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Problem of the Week #34 - January 21st, 2013

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem.

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Problem: Let $T$ be the torus generated by taking the circle of radius 1 in the $xz$-plane centered at (0,0,2) and revolving it about the $x$-axis. Find the points (or curves) on $T$ where the Gaussian curvature $K$ is zero, at a maximum, and at a minimum.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one answered this week's question. You can find my solution below:

The torus $T$ has the parameterization $f: (s,t)\mapsto(\cos t,(2+\sin t)\cos s,(2+\sin t)\sin s)$, where $t$ is the parameter used to define the circle in the $xz$ plane and $s$ is the parameter used to define the circle of revolution in the $yz$ plane. We now compute the first fundamental form $I$ and the second fundamental form $II$; but first, we need to know the derivatives of our parameterized function. We see that
\[\begin{aligned}\frac{\partial f}{\partial s} &= (0,-(2+\sin t)\sin s,(2+\sin t)\cos s)\\\frac{\partial f}{\partial t} &= (-\sin t,\cos t\cos s, \cos t\sin s)\ \\ \frac{\partial^2f}{\partial s^2} &= (0,-(2+\sin t)\cos s,-(2+\sin t)\sin s)\\ \frac{\partial^2f}{\partial t^2} &= (-\cos t,-\sin t\cos s,-\sin t\sin s)\\ \frac{\partial^2f}{\partial t\partial s} &= (0,-\cos t\sin s,\cos t\cos s)\end{aligned}\]
To compute $II$, we also need to compute the normal vector $\eta$:
\[\eta = \frac{\dfrac{\partial f}{\partial s}\times\dfrac{\partial f}{\partial t}}{\left\|\dfrac{\partial f}{\partial s}\times\dfrac{\partial f}{\partial t}\right\|}=(-\cos t,-\sin t\cos s,-\sin t\sin s).\]
With this, we now see that
\[I= \begin{bmatrix} \left\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial s} \right\rangle& \left\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle \\ \left\langle\frac{\partial f}{\partial t},\frac{\partial f}{\partial s}\right\rangle & \left\langle\frac{\partial f}{\partial t},\frac{\partial f}{\partial t}\right\rangle \end{bmatrix} = \begin{bmatrix}(2+\sin t)^2 & 0 \\ 0 & 1\end{bmatrix}\]
and
\[II= \begin{bmatrix} \left\langle\frac{\partial^2 f}{\partial s^2},\eta \right\rangle& \left\langle\frac{\partial^2f}{\partial t\partial s},\eta\right\rangle \\ \left\langle\frac{\partial f^2}{\partial s\partial t},\eta\right\rangle & \left\langle\frac{\partial^2 f}{\partial^2 t},\eta\right\rangle \end{bmatrix} = \begin{bmatrix}(2+\sin t)\sin t & 0 \\ 0 & 1\end{bmatrix}\]
where $\langle\cdot,\cdot\rangle$ is the Euclidean inner product in $\mathbb{R}^3$. Now that we know $I$ and $II$, we can now compute the Gaussian curvature $K$:
\[\begin{aligned}K &= \frac{\det II}{\det I}\\ &= \frac{(2+\sin t)\sin t}{(2+\sin t)^2}\\ &= \frac{\sin t}{2+\sin t}\end{aligned}.\]

We now observe that if $t\in[0,2\pi)$, $K=0$ whenever $t=0$ or $t=\pi$.

Note that $K$ is independent of the parameter $s$! This implies that $K=0$ on the two circles $(\pm 1, 2\cos s,2\sin s)$ (i.e. on the circle $y^2+z^2=4$ in the $x=-1$ and $x=1$ planes).

Finally, we note that if $K(t)=\dfrac{\sin t}{2+\sin t}$, then $\dot{K}(t) = \dfrac{2\cos t}{(2+\sin t)^2}$ and $\ddot{K}(t) = \dfrac{-2(2+\sin t)^2\sin t- 4\cos^2t(2+\sin t)}{(2+\sin t)^4}$. Now, for $t\in[0,2\pi)$, $\dot{K}(t)=0$ when $t=\frac{\pi}{2}$ or $t=\frac{3\pi}{2}$; we now see that $\ddot{K}\left(\frac{\pi}{2}\right)=-\frac{2}{9}<0\implies K$ is maximized at this value and $\ddot{K}\left(\frac{3\pi}{2}\right)=2>0\implies K$ is minimized at this value. Therefore, the maximum Gaussian curvature is $K\left(\frac{\pi}{2}\right) = \frac{1}{3}$ and the miniminum Gaussian curvature is $K\left(\frac{3\pi}{2}\right)=-1$.
 
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