# Gradient steepest direction and normal to surface?

From this Maths SE question, I now understand the gradient to be the directional derivative that returns the steepest slope at a point. However, reading my textbooks, they all say that the gradient is normal to the tangent plane. In the above SE question, the gradient sounds like it should be parallel to a direction in the $xy$ plane, not actually up on the surface.

My question is therefore do I have a misunderstanding of the tangent plane, gradient or directional derivative?

In addition, I question how the gradient always passes through the origin or the $xy$ plane? If for instance in the graphic, the gradient had been another direction (say horizontal), then it wouldn't pass through the origin! In addition, the gradient of other points in the graphic would satisfy this example. Many thanks!

The gradient is normal to the tangent plane of an equipotential surface. That is, $\nabla f$ is normal to the surface $f(x,y,z)=C$ (I am assuming here $f$ defined on $\mathbb{R}^3$.) Consider for instance $f(x,y,z)=x^2+y^2+z^2$. Then $f(x,y,z)=C>0$ is a sphere. The normal to the sphere at a point $(x,y,z)$ is in the direction of $(x,y,z)$, same as $\nabla f=2(x,y,z)$.
• So the normal to the tangent plane of a level curve is the direction in the $xy$ plane that the gradient points? – RedPen Jul 9 '15 at 12:22
You are having confusion because you aren't being clear about what you are taking the gradient of. If $f(x,y)$ is a function of two variables, then the gradient $\nabla f(x,y)$ is a vector in the plane which points in the direction of greatest rate of change of $f(x,y)$ at each point. Moreover, it is normal to the level curves of $f(x,y)$ (these are the equations $f(x,y)=k$ for various real numbers $k$). This makes sense because traveling along the level curves of $f(x,y)$ results in zero change.
Note that the graph of $f(x,y)$ and the level curves are very different. The graph of $f(x,y)$ is the surface $f(x,y)=z$ (here z is a dependent variable, not a constant!). We can find a normal vector to the graph of $f(x,y)$ by letting $g(x,y,z)=z-f(x,y)$ and seeing that the graph of $f(x,y)$ is a level surface to $g(x,y,z)$ (where $k=0$). Then for each $(x,y,z)$ such that $g(x,y,z)=0$ we have that $\nabla g(x,y,z)$ is normal to the level surface $g(x,y,z)=0$, which as mentioned, is the graph of $f(x,y)$. Of course $\nabla g(x,y,z)$ and $\nabla f(x,y)$ are different - they don't even have the same number of components.