Determining the Length of a Parametric Curve (Parametric Form)

Key Questions

  • 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.

    Amory W. · 3 · Aug 15 2014
  • How do you find the length of the curve x=1+3t^2, y=4+2t^3, where 0<=t<=1 ?

    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.

    Wataru · 3 · Oct 10 2014
  • How do you find the arc length of a parametric curve?
    Amory W. · · Aug 17 2014

Questions

Parametric Functions