Why does $\nabla F{(x,y,z)}$ point in the direction of greatest increase of the function, and why is $|\nabla F(x,y,z)|$ its slope?

Question

Why does $$\nabla F{(x,y,z)}$$ point in the direction of greatest increase of the function and why is $$|\nabla F{(x,y,z)}|$$ it's slope (I should actually ask what the slope would mean here as I'm not entirely sure here as we are working in $3D$ and I can't really picture what a slope would be here normally I think a line in $2D$ and it's a change in $y$) over a change in $x$ but obviously that intuition fails here?

After reading about the $\nabla$ operator severals times I understand that it has these properties but I'm not exactly sure why. I know people usually want to see effort on what you have done here but I'm not really sure what to put here, I don't see why this should be the case other than thinking there could possibly be some link between the rate of increase and the gradients but as for a proof I have no idea.

Thanks.

Answer 1

Recall that, for a differentiable function, the directional derivative of $f$ in the (unit) direction $u$, at a point $(x, y, z)$, is given by $$D_u f(x, y, z) = \nabla f(x, y, z) \cdot u.$$ Also recall the following identity: $$v \cdot w = |v||w|\cos(\theta),$$ where $\theta$ is the angle between $v$ and $w$. Since $|u| = 1$, it follows that $$D_u f(x, y, z) = |\nabla f(x, y, z)|\cos(\theta),$$ where $\theta$ is the angle between $u$ and $\nabla f(x, y, z)$ (that is, it depends on $u$). This quantity is maximised when $\cos(\theta) = 1$, that is, when the principle angle between $u$ and $\nabla f(x, y, z)$ is $0$. Thus, we want $u$ to point in the same direction as $\nabla f(x, y, z)$, so the direction of $\nabla f(x, y, z)$ is the direction of greatest directional derivative.

If we compute the directional derivative in this direction, we are taking $\theta = 0$, so the directional derivative is, $$|\nabla f(x, y, z)|\cos(0) = |\nabla f(x, y, z)|.$$