Find the length of the parametric curve $x = e^t + e^-t, y = 5 - 2t$ from $t = 0 to t = 1$ ?

Question

2234 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-20T01:32:53

$e-e^{-1}$

By differentiating w.r.t. $t$ ,

${dx}/{dt}=e^t-e^{-t}$

${dy}/{dt}=-2$

Let us simplify:

$sqrt((dx/dt)^2+(dy/dt)^2)$

$=sqrt((e^t-e^(-t))^2+(-2)^2)$

$=sqrt(e^(2t)+2+e^(-2t))$

$=sqrt((e^t+e^{-t})^2)$

$=e^t+e^{-t}$

Now, we can find the arc length.

$L=\int_0^1(e^t+e^(-t))dt=[e^t-e^{-t}]_0^1=e-e^{-1}$

I hope that this was clear.

Find the length of the parametric curve x = e^t + e^-t, y = 5 - 2tx = e^t + e^-t, y = 5 - 2t from t = 0 to t = 1t = 0 to t = 1?