Gradient Proof

We will now show that the gradient vector field $ \mathrm{grad}f(x,y,z) $ points in the direction of greatest increase of the scalar field. Consider an infinitesimal change in position within a scalar field $ f(x,y,z) $, moving $ dx $, $ dy $ and $ dz $ in the $ x $, $ y $ and $z $ directions respectively. The change $ df $ due to this change in position is given by $$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz, $$ $$ df = \Bigg(\frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k} \Bigg) \circ (dx \hat{i} + dy \hat {j} + dz \hat {k}), $$ $$ df = \vec{\nabla}f \circ d\vec{r}. $$ Where $ d\vec{r} $ is the infinitesimal change in position. Using the definition of the dot product, the change in $f$ is given by $$ df = \vert \vec{\nabla}f \vert \vert d\vec{r} \vert \cos \theta $$

Where $ \theta $ is the angle between $ \vec{\nabla}f $ and $ d\vec{r} $.

$ df $ is maximum when $ \theta = 0 $. This means that for maximum rate of change, we must move parallel to the gradient vector, therefore the gradient vectors point in the direction of greatest rate of change.

If we rearrange equation (4) when $ \theta = 0 $, we find $ \vert \vec{\nabla}f\vert = $ $ \frac{df}{dr} $ , where $ \vert d\vec{r} \vert = dr $, therefore the size of the gradient vector is the gradient in that direction.