Explanation Given: Three implicit functions are given as z = r + s 2 , x + y = s 3 + r 3 − 3 and x y = s 2 − r 2 Formula Used: To find ( ∂ x ∂ z ) s , ( ∂ x ∂ z ) r and ( ∂ x ∂ z ) y the following steps should be performed: Find d z , d x , d y , d s , and d r by taking the differential of the equations and solve for d x , d s and d r using cross multiplication method. By chain rule d x = ∂ x ∂ z d z + ∂ x ∂ y d y For ( ∂ x ∂ z ) s put d s = 0 , for ( ∂ x ∂ z ) r put d r = 0 in the obtaibned differentials. Calculation: Consider the given equations. z = r + s 2 ............ (1) x + y = s 3 + r 3 − 3 ............ (2) x y = s 2 − r 2 ............ (3) Differentiate (1), (2), and (3) as shown. d z = d r + 2 s d s ...... ( 4 ) d x + d y = 3 s 2 d s + 3 r 2 d r ...... ( 5 ) x d y + y d x = 2 s d s − 2 r d r ...... ( 6 ) Solve equation (5) and (6) for d x by cross multiplication as shown. d x = | 3 s 2 d s + 3 r 2 d r 1 2 s d s − 2 r d r x | | 1 1 y x | d x = x ( 3 s 2 d s + 3 r 2 d r ) − ( 2 s d s − 2 r d r ) x − y d x = ( 3 s 2 x − 2 s ) d s + ( 3 r 2 x + 2 r ) d r x − y d x = ( 3 s 2 x − 2 s x − y ) d s + ( 3 r 2 x + 2 r x − y ) d r ............ (7) Now, put d s = 0 for ( ∂ x ∂ z ) s in equation (4) and (7) and divide the result as shown. d x s d z s = ( 3 r 2 x + 2 r x − y ) d r d r = 3 r 2 x + 2 r x − y Hence, ( ∂ x ∂ z ) s = 3 r 2 x + 2 r x − y Substitute − 1 for r, 2 for s, 3 for x and z, and 1 for t in above result. ( ∂ x ∂ z ) s | ( − 1 , 2 , 3 , 1 , 3 ) = ( 3 r 2 x + 2 r x − y ) = [ 3 ( − 1 ) 2 ( 3 ) + 2 ( − 1 ) 3 − 1 ] = 7 2 Now, put d r = 0 for ( ∂ x ∂ z ) r in equation (4) and (7) and divide the result as shown. d x r d z r = ( 3 s 2 x − 2 s 2 s ( x − y ) ) d s d s = 3 s 2 x − 2 s 2 s ( x − y ) Hence, ( ∂ x ∂ z ) r = 3 s 2 x − 2 s 2 s ( x − y ) Substitute − 1 for r, 2 for s, 3 for x and z, and 1 for t in above result. ( ∂ x ∂ z ) r | ( − 1 , 2 , 3 , 1 , 3 ) = ( 3 s 2 x − 2 s 2 s ( x − y ) ) = [ 3 ( 2 ) 2 ( 3 ) − 2 ( 2 ) 2 ⋅ 2 ( 3 − 1 ) ] = 32 8 = 4 Solve equation (5) and (6) for d s by cross multiplication as shown

Mathematical Methods in the Physical Sciences

3rd Edition

ISBN: 9780471198260

Author: Mary L. Boas

Publisher: Wiley, John & Sons, Incorporated

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Question

Chapter 4.7, Problem 19P

To determine

To find:The value of ( ∂x ∂z)s , ( ∂x ∂z)r and ( ∂x ∂z)y at (r,s,x,y,z)=(−1,2,3,1,3) for the given functions.

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Chapter 4.7, Problem 18P

Chapter 4.7, Problem 20P

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Chapter 4 Solutions

Mathematical Methods in the Physical Sciences

Ch. 4.1 - If u=x2/x2+y2, find u/x,u/y.Ch. 4.1 - Prob. 2P Ch. 4.1 - If z=lnu2+v2+w2, find z/u,z/v,z/w.Ch. 4.1 - For w=x3y32xy+6, find 2w/ax2 and 2w/ay2 at the...Ch. 4.1 - For w=8x4+y42xy2, find 2w/x2 and 2w/y2 at the...Ch. 4.1 - For u=excosy, (a) verify that 2u/xy=2u/yx; (b)...Ch. 4.1 - Prob. 7P Ch. 4.1 - If z=x2+2y2,x=rcos,y=rsin, find the following...Ch. 4.1 - If z=x2+2y2,x=rcos,y=rsin, find the following...Ch. 4.1 - If z=x2+2y2,x=rcos,y=rsin, find the following...

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The value of ( ∂ x ∂ z ) s , ( ∂ x ∂ z ) r and ( ∂ x ∂ z ) y at ( r , s , x , y , z ) = ( − 1 , 2 , 3 , 1 , 3 ) for the given functions.

Solution Summary: The author explains the formula used to find the value of (partial z)_s, x+y=S3+r

To find:The value of ( ∂x ∂z)s , ( ∂x ∂z)r and ( ∂x ∂z)y at (r,s,x,y,z)=(−1,2,3,1,3) for the given functions.

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Chapter 4 Solutions