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Consider a catenoid $C$ parametrized by

$$r(u,v)= (u, \cosh u \cos v, \cosh u \sin v), u\in \mathbb{R}, v\in(-\pi, \pi)$$

I am required to show that the principal directions are the same as the coordiante curves, u= constant, v=constant.

The formula I want to use is $$dN_p (v)=\lambda v$$ for $v$ in the tangent space to $C$ at $p$.

I have found the the differential of the Gauss map($N$) by using $$N= \frac{r_u\times r_v}{|r_u\times r_v|}$$

But once I start calculating the differential of $N$, things get messy. Am I on the right track? Is there a better way to find the principal directions?

Thanks in advance.

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1 Answer 1

up vote 4 down vote accepted

For the catenoid $C$ parametrized by $r(u,v)= (u, \cosh u \cos v, \cosh u \sin v)$, we have $$r_u(u,v)= (1, \sinh u \cos v, \sinh u \sin v),$$

$$r_v(u,v)= (0, -\cosh u \sin v, \cosh u \cos v).$$ This implies $$r_u\times r_v=(\sinh u\cosh u,-\cosh u\cos v,-\cosh u \sin v).$$

This gives $|r_u\times r_v|=\cosh u\sqrt{(\sinh u)^2+1}=(\cosh u)^2$ and $$N=\frac{r_u\times r_v}{|r_u\times r_v|}=\frac{1}{\cosh u}(\sinh u,-\cos v,-\sin v).$$ I think this is what got for the unit normal.

Now, fixed a point $p\in C$ and assume $p=r(u_0,v_0)$. Then the coordinate curves at $p$ are given by $\alpha:(-\epsilon,\epsilon)\rightarrow C$ and $\beta:(-\epsilon,\epsilon)\rightarrow C$ such that $\alpha(t)=r(u_0+t,v_0)$ and $\alpha(t)=r(u_0,v_0+t)$ are coordinate curves at $p$, because $\alpha(0)=\beta(0)=r(u_0,v_0)=p$. Then $$\alpha'(0)=\frac{d}{dt}r(u_0,v_0+t)\Big|_{t=0}=r_u(u_0,v_0)\in T_pC,$$ $$\beta'(0)=\frac{d}{dt}r(u_0+t,v_0)\Big|_{t=0}=r_v(u_0,v_0)\in T_pC.$$

Differentiate $N(\alpha(t))$ with respect to $t$ and using $(2)$, we have $$\frac{d}{dt}N(\alpha(t))\Big|_{t=0}=\frac{d}{dt}\left(\frac{1}{\cosh(u_0+t)}(\sinh(u_0+t),-\cos v_0,-\sin v_0)\right)\Big|_{t=0}$$ $$=\left(-\frac{\sinh(u_0+t)}{\cosh^2(u_0+t)}(\sinh(u_0+t),-\cos v_0,-\sin v_0) \right)\Big|_{t=0}$$ $$+\left(\frac{1}{\cosh(u_0+t)}(\cosh(u_0+t),0,0)\right)\Big|_{t=0}$$ $$=\frac{1}{\cosh^2(u_0)}(1,\sinh u_0\cos v_0,\sinh u_0\sin v_0).$$ Using chain rule, the left hand side is $dN_{\alpha(0)}(\alpha'(0))=dN_{p}(\alpha'(0))$. On the other hand, the right hand side is equal to $\displaystyle\frac{1}{\cosh^2(u_0)}\alpha'(0)$ by $(1)$ and $(3)$. Therefore, we have shown that $$dN_{p}(\alpha'(0))=\frac{1}{\cosh^2(u_0)}\alpha'(0),$$ i.e. the coordinate curve $\alpha$ is principal direction with principal curvature $\cosh^2(u_0)$.

Similarly you can show that the coordinate curve $\beta$ is principal direction. I will let you do it.

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I am sorry for my lateness. I discussed it with a friend and I realised that I have almost solved the question. The way I did it was to find the matrix representation of $dN_p$ and to find the eigenvectors corresponding to the principal curvatures at a given point. Yours solution is also very understandable and I liked it. –  Chulumba Jan 12 '12 at 21:43
    
if you see some upvotes somewhere, do not get surprised. –  Chulumba Jan 12 '12 at 22:28

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