Multivariable Lesson 1.4: Curvature

We have already associated three vectors with every point on a space curve to indicate its directions of motion. We will close our discussion of space curves by also associating a scalar function with every point, the curvature. We will first define curvature $\kappa(t)$ as \begin{displaymath}\kappa=\left\vert\frac{d\mathbf{T}}{ds}\right\vert\end{displaymath}

If we revisit some concepts we have already discussed in previous lessons, we should be able to make sense of this definition. We recall that $\frac{d\mathbf{T}}{dt}$ is always perpendicular to $\mathbf{T}(t)$, because $\mathbf{T}$ doesn’t change in magnitude. The magnitude of $\frac{d\mathbf{T}}{dt}$ thus represents how quickly $\mathbf{T}(t)$ is changing direction. But since $\frac{d\mathbf{T}}{dt}$ depends on how quickly $t$ changes, we will always calculate curvature based on a parametrization by arc length $s$. This way, curvature only depends on the curve itself, not how we choose to represent it.

Since curvature represents how quickly $\mathbf{T}(s)$ is changing direction, it’s conceptually reasonable that higher curvature translates into more “curviness,” or at least, tighter curves. Think, for example, about the difference between a big circle and a small circle. In a small circle, $\mathbf{T}(s)$ is changing direction quickly with relation to the arc length. In a big circle, $\mathbf{T}(s)$ goes through a large $s$ without changing direction very much. Let’s try to examine real curves, though. First, let’s find equivalent and easier ways to calculate curvature: \kappa=\left\vert\frac{d\mathbf{T}}{ds}\right\vert=\left\vert......\vert=\frac{\vert\mathbf{T}'(t)\vert}{\vert\mathbf{r}'(t)\vert} The last step comes from the fact that $s(t)=\int_a^t\vert\mathbf{r}'(u)\vert du$ as discussed earlier, so differentiating both sides with respect to $t$ results in $ds/dt=\mathbf{r}'(t)$. I’ll provide one more formula for curvature without proof: \kappa(t)=\frac{\vert\mathbf{r}'(t)\times\mathbf{r}''(t)\vert}{\vert\mathbf{r}'(t)\vert^3}

Now we can calculate curvature more easily. First, let’s find the curvature of a circle with radius $a$, $\mathbf{r}(t)=(a\cos t,a\sin t)$. (Go calculate it now yourself using the second formula provided.) You should get that $\vert\mathbf{T}'(t)\vert=1$ and $\vert\mathbf{r}'(t)\vert=a$, so $\kappa=1/a$. This agrees with our previous analysis of circles, that larger radii result in smaller curvature.

Picture of the osculating circle on our favorite helix

It turns out that curvature has a lot to do with circles. At any point on a space curve, the plane defined by $\mathbf{T}$ and $\mathbf{N}$ is called the osculating plane. This plane is, loosely speaking, the plane in which the curve lies at that point: it contains both the tangent (the direction in which the curve is going at that moment) and the normal (the direction in which the curve’s motion is changing). At that point, we have already approximated the curve’s direction of motion with a vector, $\mathbf{T}$; now, we can also approximate the curve’s curve with a circle. The circle that goes through that point, shares the same tangent at that point, is on the concave side of the curve (the one that $\mathbf{N}$ is pointing toward), and has the same curvature at that point is called the osculating circle. That circle is the best approximation for the curviness of the curve at that point.

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