Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises: 38

Answer

$$\frac{\partial}{\partial x}\phi(x,y,z,t)=\frac{\alpha}{\gamma z+\delta t^2}.$$ $$\frac{\partial}{\partial y}\phi(x,y,z,t)=\frac{2\beta y}{\gamma z+\delta t^2}.$$ $$\frac{\partial}{\partial z}\phi(x,y,z,t)=-\frac{\gamma(\alpha x+\beta y^2)}{(\gamma z+\delta t^2)^2}.$$ $$\frac{\partial}{\partial t}\phi(x,y,z,t)=-\frac{2\delta t(\alpha x+\beta y^2)}{(\gamma z+\delta t^2)^2}.$$

Work Step by Step

Regard $y,z$ and $t$ as constant and differentiate with respect to $x$: $$\frac{\partial}{\partial x}\phi(x,y,z,t)=\frac{\partial}{\partial x}\left(\frac{\alpha x+\beta y^2}{\gamma z +\delta t^2}\right)=\frac{1}{\gamma z+\delta t^2}\frac{\partial}{\partial x}(\alpha x+\beta y^2)=\frac{\alpha}{\gamma z+\delta t^2}.$$ Regartd $x, z$ and $t$ as constant and differentiate with respect to $y$: $$\frac{\partial}{\partial y}\phi(x,y,z,t)=\frac{\partial}{\partial y}\left(\frac{\alpha x+\beta y^2}{\gamma z +\delta t^2}\right)=\frac{1}{\gamma z+\delta t^2}\frac{\partial}{\partial y}(\alpha x+\beta y^2)=\frac{1}{\gamma z+\delta t^2}\beta\cdot2y=\frac{2\beta y}{\gamma z+\delta t^2}.$$ Regard $x, y$ and $t$ as constant and differentiate with respect to $z$: $$\frac{\partial}{\partial z}\phi(x,y,z,t)=\frac{\partial}{\partial z}\left(\frac{\alpha x+\beta y^2}{\gamma z +\delta t^2}\right)=(\alpha x+\beta y^2)\frac{\partial}{\partial z}\left(\frac{1}{\gamma z+\delta t^2}\right)=(\alpha x+\beta y^2)\frac{-1}{(\gamma z+\delta t^2)^2}\frac{\partial}{\partial z}(\gamma z+\delta t^2)=-\frac{\gamma(\alpha x+\beta y^2)}{(\gamma z+\delta t^2)^2}.$$ Regard $x, y$ and $z$ as constant and differentiate with respect to $t$: $$\frac{\partial}{\partial t}\phi(x,y,z,t)=\frac{\partial}{\partial t}\left(\frac{\alpha x+\beta y^2}{\gamma z +\delta t^2}\right)=(\alpha x+\beta y^2)\frac{\partial}{\partial t}\left(\frac{1}{\gamma z+\delta t^2}\right)=(\alpha x+\beta y^2)\frac{-1}{(\gamma z+\delta t^2)^2}\frac{\partial}{\partial t}(\gamma z+\delta t^2)=-\frac{2\delta t(\alpha x+\beta y^2)}{(\gamma z+\delta t^2)^2}.$$
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