# Curvature

• Sep 4th 2009, 07:58 AM
chiph588@
Curvature
Let $c:[a,b] \to \mathbb{R}^2$ be a regular plane curve such that $||c(t)|| \leq 1$ for all $t \in [a,b]$. Suppose there is a point $t_0 \in [a,b]$ such that $||c(t_0)|| = 1$. Prove that the curvature at that point satisfies $|\kappa (t_0)| \geq 1$.
• Sep 4th 2009, 08:16 AM
InvisibleMan
Look at it this way, the curve is enclosed inside a circle of radius 1, there's a point in c which is tangent to this circle (at t0), now if it's the tangent circle to this point in the curve then its radius of curvature coinicide with the radius of the circle, otherwise the radius of curvature is smaller than the radius of the unit circle.
• Sep 4th 2009, 01:39 PM
ThePerfectHacker
Since $c([a,b])$ is a regular curve it means there exists a function $\gamma: I\to \mathbb{C}$ such that $\gamma(I)$ is the curve you are working with, and such that $|\dot \gamma (t) | = 1$ where $I$ is an interval. Thus, what we are doing is reparametrizing the regular curve into a unit-speed curve. You are told that somewhere on this curve the curve reaches the boundary of the unit disk, so for some $t_0\in I$ we have $|\gamma(t_0)| = 1$. The curvature is given by $\kappa = |\ddot \gamma (t_0)|$ since $\gamma$ is a unit-speed curve. We need to show that $|\ddot \gamma(t_0)| \geq 1$.

First we make a simple observation that $\dot \gamma \cdot \ddot \gamma = 0$. This is easy to show because since $|\dot \gamma | = 1 \implies \dot \gamma \cdot \dot \gamma = 1$. Now differenciate both sides to get $\dot \gamma \cdot \ddot \gamma + \ddot \gamma \cdot \dot \gamma = 0 \implies 2\dot \gamma \cdot \ddot \gamma = 0 \implies \dot \gamma \cdot \ddot \gamma = 0$. This tells us that $\dot \gamma \perp \ddot \gamma$ at any point $t\in I$.

Define the function $f: \mathbb{C} \to \mathbb{R}$ by $f(z) = z\bar z$, that is, $f(z)$ is the square of the length of the complex number.

Consider $f(\gamma(t))$, this is a real-valued function defined on $I$. But since $\gamma(t)$ never leaves the disk we have that $|f(\gamma(t))| \leq 1$, but at $t_0$ the problem says that $f(\gamma(t_0)) = 1$. Therefore, $f(\gamma(t))$ has a local maximum at $t_0$. This forces, $[ f(\gamma(t)) ]'|_{t=t_0} = 0 \text{ and }[f(\gamma(t))]''|_{t=t_0} \leq 0$. Now use the multivariable chain rule, $[f(\gamma(t))] ' = \nabla f \cdot \dot \gamma$, to get $\gamma(t_0)\cdot \dot \gamma(t_0) = 0$ and $\dot \gamma(t_0) \cdot \ddot \gamma(t_0) + 1 \leq 0$.

We have shown that $\gamma(t_0) \perp \dot \gamma(t_0)$ but we also shown above that $\dot \gamma(t_0) \perp \ddot \gamma (t_0)$, thus, $\gamma (t_0) || \ddot \gamma(t_0)$. Thus, $\ddot \gamma(t_0) = a \gamma(t_0)$ where $a\in \mathbb{R}$.

Returning back to the condition $\gamma (t_0) \cdot \ddot \gamma(t_0) + 1 \leq 0$ and substituting the result above we have that $\gamma (t_0) \cdot (a\gamma(t_0)) + 1 \leq 0 \implies a|\gamma(t_0)| + 1 \leq 0 \implies a \leq -1$. But that means $|a| \geq 1$, thus, $|\ddot \gamma (t_0)| = |a\gamma (t_0)| = |a| \geq 1$.
• Sep 4th 2009, 02:28 PM
chiph588@
Doesn't $f(\gamma(t))$ have a local maximum at $t_0$, not a minimum?
• Sep 4th 2009, 02:53 PM
ThePerfectHacker
Quote:

Originally Posted by chiph588@
Doesn't $f(\gamma(t))$ have a local maximum at $t_0$, not a minimum?

Okay I fixed it now.

Can you see how the chain rule works?
• Sep 4th 2009, 03:10 PM
chiph588@
Not quite, I'm a bit confused as to why the domain of $f$ is $\mathbb{C}$.
• Sep 4th 2009, 03:44 PM
ThePerfectHacker
Quote:

Originally Posted by chiph588@
Not quite, I'm a bit confused as to why the domain of $f$ is $\mathbb{C}$.

By definition $\mathbb{C}$ is the set $\mathbb{R}^2$. We just do not write complex numbers as $(a,b)$ we write them as $a+bi$. If it helps you can think of $f$ as $f:\mathbb{R}^2 \to \mathbb{R}$. The only reason why I used the complex number notation was that I started thinking of this problem with complex numbers so I decided to stick to complex numbers.

Here is the theorem that we used.

Theorem: Let $f:\mathbb{R}^n \to \mathbb{R}$ be a differenciable multivariable function ( $n\geq 2$) and let $\bold{g}: \mathbb{R}\to \mathbb{R}^n$ be a differenciable curve in $n$-space. Then $f\circ \bold{g}: \mathbb{R}\to \mathbb{R}$ is differenciable (in the classical sense, that is, one variable) with derivative $[ f(\bold{g}(t)) ] ' = \nabla f(\bold{g}(t))\cdot \dot{\bold{g}}(t)$.

This should be look like the classical chain rule because $\nabla f$ is the generalization of 'derivative' in $\mathbb{R}^n$. There are more general chain rules, and if you curious I can show you, but there are just not necessary here.

Now we defined (in your problem) $f:\mathbb{R}^2\to \mathbb{R}$ as $f(u,v) = u^2+v^2$, that is, the square of the distance. Thus, $\nabla f(u,v) = 2(u,v)$, so that means, $\nabla f(\gamma(t)) = 2\gamma(t)$. Therefore, the derivative of $f(\gamma(t))$ would be $2\gamma(t) \cdot \dot \gamma(t)$.