Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises: 2

Answer

$$\frac{dz}{dt}=\frac{6\pi}{(e^{\pi t}+2e^{-\pi t})^2}$$

Work Step by Step

$$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}=\frac{\partial}{\partial x}\Big(\frac{x-y}{x+2y}\Big)\frac{d}{dt}(e^{\pi t})+\frac{\partial }{\partial y}\Big(\frac{x-y}{x+2y}\Big)\frac{d}{dt}(e^{-\pi t})= \frac{\frac{\partial}{\partial x}(x-y)\cdot(x+2y)-(x-y)\frac{\partial}{\partial x}(x+2y)}{(x+2y)^2}\cdot\pi e^{\pi t}+ \frac{\frac{\partial}{\partial y}(x-y)\cdot(x+2y)-(x-y)\frac{\partial}{\partial y}(x+2y)}{(x+2y)^2}\cdot(-\pi e^{-\pi t})= \frac{1\cdot(x+2y)-(x-y)\cdot1}{(x+2y)^2}\cdot\pi e^{\pi t}-\frac{-1\cdot(x+2y)-(x-y)\cdot2}{(x+2y)^2}\cdot\pi e^{-\pi t}= \frac{3y}{(x+2y)^2}\cdot\pi e^{\pi t}-\frac{-3x}{(x+2y)^2}\cdot\pi e^{-\pi t}= 3\pi\frac{ye^{\pi t}+xe^{-\pi t}}{(x+2y)^2}$$ Expressing this in terms of $t$ we have: $$\frac{dz}{dt}=3\pi\frac{ye^{\pi t}+xe^{-\pi t}}{(x+2y)^2}= 3\pi\frac{e^{-\pi t}\cdot e^{\pi t}+e^{\pi t}\cdot e^{-\pi t}}{(e^{\pi t}+2e^{-\pi t})^2}=\frac{6\pi}{(e^{\pi t}+2e^{-\pi t})^2}$$
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