# I Proving grad operator yields perpendicular vector to contour

1. Apr 14, 2016

### henry wang

2. Apr 14, 2016

### stevendaryl

Staff Emeritus
If you have a function $f(x,y,z)$, and you have a constant $c$, then the set of points satisfying $f(x,y,z) = c$ is a two-dimensional surface. So now we let $\vec{r}(t)$ be any path that stays on that surface. In terms of components, if we write $\vec{r}(t) = (x(t), y(t), z(t))$, then to say that $\vec{r}(t)$ stays on the surface $f(x,y,z)=c$ just means that for all $t$, we have:

$f(x(t), y(t), z(t)) = c$

So far, this is just true by assumption. We're assuming that $x(t), y(t), z(t)$ are three functions such that $f(x(t), y(t), z(t)) = c$.

Since $f(x(t), y(t), z(t)) = c$, then it immediately follows that:

$\frac{d}{dt} f(x(t), y(t), z(t)) = \frac{dc}{dt} = 0$

All they're doing is defining $f(x(t), y(t), z(t)) = g(t)$. That's just the definition of $g$. So it immediately follows that $\frac{d}{dt} g(t) = 0$.

3. Apr 14, 2016

### henry wang

Thank you very much, that make a lot of sense, I understand now!