October 11th, 2017, 04:43 PM  #1 
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  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! 
October 11th, 2017, 07:07 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 621 Thanks: 392 Math Focus: Dynamical systems, analytic function theory, numerics 
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 codimension 1, then computing a normal vector is equivalent to computing $\nabla f$ which is what you have done. Last edited by SDK; October 11th, 2017 at 07:09 PM. 

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