Arc Length with Parametric Equations — Question 4

Question 4

Problem

Show that $x=e^t\cos t$, $y=e^t\sin t$, $0\le t\le\ln2$, has length $\sqrt2$.

Question 4 – Solution

Solution

1. Differentiate using the product rule: $$x'(t)=e^t(\cos t-\sin t), \qquad y'(t)=e^t(\sin t+\cos t).$$

2. Compute the squared speed: $$\begin{aligned} x'^2+y'^2 &=e^{2t}\left[(\cos t-\sin t)^2+(\sin t+\cos t)^2\right]\\ &=e^{2t}\left[2\cos^2t+2\sin^2t\right]\\ &=2e^{2t}. \end{aligned}$$ Since $e^t>0$, the speed is $\sqrt2e^t$.

3. Set up and evaluate the length: $$\begin{aligned} L&=\int_0^{\ln2}\sqrt2e^t\,dt\\ &=\sqrt2\left[e^t\right]_0^{\ln2}\\ &=\sqrt2(2-1)=\boxed{\sqrt2}. \end{aligned}$$