If the electric potential in a point $(x,y)$ of the plan $xy$ is $V(x,y)$ so the electric intensity vector in point $(x,y)$ is $E = - \nabla V(x,y)$. Suppose $V(x,y)={e^{-2x}}{\cos2y}$
Determine the electric intensity vector in $(\frac{\pi}{4},0)$ Show that in each point of the plan, the electric potential decreases faster in the same direction and sense of the $E$ vector.
So I have this question, but I do not know how answer.
I have the Gradient: $\nabla E = ({2}{\cos(2y)}{e^{-2x}},{2{e^{-2x}}\sin(2y)}$
So show the eletric intensity vector in $(\frac{\pi}{4},0)$ is simple, just use the values, but i have no ideia of how demonstrate that eletric potential decress faster when in the same direction and sense of $E$ vector.
Recall that the gradient vector direction is the direction for which the function increase faster then since $E=-\nabla V$ we have that the electric potential decreases faster in the same direction and sense of the E vector.
Refer also to Maximum rate of increase- Directional derivatives.