Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.7 Exercises - Page 933: 24

Answer

$$\eqalign{ & 4x + 4y + z = 24 \cr & {\text{Line: }}\frac{{x - 2}}{4} = \frac{{y - 2}}{4} = \frac{{z - 8}}{1} \cr} $$

Work Step by Step

$$\eqalign{ & z = 16 - {x^2} - {y^2},{\text{ }}\left( {2,2,8} \right) \cr & {x^2} + {y^2} + z - 16 = 0 \cr & {\text{Consider}} \cr & F\left( {x,y,z} \right) = {x^2} + {y^2} + z - 16 \cr & {\text{Calculating the partial derivatives}} \cr & {F_x}\left( {x,y,z} \right) = 2x,{\text{ }}{F_y}\left( {x,y,z} \right) = 2y,{\text{ }}{F_z}\left( {x,y,z} \right) = 1,{\text{ }} \cr & {\text{At the point }}\left( {2,2,8} \right) \cr & {F_x}\left( {2,2,8} \right) = 4,{\text{ }}{F_y}\left( {2,2,8} \right) = 4,{\text{ }}{F_z}\left( {2,2,8} \right) = 1,{\text{ }} \cr & {\text{Direction numbers: 4}},4,1 \cr & {\text{An equation of the tangent plane at }}\left( {2,2,8} \right){\text{ is}} \cr & {F_x}\left( {2,2,8} \right)\left( {x - 2} \right) + {F_y}\left( {2,2,8} \right)\left( {y - 2} \right) + {F_z}\left( {2,2,8} \right)\left( {z - 8} \right) = 0 \cr & 4\left( {x - 2} \right) + 4\left( {y - 2} \right) + \left( {z - 8} \right) = 0 \cr & 4x - 8 + 4y - 8 + z - 8 = 0 \cr & 4x + 4y + z = 24 \cr & {\text{The corresponding symmetric equations are:}} \cr & {\text{Line: }}\frac{{x - 2}}{4} = \frac{{y - 2}}{4} = \frac{{z - 8}}{1} \cr} $$
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