Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises - Page 974: 4



Work Step by Step

$z=\frac{x}{y^{2}}$, $(-4,2-1)$ Consider $f(x,y)= \frac{x}{y^{2}}$ $f_{x}(x,y)=\frac{1}{y^{2}}$ $f_{y}(x,y)=-\frac{2x}{y^{3}}$ At $(-4,2-1)$ $f_{x}(-4,2)=\frac{1}{2^{2}}=\frac{1}{4}$ $f_{y}(-4,2)=-\frac{2\times (-4)}{2^{3}}=1$ The equation of the tangent plane to the given surface at the specified point $(-4,2-1)$ is given by $z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$ On substituting the values, we get $z+1=\frac{1}{4}(x+4)+1(y-2)$ $z=\frac{x}{4}+1+y-2-1$ Hence, $\frac{x}{4}+y-z=2$
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