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: 2



Work Step by Step

Given: $z=(x+2)^{2}-2(y-1)^{2}-5$, $(2,3,3)$ Consider $f(x,y)= (x+2)^{2}-2(y-1)^{2}-5$ $f_{x}(x,y)=2(x+2)$ $f_{y}(x,y)=-4(y-1)$ At $(2,3,3)$ $f_{x}(2,3)=8$ $f_{y}(2,3)=-8$ The equation of the tangent plane to the given surface at the specified point $(2,3,3)$ 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 $z-3=8(x-2)-8(y-3)$ $z=8x-16-8y+24+3$ Hence, $z=8x-8y+11$
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