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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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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1411/discussion-between-richard-inglis-and-anon)2011-09-22

1 Answers 1

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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 $\nabla(z+(x^2+y^2)^{-1/2})=\left(\frac{-x}{(x^2+y^2)^{3/2}},\frac{-y}{(x^2+y^2)^{3/2}},1\right).$