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I want to find, as an exercise, an expression for the curvature of a surface given by the zero set of a function. I reached a final expression, but when I test it for a sphere I get a non-constant expression. I know I'm doing a step wrong, but I don't know why. Below is what I have.

Say we have a good enough $f:\Bbb R^3 \to \Bbb R$. This defines a surface in $\Bbb R^3$ as $$S = \{(x,y,z)\in\Bbb R^3 : f(x,y,z) = 0\}$$

Suppose that $f_z = \frac{\partial f}{\partial z}\neq 0$. Then by the implicit function theorem it's easy enough to see we have a local parametrization of $S$ (in fact, the graph of a function): $$h:\Bbb R^2\to \Bbb R^3 \atop (x,y)\mapsto (x,y,h(x,y))$$ and $h$ satisfies (found using the chain rule on $f(x,y,h(x,y)) = 0$) $$\begin{align}h_x = \frac{-f_x}{f_z} \\ h_y = \frac{-f_y}{f_z}\end{align}$$

This way we have a basis for $T_pS$ (at a point $p$ which is omitted in the expressions) $(1,0,-f_x/f_z), (0,1,-f_y/f_z)$. With the usual notation, the coefficients of the first fundamental form $I$ can be calculated as $$E = 1+(f_x/f_z)^2, F = \frac{f_xf_y}{f_z^2}, G = 1 + (f_y/f_z)^2$$ My problem, it seems, comes with the second fundamental form $II$. For example, my calculation of $h_{xx}$ results in $$h_{xx} = \frac{f_xf_{xz}-f_{xx}f_z}{f_z^2}$$ while on this page they get $$h_{xx} = \frac{2f_xf_zf_{xz}-f_x^2f_{zz}-f_z^2f_{xx}}{f_z^3}$$

so either something is horribly wrong and I've forgotten how to differentiate a quotient, or there's something else in the calculation I'm not including. Note that this is before doing anything with the normal vector, just partial derivatives of the parametrization. Could someone clear this up?

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You've forgotten to apply the chain rule again. Remember that you're evaluating the partial derivatives of $f$ at $(x,y,h(x,y))$.

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  • $\begingroup$ Thanks! I guess this is what I get for not writing $h_x(p)$... $\endgroup$ – GPerez Mar 18 '15 at 22:14

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