 2  u xy x y 2 .rejinpaul. x x y y  cos  . r x r y r x y z z x z y z z    r sin   r cos   x  y  x y PR A TH (b) YU  2u  2u 2 r   2u  2u  i.  r sin  .. we get . .www.(4)  2u   u      u u    x  y  x  y  2 r r  r   x y  x y  2 2  2u  2u 2  u 2  u x  2 xy y . www.com MA2111/ Engineering Mathematics-I  z   z     r 2 r  r    z z   cos   sin   r  x y    2 z x  2 z y   z x  2 z y   cos   2    sin   xy r  y 2 r   x r yx r    2 2 2  z  z  z  cos 2  2  sin 2  2  2sin  cos  x y xy 2 SH   2 z  2 z y  z  r cos    2  y   y   YU  r sin  A 2 z   z     2        z z    r sin   r cos      x y    2 z x  2 z y  z   r cos   r sin   2   x  x  yx    2 z 2 2 z 2 z z z  2 2       r  cos   sin   cos 2 sin cos r r 2 2 x y xy x y   2 z 1 2 z 2 z 2 z 1  z z   2  2  2   cos   sin   2 2 r x y r  x y  r   2 z  2 z 1 z   x 2 y 2 r r PR A  TH  r 2 sin 2   2 z  2 z  2 z 1  2 z 1 z     x 2 y 2 r 2 r 2 r 2 r r Jacobian Defn : from are functions of n variables the the Jacobian of the transformation is defined by 149 Get useful study materials from www.rejinpaul.rejinpaul.com . www. find Solution: 150 Get useful study materials from www. || PR A u and v are not independent. SH A 1. Find the Jacobian of u = xyz.rejinpaul.v are functionally dependent. v = xy + yz + zx. 2. || TH || YU Solution: If .rejinpaul.com . || u .com MA2111/ Engineering Mathematics-I | | | | | | and is denoted by the symbol or Problems: Prove that the functions are are functionally dependent. w = x + y + z Solution: 3. www.   x1 . y2 . x2 .com MA2111/ Engineering Mathematics-I 15. y3  x1 x2 1  x3  . if y1  1  x1  .rejinpaul. Find the value of   y1 . y2 . x3  x1 y1 x2 y1 x3 y2 x2 y2 x3 y3 x1 y3 x2 y3 x3 1  0 0 1  x2  x1 0 x2 1  x3  x1 1  x3   x1 x2   1  x1   x1 x2  A   x12 x2 PR A TH YU SH Hence the solution. y3  . x3  Solution : y1 x1   y1 . y3  y2    x1 . y2  x1 1  x2  .com . x2 . 151 Get useful study materials from www.rejinpaul. y . we get . z  . v. w  x  y  z . we first evaluate J  u x v We knowthat J   x w x   x. z .   u .rejinpaul.www. y . z    u .com . w  1 2  x  y   y  z  z  x  152 Get useful study materials from www. v.rejinpaul.com MA2111/ Engineering Mathematics-I 16. JJ   1. y . in terms of x . find J    x. y . w are exp liciltly given. v. If u  xyz . v. w  Solution : Since u . w  u y v y w y u z yz zx xy v  2x 2 y 2z z 1 1 1 w z  yz  2 y  2 z   zx  2 x  2 z   xy  2 x  2 y   2  y  z   x 2  x  y  z   yz   2  y  z  z  x  y  x   2  x  y  y  z  z  x  J   PR A By u sin g . v  x 2  y 2  z 2 . z    u . TH  2  y  z   y  z  x   x  z  x   YU  2  x 2  y  z   x  y 2  z 2   yz  y  z   SH  2  x 2 y  x 2 z  xy 2  xz 2  y 2 z  yz 2  A  2  yz  y  z   zx  x  z   xy  x  y     x. y3  1 2 .com . x2 . y3  1 2 . If Solution: and . y2  3 1 . Evaluate without actual substitution. x1 x2 x3 YU y1  x1 x2 x1 x2  x1 x2 TH Given : 1 1 1   11  1  1 1  1  1  1  022  4 Hence the proof . y3 with respect to x1 . y2 . 