Arc Length with Polar Coordinates — Question 10

Question 10

Problem

Find $a$ so that $r=a\theta$, $0\le\theta\le1$, has length $\sqrt2+\operatorname{arsinh}1$.

Question 10 – Solution

Solution

1. Differentiate the polar radius to obtain $r'=dr/d\theta$ and choose an interval that traces the requested arc exactly once.

2. Use the polar arc-length formula $$L=\int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta.$$ Simplify the expression under the square root before evaluating or reporting the integral.

3. Its length is $|a|\int_0^1\sqrt{1+\theta^2}d\theta=\tfrac{|a|}{2}(\sqrt2+\operatorname{arsinh}1)$.

4. Thus $\boxed{a=\pm2}$.