# What does the definition of curvature mean?

First question, am I right in saying that curvature measure how quickly the direction of a cruve changes?

Also, we have been given the "definition":

$$T'(t) = \kappa(t) | \gamma'(t) | U(t)$$

where $T(t)$ is the unit tangent of the curve $\gamma(t)$ (so $T'(t)$ is its derivative), $\kappa(t)$ is the curvature, $|\gamma ' (t)|$ is the speed vector and $U(t)$ is the unit normal of the curve $\gamma (t)$.

My lecturer keeps saying that this is more of a "definition" rather than a formula to work out the curvature, how is it? what does this definition then say/mean?

Also, from this definition, somehow we get the formula

$$\kappa = \frac{T' \cdot U}{|\gamma '|}$$

where $T' \cdot U$ is the dot product between $T'$ and $U$. How do we get this formula? I thought that we would have to divide by $|\gamma'|$ and $U$ to get $\kappa$, so where does the dot product come from?

EDIT: I figured out the last bit. Using the definition, we get

$$T' \cdot U = (\kappa |\gamma '| U) \cdot U = \kappa |\gamma ' | (U \cdot U) = \kappa |\gamma '|$$

which we can then rearrange. You get $U \cdot U = 1$ as they are both unit normal vectors (i.e of length 1) and there is a formula that says we can do $| \alpha| = \alpha \cdot \alpha$.

• Started editing before I saw the answer was posted. Not sure if I should delete or not? – Kaish Jan 30 '13 at 14:22
• You shouldn't delete. Part of the goal of this site is to provide future reference to people who have the same problems. So with upvotes and accepted answers someone who has the same problem as you can quickly and easily find what they want. – guest196883 Jan 30 '13 at 14:37

## 1 Answer

There are a number of characterizations of curvature. The "most intuitive" and geometric one, in my opinion, is the inverse of the radius of curvature. That is, intuitively, the radius of the circle that lies tangent to the curve in a neighborhood of the point (more formally, the osculating circle.

You may think of it as the limit of the distance to the intersection pont of two normals of arbitrarily close points on the curve. Your lecturer's definition coincides nicely with this one.

I'll first explain where the dot product comes from. If $$\vec v = \alpha \vec w$$ Then taking the definition of $\vec v \cdot \vec w$ as $|v||w|\cos\theta$ where $\theta$ is the angle between them. We have that since $|w| = \alpha|v|$, and $\theta = 0$ because they are scalar multiples, and therefore in the same direction,$$\vec v \cdot \vec w = \alpha|v|^2$$ and since $|v|$ is 0 for a unit vector (as in your case): $$T'\cdot U = \kappa |\gamma'|$$ And division gives your second formula.