Explanation Given: w = x + y 2 − z And, x = 2 r s , y = sin r t , z = s t 2 Concept Used: Chain rule for three independent parameters: Let w = f ( x , y , z ) is differentiable at ( x , y , z ) and the partial derivative of x = x ( s , u , t ) and y = y ( s , u , t ) exist at ( s , u , t ) then the function is differentiable with ∂ w ∂ s = ∂ w ∂ x ∂ x ∂ s + ∂ w ∂ y ∂ y ∂ s + ∂ w ∂ z ∂ z ∂ s , ∂ w ∂ u = ∂ w ∂ x ∂ x ∂ u + ∂ w ∂ y ∂ y ∂ u + ∂ w ∂ z ∂ z ∂ u , and ∂ w ∂ t = ∂ w ∂ x ∂ x ∂ t + ∂ w ∂ y ∂ y ∂ t + ∂ w ∂ z ∂ z ∂ t Calculation: To find ∂ w ∂ r , and ∂ w ∂ t Where, w = x + y 2 − z And, x = 2 r s , y = sin r t , z = s t 2 , apply chain rule with partial derivative, since x, y and z are function of more than one variable ( s, r and t ) The formula for ∂ w ∂ r is written by treating s, and t as constant, then the function depends upon r alone, as ∂ w ∂ r = ∂ w ∂ x ∂ x ∂ r + ∂ w ∂ y ∂ y ∂ r + ∂ w ∂ z ∂ z ∂ r This, can be solved as ∂ w ∂ r = ∂ w ∂ x ∂ x ∂ r + ∂ w ∂ y ∂ y ∂ r + ∂ w ∂ z ∂ z ∂ r ∂ w ∂ r = ∂ ( x + y 2 − z ) ∂ x ∂ ( 2 r s ) ∂ r + ∂ ( x + y 2 − z ) ∂ y ∂ ( sin r t ) ∂ r + ∂ ( x + y 2 − z ) ∂ z ∂ ( s t 2 ) ∂ r = 1 2 − z ∂ ( x + y ) ∂ x ( 2 s ) + 1 2 − z ∂ ( x + y ) ∂ y cos r t ∂ ( r t

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Chapter 11.5, Problem 21PS

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To find ∂w∂r , and ∂w∂t

Where, w=x+y2−z

And, x=2rs,y=sinrt,z=st2

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Chapter 11.5, Problem 20PS

Chapter 11.5, Problem 22PS

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To find ∂ w ∂ r , and ∂ w ∂ t Where, w = x + y 2 − z And, x = 2 r s , y = sin r t , z = s t 2

Solution Summary: The author explains how w=f(x,y,z) is differentiable, and the partial derivative of

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To find ∂w∂r , and ∂w∂t Where, w=x+y2−z And, x=2rs,y=sinrt,z=st2

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Chapter 11 Solutions

To find ∂w∂r , and ∂w∂t

Where, w=x+y2−z

And, x=2rs,y=sinrt,z=st2