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: 14

Answer

$\partial w/\partial s=-6e^{2s}e^t, \partial w/\partial t=3e^t(e^{2t}-e^{2s})$ For $s=-1$ and $t=2$: $\partial w/\partial s=-6$ and $\partial w/\partial t=3(e^6-1)$

Work Step by Step

We will find both partial derivatives using the Chain Rule. The partial derivative with respect to $s$ is: $$\frac{\partial w}{\partial s}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial}{\partial x}(y^3-3x^2y)\frac{\partial}{\partial s}(e^s)+\frac{\partial}{\partial y}(y^3-3x^2y)\frac{\partial}{\partial s}(e^t)= -3\cdot2xy\cdot e^s+(3y^2-3x^2)\cdot0=-6xye^s$$ Expressing this in terms of $s$ and $t$ we have: $$\frac{\partial w}{\partial s}=-6xye^s=-6e^{2s}e^t$$ The partial derivative with respect to $t$ is: $$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial}{\partial x}(y^3-3x^2y)\frac{\partial}{\partial t}(e^s)+\frac{\partial}{\partial y}(y^3-3x^2y)\frac{\partial}{\partial t}(e^t)= -3\cdot2xy\cdot0+(3y^2-3x^2)\cdot e^t=3e^t(y^2-x^2)$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial w}{\partial t}=3e^t(y^2-x^2)=3e^t(e^{2t}-e^{2s})$$ Now we will evaluate these partial derivatives at the given values $s=-1$ and $t=2$: $$\frac{\partial w}{\partial s}=-6e^{2s}e^t=-6e^{-1\cdot2}e^2=-6$$ $$\frac{\partial w}{\partial t}=3e^t(e^{2t}-e^{2s})=3e^2(e^{2\cdot2}-e^{2\cdot(-1)})=3e^2(e^4-e^{-2})=3e^6-3\cdot1=3(e^6-1)$$
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