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Homework Help: Finding total length of a parametric curve

  1. Oct 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the total length of the curve t --> (cos^3(t), sin^3(t)), and t is between 0 and ∏/2 where t is in radians. Find also
    the partial arc length s(t) along the curve between 0 and ∏/2

    2. Relevant equations

    The length is given by:

    S = ∫[itex]\sqrt{xdot^2 + ydot^2}[/itex] dt

    3. The attempt at a solution

    xdot = -3sin(t)cos^2(t) and ydot = 3sin^2(t)cos(t)

    so subst. these into the formula gave me:

    S = ∫[itex]\sqrt{9sin^2(t)cos^4(t)+9sin^4(t)cos^2(t)}[/itex]

    I then used integration by subst. using u^2 = 9sin^2(t)cos^4(t)+9sin^4(t)cos^2(t)

    The method I have been shown tells me to differentiate both sides and use this to subst. back into the intregal to solve.

    After differentiation and some cancelling I have arrived at the following:

    2u du = 18sin(t)cos^5(t) - 18cos(t)sin^5(t) dt

    now I'm not sure what to do. Can anyone please help?
  2. jcsd
  3. Oct 26, 2013 #2


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    Try simplifying that before trying to integrate it. One way is to convert everything to sines and see what cancels.
  4. Oct 26, 2013 #3
    Thanks for the reply.

    I realised that I could use the substitution u^2 = 9cos^4(t)sin^2(t) + 9sin^4(t)cos^2(t)

    Now taking out a factor of 9sin^2(t)cos^2(t) gave me:

    u^2 = 9sin^2(t)cos^2(t) (cos^2(t) + sin^2(t))

    and now you can use that cos^2(t) + sin^2(t) = 1

    which now leaves that:

    u^2 = 9sin^2(t)cos^2(t)

    so now you have:

    u = 3sin(t)cos(t) and using the double angle formula 2sin(t)cos(t) = sin(2t)

    this can become:

    u = 3/2 * sin(2t)

    Now you have to integrate this to solve. The question that I have now is, what is the difference between finding the total length and the partial arc length between the limits given?

    Would the total length be:

    -0.75cos(2t) + C

    and then you apply the limits of 0 and ∏/2 to give the partial arc length?

    Thanks again.
  5. Oct 26, 2013 #4


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    That's the indefinite integral, yes. Not sure what you mean by 'total' length.
    Yes, but I don't know why you call it the partial length. It's the length of the specified arc.
    Note that if you try to take t beyond that range the length appears to go down. This is because, when you took the square root, you should have taken the positive value only. Thus, your integration step was only valid where sine is positive. To go beyond that range you'd need to split the integral into a sum of as many pieces as necessary to cope with the reversing sign.
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