Explanation Chain Rule (one independent variable): “Let z = f ( x , y ) be a differentiable function of x and y , where x and y are differentiable functions of t on an interval I . Then d z d t = ∂ z ∂ x ⋅ d x d t + ∂ z ∂ y ⋅ d y d t ”. Calculation: The given function is, Q = x 2 + y 2 + z 2 (1) Take partial derivative with respect to x of the equation (1), ∂ Q ∂ x = ∂ ∂ x ( x 2 + y 2 + z 2 ) = 1 2 x 2 + y 2 + z 2 ( 2 x ) = x x 2 + y 2 + z 2 Thus, the partial derivate, ∂ Q ∂ x = x x 2 + y 2 + z 2 . Take partial derivative with respect to y of the equation (1), ∂ Q ∂ y = ∂ ∂ y ( x 2 + y 2 + z 2 ) = 1 2 x 2 + y 2 + z 2 ( 2 y ) = y x 2 + y 2 + z 2 Thus, the partial derivate, ∂ Q ∂ y = y x 2 + y 2 + z 2 . Take partial derivative with respect to z of the equation (1), ∂ Q ∂ z = ∂ ∂ z ( x 2 + y 2 + z 2 ) = z x 2 + y 2 + z 2 = z x 2 + y 2 + z 2 Thus, the partial derivate, ∂ Q ∂ z = z x 2 + y 2 + z 2 . Obtain the derivative of x with respect to t . x = sin t d x d t = ( cos t ) ( 1 ) = cos t Thus, the derivate, d x d t = cos t . Obtain the derivative of y with respect to t . y = cos t d y d t = ( − sin t ) ( 1 ) = − sin t Thus, the derivate, d y d t = − sin t

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Author: Briggs

Publisher: PEARSON

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Chapter 12.5, Problem 14E

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To find: The value of dQdt by using Chain Rule if Q=x2+y2+z2 , x=sint , y=cost and z=cost .

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Chapter 12.5, Problem 13E

Chapter 12.5, Problem 15E

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The value of d Q d t by using Chain Rule if Q = x 2 + y 2 + z 2 , x = sin t , y = cos t and z = cos t .

Solution Summary: The author explains how to find the value of qdt by using Chain Rule.

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To find: The value of dQdt by using Chain Rule if Q=x2+y2+z2 , x=sint , y=cost and z=cost .

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Chapter 12 Solutions