\(f\left( x,y \right) ={ x }^{ 2 }+2{ y }^{ 2 }-xy-7x,\quad p=(4,1)\)
Find the directional derivative in the indicated directions
\({ f }_{ x }=2x-y-7,\quad { f }_{ y }=4y-x\)
\(\nabla=\begin{bmatrix} 2x-y-7\quad \\ 4y-x \end{bmatrix}\)
\(\nabla =\begin{bmatrix} 0 \\ 0 \end{bmatrix}\)
Since these will both be 0, we can try the point (3,2) instead: \(\nabla =\begin{bmatrix} -3 \\ 5 \end{bmatrix}\)
a)In the direction of the vector (2,5)
\(u=\begin{bmatrix} \frac { -2 }{ \sqrt { 29 } } \\ \frac { 5 }{ \sqrt { 29 } } \end{bmatrix}\) \(Du=\begin{bmatrix} \frac { -2 }{ \sqrt { 29 } } \\ \frac { 5 }{ \sqrt { 29 } } \end{bmatrix}\cdot \begin{bmatrix} -3 \\ 5 \end{bmatrix}=\frac { 30 }{ \sqrt { 29 } }\)
b)Toward the point (4,0)
\(Du=\begin{bmatrix} 0 \\ 1 \end{bmatrix}\cdot \begin{bmatrix} -3 \\ 5 \end{bmatrix}=5\)