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 $z+(x^2+y^2)^{-1/2}=0$...

Ok, but how do I work out $\nabla F(a)$? Sorry, my maths is very rusty...

Is it $\nabla F(a)=(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z})$?

Richard: Yes, that's the gradient of $F$. And here $F$ is..?

@anon: um, $z+(x^2+y^2)^{-1/2}=0$, but it may take me several days to differentiate it... :)

Now, now. I know you can differentiate it, just take your time. :)

I'm getting something like $(x^{-2},y^{-2},1)$...

Nope. Use the chain rule: $f(g)=g^{-1/2}$ and $g(x)=x^2+y^2$.

Brain and laptop both about to run out of batteries... thanks for your help anon - I'll have to try to work out the rest tomorrow.

you here?

on my phone now... sorry: battery failure.

that's cool

so, the chain rule says d/dx f(g(x)) = f'(g) * g'(x)

(1) What's f' ? (2) What's g' ?

(1) -1/2g^-3/2 (2) 2x

Yes. So what is f'(g(x)) * g'(x) ?

So f'.g' should be -1/2(x^2+y^2).2x...

Oops...

-1/2(x^2+y^2)^-3/2.2x

(sorry, slow phone typing...)

yep! You can simplify it a bit too

Well, the 2's cancel...

so it's x/r^3

-x/r^3 I mean...

that's correct! and similarly, for the we get -y/r^3 for the d/dy derivative. so then what is your gradient?

So the normal is (-x/r^3,-y/r^3,1) ?

That's more symmetrical than my first answer, which only had even powers of x...

that's correct

My last calculus lesson was 30 years ago! Thanks so much for helping with this...

but remember that r only stands for x and y, not all three vectors

im posting our process as an answer :)

I'm happy to accept it...

also, keep in mind this isn't a unit vector, but it is normal

do you need it to be a unit normal vector?

No, I don't think it needs to be a unit vector.

well, then it's posted

...and accepted. Thanks again. Goodnight.

night