18.www.com MA2111/ Engineering Mathematics-I 17. and 153 Get useful study materials from www. Pr oof : We knowthat   y1 . y2 .rejinpaul. x3 is 4. y2  3 1 .rejinpaul. x3  x1 y1 x2 y1 x3 y2 x2 y2 x3 y3 x1 y3 x2 y3 x3   x2 x3 x12 x3 x2 x2 x3  x3 x1 x2 x1 x3 x1 x22 x1 x2 x1 x3  x2 x3 1  2 2 2 x2 x3 x1 x2 x3 x2 x3 x1 x2 x32 x3 x1  x3 x1 x3 x1 1 1 1 1 1 1 PR A x2 x2 x2  12 22 32 x1 x2 x3  SH y1 x1 A x2 x3 xx xx . y3  y2    x1 . x2 . If y1  x2 x3 xx xx . x1 x2 x3 Showthat the Jacobian of y1 . PR A TH YU SH A Solution: 154 Get useful study materials from www.com .rejinpaul. Find the Jacobian . If .com MA2111/ Engineering Mathematics-I 19.www.rejinpaul. find MA2111/ Engineering Mathematics-I  ( x. To find the relation between them: Given and Now. PR A will not be independent if Hence are not independent.rejinpaul. ) Solution Given x  r cos  y  r sin  x  cos  r x   r sin   y  sin  r y  r cos   x x  ( x. y ) r  We knowthat  y y  (r . If x  r cos  .www.rejinpaul.com . y )  (r . . = = 0. 155 Get useful study materials from www.com 23. And Sol: Hence.If TH also find the relation between them. show that they are not independent. ) r  cos  r sin   sin  r cos   A  r cos 2   r sin 2   YU SH  r cos 2   sin 2   r 24. y  r sin  . (1) ..rejinpaul. A Problems: +….www.…. in the power of and is PR A TH YU Sol: The Taylor’s series expansion of SH 1. = = = =1 = . .com .rejinpaul. which is the required relation. =1 156 Get useful study materials from www.com MA2111/ Engineering Mathematics-I . Taylor`s Series and Laurent Series The Taylor’s series expansion of in the power of and is Find the Taylor’s series expansion of in the power of x and y upto third degree terms. in powers of x and y upto second Solution: 157 Get useful study materials from www. =0 2.rejinpaul.www.com ..com MA2111/ Engineering Mathematics-I . PR A = TH YU SH A =2 Using Taylors series expansion express degree terms at + …….rejinpaul. Expand as Taylors series up to second degree terms Taylors series is PR A TH YU SH A Solution: 158 Get useful study materials from www.www.rejinpaul.com .com MA2111/ Engineering Mathematics-I Taylors series is 3.rejinpaul. 2)  2 f xyy  0 f xyy (1.(1)  f x (1. y )  x 2  3 y  2 A Here a  1.com . b)  ( y  b) 2 f yy (a. b )  ( x  a ) f x ( a . 2)  0 f yyy  0 f yyy (1. b  2 x 2 y  3 y  2  6  ( x  1)( 4)  ( y  2)(4)  1 [( x  1) 2 (4)  2( x  1)( y  2)( 2)  ( y  2) 2 (0)] 2! 1 [0  3( x  1) 2 ( y  2)(2)  0  0] 3!  6  4( x  1)  4( y  2)  2( x  1) 2  2( x  1)( y  2)  2( x  1) 2 ( y  2)  Hence the solution. 2)  2 f yy  0 f yy (1. 2)  4 f xy  2 x f xy (1. 2)  4 fy  x  3 f y (1. b)  2( x  a )( y  b) f xy (a. y ) in powers x  a and y  b is given by f ( x. b ) 1 ( x  a ) 2 f xx (a. 2)  0 2 PR A Substituting these value s in (1) we get ..rejinpaul.com MA2111/ Engineering Mathematics-I 16. 2)  4 f xx  2 y f xx (1..rejinpaul.. b)  2! 1  [( x  a )3 f xxx  3( x  a ) 2 ( y  b) f xyy  3( x  a )( y  b) 2 f xyy 3!  ( y  b)3 f yyy ] . 159 Get useful study materials from www. 2)  0 f xxx  0 f xxx (1. y )  f ( a . SH f x  2 xy YU f (1. 2)  6 TH Now f ( x. b )  ( y  b ) f y ( a . Solution We knowthat the exp ansion of f ( x. 2)  0 f xxy  2 f xxy (1. Use Taylor ' s series oftwo var iables to exp and x y  3 y  2 in powers of 2 x  1 and y  2.www. 1)  ( y  1) f y (1.  2 x  y2 1 1 f y ( x.. 2 2 SH 2 xy   2 2 f yy (1.com . y )  . y )   y (1)( x 2  y 2 ) 2 ..1)   . x 4 2 2 4 4 Hence the solution  160 Get useful study materials from www.1)  ( y  1) 2 f yy (1...1)    f (1. y )  tan 1 y x 1  y .rejinpaul. y )  x(1)( x 2  y 2 ) 2 .1)   1 2 1 ( x  1) 2 f xx (1. x solution Let f ( x. y )  x  y  2 2  y2  f xy (1.2 x x  y   x  y 1  x 2 x ( x.www.2 y  YU 2 2 y 2  x2 x 1 2 2 2  f xx (1.  y 2  x 2  1 2 x y .1)  TH 2 f xy .1)  0 2 tan 1 y  f ( x.  2( x  1)( y  1)0  ( y  1) 2      .  