Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.4 - Tangent Planes and Linear Approximation - 14.4 Exercise - Page 934: 1

Answer

$z = 4x-y-6$

Work Step by Step

Question: $z = 2x^{2} + x^{2} -5y$ $(1,2,-4)$ The equation of the tangent plane to the surface is: $z - z_{0} = f_{x}(x_{0},y_{0})(x - x_{0}) + f_{y}(x_{0},y_{0})(y - y_{0})$ $f_{x} = 4x$ so $f_{x}(1,2) = 4$ $f_{y} = 2y - 5$ so $f_{y}(1,2) = -1$ So the equation is: $z + 4 = 4*(x-1) - 1(y-2)$ And the final answer is: $z = 4x-y-6$
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