Question

Find the length of the curve x=e^t e^{-t},\;\;y=5-2t,\;\;0 \le t \le 3.

Answer 1

`\begin{cases}x(t)=e^t+e^(-t)\\y(t)=5-2t\end{cases}\implies\begin{cases}x'(t)=e^t-e^(-t)\\y'(t)=-2\end{cases}`

The length of the curve is given by the integral


`\displaystyle\int_0^3\sqrt{(e^t-e^(-t))^2+(-2)^2}\,\mathrm dt`

Expand and rewrite the integrand:


`(e^t-e^(-t))^2+(-2)^2=e^(2t)+2+e^(-2t)`

`=e^(-2t)(e^(4t)+2e^(2t)+1)`

`=e^(-2t)(e^(2t)+1)^2`

`\implies\sqrt{(e^t-e^(-t))^2+(-2)^2}=(e^(2t)+1)/(e^t)`

Now the integral is


`\displaystyle\int_0^3(e^(2t)+1)/(e^t)\,\mathrm dt=\int_0^3(e^t+e^(-t))\,\mathrm dt`

`=2\displaystyle\int_0^3\cosh t\,\mathrm dt`

`=2\sinh t\bigg|_(t=0)^(t=3)`

`=2\sinh 3`