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: 1

Answer

$z=4x-y-6$ or $4x-y-z=6$

Work Step by Step

$z=2x^{2}+y^{2}-5y$, $(1,2,-4)$ Consider $f(x,y)= 2x^{2}+y^{2}-5y$ $f_{x}(x,y)=4x$ $f_{y}(x,y)=2y-5$ At $(1,2,-4)$ $f_{x}(-4,2)=4\times1=4$ $f_{y}(-4,2)=2\times 2 -5=-1$ The equation of the tangent plane to the given surface at the specified point $(1,2,-4)$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 $z+4=4(x-1)+(-1)(y-2)$ $z=4x-4-y+2-4$ Hence, $z=4x-y-6$ or $4x-y-z=6$
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