Determining the Length of a Parametric Curve (Parametric Form)

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Arc Length Using Parametric Curves
8:18 — by patrickJMT

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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=3t-t^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)=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.

  • 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 square-root,

    #=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.

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