4 2  2!  2  2  2  y  1 1 1 1 tan 1   ( x  1)  ( y  1)  ( x  1) 2  ( y  1) 2  .. 2!   u sin g cor : 2   1 1  1  1  1  ( x  1)     ( y  1)   ( x  1) 2 . Expand tan 1 y in the neighbourhood of (1.. y ) x x 2 xy 2 y PR A f yy ( x..1)  ( x  1) f x (1.1)   1 2 A f y (1.1)  2( x  1)( y  1) f xy (1.  2 x  y2 f x ( x. y )  f x (1. y2 x 1 2 x x .1)  1 2 f xx ( x..1).rejinpaul.com MA2111/ Engineering Mathematics-I 15. z Problems: Examine for its extreme values Solution: Given 161 Get useful study materials from www.rejinpaul.www.com .z are subject to a constraint equation We define a function where Which is independent of x.rejinpaul.y.y.com MA2111/ Engineering Mathematics-I Maxima and Minima and Lagrangian Multiplier Defn: Maximum Value YU SH A Defn: Minimum Value PR A TH Defn: Extremum Value Defn: Lagrangian Multiplier Suppose we require to find the maximum and minimum values of where x. com MA2111/ Engineering Mathematics-I At maximum point and minimum point The points may be maximum points or minimum points. At the maximum point and minimum point 162 Get useful study materials from www. .find the maximum value of .rejinpaul.rejinpaul.www.com . At and r = 12 > 0 is a minimum point Minimum value = are saddle points. In a plane triangle ABC . and r = -12 < 0 YU At SH The points A At maximum value = TH is a maximum point Solution: PR A 8. www. TH The point .rejinpaul.rejinpaul.com . Find the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid Solution: The given ellipsoid is (1) The volume of the parallelepiped is (2) At the max point or min point (3) (4) 163 Get useful study materials from www.com MA2111/ Engineering Mathematics-I Solving these equations A+B+C = π . Maximum value = . SH A At YU and r < 0 is a maximum point. PR A 9. .rejinpaul. A Max volume = SH Put in (2) 10.(4)y 164 Get useful study materials from www. cm PR A TH Solution: Given Surface area The volume is (1) (2) At the max point or min point (3) (4) (5) Solve the equation (3)x .rejinpaul.com MA2111/ Engineering Mathematics-I (5) Solve the equation (3)x+ (4)y+(5)z  Put in (3) Similarly. whose surface is YU 108 sq.www. Find the dimensions of the rectangular box without a top of maximum capacity.com . Find the highest temperature on the SH A surface of the unit sphere Solution: YU (1) PR A At the max point or min point TH (2) (3) (4) (5) Solve the equation (3)x+ (4)y+(5)z 1600 Put in (3) and (4) we get 165 Get useful study materials from www.(5)z Put in (1) y =6. y. Breadth = 6cm. 11. z =3 The dimension of the box.www. The temperature T at any point (x.com . having max capacity is Length=6cm.rejinpaul.rejinpaul. z) in space is T = 400xy . Height = 3cm.com MA2111/ Engineering Mathematics-I (3)x . com .rejinpaul. showthat 166 Get useful study materials from www.rejinpaul. u 2 v 2 x 2 y 2 PR A TH YU SH A y  u sin   v cos  .www. 13.com MA2111/ Engineering Mathematics-I The highest temperature is = 50. y  u sin   v cos  x 2 y 2 u 2 v 2 (OR) By changing independent var iables u and v to x and y by means of the relations x  u cos   v sin  . 