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 935: 12

Answer

$$\eqalign{ & 3x + 5y - 5z - 5 = 0 \cr & - 4x - 3y - 5z = 0 \cr} $$

Work Step by Step

$$\eqalign{ & {x^2} + {y^2} - {z^2} = 0;\,\,\,\left( {3,4,5} \right){\text{ and }}\left( { - 4, - 3,5} \right) \cr & {\text{Let }}F\left( {x,y,z} \right) = {x^2} + {y^2} - {z^2} \cr & {\text{Calculate the partial derivatives }}{F_x}\left( {x,y,z} \right){\text{, }}{F_y}\left( {x,y,z} \right){\text{ }} \cr & {\text{and }}{F_z}\left( {x,y,z} \right){\text{ }} \cr & {F_x}\left( {x,y,z} \right) = 2x \cr & {F_y}\left( {x,y,z} \right) = 2y \cr & {F_z}\left( {x,y,z} \right) = - 2z \cr & {\text{Evaluate at the point }}\left( {3,4,5} \right){\text{ and }}\left( { - 4, - 3,5} \right) \cr & {\text{at }}{F_x}\left( {x,y,z} \right),\,{F_y}\left( {x,y,z} \right){\text{ and }}{F_z}\left( {x,y,z} \right) \cr & {F_x}\left( {3,4,5} \right) = 6 \cr & {F_y}\left( {3,4,5} \right) = 8 \cr & {F_z}\left( {3,4,5} \right) = - 10 \cr & and \cr & {F_x}\left( { - 4, - 3,5} \right) = - 8 \cr & {F_y}\left( { - 4, - 3,5} \right) = - 6 \cr & {F_z}\left( { - 4, - 3,5} \right) = - 10 \cr & \cr & {\text{An equation of the plane tangent to the surface is}} \cr & {F_x}\left( {a,b,c} \right)\left( {x - a} \right) + {F_y}\left( {a,b,c} \right)\left( {y - b} \right) + {F_z}\left( {a,b,c} \right)\left( {z - c} \right) = 0 \cr & {\text{For the point }}\left( {3,4,5} \right) \cr & 6\left( {x - 3} \right) + 8\left( {y - 4} \right) - 10\left( {z - 4} \right) = 0 \cr & 6x - 18 + 8y - 32 - 10z + 40 = 0 \cr & 6x + 8y - 10z - 10 = 0 \cr & 3x + 5y - 5z - 5 = 0 \cr & {\text{For the point }}\left( { - 4, - 3,5} \right) \cr & - 8\left( {x + 4} \right) - 6\left( {y + 3} \right) - 10\left( {z - 5} \right) = 0 \cr & - 8x - 32 - 6y - 18 - 10z + 50 = 0 \cr & - 8x - 6y - 10z = 0 \cr & - 4x - 3y - 5z = 0 \cr} $$
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