Arc Length with Parametric Equations — Question 8

Question 8

Problem

Find the length of the tractable curve $x=t-\tanh t$, $y=\operatorname{sech}t$, $0\le t\le1$.

Question 8 – Solution

Solution

1. Differentiate and use $1-\operatorname{sech}^2t=\tanh^2t$: $$x'(t)=1-\operatorname{sech}^2t=\tanh^2t,$$ $$y'(t)=-\operatorname{sech}t\tanh t.$$

2. Simplify the squared speed: $$\begin{aligned} x'^2+y'^2 &=\tanh^4t+\operatorname{sech}^2t\tanh^2t\\ &=\tanh^2t\left(\tanh^2t+\operatorname{sech}^2t\right)\\ &=\tanh^2t. \end{aligned}$$ Thus the speed is $|\tanh t|$.

3. On $0\le t\le1$, $\tanh t\ge0$, so $$L=\int_0^1\tanh t\,dt.$$ Since $d(\ln\cosh t)/dt=\tanh t$, $$L=\left[\ln(\cosh t)\right]_0^1 =\ln(\cosh1)-\ln1.$$

4. Therefore, $$\boxed{L=\ln(\cosh1)}.$$