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: 41

Answer

$$\left( {0,3,12} \right)$$

Work Step by Step

$$\eqalign{ & z = 3 - {x^2} - {y^2} + 6y \cr & z - 3 + {x^2} + {y^2} - 6y = 0 \cr & {\text{Consider}} \cr & F\left( {x,y,z} \right) = z - 3 + {x^2} + {y^2} - 6y \cr & {\text{Calculating the partial derivatives}} \cr & {F_x}\left( {x,y,z} \right) = 2x \cr & {F_y}\left( {x,y,z} \right) = 2y - 6 \cr & {F_z}\left( {x,y,z} \right) = 1 \cr & {\text{Find the gradient }}\nabla F\left( {x,y,z} \right) \cr & \nabla F\left( {x,y,z} \right) = 2x{\bf{i}} + \left( {2y - 6} \right){\bf{j}} + {\bf{k}} \cr & {\text{Find the point}}\left( {\text{s}} \right){\text{ on the surface at which the tangent plane }} \cr & {\text{is horizontal}}. \cr & 2x = 0 \to x = 0 \cr & 2y - 6 = 0 \to y = 3 \cr & z = 3 - {x^2} - {y^2} + 6y \cr & z = 3 - {\left( 0 \right)^2} - {\left( 3 \right)^2} + 6\left( 3 \right) \cr & z = 12 \cr & {\text{The point is}} \cr & \left( {0,3,12} \right) \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.