Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.5 Exercises - Page 913: 18

Answer

$$\frac{\partial w}{\partial s}=2st^4,\frac{\partial w}{\partial t}=2t+4s^2t^3$$

Work Step by Step

(a) We will use the Chain Rule. The partial derivative with respect to $s$ is: $$\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial s}=\frac{\partial }{\partial x}(x^2+y^2+z^2)\frac{\partial }{\partial s}(t\sin s)+\frac{\partial }{\partial y}(x^2+y^2+z^2)\frac{\partial }{\partial s}(t\cos s)+\frac{\partial }{\partial z}(x^2+y^2+z^2)\frac{\partial }{\partial s}(st^2)=2x\cdot t\cos s+2y\cdot(-t\sin s)+2z\cdot t^2=2xt\cos s-2yt\sin s+2zt^2$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial w}{\partial s}=2xt\cos s-2yt\sin s+2zt^2=2t\sin s\cdot t\cos s-2t\cos s\cdot t\sin s+2st^2\cdot t^2= 2t^2\sin s\cos s-2t^2\sin s\cos s+2st^4=2st^4$$ The partial derivative with respect to $t$ is: $$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}= \frac{\partial }{\partial x}(x^2+y^2+z^2)\frac{\partial }{\partial t}(t\sin s)+\frac{\partial }{\partial y}(x^2+y^2+z^2)\frac{\partial }{\partial t}(t\cos s)+\frac{\partial }{\partial z}(x^2+y^2+z^2)\frac{\partial }{\partial t}(st^2)= 2x\sin s+2y\cos s+2z\cdot2st=2x\sin s+2y\cos s+4zst$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial w}{\partial t}=2x\sin s+2y\cos s+4zst=2t\sin s\cdot\sin s+2t\cos s\cdot\cos s+4st^2\cdot st= 2t(\sin^2s+\cos^2s)+4s^2t^3=2t+4s^2t^3$$ (b) We will first convert $w$ to a function of $s$ and $t$ and then we will differentiate. $$w=x^2+y^2+z^2=t^2\sin^2s+t^2\cos^2s+s^2t^4=t^2+s^2t^4$$ The partial derivative with respect to $s$ is: $$\frac{\partial w}{\partial s}=\frac{\partial }{\partial s}(t^2+s^2t^4)=2st^4$$ The partial derivative with respect to $t$ is: $$\frac{\partial w}{\partial t}=\frac{\partial }{\partial t}(t^2+s^2t^4)=2t+4s^2t^3$$
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