Calculate the length of the curve:
$\vec r(t)= a(\sinh t-t)\vec i+a(1-\cosh t)\vec j, 0\leq t\leq T, a>0$
Answer: $2a\left(\cosh \frac T2 \sqrt{\cosh T}-1\right)-\sqrt2a\ln\left(\frac{\sqrt2\cosh\frac T2+\sqrt\cosh T}{1+\sqrt2}\right)$
My attempt:
$\vec r'(t)=a(\cosh t-1)\vec i+a(-\sinh t)\vec j$
$|\vec r'(t)|=\sqrt{a^2[(\cosh t-1)^2+(-\sinh t)^2]}$
$=a\sqrt{\cosh^2t +\sinh^2 t-2\cosh t+1}$
$L=\int_0^Ta\sqrt{\cosh^2t +\sinh^2 t-2\cosh t+1}dt$
The problem is that I don't have any experience dealing with hyperbolic functions, so how can I simplify $|\vec r'(t)|$ to be able to solve this integral?
Best Answer
HINT
Let us start with the definition involved: \begin{align*} L & = \int_{0}^{T}\|r'(t)\|\mathrm{d}t\\\\ & = \int_{0}^{T}\sqrt{a^{2}(\cosh(t) - 1)^{2} + a^{2}\sinh^{2}(t)}\mathrm{d}t\\\\ & = \int_{0}^{T}a\sqrt{\cosh^{2}(t) - 2\cosh(t) + 1 + \sinh^{2}(t)}\mathrm{d}t\\\\ & = \int_{0}^{T}a\sqrt{2\cosh^{2}(t) - 2\cosh(t)}\mathrm{d}t\\\\ & = \int_{0}^{T}a\sqrt{2}\sqrt{\cosh^{2}(t) - \cosh(t)}\mathrm{d}t\\\\ & = \int_{0}^{T}a\sqrt{2}\sqrt{\left(\cosh(t) - \frac{1}{2}\right)^{2} - \frac{1}{4}}\mathrm{d}t\\\\\ & = \int_{0}^{T}\frac{a\sqrt{2}}{2}\sqrt{\left(2\cosh(t) - 1\right)^{2} - 1}\mathrm{d}t \end{align*}
Now you can apply the change of variable $\cosh(x) = 2\cosh(t) - 1$.
Can you take it from here?