2 z 2 z 2 z 2 z  transforms in t o  . where x  u cos   v sin  . Pr ovethat 2 z 2 z 2 z 2 z    . .rejinpaul.com .www.com MA2111/ Engineering Mathematics-I Solution : Here z is a composite function of u and v.     v x v y v z z   sin   cos  x y or A       z     sin   cos   z v x y         sin   cos   (2) v x y Now we shall make use of the equivalance of operators as given by (1) and (2). TH  2 z   z     u 2 u  u  YU SH or ie. z z x z y     u x u y u z z  cos   sin  x y       z    cos   sin   z x x y        cos   sin   (1) u x y z z x z y Also. PR A     z z    cos   sin    cos   sin   (u sin g (1) and (2) ) u y   x y   2 z 2 z 2 z 2 z  cos 2  2  cos  sin   sin  cos   sin 2  2 x xy yx y 2 z 2 z 2 z 2 z 2 2  cos   2 cos  sin   sin  u 2 x 2 xy y 2 Similarly  (3)  2 z   z     v 2 v  v      z z     sin   cos     sin   cos   (u sin g (3) and (4) ) x y   x y   167 Get useful study materials from www.rejinpaul. 2 x 3  x 2 y  1  0 . y   x 2  xy  y 2  TH PR A 2 f 2  2 3 2 x x 2 2 f  2 3 2 y y YU SH f 1  2x  y  2 x x f 1  x  2y  2 y y 2 f 1 xy Step1: For a max imum or min imum. 2 y 3  xy 2  1  0 .. 17.(4) From (3) we get x y  y  2 x y 2 2 3 .. x y Solution : 1 1  x y A Given f  x..(5) Substituting (5) in (4) we get 168 Get useful study materials from www.e.(2) (1)  y  2 x3 y  x 2 y 2  y  0 . we must have f  0  2x  y  x f  0  x  2y  y 1 0 x2 1 0 y2 i....(1) i...e. Investigatethe max ima of the functions f  x...www..  (4) 2 z 2 z 2 z 2 z    u 2 v 2 x 2 y 2 Hence the proof . y   x 2  xy  y 2  1 1  .(3) (2)  x  2 xy 3  x 2 y 2  x  0 .rejinpaul.rejinpaul..com .com MA2111/ Engineering Mathematics-I  sin 2   z  z  z 2 z 2  sin  cos   cos  sin   cos  x 2 xy yx y 2 2 2 2 2 z 2 z 2 z 2 z 2 2  sin   2 cos  sin   cos  v 2 x 2 xy y 2 Adding (3) and (4).    8 8 1  0 x 2 y 2  xy  2 f and 2  8  0 x  1 1 f ( x. 2 xy ( x  y )( x  y )  ( x  y )  0 ( x  y ) 2 xy ( x  y )  1  0 x  y (or )2 xy ( x  y )  1  0 when x  y.  3 3 2 PR A 2 2 f  2 8 2 1 x 3 2 f 2 f 8 .com . 2 x 3 y  2 xy 3  y  x  0 i.  1 y 2 xy TH YU SH  1 1 Hence  3 ..e. 169 Get useful study materials from www. 3  and  3 3 4 3 the min imum valueis 3 .3 x 3  1 1 3 1 x3 3 x3  3  1 1 Step 2 : At  3 .rejinpaul.www.rejinpaul. 3  .com MA2111/ Engineering Mathematics-I 2 xy  y  2 x y  x  0 3 3 i.. y ) has a min imum at the po int  3 . the equation 2 x 3  x 3 y  1  0 gives 2 x 3  x 3  1  0 i. 3  is a critical po int  3 3 1 3 A y 2 f 2 f  2 f  .e..e..e. 2 xy ( x 2  y 2 )  ( x  y )  0 i. .. Find the min imum value of the function f ( x.(4) .....com . y )  x 2  y 2  xy  ax  by....www.rejinpaul...(3) 2 f 2 x 2 2 f 2 y 2 .(2) .. Solution Given f ( x..com MA2111/ Engineering Mathematics-I 18.rejinpaul.. y )  x 2  y 2  xy  ax  by f  2x  y  a x f  2y  x  b y ...(5) 2 f 1 xy SH A .(7) YU Step 1: For miimum value of the function f  0  2x  y  a x .(1) .(6) 170 Get useful study materials from www.(8) TH  2y  x  b PR A f 0 y . rejinpaul..com .  2 and 1 x 2 y 2 xy YU SH A x 2 f 2 f  2 f  .e.com MA2111/ Engineering Mathematics-I Solving (7) and (8) we get 2 x  y  a x  2 y  b i. a  2b   3   b  2a  Step 2 : At  . a  2b   3  171 Get useful study materials from www.www. 4 x  2 y  2a x  2 y  b 3x  b  2a b  2a 3 b  2a Substituting x  in (8) we get 3  b  2a    b 3  y  2 (4b  2a )  2  a  2b 2 PR A TH  b  2a  The critical po int is  . a  2b   3  2 f 2 f 2 f  2 .rejinpaul.    2  2 1 x 2 y 2  xy  30  b  2a  We have min imum value of the function at  . .(2)    2 1 6 Any po int on the first line is P (3  2.com MA2111/ Engineering Mathematics-I Step 3 :The min imum value of f ( x. 2  6. PQ 2 17 2  41 2  32  66  114   99 Let f ( .e.rejinpaul.www..com .