How can ∇f(x,y) be perpendicular to the tangent plane but also point in the direction of greatest ascent

Question

I know ∇f(x,y) is perp. to a graphs contour lines and points in the direction of greatest ascension implying it lies along the surface and pointing up the slope at each point. So I cant understand how ∇f(x,y) can be the normal vector to the surface at a point as well as shown in image 2.

Image of a graph with contour lines

Image of a graph with ∇f(x,y) sketched on

Answer 1

The gradient is perpendicular to level surfaces (or curves).

For example, consider the parabloid $z=x^2+y^2$.

There are two ways to think about this parabola. The first is the function $\mathbb R^2 \to \mathbb R^3$, where it is parametrized by $r(s,t)=(r,s,t^2+s^2)$

However, there is another way to think about it as the zero set of a function $\mathbb R^3 \to \mathbb R$ given by $f(x,y,z)=x^2+y^2-z$.

Note that $f \circ r=0$, by assumption.

The chain rule implies that $0=(f \circ r)^{\prime}=f^{\prime}(r^{\prime}(s,t))=\vec{0}$

which implies that $$\begin{pmatrix} \frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}&\frac{\partial f}{\partial z}\end{pmatrix} \cdot \begin{pmatrix}\frac{\partial x }{\partial r} &\frac{\partial x }{\partial t}\\\frac{\partial x }{\partial r}&\frac{\partial x }{\partial t}\\ \frac{\partial x }{\partial r}&\frac{\partial x }{\partial t}\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}$$

in other words, \begin{pmatrix} \frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}&\frac{\partial f}{\partial z}\end{pmatrix} is orthogonal to the surface $f(x,y,z)=0$ which is an example of a level surface.