Explanation Given: The speed of fluid at ( x , y ) is s ( x , y ) = u ( x , y ) 2 + v ( x , y ) 2 , where u ( x , y ) = x ( 1 − x ) ( 1 − 2 y ) and v ( x , y ) = y ( y − 1 ) ( 1 − 2 x ) ; 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 . Chain Rule(Two Independent Variables): “Let z = f ( x , y ) is a differentiable function of x and y , where x and y are differentiable functions of s and t . Then ∂ z ∂ s = ∂ z ∂ x ⋅ ∂ x ∂ s + ∂ z ∂ y ⋅ ∂ y ∂ s and ∂ z ∂ t = ∂ z ∂ x ⋅ ∂ x ∂ t + ∂ z ∂ y ⋅ ∂ y ∂ t ”. Calculation: The speed of fluid at ( x , y ) is s ( x , y ) = u ( x , y ) 2 + v ( x , y ) 2 (1) Take the partial derivative with respect to u of the equation (1), ∂ s ∂ u = ∂ ∂ u ( u 2 + v 2 ) = 1 2 u 2 + v 2 ( 2 u ) = u u 2 + v 2 Thus, the partial derivate, ∂ s ∂ u = u u 2 + v 2 . Take the partial derivative with respect to v of the equation (1), ∂ s ∂ v = ∂ ∂ v ( u 2 + v 2 ) = 1 2 u 2 + v 2 ( 2 v ) = v u 2 + v 2 Thus, the partial derivate, ∂ s ∂ v = v u 2 + v 2 . Obtain the derivative of u with respect to x . u = x ( 1 − x ) ( 1 − 2 y ) ∂ u ∂ x = ( 1 − 2 y ) [ x ( 0 − 1 ) + ( 1 − x ) 1 ] = ( 1 − 2 y ) [ − x + 1 − x ] = ( 1 − 2 x ) ( 1 − 2 y ) Thus, the derivate, ∂ u ∂ x = ( 1 − 2 x ) ( 1 − 2 y ) . Obtain the derivative of v with respect to x . v = y ( y − 1 ) ( 1 − 2 x ) ∂ v ∂ x = y ( y − 1 ) [ 0 − 2 ] = − 2 y ( y − 1 ) Thus, the derivate, ∂ v ∂ x = − 2 y ( y − 1 ) . Substitute the respective values in Chain Rule, ∂ s ∂ x = ∂ s ∂ u ⋅ ∂ u ∂ x + ∂ s ∂ v ⋅ ∂ v ∂ x = ( u u 2 + v 2 ) [ ( 1 − 2 x ) ( 1 − 2 y ) ] + ( v u 2 + v 2 ) [ − 2 y ( y − 1 ) ] Thus, the derivative, ∂ s ∂ x is ( u u 2 + v 2 ) [ ( 1 − 2 x ) ( 1 − 2 y ) ] + ( v u 2 + v 2 ) [ − 2 y ( y − 1 ) ] . Substitute u ( x , y ) = x ( 1 − x ) ( 1 − 2 y ) and v ( x , y ) = y ( y − 1 ) ( 1 − 2 x ) in ∂ s ∂ x and obtain the required value. ∂ s ∂ x = ( u u 2 + v 2 ) [ ( 1 − 2 x ) ( 1 − 2 y ) ] + ( v u 2 + v 2 ) [ − 2 y ( y − 1 ) ] = { x ( 1 − x ) ( 1 − 2 y ) [ ( 1 − 2 x ) ( 1 − 2 y ) ] [ x ( 1 − x ) ( 1 − 2 y ) ] 2 + [ y ( y − 1 ) ( 1 − 2 x ) ] 2 + [ y ( y − 1 ) ( 1 − 2 x ) ] [ − 2 y ( y − 1 ) ] [ x ( 1 − x ) ( 1 − 2 y ) ] 2 + [ y ( y − 1 ) ( 1 − 2 x ) ] 2 } = { x ( 1 − x ) ( 1 − 2 y ) [ ( 1 − 2 x ) ( 1 − 2 y ) ] + [ y ( y − 1 ) ( 1 − 2 x ) ] [ − 2 y ( y − 1 ) ] [ x 2 ( 1 − x ) 2 ( 1 − 2 y ) 2 ] + [ y 2 ( y − 1 ) 2 ( 1 − 2 x ) 2 ] } = { x ( 1 − x ) ( 1 − 2 x ) ( 1 − 2 y ) 2 − 2 y 2 ( 1 − 2 x ) ( y − 1 ) 2 [ x 2 ( 1 − x ) 2 ( 1 − 2 y ) 2 ] + [ y 2 ( y − 1 ) 2 ( 1 − 2 x ) 2 ] } Therefore, the value of ∂ s ∂ x is { x ( 1 − x ) ( 1 − 2 x ) ( 1 − 2 y ) 2 − 2 y 2 ( 1 − 2 x ) ( y − 1 ) 2 [ x 2 ( 1 − x ) 2 ( 1 − 2 y ) 2 ] + [ y 2 ( y − 1 ) 2 ( 1 − 2 x ) 2 ] }

3rd Edition

ISBN: 9780135182543

Author: Briggs

Publisher: PEARSON

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Question

Chapter 15.4, Problem 42E

To determine

To find: The values of ∂s∂x and ∂s∂y using Chain Rule.

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Chapter 15.4, Problem 41E

Chapter 15.4, Problem 43E

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