The Curvature of Straight Lines and Circles

We are now going to apply the concept of curvature to the classic examples of computing the curvature of a straight line and a circle. Before we look at the following theorems, recall from the Curvature at a Point on a Single Variable Real Valued Function page that the curvature of a twice differentiable single-variable function $y = f(x)$ is given by $\kappa (x) = \frac{\mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}}$.

Also recall that in general, for a vector-valued function $\vec{r}(t)$, the curvature can be calculated as $\kappa (t) = \frac{ \| \hat{T'}(t) \|}{\| \vec{r'}(t) \|}$ where $\hat{T'}(t)$ is the derivative of the unit tangent vector, and $\vec{r'}(t)$ is the derivative of the vector-valued function.

The Curvature of Straight Lines

Theorem 1: A straight line has curvature $\kappa_{\mathrm{straight\: line}} = 0$.

• Proof: Consider the line $f(x) = mx + b$. Notice that $f'(x) = m$ and $f''(x) = 0$. Therefore $\mid f''(x) \mid = 0$ and since $\kappa (x) = \frac{\mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}}$, we have that $\kappa (x) = 0$ for all $x \in \mathbb{R}$.

• More generally, consider a generic line with parametric equations $x(t) = x_0 + at$, $y(t) = y_0 + bt$, and $z(t) = z_0 + ct$, and so define $\vec{r}(t) = (x_0 + at, y_0 + bt, z_0 + ct)$.

• First compute $\vec{r'}(t) = (a, b, c)$. Therefore $\| \vec{r'}(t) \| = \sqrt{a^2 + b^2 + c^2}$. Now we have that $\hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \left ( \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}} \right )$. Notice that each component is a scalar, so differentiating $\hat{T}(t)$, we get $\hat{T'}(t) = (0, 0, 0)$, and $\| \hat{T'}(t) \| = 0$. Therefore since $\kappa = \frac{ \| \hat{T'}(t) \|}{\| \vec{r'}(t) \|}$ we have $\kappa = 0$. $\blacksquare$

Intuitively, it should make sense that a straight line has curvature $0$ as the tangent vector is constant along the line, and so its rate of change is $0$.

The Curvature of Circles

Theorem 2: A circle with radius $R$ has curvature $\frac{1}{R}$, that is $\kappa_{\mathrm{circle}} = \frac{1}{R}$.

• Proof: Consider an arbitrary circle with radius $R > 0$ that is centered at the origin and given by the vector equation $\vec{r}(t) = (R \cos t, R \sin t)$. In order to calculate $\kappa$ we must calculate $\| \hat{T'}(t) \|$ and $\| \vec{r'}(t) \|$.

• Let's first calculate $\vec{r'}(t)$. To calculate $\vec{r'}(t)$ we take the derivative of $\vec{r}(t)$ and get $\vec{r'}(t) = (-R \sin t, R \cos t)$. Taking the magnitude, we have $\| \vec{r'}(t) \| = \sqrt{R^2 \sin ^2 t + R^2 \cos ^2 t} = \sqrt{R^2} = R$.

• Now let's calculate $\vec{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \frac{1}{R} (-R \sin t, R \cos t) = (-\sin t, \cos t )$. Now take the derivative of $\vec{T}(t)$ to get $\vec{T'}(t) = (-\cos t, -\sin t)$. Lastly, compute $\| \vec{T'}(t) \| = \sqrt{ \cos ^2 t + \sin ^2 t} = \sqrt{1} = 1$.

• Using the formula for the curvature, we have that $\kappa_{\mathrm{circle}} = \frac{\| \hat{T'}(t) \|}{\| \vec{r'}(t)\|} = \frac{1}{R}$. $\blacksquare$

Corollary 1: A circle with a large radius has a small curvature, and a circle with a small radius has a large curvature.

• Proof: Since $\kappa_{\mathrm{circle}} = \frac{1}{R}$, then as $R \to 0$, $\kappa \to \infty$ and as $R \to \infty$, $\kappa \to 0$. $\blacksquare$

Once again, it should be rather clear that a larger circle has a smaller curvature. If we were to enlarge a circle and look at a portion of the arc of the circle, then as the circle gets larger, the arc would appear to look more and more like a straight line (which we saw had curvature zero), similarly to what we acknowledged with Linear Approximation. The converse for small circles is also sensible.