(1) 3 2 2 x 5 y 3 z  4 . 6   4) PQ  (2  3  5  2  ) 2  (6  2  3   ) 2  (5  2  4  6  ) 2 i. y ). y) is obtained by putting x b  2a and y  a  2b in f ( x.  )  17 2  41 2  32  66  114   99 f  34  32   66  f  32  82   114  2 f 2 f 2 f    32 34...  2  2  172 Get useful study materials from www. 2  5) and PR A Let any po int on the sec ond line is Q (2   5..rejinpaul. Find the shortest dis tan ce between the lines SH TH x2 y 6 z 5 x5 y 3 z  4   and   3 2 2 2 1 6 Solution YU  b  2a   4a  5b      (a  2b)(a  b0  3  3  4ab  5b 2  8a 2  10ab   a 2  ab  2ab  2b 2 9 2 4ab  5b  8a 2  9a 2  27ab  18b 2  9 2 2 13b  a  23ab  9 A 2 x 2 y 6 z 5    .   3. 82. 3 2  b  2a   b  2a   b  2a  Minimum value      a  2b      a  2b   a    b  a  2b   3   3   3    b  2 a   b  2a       a  2b  a    a  2b   a  2b  b   3   3   19. PQ 2  17  41  32  66  114  99 9 PR A TH YU Shortest length  PQ  9  3 A  2 f    0    At (1.  ) has min imum. PQ 2 has min imum which gives the shortest length.. 2 f 2 f .  2  2 173 Get useful study materials from www. 1). 0   i.rejinpaul.rejinpaul.e. 1) .www. 1) the function f ( .   1 At (1.e. 1) 2 f  34  0  2 At (1. At (1. 34  32   66  0  32  82   114  0 Solving thesetwo equations we get   1. SH i.com MA2111/ Engineering Mathematics-I For a max imum or a min imum of ' f ' we should have f f  0..com . (3) TH x3  y 3  z 3  ..... y .rejinpaul...e.(1) F ( x.rejinpaul. x y  z   2 Substituting (6) in (5) we get 3  1 or x  3 x 3  1 or y  3 y 3  1 or z  3 z (3.3) is the po int where min imum value occur.. The min imum value of x 2  y 2  z 2 is 32  32  32  27 174 Get useful study materials from www. (3) and (4) we get 2 1 A SH . z )  ( x 2  y 2  z 2 )       1  x y z  By Lagranges method the values of x...(6) i..(5) YU  .(4) .com . y ...3.www..com MA2111/ Engineering Mathematics-I 1 1 1 20.(2) PR A   3 . Find the min imum value of x 2  y 2  z 2 subject to the condition    1 x y z Solution Let f  x 2  y 2  z 2 1 1 1 g    1 x y z Let the auxillary function ' F ' be 1 1 1  . z for which ' f ' is min imum are obtained by the following eqations F    0  2 x  2  0   x3 x x 2 F    0  2 y  2  0   y3 y y 2 F    0  2z  2  0   z3 z z 2 F 1 1 1  0    1  0  x y z From (2) . . y )  4 x 2  6 xy Now we have to find the min imum surface area .rejinpaul.(7)  Substituting x y 6  in (6) we get 36  2 2 x . x 2 y  36 .e. y  72 i.x)  2(2 x. Solution Let the sides of the rec tan gular box be 2 x. PR A By Lagranges method the values of x. y.x. and a cons tan t valume 72m3 ..(4) x F 0 6x   x2  0 .. Then volume is 2 x.(8) 175 Get useful study materials from www. y )  4 x 2  6 xy   x 2 y  36 YU Let f  4 x 2  6 xy.. Find theleast surface area of thebox.e.......(3) TH F ( x. Athin closed rec tan gular box is to have one edge equal to twice the other .(5) y F 0 00 z F 0 x 2 y  36  0  x 2 y  36 . y..(6)  6 From (5) we get x   .. y )  2( x...(2) S  4 x 2  6 xy under the condition SH A x 2 y  36 g  x 2 y  36 Let the auxillary function ' F ' be F  f   g     ..(1) The surface area is given by S  2(2 x..rejinpaul.. x. z for which ' F ' is min mum is obtained from the following equations F  0  8 x  6 y  2 xy  0 .www.com . 2 x 2 y  72 i..com MA2111/ Engineering Mathematics-I 21.   x y z mn p mn p   x yz a 176 Get useful study materials from www. z )  x m y n z p    x  y  z  a  A Let f  x m y n z p and g  x  y  z  a Let the auxillary function F  f   g .e. F ( x....com ...