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What would be a simple way to prove that $\nabla F(x,y,z)$ is normal to the surface $F(x,y,z)=0$? I was wondering if anyone had a simple way to do this.

Thanks in advance

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Take any curve $p(t)$ on $F(x,y,z) = 0$, then $\dot{p} \in \nabla F(x,y,z)$, and $\langle p(t), \dot{p(t)} \rangle = 0$. Then take a look at what the tangent spaces look like and conclude they are the same. – Andy Nov 26 '12 at 14:02
The idea that Andy suggests is a nice way to go. To elaborate a bit, call the surface $S$, and let $q \in S$, and recall that any vector $v \in T_qS$ is the tangent vector to some curve $p(t)$ on $S$ with $p(0) = q$. Then consider the function $F \circ p$; it is constant. Differentiate it to conclude that $v= \dot{p}(0)$ is perpendicular to $\nabla F(x,y,z)$. – yasmar Nov 26 '12 at 20:33
@yasmar, thanks for explaining better; I was on the phone and it was hard to type. – Andy Nov 27 '12 at 10:25
up vote 4 down vote accepted

Let $u,v$ parametrise the surface $S$ where $F(x,y,z) = 0$ on $S$. Then we can write $F(S(u,v)) = 0$. What happens when we take derivatives with respect to $u$ and $v$?

(Hint: apply chain rule)

Edit: Taking partial derivatives of $F$ over the surface with respect to, for example, u, gives $\frac{\partial F(S(u,v))}{\partial u} = \frac{\partial F}{\partial x_i} \frac{\partial x_i}{\partial u}$. $\frac{\partial F}{\partial x_i} = \nabla F$, and since this derivative is along the curve, $\frac{\partial x_i}{\partial u} = \frac{\partial S}{\partial u} $. So we're left with $\frac{\partial F(S(u,v))}{\partial u} =\nabla F \bullet \frac{\partial S}{\partial u} $.

But since this is along the surface, the left hand side is zero (since changing u keeps us on the surface, so $F$ doesn't change.) so $\nabla F$ is normal to $\frac{\partial S}{\partial u}$. Similarly for $\frac{\partial S}{\partial v}$. But the partial derivatives are the two tangent vectors that span the plane normal to the surface, so $\nabla F$ mus be normal to the plane!

Note that this result is true for $F(x,y,z) = c$ on a surface for any constant $c$, not just $0$.

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We get $F_u(S(u,v))=F'S_u$? Sorry having a bit of a mind block, could you explain a little more? Thanks :) – Freeman Nov 26 '12 at 14:27
@LHS edited it explain in further detail. Hope that helps! Let me know if there's anything more you want clarified. :) – Tom Oldfield Nov 26 '12 at 14:46
Thank you!! that's great – Freeman Dec 3 '12 at 17:51

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