# Gradient vector perpendicular to level curves?

1. Jul 28, 2014

### princejan7

1. The problem statement, all variables and given/known data

can anyone explain/prove why the gradient vector is perpendicular to level curves?

2. Relevant equations

3. The attempt at a solution

2. Jul 28, 2014

I think this could be a good exercise for you to prove yourself. I will give some pointers. You have a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ and level sets such that $f(x_1,..., x_n) = c.$

Let $\gamma(t) = (x_1(t),...,x_n(t))$ be a curve on the level surface and consider $g(t) = f(x_1(t),...,x_n(t))$. Now compute $\frac{d}{dt}g(t)$.

Last edited: Jul 28, 2014
3. Jul 28, 2014

### slider142

The directional derivative DuF(P) of a function F in the direction of the unit vector u at the point P is equivalent to the dot product of the gradient vector ∇F(P) with the unit vector u (this is something you should prove if you have not already done so).
The value of F does not change in the direction of the tangent vector to the level curve that passes through P, since a level curve is a set of points where the value of F is constant. Therefore, the directional derivative of F in that direction is 0. Since this is equivalent to the dot product between ∇F(P) and any unit tangent vector to the level curve, and assuming neither vector is the zero vector, the two vectors must be perpendicular.
You can put each of these intuitive arguments into rigorous mathematical form in order to get a more rigorous picture.

Edit: I agree with Quesadilla (it looks like we posted at the same time). This is something you may like to prove rigorously yourself, if you understand the definition of the gradient, dot product, and directional derivative. It is a walk through a series of definitions.