Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 12 - Functions of Several Veriables - 12.7 Tangent Planes and Linear Approximation - 12.7 Exercises - Page 936: 18

Answer

$$z = - 2x + y + 1{\text{ and }}z = 12x - 2y - 18$$

Work Step by Step

$$\eqalign{ & z = 2 + 2{x^2} + \frac{{{y^2}}}{2};\,\,\,\left( { - \frac{1}{2},1,3} \right){\text{ and }}\left( {3, - 2,22} \right) \cr & f\left( {x,y} \right) = 2 + 2{x^2} + \frac{{{y^2}}}{2} \cr & {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = 4x \cr & {f_y}\left( {x,y} \right) = y \cr & {\text{Evaluate at the point }}\left( { - \frac{1}{2},1,3} \right){\text{ and }}\left( {3, - 2,22} \right) \cr & {\text{at }}{f_x}\left( {x,y} \right),\,{f_y}\left( {x,y} \right) \cr & {f_x}\left( { - \frac{1}{2},1} \right) = - 2 \cr & {f_y}\left( { - \frac{1}{2},1} \right) = 1 \cr & and \cr & {f_x}\left( {3, - 2} \right) = 12 \cr & {f_y}\left( {3, - 2} \right) = - 2 \cr & \cr & {\text{An equation of the plane tangent to the surface is}} \cr & z = f\left( {x,y} \right){\text{ at the point }}\left( {a,b,f\left( {a,b} \right)} \right){\text{ is}} \cr & {\text{For the point }}\left( { - \frac{1}{2},1,3} \right) \cr & z = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_x}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr & z = - 2\left( {x + \frac{1}{2}} \right) + \left( {y - 1} \right) + 3 \cr & z = - 2x - 1 + y - 1 + 3 \cr & z = - 2x + y + 1 \cr & {\text{For the point }}\left( {3, - 2,22} \right) \cr & z = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_x}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr & z = 12\left( {x - 3} \right) - 2\left( {y + 2} \right) + 22 \cr & z = 12x - 36 - 2y - 4 + 22 \cr & z = 12x - 2y - 18 \cr} $$
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