The Curvature of Straight Lines and Circles

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 $\| \vec{r'}(t) \| = \sqrt{R^2 \sin ^2 t + R^2 \cos ^2 t} = \sqrt{R^2} = R^2$.
  • 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.

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