I need to calculate the Gaussian and mean curvatures of a sphere of radius a.
Writing the equation of the sphere in the form $$p(u,v)= \begin{pmatrix}f(u)\cos(v)\\f(u)\sin(v)\\g(u)\\\end{pmatrix}$$ which in the case of my sphere is
$$\begin{pmatrix}a\cos(u)\cos(v)\\a\cos(u)\sin(v)\\a\sin(u)\\\end{pmatrix}$$
I see that $f(u)=a\cos(u)$ and $g(u)=a\sin(u)$.
I have been given that the Gaussian curvature can be calculated by $K=\frac{-f''(u)}{f(u)}$ and the mean curvature by $H=\frac12(\frac{g'(u)}{f(u)}-\frac{f''(u)}{g'(u)})$
(with $'$ meaning differentiation with respect to $u$).
However, when I differentiate my values for $f(u)$ and $g(u)$ and substitute them in, I keep getting both Gaussian and mean curvatures to be $1$, when they should be $K=\frac{1}{a^2}$ and $H=\frac{1}{a}$.
Where am I going wrong?
Best Answer
Gaussian and Mean curvature formulas you've written are correct only if $\alpha(u)=(f(u),g(u))$ has unit-speed i.e. $\|\alpha'(u)\|=1$ that means $u$ is the arc-length parameter. But, in your case, it seems that $(a\cos u, a\sin u)$ is not a unit-speed curve. You can use the formulas $$K=\dfrac{-{g'}^2f''+f'g'g''}{f({f'}^2+{g'}^2)^2}\quad \text{and}\quad H=\dfrac{f(f''g'-f'g'')-g'({f'}^2+{g'}^2)}{2|f|({f'}^2+{g'}^2)^{3/2}}, $$ which yield the correct results. Moreover, if you consider a unit-speed curve i.e. ${f'}^2+{g'}^2=1$, you can derive the formulas you've written. It is easy to see that $f'f''+g'g''=0$ in this case. Then, you can find $$K=\dfrac{-{g'}^2f''+f'g'g''}{f({f'}^2+{g'}^2)^2}=\dfrac{-{g'}^2f''+f'g'g''}{f}= \dfrac{-{g'}^2f''+f'(-f'f'')}{f}=\dfrac{-f''}{f}.$$