Gradient vector interpretation I have trouble understanding the gradient vector $\nabla$. 
For a surface, when we find it's $\nabla$, why is the resultant vector the normal vector when $\nabla$ means gradient vector??
Thanks
 A: From my understanding, a general 2D surface $S$ sitting in $\mathbb{R}^3$ can be defined as the solution set of a single equation
$$
f(\mathbf{x}) = c
$$
Where $f : \mathbf{x} \in \mathbb{R}^3 \mapsto f(\mathbf{x}) \in \mathbb{R}$ and $c$ here is a real number. Since $f$ is a scalar function, $\nabla f$ is its gradient, which can be seen as a gradient vector. 
Here’s my understanding of why the gradient vector $\nabla f$ is orthogonal to the surface $S$. (This is more of an intuitive explanation than a proof.)
Consider a separate surface $S^\prime$ defined by the equation 
$$
f(\mathbf{x}) = c^{\prime}\, , \,\,\,\, c^\prime > c.
$$
Let $c^\prime$ approach $c$. Since $f$ is differentiable (and thus continuous), all the points in the surface $S^\prime$ will approach the points in the surface $S$. So for $c^\prime$ extremely close to $c$, you can imagine the surface $S^\prime$ as locally tangent (or “parallel”) to the surface $S$ everywhere. 
Now, imagine the gradient vector $\nabla f$ sitting on some point of $S$. You should know that $\nabla f$ points in the direction of highest rate of increase of $f$. Since, $c^\prime > c$, in the context of the two surfaces, this means that $\nabla f$ points in the direction of shortest path between $S$ and $S^\prime$. But as we said, for $c^\prime$ approaching $c$, $S^\prime$ becomes tangent to $S$. Thus $\nabla f$ becomes orthogonal to $S$, i.e. the gradient vector is normal to the surface. 
