Arc Length with Parametric Equations — Question 6

Question 6

Problem

Find the length of $x=t^2/2$, $y=t^3/3$, $0\le t\le2$.

Question 6 – Solution

Solution

1. Differentiate the coordinates: $$x'(t)=t, \qquad y'(t)=t^2.$$

2. The speed is $$\sqrt{t^2+t^4} =\sqrt{t^2(1+t^2)} =|t|\sqrt{1+t^2}.$$ Since $0\le t\le2$, $|t|=t$.

3. Therefore, $$L=\int_0^2t\sqrt{1+t^2}\,dt.$$ Let $u=1+t^2$, so $du=2t\,dt$. Then $$\begin{aligned} L&=\frac12\int_1^5u^{1/2}\,du\\ &=\frac13\left[u^{3/2}\right]_1^5\\ &=\frac13(5^{3/2}-1). \end{aligned}$$

4. Hence $$\boxed{L=\frac{5\sqrt5-1}{3}}.$$