# If $\| \alpha(s) \| \leq \| \alpha(s_0) \| = R$, then the curvature $k(s_0)$ is greater than $1/R$

This question is in Ted Shifrin's A first course in curves and surfaces, page 18, exercise 7:

Suppose $\alpha$ is an arclength-parametrized space curve with the property that $\| \alpha(s) \| \leq \| \alpha(s_0) \| = R$ for all $s$ sufficiently close to $s_0$ . Prove that $k(s_0) \geq 1/R$. (Hint: Consider the function $f(s)=\| \alpha(s)\|^2$. What do you know about $f''(s_0)$?)

Here's what I have tried:

$$f(s) = \| \alpha(s)\|^2 = \langle \alpha(s), \alpha(s) \rangle \implies f'(s) = 2 \langle T(s), \alpha(s) \rangle$$

And using the Cauchy-Schwartz inequality :

$$f'(s_0) = 2 \langle T(s_0), \alpha(s_0) \rangle \leq 2 \|T(s_0)\|\|\alpha(s_0)\|=2R \implies \langle T(s_0), \alpha(s_0) \rangle \leq R$$

Now we find $f''(s)$:

$$f''(s) = 2 \langle k(s)N(s), \alpha(s) \rangle + 2 (\langle T(s), \langle T(s \rangle )= 2 k(s) \langle N(s), \alpha(s) \rangle + 2$$

And because $s_0$ is maximum of $\| \alpha(s)\|$:

$$2 k(s_0) \langle N(s_0), \alpha(s_0) \rangle + 2 \leq 0 \implies -1/k(s_0) \geq \langle N(s_0), \alpha(s_0) \rangle$$

I don't know how to continue from here.

First of all, what is $f'(s_0)$, precisely? No inequalities here. You should be thinking about Cauchy-Schwarz in the analysis of the second derivative. Several warnings: First, $s_0$ is a maximum, not a minimum; but $f''(s_0)\le 0$ (not strict inequality). Second, think about the sign of $\langle N(s_0),\alpha(s_0)\rangle$. Last, don't divide by $k(s_0)$ if you want to solve for $k(s_0)$ :)

• Incomplete "hint answers" like this do not help anyone who wants the actual answer. As someone with a very poor understanding of differential geometry, this answer is useless to me. Stack Exchange is about curating knowledge, not about helping the person who asked the question and them alone. – forest Sep 25 '18 at 10:19
• @forest We'll have to agree to disagree. Witness the huge number of duplicates and near duplicates. – Ted Shifrin Sep 25 '18 at 13:59
• That is a thought-terminating cliché. Referring to duplicates may be valid (I have not seen any), but does not change my opinion that this answer to this question is incomplete. – forest Sep 25 '18 at 14:04
• @forest if you have a very poor understanding of differential geometry, then you should start with easier problems. SE is not an encyclopedia, it is not a list of solved questions that your textbook is "missing." Curating knowledge in mathematics involves providing insight, and Professor Shifrin does this by offering a thought process that would lead OP to the answer. That is probably how you curate knowledge. Have you taught a class? Do you have a degree in math education? Have you taught any classes? Why would you claim to know more about curating knowledge than a lifelong professor. Sheesh – Andres Mejia Sep 27 '18 at 22:37
• @TedShifrin +1 by the way. Sorry for the rant on what I thought was a nice answer, that helped me answer the question (with some work), which I stumbled upon. – Andres Mejia Sep 27 '18 at 22:41

This question is a classic from curve and surface geometry. Good to run into it here, like an old friend one hasn't see in awhile.

These things being said:

We note the hypothesis that

$\Vert \alpha(s) \Vert \le \Vert \alpha(s_0) \Vert \tag 1$

essentially tells us that $s_0$ is a local maximum of the function $\Vert \alpha(s_0) \Vert$; since the real function $w \to w^2$ is strictly monotonically increasing for $w \ge 0$, $s_0$ will also be a local maximum for $\Vert \alpha(s) \Vert^2$; we shall work with this function in lieu of $\Vert \alpha(s) \Vert$.

