Arc Length — Question 7

Question 7 – Solution

Step 1: Correct the geometric formula. A small arc segment is the hypotenuse of a right triangle: $$ds^2=dx^2+dy^2.$$ Since $dy=y'\,dx$, $$ds=\sqrt{dx^2+(y'\,dx)^2}=\sqrt{1+(y')^2}\,dx.$$ The student’s expression adds horizontal and vertical changes instead of using the Pythagorean theorem.

Step 2: Differentiate and set up the length. $$y'=2x,\qquad L=\int_0^1\sqrt{1+4x^2}\,dx.$$ Step 3: Evaluate. $$\begin{align*} L&=\left[\frac{x}{2}\sqrt{1+4x^2} +\frac14\ln\left(2x+\sqrt{1+4x^2}\right)\right]_0^1\\ &=\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5). \end{align*}$$ $$\boxed{L=\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5)}$$