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Take a curve $\vec{r} = \vec{r}(t)$ that stays on the level $w=c$ where $c$ is a constant. Velocity is $\vec{v} = \frac{d\vec{r}}{dt}$ and is tangent to the level $w=c$ because it's tangent to the level, because it's tangent to the curve and the curve is inside in the level.

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By the chain rule

$\frac{dw}{dt} = \nabla w \cdot \frac{d\vec{r}}{dt}$

which also is

$ \frac{dw}{dt} = \nabla w \cdot \vec{v}$

Then the teacher says

$\nabla w \cdot \vec{v} = 0 $

because $w=c$ hence there is no change in $w$ (we are on a contour). So I guess that's why $\nabla w \cdot \vec{v} = 0$? Because $\nabla w = 0$, and $0 \cdot \vec{v} = 0$?

Then he gives another conclusion that since $\nabla w \cdot \vec{v} = 0$, hence we should have $\nabla w \perp \vec{v}$ -- the two vectors are perpendicular. But if we are taking the gradient of a function that is a constant, hence the result is the zero vector, arent we saying that any vector is always perpendicular to a zero vector? That seems kind of backwards to me, even though I intuitively understand what he is trying to do, algebraically it doesnt really make sense.

Any ideas?

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    It seems rather backwards to me, too. But it's true that the zero vector is perpendicular to _any_ vector at all.2011-09-18

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$\nabla w$ is not "the gradient of a constant", because the gradient is taken in the entire space that the surface is embedded in, not just inside the surface itself. So $\nabla w$ is in general a vector that points away from the surface. (The case where the gradient is the zero vector will usually be excluded as a degenerate case anyway).

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    $w$ itself does not have $r$ inside. It is just _some_ smooth function from $\mathbb R^3$ to $\mathbb R$. The argument then goes that there are two different ways to compute $\frac{d}{dt}w(r(t))$. One is to mindlessly apply the chain rule, which produces $\nabla w\cdot v$; the other is to remember that we have assumed that $w(r(t))=c$ for all $t$, and since $c$ is just a constant, $\frac{d}{dt}w(r(t))=\frac{d}{dt}c=0$. Because those two calculations are calculations of _the same thing_, they must give the same result, and therefore we can conclude $\nabla w\cdot v=0$2011-09-18