# Proving Grad(F) is perpendicular to level curve - question

1. Nov 8, 2013

Hi -

I was reading Adams' "Calculus: A Complete Course" (6th edition) and he offers the following proof that the gradient of a function:

I'm just wondering why the emphasis on specifying the values of x(0) and y(0), or even the need to specify t = 0 in the last statement? If $\mathbf{r}$ is a parameterization of the level curve, then isn't the derivative of f zero for all values of t? Why not go for the more general statement?

I guess I just don't find the proof very satisfying. Obviously I don't doubt it, but does anyone know a more intuitive proof? I'm teaching the concept shortly and always appreciate bringing an intuitive approach. Usually I try to make connections to single variable, but that's not possible here …

Rax

2. Nov 8, 2013

### MisterX

I agree; there seems to be no reason to have a, b, or t = 0. If x(t), y(t) is parameterization of the level curve for some values of t, then the value of f(x(t), y(t)) is a constant for all t that correspond to being on the level curve. It should be okay to write

$\frac{df}{dt} = \nabla f \cdot \frac{d\vec{r}}{dt} = 0$

3. Nov 8, 2013

### arildno

How is the book's presentation a restriction of the statement's generality?

Having t=0 being a point on the parametrized curve is no restriction on the types of curves studied.
Having x(0) equal some arbitrary number "a" and y(0) some arbitrary number "b" do not represent restrictions either.

4. Nov 9, 2013

@arildno thanks for your comment. I realize that they do not represent restrictions, I guess I just feel that they are **suggestive** of restrictions because (ISFAIA) it is unnecessary. I also just don't understand why he would bother with those details?

5. Nov 9, 2013