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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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  • $\begingroup$ It seems rather backwards to me, too. But it's true that the zero vector is perpendicular to any vector at all. $\endgroup$ – leftaroundabout Sep 18 '11 at 19:08
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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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  • $\begingroup$ Actually I meant to say, $w$ is a constant, since there is no change in $w$, the gradient is zero. Right? Then it's $0 \cdot \vec{v}$. $\endgroup$ – BBDynSys Sep 18 '11 at 19:21
  • $\begingroup$ Why do you think $w$ is a constant? It is not. It cannot be a constant, because it is $c$ at the surface and $\neq c$ at every point not on the surface. $\endgroup$ – hmakholm left over Monica Sep 18 '11 at 19:36
  • $\begingroup$ I see. Then how does the teacher make the conclusion $\nabla w \cdot \vec{v} = 0$? $\endgroup$ – BBDynSys Sep 18 '11 at 19:45
  • $\begingroup$ Um, that is what you quote in your question. By the chain rule $dw/dt=\nabla w\cdot v$, but $dw/dt$ is 0 because by assumption the curve stays within the surface. That is, $r(t)$ always has the property $w(r(t))=c$ -- this is a particular assumed property of $r$, not one of $w$ itself. $\endgroup$ – hmakholm left over Monica Sep 18 '11 at 19:50
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    $\begingroup$ $w$ 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 $\nabla w\cdot v=0$ $\endgroup$ – hmakholm left over Monica Sep 18 '11 at 20:20

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