## 275- Constant curvature

The text we are using for Calculus III introduces the notion of unit tangent vector, principal unit normal vector, and curvature, for smooth curves $\vec r(t)$, and it also mentions that circles and lines are planar curves of constant curvature. (A curve is planar if its image is contained in a plane, i.e., if it describes a two-dimensional trajectory.)

Surprisingly, though, the text does not explain (not even in the exercises) that circles and lines are the only smooth planar curves of constant curvature. The argument for this is simple enough, so I will show it here:

Recall that if $\vec r(t)$ is a smooth curve, we set

$\displaystyle\vec T=\frac{d\vec r}{ds}$

where $s$ denotes arc length. We can define the curvature $\kappa$ of $\vec T$ as

$\displaystyle\kappa=\left\|\frac{d\vec T}{ds}\right\|.$

Assume that $\kappa$ is constantly equal to zero. Then $\frac{d\vec T}{ds}=\vec 0$, so $\vec T$ is constant, so $\vec r=s\vec T+\vec r_0$ for some constant vector $\vec r_0$. This is the parametric equation of a line, so we have shown that the only curves with zero curvature are straight lines. Notice that for this we did not need to assume that $\vec r$ is a planar curve. From now on we may assume that this is the case and that $\vec\kappa$ is constant and different from zero.

In terms of $t$, we can equivalently compute $\vec T$ as

$\displaystyle\vec T=\frac{d\vec r/dt}{\|d\vec r/dt\|}=\frac{\vec v}{\|\vec v\|}.$

We then set $\vec N$ to be the unit vector in the direction of $\frac{d\vec T}{ds}$, so

$\displaystyle\kappa\vec N=\frac{d\vec T}{ds}$

or, equivalently in terms of $t$,

$\displaystyle\vec N=\frac{d\vec T/dt}{\|d\vec T/dt\|},$

so $\displaystyle \vec T'=\kappa\|\vec v\|\vec N$. Recall that $\vec N\cdot\vec T=0$, since $\vec T\cdot\vec T=1$ is constant, so its derivative is zero.

Since $\vec N\cdot\vec N=1$ is also constant, again we have that $\vec N\cdot \vec N'=0$. Since $\vec T$ is also perpendicular to $\vec N$ and we are assuming that $\vec r$ moves on a plane, this means that $\vec N'$ is a multiple of $\vec T$,

$\vec N'(t)=a(t)\vec T(t),$

where $a(t)$ is some scalar function. To compute $a$, notice that $\vec N'\cdot \vec T=a\vec T\cdot\vec T=a$. But also $\vec T\cdot \vec N=0$, so $\vec T'\cdot\vec N+\vec T\cdot\vec N'=0$, or $a=-\vec T'\cdot\vec N$. Since $\vec T'=\kappa\|\vec v\|\vec N$, we finally find that

$a=-\kappa\|\vec v\|\mbox{ and }\vec N'=-\kappa\|\vec v\|\vec T.$

Recall also that the osculating circle to the curve $\vec r$ at the point $t$ is the circle of the same curvature as $\vec r$ at $t$ and centered in the direction of the vector $\vec N$ at $t$, so its center is given by

$\displaystyle \vec c(t)=\vec r(t)+\frac1{\kappa}\vec N(t),$

since the curvature of a circle is the inverse of its radius. In particular, since $\kappa$ is constant, all the osculating circles to $\vec r$ have the same radius. Now we prove that $\vec c$ is constant (from the assumptions that $\vec r$ is planar and $\kappa$ is constant). To see this, compute $\vec c\,'(t)$ to find

$\displaystyle \vec c\,'(t)=\vec r\,'(t)+\frac1{\kappa}\vec N'(t)=\vec v-\frac{\kappa\|\vec v\|}{\kappa}\vec T,$

or $\vec c\,'=\vec v-\|\vec v\|\vec T=\vec 0$.

But this completes the proof, because then $\displaystyle\|\vec r(t)-\vec c\|=\frac1{\kappa}$, since $\|\vec N\|=1$, i.e., the curve $\vec r$ lies on a circle of radius $1/\kappa.$ ${\sf QED}$

On the other hand, there are curves of constant curvature other than lines and circles, if the curve is not confined to a plane. The helix $\vec r(t)=\cos(t)\vec i+\sin(t)\vec j+t\vec k$ is an example.