Mathematical Immaturity

14.13 Exercises

• Two points $A$ and $B$ on a unit circle with center at $O$ determine a circular sector $AOB.$ Prove that the arc $AB$ has a length equal to twice the area of the sector.
• Recall from section 2.10 that for a polar function with radius $r = f(\theta),$ the area of the sector $R$ from $0$ to $\theta_0,\, (0 \leq \theta_0 \leq 2\pi)$ is $$ a(R) = \frac{1}{2}\int_0^{\theta_0}f^2(\theta)\,d\theta $$
• Since the curve is a circle of radius $1$ centered at the origin, we can write position $\mathbf{r}$ as a parameter of angle $\theta:$ $$ \mathbf{r}(\theta) = \cos \theta\,\mathbf{i} + \sin \theta\,\mathbf{j} $$ Then, velocity $\mathbf{v}(\theta)$ and speed $v(\theta)$ are: $$ \begin{align*} \mathbf{v}(\theta) &= -\sin \theta\,\mathbf{i} + \cos \theta\,\mathbf{j} \\ \\ v(\theta) &= \sqrt{\sin^2 \theta + \cos^2 \theta} \\ &= 1 \end{align*} $$ Arc length $s$ for angle $\theta = \theta_0$ is $$ \begin{align*} s &= \int_0^{\theta_0}v(\theta)\,d\theta \\ &= \int_0^{\theta_0}d\theta \\ &= \theta_0 \end{align*} $$ Now, we recall the formula for the area of a region $R$ from $\theta = 0$ to $\theta = \theta_0$ of a polar function with radius $r(\theta):$ $$ \begin{align*} a(R) &= \frac{1}{2}\int_0^{\theta_0}r^2(\theta)\,d\theta \end{align*} $$ But for a circle with radius $1$ centered at the origin, $r(\theta) = 1$ for all $\theta,$ thus for the sector drawn from $\theta = 0$ to $\theta = \theta_0,$ its area a(R) is: $$ \begin{align*} a(R) &= \frac{1}{2}\int_0^{\theta_0}\,d\theta \\ &= \frac{1}{2}\theta_0 \end{align*} $$ But this is half of the arc length $s.$ This completes the proof.