Determining the Length of a Parametric Curve (Parametric Form)
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Key Questions

Please see how I solved this problem, it has a full explanation.
How do you find the length of the curve
#x=3tt^3# ,#y=3t^2# , where#0<=t<=sqrt(3)# ? 
Answer:
#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)=33t^2#
#(dy)/(dt)=6t# And we substitute these into the integral:
#L=int_0^(sqrt3)sqrt((33t^2)^2+(6t)^2)dt# And solve:
#=int_0^(sqrt3)sqrt(918t^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.

By taking the derivative with respect to
#t# ,#{(x'(t)=6t),(y'(t)=6t^2):}# Let us now find the length
#L# of the curve.#L=int_0^1 sqrt{[x'(t)]^2+[y'(t)]^2}dt# #=int_0^1 sqrt{6^2t^2+6^2t^4} dt# by pulling
#6t# out of the squareroot,#=int_0^1 6t sqrt{1+t^2} dt# by rewriting a bit further,
#=3int_0^1 2t(1+t^2)^{1/2}dt# by General Power Rule,
#=3[2/3(1+t^2)^{3/2}]_0^1=2(2^{3/2}1)# I hope that this was helpful.