Provide an appropriate response.
-Find the direction in which the function is increasing most rapidly at the point $\mathrm { P } _ { 0 }$.
$$f ( x , y ) = x e ^ { y } - \ln ( x ) , P _ { 0 } ( - 3,0 )$$
A) $\left( - \frac { 9 } { \sqrt { 97 } } \right) i + \left( \frac { 4 } { \sqrt { 97 } } \right) j$
B) $\left( - \frac { 4 } { \sqrt { 97 } } \right) \mathrm { i } - \left( \frac { 9 } { \sqrt { 97 } } \right) \mathrm { j }$
C) $\left( - \frac { 4 } { \sqrt { 97 } } \right) \mathrm { i }$
D) $\left( \frac { 4 } { \sqrt { 97 } } \right) \mathrm { i } - \left( \frac { 9 } { \sqrt { 97 } } \right) \mathrm { j }$

Question

Quizplus · Accepted Answer

The direction of maximum increase of a function is given by the direction of the gradient vector. The gradient vector of the function is $\nabla f ( x , y ) =\left( e ^ { y } - \frac { 1 } { x } \right) \boldsymbol { i } + x e ^ { y } \boldsymbol { j }$. At the point $\mathrm { P } _ { 0 } ( - 3,0 )$, we have $\nabla f ( - 3,0 ) = - \frac { 1 } { 3 } \boldsymbol { i }$. Thus, the direction of maximum increase is the direction of $-\nabla f ( - 3,0 ) = \frac { 1 } { 3 } \mathrm { i }$ which is equivalent to the vector $\left( \frac { 4 } { \sqrt { 97 } } \right) \mathrm { i } - \left( \frac { 9 } { \sqrt { 97 } } \right) \mathrm { j }$ after normalizing. Therefore, the answer is D.

Provide an Appropriate Response P0\mathrm { P } _ { 0 }P0​

Provide an Appropriate Response $\mathrm { P } _ { 0 }$