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 - 13.6 Exercises - Page 924: 7

Answer

$${D_u}g\left( {3,4} \right) = - \frac{7}{{25}}$$

Work Step by Step

$$\eqalign{ & g\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} ,{\text{ }}P\left( {3,4} \right),{\text{ }}{\bf{v}} = 3{\bf{i}} - 4{\bf{j}} \cr & {\text{Calculate }}\nabla g\left( {x,y} \right) \cr & \nabla g\left( {x,y} \right) = {g_x}\left( {x,y} \right){\bf{i}} + {g_y}\left( {x,y} \right){\bf{j}} \cr & {g_x}\left( {x,y} \right) = \frac{{2x}}{{2\sqrt {{x^2} + {y^2}} }} = \frac{x}{{\sqrt {{x^2} + {y^2}} }} \cr & {g_y}\left( {x,y} \right) = \frac{y}{{\sqrt {{x^2} + {y^2}} }} \cr & \nabla g\left( {x,y} \right) = \frac{x}{{\sqrt {{x^2} + {y^2}} }}{\bf{i}} + \frac{y}{{\sqrt {{x^2} + {y^2}} }}{\bf{j}} \cr & {\text{Evaluate }}\nabla g\left( {3,4} \right) \cr & \nabla g\left( {3,4} \right) = \frac{3}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} }}{\bf{i}} + \frac{4}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} }}{\bf{j}} \cr & \nabla g\left( {3,4} \right) = \frac{3}{5}{\bf{i}} + \frac{4}{5}{\bf{j}} \cr & {\text{Calculate }}\left| {\bf{v}} \right| \cr & \left| {\bf{v}} \right| = \left| {3{\bf{i}} - 4{\bf{j}}} \right| = \sqrt {9 + 16} = 5 \cr & {\bf{v}}{\text{ is not a unit vector; the unit vector in the direction of }}{\bf{v}}{\text{ is:}} \cr & {\bf{u}} = \frac{{\bf{v}}}{{\left| {\bf{v}} \right|}} = \frac{{3{\bf{i}} - 4{\bf{j}}}}{5} = \frac{3}{5}{\bf{i}} - \frac{4}{5}{\bf{j}} \cr & {\text{The directional derivative at }}\left( {3,4} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr & {D_u}g\left( {x,y} \right) = \nabla g\left( {x,y} \right) \cdot {\bf{u}} \cr & {D_u}g\left( {3,4} \right) = \nabla g\left( {3,4} \right) \cdot \left( {\frac{3}{5}{\bf{i}} - \frac{4}{5}{\bf{j}}} \right) \cr & {D_u}g\left( {3,4} \right) = \left( {\frac{3}{5}{\bf{i}} + \frac{4}{5}{\bf{j}}} \right) \cdot \left( {\frac{3}{5}{\bf{i}} - \frac{4}{5}{\bf{j}}} \right) \cr & {D_u}g\left( {3,4} \right) = {\left( {\frac{3}{5}} \right)^2} - {\left( {\frac{4}{5}} \right)^2} \cr & {D_u}g\left( {3,4} \right) = - \frac{7}{{25}} \cr} $$
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