Question

How do you find the length of the curve `x=3t-t^3`, `y=3t^2`, where `0<=t<=sqrt(3)` ?

Answer 1

`6sqrt3`.

Explanation:

The answer is `6sqrt3`.

The arclength of a parametric curve can be found using the formula: `L=int_(t_i)^(t_f)sqrt(((dx)/(dt))^2+((dy)/(dt))^2)dt`. Since `x` and `y` are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function: `L=int_a^b sqrt(1+((dy)/(dx))^2)dx`. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:
`(dx)/(dt)=3-3t^2`
`(dy)/(dt)=6t`

And we substitute these into the integral:
`L=int_0^(sqrt3)sqrt((3-3t^2)^2+(6t)^2)dt`

And solve:
`=int_0^(sqrt3)sqrt(9-18t^2+9t^4+36t^2)dt`
`=int_0^(sqrt3)sqrt(9+18t^2+9t^4)dt`
`=int_0^(sqrt3)sqrt((3+3t^2)^2)dt`
`=int_0^(sqrt3)(3+3t^2)dt`
`=3t+t^3|_0^(sqrt3)`
`=3sqrt3+3sqrt3`
`=6sqrt3`

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.