Arc Length — Question 5

Question 5

Find $a>0$ so that $$y=\frac23x^{3/2},\qquad0\le x\le a,$$ has length $14/3$.

Question 5 – Solution

Step 1: Differentiate. $$y'=\sqrt{x}.$$ Step 2: Write and evaluate the length. $$\begin{align*} L&=\int_0^a\sqrt{1+x}\,dx\\ &=\left[\frac23(1+x)^{3/2}\right]_0^a\\ &=\frac23\left((1+a)^{3/2}-1\right). \end{align*}$$ Step 3: Set the length equal to $14/3$ and solve. $$\begin{align*} \frac23\left((1+a)^{3/2}-1\right)&=\frac{14}{3},\\ (1+a)^{3/2}-1&=7,\\ (1+a)^{3/2}&=8,\\ 1+a&=8^{2/3}=4,\\ a&=3. \end{align*}$$ $$\boxed{a=3}$$