Find the exact length of the parametric curve: $x={e^t}+{e^{-t}}, y=5-2t,0\le t \le 3$.

Question

Find the exact length of the parametric curve: $$x={e^t}+{e^{-t}}, y=5-2t,0\le t \le 3$$ Solution: $$\frac{dx}{dt}=e^t -e^{-t}, \frac{dy}{dt}=-2$$ $$\int_{0}^{3} \sqrt{(e^t-e^{-t})^2+(-2)^2}dt$$ $$= \int_{0}^{3} \sqrt{e^{2t}+e^{-2t}+2}dt$$ Let$$e^{2t}=u, e^{-2t}=\frac{1}{u}$$ and then$$du=2u\ dt$$ Therefore, $$\int \sqrt{u+\frac{1}{u}+2}\left(\frac{1}{2u}\right)du$$ $$=\int \sqrt{\frac{(u+1)^2}{u}}\left(\frac{1}{2u}\right)du$$ $$=\int \frac{u+1}{2u\sqrt u}du$$ $$=\int \frac{1}{2\sqrt u}+\frac{1}{2u\sqrt u}du$$ $$=\left[\sqrt {e^{2t}} - \frac{1}{\sqrt {e^{2t}}}\right]_{0}^{3}$$ $$=e^3-\frac{1}{e^3}-1+1$$ $$=e^3+e^{-3}$$ Is my answer correct?

Answer 1

Hint:

It's better to utilize $$e^{2t}+e^{-2t}+2=(e^t+e^{-t})^2$$

and $\displaystyle\dfrac{d(e^{ax})}{dx}=ae^{ex}\implies\int e^{ax}\ dx=\dfrac{e^{ax}}a+K$

Answer 2

It is faster to use some hyperbolic trigonometry: as $x(t)=2\cosh t$, we have $\;x'(t)=2\sinh t$, so \begin{align} \ell&=\int_0^3\sqrt{4(\sinh^2t+1)}\,\mathrm d t=2\int_0^3\cosh t\,\mathrm d t=2\sinh t\Biggm\vert_0^3=2\sinh 3=\mathrm e^3-\mathrm e^{-3}. \end{align}