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October 11th, 2017, 04:43 PM   #1
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Gradient question

Hello forum,

A couple friends and I have been discussing a problem.

Consider a field $\phi(\vec{r}) = x^2 + sin (y) - xz$

I want to find a unit vector normal to the surface $\phi(\vec{r}) = 5$ at the point (x,y,z) = (1, $\frac{\pi}{2}$, -3)

My plan of attack was to $\nabla \phi(\vec{r})$ at the points (1, $\frac{\pi}{2}$, -3). = $<2,0,0>$

After I normalized the vector I got $<1,0,0>$

Then we realized we never utilized the information $\phi(\vec{r}) = 5$ given in the problem. Where did I go wrong?

Thanks!
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October 11th, 2017, 07:07 PM   #2
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You did use the information whether or not you realize it. Its a simple fact that if one $f$ is a smooth funcion, then $f(x) = c$ is locally a graph as long as $x$ is a regular point (i.e. as long as $\nabla f(x) \neq 0$.

Now, let $M = \{x : f(x) = c\}$ and let $\gamma(s)$ be a parameterized curve lying in $M$. Then we have $f(\gamma(s)) = c$ so by differentiating we obtain
\[
\nabla f(\gamma(s)) \cdot \frac{d}{ds}\gamma(s) = 0.
\]

Since $\frac{d}{ds}\gamma(s)$ must be a vector in the tangent space, $TM_{\gamma(s)}$. Since $\gamma$ was an arbitrary curve, one immediately recovers that if $x \in M$, then for any vector $v \in TM_x$ we have $\nabla f(x) \cdot v = 0$.

Or to put it in English, the gradient vector at a regular point is always orthogonal to the tangent space of the level surface (if the surface is sufficiently smooth). When this tangent space has co-dimension 1, then computing a normal vector is equivalent to computing $\nabla f$ which is what you have done.
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Last edited by SDK; October 11th, 2017 at 07:09 PM.
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