Explanation Given: x = v u 2 + v 2 , y = u u 2 + v 2 Concept used: Jacobian = ∂ ( x , y ) ∂ ( u , v ) = ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v Calculation: x = v u 2 + v 2 = v ( u 2 + v 2 ) − 1 ⇒ ∂ x ∂ u = v ( − 1 ) ( u 2 + v 2 ) − 2 × ( 2 u ) [ Treating v as constant ] ⇒ ∂ x ∂ u = − 2 u v ( u 2 + v 2 ) − 2 ⇒ ∂ x ∂ u = − 2 u v ( u 2 + v 2 ) 2 ⇒ ∂ x ∂ v = ( u 2 + v 2 ) − 1 + v ( − 1 ) ( u 2 + v 2 ) − 2 × ( 2 v ) [ Treating u as constant and applying product rule ] ⇒ ∂ x ∂ v = ( u 2 + v 2 ) − 1 − 2 v 2 ( u 2 + v 2 ) − 2 ⇒ ∂ x ∂ v = 1 ( u 2 + v 2 ) − 2 v 2 ( u 2 + v 2 ) 2 ⇒ ∂ x ∂ v = ( u 2 + v 2 ) − 2 v 2 ( u 2 + v 2 ) 2 ⇒ ∂ x ∂ v = u 2 − v 2 ( u 2 + v 2 ) 2 y = u u 2 + v 2 = u ( u 2 + v 2 ) − 1 ⇒ ∂ y ∂ u = ( u 2 + v 2 ) − 1 + u ( − 1 ) ( u 2 + v 2 ) − 2 × ( 2 u ) [ Treating v as constant and applying product rule ] ⇒ ∂ y ∂ u = ( u 2 + v 2 ) − 1 − 2 u 2 ( u 2 + v 2 ) − 2 ⇒ ∂ y ∂ u = 1 ( u 2 + v 2 ) − 2 u 2 ( u 2 + v 2 ) 2 ⇒ ∂ y ∂ u = ( u 2 + v 2 ) − 2 u 2 ( u 2 + v 2 ) 2 ⇒ ∂ y ∂ u = v 2 − u 2 ( u 2 + v 2 ) 2 ⇒ ∂ y ∂ v = u ( − 1 ) ( u 2 + v 2 ) − 2 × ( 2 v ) [ Treating u as constant ] ⇒ ∂ y ∂ v = − 2 u v ( u 2 + v 2 ) − 2 ⇒ ∂ y

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 12.8, Problem 10PS

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To find the Jacobian when x=vu2+v2,y=uu2+v2

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Chapter 12.8, Problem 9PS

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

Ch. 12.1 - Prob. 1PS Ch. 12.1 - Prob. 2PS Ch. 12.1 - Prob. 3PS Ch. 12.1 - Prob. 4PS Ch. 12.1 - Prob. 5PS Ch. 12.1 - Prob. 6PS Ch. 12.1 - Prob. 7PS Ch. 12.1 - Prob. 8PS Ch. 12.1 - Prob. 9PS Ch. 12.1 - Prob. 10PS

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To find the Jacobian when x = v u 2 + v 2 , y = u u 2 + v 2

Solution Summary: The author explains how to find the Jacobian when x=vu2+.

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To find the Jacobian when x=vu2+v2,y=uu2+v2

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Chapter 12 Solutions