Arc Length — Question 1

Question 1

A cable follows $$y=\frac{x^2}{4}-\frac12\ln x,\qquad1\le x\le e.$$ Find its exact length.

Question 1 – Solution

Step 1: Differentiate. $$y'=\frac{x}{2}-\frac1{2x}.$$ Step 2: Simplify the arc-length integrand. $$\begin{align*} 1+(y')^2 &=1+\left(\frac{x}{2}-\frac1{2x}\right)^2\\ &=\frac{x^2}{4}+\frac12+\frac1{4x^2}\\ &=\left(\frac{x}{2}+\frac1{2x}\right)^2. \end{align*}$$ Since $x>0$, $$\sqrt{1+(y')^2}=\frac{x}{2}+\frac1{2x}.$$ Step 3: Integrate and evaluate. $$\begin{align*} L&=\int_1^e\left(\frac{x}{2}+\frac1{2x}\right)dx\\ &=\left[\frac{x^2}{4}+\frac12\ln x\right]_1^e\\ &=\left(\frac{e^2}{4}+\frac12\right)-\frac14 =\frac{e^2+1}{4}. \end{align*}$$ $$\boxed{L=\frac{e^2+1}{4}}$$