Proof That Grad f is in the Direction of Greatest Rate of Increase of f
TheoremA funvtion increasing most rapidly in the direction of
\[\nabla f\]
, The gretest rate of increase of \[\nabla f\]
at a point is \[ |\nabla f|\]
at that point.Proof
Suppose
\[f\]
is a parametrised in term of the distance \[s\]
along the curve.\[\frac{df}{ds} = \frac{\partial f}{\partial x} \frac{dx}{ds}+\frac{\partial f}{\partial y} \frac{dy}{ds}+\frac{\partial f}{\partial z} \frac{dz}{ds}=\nabla f \cdot \frac{d \mathbf{r}}{ds}=\nabla f \cdot \mathbf{v}\]
since
\[\mathbf{v}=\frac{d \mathbf{r}}{ds}\]
is a unit vector.But
\[\frac{df}{ds}=\nabla f \cdot \mathbf{v}=|\nabla f| \cos \theta <=|\nabla f|\]
since \[\cos \theta <=1\]
.