Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises: 6

Answer

$$\frac{dw}{dt}=\tan t$$

Work Step by Step

$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt}= \frac{\partial}{\partial x}\big(\ln{\sqrt{x^2+y^2+z^2}}\big)\frac{d}{dt}(\sin t)+\frac{\partial}{\partial y}\big(\ln{\sqrt{x^2+y^2+z^2}}\big)\frac{d}{dt}(\cos t)+\frac{\partial}{\partial z}\big(\ln{\sqrt{x^2+y^2+z^2}}\big)\frac{d}{dt}(\tan t)= \frac{1}{\sqrt{x^2+y^2+z^2}}\frac{\partial}{\partial x}\big(\sqrt{x^2+y^2+z^2}\big)\cdot \cos t+ \frac{1}{\sqrt{x^2+y^2+z^2}}\frac{\partial}{\partial y}\big(\sqrt{x^2+y^2+z^2}\big)(-\sin t)+\frac{1}{\sqrt{x^2+y^2+z^2}}\frac{\partial}{\partial z}\big(\sqrt{x^2+y^2+z^2}\big)\frac{1}{\cos ^2t}= \frac{1}{\sqrt{x^2+y^2+z^2}}\frac{1}{2\sqrt{x^2+y^2+z^2}}\frac{\partial}{\partial x}(x^2+y^2+z^2)\cdot \cos t-\frac{1}{\sqrt{x^2+y^2+z^2}}\frac{1}{2\sqrt{x^2+y^2+z^2}}\frac{\partial}{\partial x}(x^2+y^2+z^2)\cdot \sin t+ \frac{1}{\sqrt{x^2+y^2+z^2}}\frac{1}{2\sqrt{x^2+y^2+z^2}}\frac{\partial}{\partial x}(x^2+y^2+z^2)\sec^2t= \frac{2x\cos t}{2(x^2+y^2+z^2)}-\frac{2y\sin t}{2(x^2+y^2+z^2)}+\frac{2z\sec^2t}{2(x^2+y^2+z^2)}= \frac{x\cos t-y\sin t+z\sec^2t}{x^2+y^2+z^2}$$ Now we have to express solution in terms of $t$: $$\frac{dw}{dt}=\frac{x\cos t-y\sin t+z\sec^2t}{x^2+y^2+z^2}= \frac{\sin t\cos t-\cos t\sin t+\sec^2t\tan t}{\cos^2t+\sin^2t+\tan^2t}=\frac{\sec^2t\tan t}{1+\tan^2t}=\frac{\tan t}{\cos^2t+\sin^2t}=\tan t$$
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