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I'm reading by myself the classical Do Carmo's differential geometry book and I couldn't solve this question (page 25):

The hint on the back of the book says:

My attempt

$(|\alpha(s)|^2)'=0\implies (\langle \alpha(s),\alpha(s)\rangle)'=0\implies \langle\alpha'(s),\alpha(s)\rangle=0$

Differentiating again the last equation above, we have: $\langle\alpha''(s),\alpha(s)\rangle=-1$, in other words $\langle k(s)n(s),\alpha(s)\rangle=-1$

Differentiating a third time, we have:

$\langle k'(s)n(s)+k(s)n'(s),\alpha(s)\rangle+\langle k(s)n(s),\alpha'(s)\rangle=0\implies \langle k'(s)n(s)+k(s)n'(s),\alpha(s)\rangle=0$

So my last result was

$$\langle k'(s)n(s)+k(s)n'(s),\alpha(s)\rangle=0$$

I'm stuck there, any help would be very useful.

EDIT

Using the Frenet-serret formulas for the $n'$ in my last result above gave me:

$$\langle b(s),\alpha(s)\rangle=\frac{1}{R'(s)\tau(s)}$$

However, I couldn't prove $\alpha(s)=-Rn+R'Tb$ using this fact. I'm stuck again.

ziggurism
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user42912
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1 Answers1

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If you substitute the Frenet-Serret equation $\displaystyle n'(s)=-k(s)\alpha'(s)+\tau(s) b(s)$ into your third equation, and also substitute your second equation $n(s)\cdot\alpha(s)=-\frac{1}{k(s)},$ you're almost there. Finally just express $\alpha^2$ in terms of the Frenet-Serret frame.

Edit: Here it is step by step.

  1. $\alpha^2 = r^2$
  2. Take derivative: $\alpha\cdot\alpha' = 0.$
  3. Take derivative: $(\alpha')^2 + \alpha\cdot\alpha''=0.$
  4. Take the derivative: $3\alpha'\cdot\alpha'' + \alpha\cdot\alpha''' = 0.$
  5. Arc length is by definition the parametrization which makes $\alpha'$ the unit tangent vector, so $(\alpha')^2=1$ in step 3 or $\alpha'\cdot\alpha'' = 0 $ in step 4. Leaving $\alpha\cdot\alpha''' = 0.$
  6. $\alpha''=kn,$ so now step 3 can be written $\alpha\cdot n=-\frac{1}{k}.$
  7. Also $\alpha'''=k'n+kn'.$
  8. From steps 5 and 7, $k'\alpha\cdot n+k\alpha\cdot n'=0.$
  9. By the Frenet-Serret formula $n'=-k\alpha'+\tau b,$ so we now have $k'\alpha\cdot n+k\tau\alpha\cdot b=0.$
  10. By step 6, this is $-\frac{k'}{k}+k\tau(\alpha\cdot b)=0$ or $\alpha\cdot b=\frac{k'}{k^2\tau}.$
  11. Writing $\alpha$ in the Frenet-Serret frame, $\alpha = (\alpha\cdot\alpha')\alpha' + (\alpha\cdot n)n + (\alpha\cdot b)b,$ we see that $\alpha^2=r^2=(\alpha\cdot n)^2+(\alpha\cdot b)^2$ by the Pythagorean theorem.
  12. So $\alpha = -\frac{n}{k} + \frac{k'b}{k^2\tau}$ and $r^2=\left(-\frac{1}{k}\right)^2+\left(\frac{k'}{k^2\tau}\right)^2=\left(\frac{1}{k}\right)^2+\left(\left(\frac{1}{k}\right)'\right)^2\left(\frac{1}{\tau}\right)^2,$ which was to be proved.
ziggurism
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  • Thank you for your answer. What do you mean by $\alpha^2$? – user42912 Nov 11 '17 at 03:48
  • Why downvotes? If this answer is not useful I am happy to delete... – ziggurism Nov 11 '17 at 04:45
  • It wasn't me who gave the downvote! – user42912 Nov 11 '17 at 11:08
  • @user42912 $\alpha^2=\alpha\cdot\alpha.$ – ziggurism Nov 11 '17 at 13:35
  • I have added a step-by-step solution. I hope it will be more helpful. Also I hope the downvoter will reveal why it is deficient? – ziggurism Nov 11 '17 at 13:36
  • We are in the same boat. I hope the downvoter also reveal why my question is deficient. Thank you for your answer again. My doubt is why the radius of the sphere is $1/k$ – user42912 Nov 11 '17 at 15:45
  • @user42912 the radius of the sphere is not $1/k,$ unless $k$ is constant. Sorry if I suggested otherwise by reusing the symbol $R$. We just proved that the radius of the sphere is $\sqrt{\left(\frac{1}{k}\right)^2+\left(\left(\frac{1}{k}\right)'\right)^2\left(\frac{1}{\tau}\right)^2}$ – ziggurism Nov 11 '17 at 15:50
  • Great answer! I upvoted and accepted your answer. I hope our downvoters come here to correct these unfair votes – user42912 Nov 11 '17 at 21:52
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    I am very sorry. I am responsible for downvoting, but I misclicked. I actually find your answer well written, and I have upvoted it. Sorry again. – Salvatore Baldino Nov 12 '17 at 23:19
  • @SalvatoreBaldino thanks Sal. It's a little disheartening to see a negative score on an honest attempt at an answer. But it eventually got back up into the positive, so I got over it. Anyway of course it's just for fun. Cheers. – ziggurism Nov 13 '17 at 00:22