We set

$f(s) = \Vert \alpha(s) \Vert^2 = \langle \alpha(s), \alpha(s) \rangle; \tag 2$

then

$f'(s) = 2\langle \alpha'(s), \alpha(s) \rangle = 2\langle T(s), \alpha(s) \rangle, \tag 3$

where

$T(s) =\alpha'(s) \tag 4$

is the unit tangent vector field to the curve $\alpha(s)$; we further have

$f''(s) = 2\langle T'(s), \alpha(s) \rangle + 2\langle T(s), T(s) \rangle; \tag{5}$

we may take this equation a step further by recalling that $T(s)$ is a unit vector and

$T'(s) = \kappa(s) N(s), \tag 6$

where $N(s)$ is the unit normal field to, and $\kappa(s) > 0$ is the curvature of, $\alpha(s)$. Thus,

$f''(s) = 2\kappa(s) \langle N(s), \alpha(s) \rangle + 2; \tag 7$

since $s_0$ is a local maximum of $f(s)$, we must have

$f''(s) \le 0; \tag 8$

thus,

$\kappa(s_0) \langle N(s_0), \alpha(s_0) \rangle + 1 \le 0, \tag 9$

or

$\kappa(s_0)\langle N(s_0), \alpha(s_0) \rangle \le -1; \tag{10}$

taking absolute values of both sides, we find

$\kappa(s_0)\vert \langle N(s_0), \alpha(s_0) \rangle \vert \ge 1; \tag{11}$

by Cauchy-Schwarz,

$\vert \langle N(s_0), \alpha(s_0) \rangle \vert \le \Vert N(s_0) \Vert \Vert \alpha(s_0) \Vert = \Vert \alpha(s_0) \Vert \tag{12}$

since

$\Vert N(s_0) \Vert = 1; \tag{13}$

combining (11) and (12), we obtain

$\kappa(s_0) R \ge 1, \tag{14}$

whence

$\kappa(s_0) \ge \dfrac{1}{R}. \tag{15}$

• Robert: Please don't ruin my textbook by posting complete solutions to the homeworks. You realize you're likely doing lots of students' homework this way. (This person is asking for guidance, not a professional solution.) – Ted Shifrin Sep 20 '18 at 0:01
• @TedShifrin: with all due respect, and I do admire your book and your work, have I done so before? If so, can you cite links to such answers? Also, most of my post re-iterated what the OP already had stated; in light of that, I merely added a few notes to the end of his attempt. "Ruin my textbook", which I said I admired, seems pretty strong. Best Regards. – Robert Lewis Sep 20 '18 at 0:08
• @TedShifrin: It seems to me a legitimate question as to how is myself or any other user of this site to know if the poster is taking a class using your book or just reading it? I don't think I'm wrong in asserting that many questions posted on this site occur in textbooks, whether they are quoted as such or not. Also, if you look carefully at the last line in the question, you will see that the poster was right on top of the answer, he just didn't get how to use Cauchy-Schwarz in this context. I did, and so posted. By the way, am I wrong in assuming that you downvoted my answer? – Robert Lewis Sep 20 '18 at 0:19
• You too are free to not give textbook problems as graded homework. With the rise of the internet (and Yahoo Answers being a much worse culprit than Math SE), the only way to prevent student cheating is to use proctored assessment. Attempting to impose rules on online communities would be like a Whack-A-Mole game that never actually works. Maybe sad, but that's the way it is... – user21820 Sep 21 '18 at 8:32
• @TedShifrin Respectfully, this answer at most ruins your graded homework, if anything, not your book. Secondly, students that receive graded homework have hundreds of ways to cheat. They can, for example, copy the solutions from each other. They can pay someone to do it (I have seen this happen several times unfortunately). A well written answer on Math.SE should be the least of your concerns. Congratulations on giving graded homework assignments, that is I great thing in my opinion, but respectfully I think your reaction doesn't make sense. – Pedro A Sep 21 '18 at 21:50