Arc Length — Question 4

Question 4 – Solution

Step 1: Differentiate. $$y'=\frac{\sec x\tan x+\sec^2x}{\sec x+\tan x}=\sec x.$$ Thus $$L=\int_0^{\pi/3}\sqrt{1+\sec^2x}\,dx.$$ Step 2: Substitute $u=\tan x$. Then $du=\sec^2x\,dx$ and the bounds become $0$ and $\sqrt3$: $$L=\int_0^{\sqrt3}\frac{\sqrt{u^2+2}}{1+u^2}\,du.$$ Step 3: Substitute $u=\sqrt2\sinh t$. Then $$du=\sqrt2\cosh t\,dt,\qquad\sqrt{u^2+2}=\sqrt2\cosh t.$$ The upper bound is $T=\operatorname{arsinh}\sqrt{3/2}$, and $$\begin{align*} L&=\int_0^T\frac{2\cosh^2t}{1+2\sinh^2t}\,dt\\ &=\int_0^T\left(1+\operatorname{sech}(2t)\right)dt\\ &=\left[t+\frac12\arctan(\sinh2t)\right]_0^T. \end{align*}$$ Since $\sinh(2T)=\sqrt{15}$, $$\boxed{L=\operatorname{arsinh}\sqrt{\frac32} +\frac12\arctan\sqrt{15}}$$