Arc Length — Question 6

Question 6

Rank the arc lengths on $0\le x\le1$: $$y=0,\qquad y=x,\qquad y=x^2.$$

Question 6 – Solution

Step 1: Find the first two lengths. $$\begin{align*} L_0&=\int_0^1\sqrt{1+0^2}\,dx=1,\\ L_x&=\int_0^1\sqrt{1+1^2}\,dx=\sqrt2\approx1.414. \end{align*}$$ Step 2: Set up the third length. For $y=x^2$, $y'=2x$, so $$L_{x^2}=\int_0^1\sqrt{1+4x^2}\,dx.$$ Use the standard antiderivative $$\int\sqrt{1+4x^2}\,dx =\frac{x}{2}\sqrt{1+4x^2}+\frac14\operatorname{arsinh}(2x).$$ Therefore, $$\begin{align*} L_{x^2} &=\left[\frac{x}{2}\sqrt{1+4x^2} +\frac14\operatorname{arsinh}(2x)\right]_0^1\\ &=\frac{\sqrt5}{2}+\frac14\operatorname{arsinh}2\\ &\approx1.479. \end{align*}$$ Step 3: Rank the values. $$\boxed{L_0<L_x<L_{x^2}}$$