(3) PR A F  0  mx m 1 y n z p    0 x F  0  ny n 1 x m z p    0 y F  0  p x m y n z p 1    0 z F 0  x y za 0  From (2) ..www. we get SH i. mx m 1 y n z p  ny n 1 x m z p  p x m y n z p 1 m n p i... S has a min imum value at (3.(5)    p x m y n z p 1 i. Find the max imum value of x m y n z p when x  y  z  a Solution    mx m 1 y n z p    ny n 1 x m z p ... 4) The min imum value of S  4(3) 2  6(3)(4)  108 22. (3) and (4) .(4) ..com MA2111/ Engineering Mathematics-I Substituting  48  x 6  ..e.rejinpaul. Substituting   2   2 in (7) and (8) we get x  3. y..rejinpaul.e.(2) TH .(1) YU . y   2 in (4) we get  6 2  12 2  0   3   8 i.e.. y  4. www.com .com MA2111/ Engineering Mathematics-I Hence max imum value of f occurs when am mn p an y mn p ap z mn p The max imum value of x mn nn p p m  n  p mn p PR A TH YU SH A f  a mn p 177 Get useful study materials from www.rejinpaul.rejinpaul. www. Evaluate: Solution: 4.com MA2111/ Engineering Mathematics-I UNIT – V Multiple Integrals Part – A 1.rejinpaul. Evaluate 2.rejinpaul. SH A Solution: Let I = Evaluate: 3. PR A TH YU Solution: Let I = Evaluate: Solution: Let I = 178 Get useful study materials from www.com . com . Evaluate: Solution: I= 7.rejinpaul. PR A TH YU SH A Solution: Let I = Evaluate: Solution: I= 179 Get useful study materials from www.rejinpaul. Evaluate: 6.www.com MA2111/ Engineering Mathematics-I 5. After changing the order. Change the order of integration PR A Solution: The given region of integration is bounded by y=0. Evaluate: Find the limits of integration in the double integral YU 9.com MA2111/ Engineering Mathematics-I 8.rejinpaul.com . y=a. I = 11. I= 180 Get useful study materials from www.www. x=y & x=a. Solution: The limits are: x varies from 0 to 1 and y varies from x2 to x. Change the order of integration for the double integral Ans: 1. Evaluate: Part – B over the area between y = x2 and y = x.rejinpaul. SH A Solution: I = TH Solution: The limits are: y varies from 0 to 1 and x varies from 0 to 1-y. 10. we have. www.rejinpaul.com MA2111/ Engineering Mathematics-I 2. Evaluate: over the cardioids r = a (1+cosθ). Solution: The limits of r: 0 to a (1+cosθ) and The limits of θ: 0 to π. I= = 3. YU SH A Put 1+cosθ = t then –sinθ dθ = dt When θ = 0, t = 2 When θ = π, t = 0. Evaluate Ans: PR A TH are the polar coordinates for the above integral 4. Evaluate: over the region in the first quadrant of the circle x2+y2=1. Solution: In the given region, y varies from 0 to and x varies from 0 to 1. I= dx Put x = sinθ. Then dx = cosθdθ.  varies from 0 to π/2. 181 Get useful study materials from www.rejinpaul.com www.rejinpaul.com MA2111/ Engineering Mathematics-I 5. and hence evaluate it. Change the order of integration in I = Solution: I = The given region of integration is bounded by x=0, x=1, y=x2 and x+y=2. In the given integration x is fixed and y is varying. So, after changing the order we have to keep y fixed and x should vary. A After changing the order we’ve two regions R1 & R2 SH I = I1+I2 TH YU I= 6. PR A = Evaluate by changing the order of integration. Solution: The given region is bounded by x=0, x=1, y=x and x2+y2=2. I= After changing the order we’ve, The region R is splinted into two regions R1& R2. In R1: limits of