# Gradient: Normal vs Direction of Increase

1. Mar 10, 2010

### thepopasmurf

Hi,

I'm having trouble understanding what exactly the gradient of a scalar field represents. According to wikipedia and the textbooks I have it points in the direction of greatest increase and has a magnitude of greatest increase. This by itself seems fine. However, I have also been using it to find the normal of a surface and I don't understand how it can be both.

Also, do the properties( eg, what it represents) of the gradient change as you change dimensions and how does it work on a simple 2d graph?

2. Mar 10, 2010

### Tinyboss

The gradient is perpendicular to the level sets of a function. So if your surface is given by f(v)=c, the normal at v is given by the direction of the gradient at v.

3. Mar 10, 2010

### quasar987

It all comes from the formula <v,w>=|v||w|cosθ where θ is the angle between v and w.

Let F:R³-->R be a function. By looking at the definition of DF(x)v for |v|=1 (the derivative of F at a point x in the direction of the vector v), we agree that this number represents the rate of change of F at x in the direction v. On the other hand, the gradient of F at x is defined as the (unique) vector ∇F(x) such that DF(x)v=<∇F(x),v> for all vectors v.

So to ask in which direction is F increasing the most rapidly at x is to ask which vector v of unit lenght (|v|=1) maximizes the value of DF(x)v. But DF(x)v=<∇F(x),v>=|∇F(x)||v|cosθ=|∇F(x)|cosθ, with cosθ taking values between -1 and 1. Clearly, |∇F(x)|cosθ is largest when cosθ=1; i.e. when θ=0. That is, when v points in the direction of ∇F(x)!

Now, consider S a surface in R³ that is realized as the level set F=c of F. That is, $S=F^{-1}(c)$ for some constant c. Take x a point in S. By definition, a tangent vector to S at x is a vector v of the form $v=\gamma'(0)$ for some curve $\gamma:]-1,1[\rightarrow S$ on S with $\gamma(0)=x$. Notice that for $v=\gamma'(0)$ a tangent vector to S at x, the derivative of F at x in the direction v vanishes:

$$DF(x)v=DF(x)\gamma'(0)=\frac{d}{dt}_{t=0}(F\circ\gamma)(t)=0$$

The second equality is the chain rule and the third equality is because the map $(F\circ\gamma)(t)$ is the map $t\mapsto c$.
Ok, so in terms of the gradient, what does this tells us? It tells us that 0=DF(x)v=<∇F(x),v>=|∇F(x)||v|cosθ, so cosθ=0, so θ=±90°. That is, ∇F(x) and v are perpendicular. By definition, this means ∇F(x) is perpendicular (or normal) to the surface S at the point x.