Arc Length — Question 9

Question 9

For $$x=\frac{y^2}{2},\qquad0\le y\le2,$$ find the length using $y$. Then write the equivalent $x$-integral without evaluating it.

Question 9 – Solution

Step 1: Differentiate with respect to $y$. $$\frac{dx}{dy}=y.$$ Step 2: Set up and evaluate the $y$-integral. $$\begin{align*} L&=\int_0^2\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy\\ &=\int_0^2\sqrt{1+y^2}\,dy\\ &=\left[\frac y2\sqrt{1+y^2}+\frac12\operatorname{arsinh}y\right]_0^2\\ &=\sqrt5+\frac12\operatorname{arsinh}2. \end{align*}$$ $$\boxed{L=\sqrt5+\frac12\operatorname{arsinh}2}$$ Step 3: Rewrite using $x$. Since $y=\sqrt{2x}$, $$\frac{dy}{dx}=\frac1{\sqrt{2x}}.$$ The $x$-values run from $0$ to $2$, so $$\boxed{L=\int_0^2\sqrt{1+\frac1{2x}}\,dx}.$$