x: 0 to y & limits of y: 0 to 1 In R2: limits of x: 0 to & limits of y: 1 to I = I1+I2 182 Get useful study materials from www.rejinpaul.com www.rejinpaul.com MA2111/ Engineering Mathematics-I I1 = A I2 = Evaluate by changing the order of integration in TH 7. YU SH I= Solution: I = PR A The given region of integration is bounded by x=0, x=4, y = , y2 = 4x After changing the order we’ve Limits of x: y2/4 to 2√y Limits of y: 0 to 4 I= 8. = 16/3. Find the area enclosed by the ellipse Solution: Area of the ellipse = 4 x area of the first quadrant =4 183 Get useful study materials from www.rejinpaul.com I= 13.rejinpaul. r varies from r=2sinθ& r= 4sinθ and θ varies from0to π. Solution: Transforming Cartesian in Polar coordinates (i.e) x=rcosθ & y=rsinθ. Solution:In the region of integration. Find the area of the region outside the inner circle r=2cos and inside the outer circle r=4 cos by double integration.www. Solution: Required area = 10.com .com MA2111/ Engineering Mathematics-I 9. Find PR A = TH =2 over the area bounded between the circles r = 2sinθ & r= 4sinθ. Find the volume bounded by the cylinder x2+y2=4 and the planes y+z=4 and z=0.rejinpaul. 2 2 Find the smaller area bounded by y = 2-x and x +y =4. Solution: Required Area = =2 SH A 11. Solution: The limits are: Z varies from: 0 to 4-y 184 Get useful study materials from www. Find the area of the circle of radius ‘a’ by double integration. Then dxdy = rdrdθ limits of θ: 0 to and limits of r: o to YU Required Area = 2xupper area 12. Find the volume of the tetrahedron bounded by the plane SH A the coordinate plane. Solution: The limits are: X varies from 0 to a Y varies from 0 to b Required Volume = = TH =c = =c PR A =c = YU Z varies from o to c = = 15. Solution: Required Volume = 8 x volume in the positive octant = 8 Limits of integration are: Z varies from 0 to Y varies from 0 to X varies from 0 to a Volume = 8 =8 =8 185 Get useful study materials from www.com . Find the volume of the sphere x2+y2+z2=a2 using triple integral.www. Required volume = 2 =2 =2 dy= 2 =2 =8 since y = 16 is an odd function.rejinpaul. =16 = 16x2x = 16π and 14.com MA2111/ Engineering Mathematics-I X varies from: - to Y varies from: -2 to 2.rejinpaul. rejinpaul. Find the Volume of the ellipsoid Solution: Required Volume = 8 x Volume in the first octant Limits of Integration are: Z varies from 0 to c A Y varies from 0 to b Volume = 8 SH X varies from 0 to a.com MA2111/ Engineering Mathematics-I =8 =8 = 2π = 2π = = 2π cu.com .www. Evaluate   e 0   e 0 x y  z = =2πbc = dzdydx 0 0 log a x x  y Solution : PR A = TH = 8c x y  z log a x   [e dzdydx= 0 0 0  0 log a = ] log a x dydx= 0 log a = x y  z x y 0  0   (e 0 x  e 2( x y )    e x  y  dx =  2 0 2( x y )  e x  y )dydx 0 log a  0  1 4 x  e2x  2x  e  e   e x  dx      2   2 log a  e4x 3 2x   e4x 3 2x    e  ex   e  e x dx =  2  2 0  2 2  186 Get useful study materials from www. YU =8 where = =2πbc =2πbc log a x x  y 16.units 16.rejinpaul. drd SH 0 y = YU   / 4 a sec  x 2 dxdy TH a a A The limits are r=0 to r= asec and  = o to /4 = a3   a3 [log(sec  tan )  log(sec  tan ] = [log( 2  1)  0] 3 4 4 3 = a3 log( 2  1) 3 18. sec2  d = 3 3 =   0 r 3 cos2 drd r 0 cos2 d = a3 3  /4  sec 3  . (By changing) express  0 y 1 4 3 2 3 a  a a 8 4 8 x 2 dxdy x2  y2 Solution:The limits for x are x=y to x=a and