a. Explanation Given: z = x 2 + y 2 Where, x = u sin v , y = u − 2 v Calculation: To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = x 2 + y 2 Where, x = u sin v , y = u − 2 v , first lets express z in terms of u and v , as z = x 2 + y 2 = ( u sin v ) 2 + ( u − 2 v ) 2 = u 2 sin 2 v + u 2 + 4 v 2 − 4 u v Now, to find ∂ z ∂ u differentiate above with respect to u treating v as constant, as ∂ z ∂ u = ∂ ∂ u ( u 2 sin 2 v + u 2 + 4 v 2 − 4 u v ) = sin 2 v ( 2 u ) + ( 2 u ) + 0 − 4 v b. To determine To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = x 2 + y 2 Where, x = u sin v , y = u − 2 v

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 11.5, Problem 9PS

To determine

To find the partial derivatives ∂z∂u,∂z∂v , where

z=x2+y2

Where, x=usinv,y=u−2v

To determine

To find the partial derivatives ∂z∂u,∂z∂v , where

z=x2+y2

Where, x=usinv,y=u−2v

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Chapter 11.5, Problem 8PS

Chapter 11.5, Problem 10PS

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Chapter 11 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

Ch. 11.1 - Prob. 1PS Ch. 11.1 - Prob. 2PS Ch. 11.1 - Prob. 3PS Ch. 11.1 - Prob. 4PS Ch. 11.1 - Prob. 5PS Ch. 11.1 - Prob. 6PS Ch. 11.1 - Prob. 7PS Ch. 11.1 - Prob. 8PS Ch. 11.1 - Prob. 9PS Ch. 11.1 - Prob. 10PS

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To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = x 2 + y 2 Where, x = u sin v , y = u − 2 v

Solution Summary: The author explains how to find the partial derivatives of partial z, v, and u.

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To find the partial derivatives ∂z∂u,∂z∂v , where z=x2+y2 Where, x=usinv,y=u−2v

To find the partial derivatives ∂z∂u,∂z∂v , where z=x2+y2 Where, x=usinv,y=u−2v

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Chapter 11 Solutions

To find the partial derivatives ∂z∂u,∂z∂v , where

z=x2+y2

Where, x=usinv,y=u−2v

To find the partial derivatives ∂z∂u,∂z∂v , where

z=x2+y2

Where, x=usinv,y=u−2v