Question

How do you find the length of the curve `x=e^t+e^-t`, `y=5-2t`, where `0<=t<=3` ?

Answer 1

The answer is `e^3-e^(-3)`.

Recall that the arclength for parametric curves is:

`L=int_a^b sqrt(((dx)/(dt))^2+((dy)/(dt))^2)dt`

So,

`(dx)/(dt)=e^t-e^(-t)`
`(dy)/(dt)=-2`

Now substituting:

`L=int_0^3 sqrt((e^t-e^(-t))^2+(-2)^2)dt`
`=int_0^3 sqrt(e^(2t)-2+e^(-2t)+4)dt` expand
`=int_0^3 sqrt(e^(2t)+2+e^(-2t))dt` simplify
`=int_0^3 sqrt((e^t+e^(-t))^2)dt` factor
`=int_0^3 (e^t+e^(-t))dt` simplify
`=e^t-e^(-t)|_0^3` integrate
`=e^3-e^(-3)-(e^0-e^0)` evaluate
`=e^3-e^(-3)`

Note that there aren't many questions that can be solved algebraically. Please note the pattern of this problem because most algebraic solutions have this form.