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Let $\textbf{$\gamma$ }(t)$ be a unit-speed curve with $\kappa (t) > 0$ and $\tau (t) \neq 0$ for all $t$. Show that, if $\textbf{$\gamma$}$ is spherical, i.e., if it lies on the surface of a sphere, then $$\frac{\tau }{\kappa }=\frac{d}{ds}\left (\frac{\dot \kappa}{\tau \kappa^2}\right ) \tag 1$$

Conversely, show that if Eq. $(1)$ holds, then $$\rho^2+(\dot \rho \sigma )^2=r^2$$ for some (positive) constant $r$, where $\rho = \frac{1}{\kappa}$ and $\sigma = \frac{1}{\tau}$, and deduce that $\textbf{$\gamma$}$ lies on a sphere of radius $r$. Verify that Eq. $(1)$ holds for Viviani’s curve.

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At the second part, I am facing some difficulties at showing that $\textbf{$\gamma$}$ lies on a sphere of radius $r$.

Could you give me some hints how to show it?

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EDIT:

I have done everything besides the verification that Eq. $(1)$ holds for Viviani’s curve.

I have done the following:

The Viviani's curve is $$\gamma (t)=\left (\cos^2 t -\frac{1}{2}, \sin t\cos t , \sin t\right )$$

The curvature is given by the formula $$\kappa =\frac{ \| \gamma '' \times \gamma '\|}{\|\gamma '\|^3}$$

I found that it is equal to $$\kappa =\frac{\sqrt{5+3\cos^2 t}}{(1+\cos^2 t)^{\frac{3}{2}}}$$

Its derivative is $$\kappa '=\frac{6 \cos \sin t (\cos^2 t+2)}{\sqrt{5+3\cos^2 t}(1+\cos^2 t)^{\frac{5}{2}}}$$

The torsion is given by the formula $$\tau =\frac{(\gamma ' \times \gamma '' ) \cdot }{\|\gamma ' \times \gamma '' \|^2}$$

I found that it is equal to $$\tau =\frac{6 \cos t}{5+3 \cos^2 t}$$

Then $$\frac{\kappa '}{\tau \kappa^2}=\frac{\sin t(\cos^2 t+2)(1+\cos^2 t)^{\frac{1}{2}}}{\sqrt{5+3 \cos^2 t}}$$

I calculated $$\left ( \frac{\kappa '}{\tau \kappa^2} \right ) '$$ at Wolfram but this is not equal to $$\frac{\tau }{\kappa}$$ What have I done wrong?

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    $\begingroup$ If $\tau = 0$, what does $\frac{\ddot{\kappa}}{\tau \kappa^2}$ mean? $\endgroup$ Commented Oct 15, 2015 at 20:00
  • $\begingroup$ It should be $\tau (t) \neq 0, \forall t$. I edited it. @JasonDeVito $\endgroup$
    – Mary Star
    Commented Oct 15, 2015 at 20:58
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    $\begingroup$ I just found this:math.stackexchange.com/questions/455661/…, which I think makes this question a duplicate. $\endgroup$ Commented Oct 15, 2015 at 21:55
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    $\begingroup$ Could you clarify exactly what in Ted's answer (and the comments following it) that you didn't understand? $\endgroup$ Commented Oct 22, 2015 at 20:58
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    $\begingroup$ Yes, they are an orthonormal basis of $\mathbb{R}^3$. In addition, there are known formulas relating $t',n'$ and $b'$ to $t,n,b$ (called the Frenet-Serre formulas.) $\endgroup$ Commented Oct 22, 2015 at 21:12

1 Answer 1

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I think this is a lot easier than what you think: Take (1) above: $$\frac{\tau }{\kappa }=\frac{d}{ds}\left (\frac{\dot \kappa}{\tau \kappa^2}\right ) \tag 1$$

Rewrite this in terms of $\sigma$ and $\rho$ $$\frac{\rho}{\sigma }=\frac{d}{ds}\left (\sigma (-\dot \rho)\right ) = -\frac{d}{ds}\left (\dot{\rho} \sigma \right ) \tag 2$$

Now you're almost done.

Take the derivative of the LHS of your expression:

$$\frac{d}{ds}\left(\rho^2 + (\dot{\rho} \sigma)^2 \right) = 2\rho \dot{\rho} + 2(\dot{\rho} \sigma)\frac{d}{ds}(\dot{\rho} \sigma)$$ But we just figured out what that last term is in terms of $\rho$ so substitute: $$\frac{d}{ds}\left(\rho^2 + (\dot{\rho} \sigma)^2 \right) = 2\rho \dot{\rho} + 2(\dot{\rho} \sigma)\left(\frac{-\rho}{\sigma}\right)$$ You can now see that $$\frac{d}{ds}\left(\rho^2 + (\dot{\rho} \sigma)^2 \right) = 0$$

Or in other words $\rho^2 + (\dot{\rho} \sigma)^2$ is a constant (it must be positive, do you see why?). And if you call that constant $r^2$ you are done.

Put $a = \rho N + \sigma \dot{\rho} B$

We have just shown that this vector has a constant length of $r$. You can use the Frenet equations to show that $\dot{a} = -T = -\dot{\gamma}$

Then it is clear that $\gamma + a$ is a constant vector (the center of the circle) and the path $\gamma$ is at a constant distance $r$ from it.

For Viviani's curve put $$ \gamma(t) = \left(\frac{r}{2}(1 + \cos t), \frac{r}{2} \sin(t), r \sin(t/2)\right)$$ Which you can get from Wikipedia or Mathworld (remember we have a constant $r$ for our radius). then $$ \gamma'(t) = \frac{r}{2}\left(-\sin (t), \cos(t), \cos(t/2)\right)$$ $$ \gamma''(t) = -\frac{r}{2}\left(\cos (t), \sin(t), \sin(t/2)/2\right)$$ This is different from what you have. You need to calculate this out to finish.

For example I get $||\gamma'' \times \gamma'|| = \frac{r^2}{8\sqrt{2}}\sqrt{13 + \cos (t)}$

Can you take it from there?

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    $\begingroup$ OK sorry for the confusion it's been a few years since I did this stuff. I think I got it all correct now. Construct the vector $a$ and see that (according to Frenet) it's derivative is $-T$. This should be all you need for the last bit. $\endgroup$
    – amcalde
    Commented Oct 23, 2015 at 17:42
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    $\begingroup$ Because it works! We just saw that this vector has a constant norm. You use that particular combination of $N$ and $B$ to make the Frenet equations work out the right way so that $\dot{a} = -T$. Then everything works out nicely! $\endgroup$
    – amcalde
    Commented Oct 23, 2015 at 18:34
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    $\begingroup$ That is what you conclude. $\gamma$ is a constant distance away from a constant vector in space. That is what it means to be on the surface of a sphere. $\endgroup$
    – amcalde
    Commented Oct 24, 2015 at 10:44
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    $\begingroup$ The center of the circle is fixed in space (that's $\gamma + a$ for us). It doesn't change. The path $\gamma$ is always at a fixed distance from this center. This means that $|a|$ is constant (and equal to $r$) which we have already shown. This is all you need to show. Your path is at a fixed distance from some fixed point in space. QED. $\endgroup$
    – amcalde
    Commented Oct 24, 2015 at 11:11
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    $\begingroup$ @MaryStar I'll take a look and add to the post. $\endgroup$
    – amcalde
    Commented Oct 26, 2015 at 18:50

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