the limits for y are y=0 to y=a change to polar co-ordinate we have x= rcos.rejinpaul. y = rsin dxdy = rdrd x = a = rcos  r  a  a sec  cos  x2  y2   0 r 2 cos2   r 2 sin 2  0  / 4 a sec   0 0 a3 = 3 2  /4  0  /4 a sec   r3     3 0 a3 1 0 sec  .www. Find the area inside the circle r=asin but lying outside the cardiod r=a(1-cos) Solution : Given curves are r=asin and r =a(1-cos) The curves intersect where a sin  = a (1-cos)  a sin  = a –a cos  a sin  + a cos = a  sin  + cos =1 187 Get useful study materials from www.r.com . cos2 d 0  /4 a3  /4 0 sec  d = 3 [log(sec  tan )]0 PR A =  r cos drd = 3  / 4 a sec  r 2 cos2  .com MA2111/ Engineering Mathematics-I 3 1  1 3  =  e 4 log a  e 2 log a  e log a      1 4 8  8 4  = a a 17.rejinpaul. rejinpaul.com .com  sin(   4 2 )    0(or )     0(or )  1 cos  1 2  4 2  sin    sin  cos    4   4      4  4  4  cos (or )   4 cos  1 2  4 2      4 2 2  2  /4 The required area =   /2 a sin    rdrd  = 0 a (1 cos ) a2 = 2  /2 a2 = 2  /2 0 2  2  /2  sin 2   (1  cos2   2 cos d 0 0  cos   cos  d 2 0 TH   / 2  1  cos2   = a 2 1    d  2   0   PR A  /2  1 sin 2   = a 1     2  0   2    /2  2 cos  2 cos  d  /2    /2 = a 2 sin  0   cos2 d  0    a2 . Find by double integration. the area enclosed by the curves Ans: Sub (1) in (2) we get 188 Get useful study materials from www.2 = 2 2 2 a2 = 2  /2  a2  2 2 sin   cos   1  2 cos d =   1  cos   cos   1  2 cos d  2  0   0  a sin   r2    d  2  a (1cos ) A 2 1 sin   SH 1 YU  MA2111/ Engineering Mathematics-I  1   = a 2 1    0   0    2 2 2    a (4   ) = a 2 1    4  4 17.www.rejinpaul. rejinpaul.4a) x Varies from 0 to 4a and y varies from The required Area = 2   y 2  z 2 dxdydz A  x a b c 19. Evaluate a  c 3b cb3  =    cbz2  dz 3 3  0 PR A a TH b  c 3 y cy 3    cyz2  dz =  3 3 0 0  a  c3  =    cy 2  cz 2 dydz 3  0 0 a b YU c  x3  Solution : I      y 2 x  z 2 x  dydz 3 0 0 0  a b SH 0 0 0  c 3bz cb3 z cbz3  c 3ba cb3 a cba3 abc 2   [c  b 2  a 2 ]   = =  = 3 3 3 0 3 3 3  3 1 2 x 20.www. y=x2 and x+y=2 In the given integration x is fixed and y is varying So. Change the order of integration in I    f ( x. after changing the order we have to keep y fixed and x should vary. y)dydx 0 x2 1 2 x Solution :Given I    f ( x. After changing the order we have two regions R1 & R2 189 Get useful study materials from www.com MA2111/ Engineering Mathematics-I Therefore the point of intersection of (1)&(2) is (0.0) and (4a. x=1. y)dydx 0 x2 The given region of integration is bounded by x=0.rejinpaul.com . Evaluate over annular region between the PR A TH circles Ans: Putting A 1 190 Get useful study materials from www.xy3 / 2   x 4 x 4  x7 / 2 5  x5        dx =      =    2 3   2 3   8 ( 7 )(3) 6  5  0  2  0 SH 9 3  21 16  28   = = 168   168 56 YU 1 2 1 =      (0)  8 21 6  21. By Transforming into polar coordinates .rejinpaul.com MA2111/ Engineering Mathematics-I I = I1 + I2 1 I  2 2 x y  f ( x.com . y )dxdy   0 0  f ( x.rejinpaul.Evaluate :    ( x 2 y  xy 2 )dydx x 0   x  I=    ( x 2 y  xy 2 )dydx =  0 x 1 Solution : Let  x 2  2 ( x y  xy ) dydx  = 0  x   1 x  x 2 y 2 xy 3  0  2  3  dx x 1 1  x4  x 3 x. y)dxdy 1 1 1 x  21.www. www.com through the positive spherical octant for which YU 22.rejinpaul.com .rejinpaul. Find the value of SH A MA2111/ Engineering Mathematics-I PR A TH Ans: 191 Get useful study materials from www.