A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency

Question

Suppose $a$ is an arc length-parametrized space curve with the property that $\|a(s)\| \leq \|a(s_0)\| = R$ for all $s$ sufficiently close to $s_0$. Prove that $k(s_0) \geq 1/R$.

So, I was going to consider the function $f(s) = \|a(s)\|^2$ but then the hint asks what we know about $f''(s_0)$? I am confused on how to link the two together and show from there that the curvature of $k(s_0) \geq 1/R$.

I know that we should use the second derivative test to see the local minima or maxima thus giving us if the curve points upward or downward, but I am a little lost at where to begin. I was told that $f(s_0)$ is constant, why? How come $f(s)$ is not a constant function here?

Answer 1

Here's what I think is going on, I'll let you fill in the details. The relevant definitions are all included here: http://en.wikipedia.org/wiki/Curvature#Curvature_of_space_curves

Because $a$ is parametrized by the arc-length, the curvature at a point $s$ is simply $\kappa(s) = ||a''(s)||$. Now imagine a curve $\gamma$ (also parametrized by the arc-length) whose image is the intersection of a sphere of radius $R$, and a plane passing through the origin in $\mathbb{R}^3$, and such that $\gamma(s_{0}) = a(s_{0})$ (and if you want $\gamma'(s_{0})= a'(s_{0})$); in this case the curvature of $\gamma$ at $s_{0}$ (and at any other point for that matter) is simply $\kappa(s_{0}) = 1/R$.

Now we know that $||a(s)|| \leq ||a(s_{0})|| = R$ for $s$ close to $s_{0}$. If you think about this geometrically, this means that near $s_{0}$, the curve $a$ does not leave the sphere of radius $R$; in fact it may even curve inward. Therefore, it is not too hard to see that $||a''(s_{0})|| \geq ||\gamma''(s_{0})||$, or in other words, $\kappa(s_{0}) \geq 1/R$.

Answer 2

Some intermediate hints:

1. We are given that $\Vert a(s) \Vert \leq \Vert a(s_0)\Vert$ for all $s$ near $s_0$. Squaring both sides tells us that $f(s) \leq f(s_0)$ for all $s$ near $s_0$. In other words, $s = s_0$ is a local max for the function $f$. What does that imply about $f''(s_0)$?

2. You should compute $f''(s_0)$ as the hint says. This means that you'll have to compute $f'(s)$, then its derivative $f''(s)$, then finally plug in $f''(s_0)$. To compute these derivatives, you should use that $f(s) = \Vert a(s) \Vert^2 = \langle a(s), a(s) \rangle$, and use the product rule for inner products: $$\frac{d}{ds}\langle F(s), G(s) \rangle = \langle F'(s), G(s) \rangle + \langle F(s), G'(s) \rangle.$$

3. Finally, once you've computed $f''(s)$ and also used Hint 1, you should get an inequality. At this point, you might want to apply the Cauchy-Schwarz Inequality $$\left|\langle v,w \rangle\right| \leq \Vert v \Vert \Vert w \Vert.$$ You'll also want to remember that since $a(s)$ is unit speed, we have $\Vert a'(s) \Vert = 1$ for all $s$.