# normal vector to surface $z=\frac{-1}{\sqrt{x^2+y^2}}$

I have a surface defined as $z=\frac{-1}{r}$ where $r=\sqrt{x^2+y^2}$ and I'd like to know how to calculate a surface normal vector at a point $(x,y)$.

An approximation would be acceptable.

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The normal vector of a surface implicitly defined by $F(x,y,z)=C$ (i.e. a level sets of $F$) is $\nabla F(a)$ at the point $x=a$. Your surface is $z+(x^2+y^2)^{-1/2}=0$... – anon Sep 22 '11 at 22:52
Ok, but how do I work out $\nabla F(a)$? Sorry, my maths is very rusty... – Richard Inglis Sep 22 '11 at 23:06
Is it $\nabla F(a)=(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z})$? – Richard Inglis Sep 22 '11 at 23:09
Richard: Yes, that's the gradient of $F$. And here $F$ is..? – anon Sep 22 '11 at 23:13
@anon: um, $z+(x^2+y^2)^{-1/2}=0$, but it may take me several days to differentiate it... :) – Richard Inglis Sep 22 '11 at 23:15

We went over the answer in chat. To review, in order to differentiate $(x^2+y^2)^{-1/2}$ we used the chain rule after setting $g(x)=x^2+y^2$ and $f(g)=g^{-1/2}$, so that $f'(g)=-(1/2)g^{-3/2}$ and $g_x=2x$ (similarly for $d/dy$) so the gradient (and therefore the normal vector, though not the unit normal) can be expressed as $$\nabla(z+(x^2+y^2)^{-1/2})=\left(\frac{-x}{(x^2+y^2)^{3/2}},\frac{-y}{(x^2+y^2)^{3/2}},1\right).$$