Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 37

Answer

$$\nabla w = 3\left( { - 3{x^8}{\bf{i}} - {y^2}{\bf{j}} + 4{z^{11}}{\bf{k}}} \right)$$

Work Step by Step

$$\eqalign{ & w = - {x^9} - {y^3} + {z^{12}} \cr & \cr & {\text{calculate the partial derivatives }}{w_x}{\text{,}}\,\,{w_y}{\text{, and }}{w_z} \cr & {w_x} = \frac{\partial }{{\partial x}}\left[ { - {x^9} - {y^3} + {z^{12}}} \right] \cr & {\text{treat }}y{\text{ and }}z{\text{ as constants}} \cr & {w_x} = - 9{x^8} + 0 + 0 \cr & {w_x} = - 9{x^8} \cr & \cr & {w_y} = \frac{\partial }{{\partial y}}\left[ { - {x^9} - {y^3} + {z^{12}}} \right] \cr & {\text{treat }}x{\text{ and }}z{\text{ as constants}} \cr & {w_y} = 0 - 3{y^2} \cr & {w_y} = - 3{y^2} \cr & \cr & {w_z} = \frac{\partial }{{\partial z}}\left[ { - {x^9} - {y^3} + {z^{12}}} \right] \cr & {\text{treat }}x{\text{ and }}y{\text{ as constants}} \cr & {w_z} = 0 + 0 + 12{z^{11}} \cr & {w_z} = 12{z^{11}} \cr & \cr & {\text{The gradient of the function }}w\left( {x,y,z} \right){\text{ is defined by }}\left( {{\text{see page 963}}} \right) \cr & \nabla w = {z_x}{\bf{i}} + {z_y}{\bf{j}} + {z_z}{\bf{k}} \cr & {\text{substituting the partial derivatives, we obtain}} \cr & \nabla w = - 9{x^8}{\bf{i}} - 3{y^2}{\bf{j}} + 12{z^{11}}{\bf{k}} \cr & {\text{factoring}} \cr & \nabla w = 3\left( { - 3{x^8}{\bf{i}} - {y^2}{\bf{j}} + 4{z^{11}}{\bf{k}}} \right) \cr} $$
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