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

1 Answer
Aug 15, 2014

#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.