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Arc length of vector function with trigonometric components

  1. Nov 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the length of the path traced out by a particle moving on a curve according to the given equation during the time interval specified in each case.
    r(t) = (c2/a)cos3t i + (c2/b)sin3t j

    where i and j are the usual unit vectors, 0 [itex]\leq[/itex] t [itex]\leq[/itex] 2[itex]\pi[/itex], c2 = a2 - b2, and 0 < b < a

    Also, r(t) is the position vector-valued function of t. Scalar functions will not be bolded.

    2. Relevant equations
    I know that the arc length in this case is the integral of ||r'(t)|| = v(t) from 0 to 2[itex]\pi[/itex].


    3. The attempt at a solution
    Differentiating gives: r'(t) = (-3c2/a)cos2t sin t i + (3c2/b)sin2t cos t j.

    Therefore, v(t) = √{(9c4/a2)cos4t sin2t + (9c4/b2)sin4t cos2t)} = √{(9c4/a2)cos2t(cos2t sin2t) + (9c4/b2)sin2t(sin2t cos2t)} =

    √{(9c4/a2)cos2t(0.25sin22t) + (9c4/b2)sin2t(0.25sin22t)} = √{(9c4/a2)cos2t(0.25sin22t) + (9c4/b2)sin2t(0.25sin22t)} =

    (3c2/2)(sin2t)*√{(1/a2)cos2t + (1/b2)sin2t} = (3c2/(2ab))(sin2t)*√{(b2)cos2t + (a2)sin2t}

    And that is where I get stuck. I'm not sure what the integral of that expression is, nor do I know what I can do to simplify it further. I would say that it has no simple solution, since the argument under the square root looks an awfully lot like an integral for an ellipse, and I have heard one needs something called an "elliptic integral" to solve. However, the answer for this problem is quite simple: (4a3-4b3)/(ab). Will someone please help me with this?
     
    Last edited: Nov 25, 2009
  2. jcsd
  3. Nov 25, 2009 #2

    tiny-tim

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    Hi Subdot! :smile:

    (have a square-root: √ :wink:)
    Use standard trigonometric identities to write it as sin2t f(cos2t) :smile:
     
  4. Nov 25, 2009 #3
    Hello! Oh! I get it now. I just use the identities: sin2 = .5 - .5cos2t and cos2 = .5 + .5cos2t after which I do a simple substitution. Thanks for that! On another related note, now it's obvious that I'm going to need to integrate it a step over the interval at a time.

    [(1/(ab)) (.5b2 + (.5b2 - .5a2)cos(2t) + .5a2)1.5] from 0 to 2[itex]\pi[/itex]. If I try to integrate it from 0 to .5[itex]\pi[/itex] then from .5[itex]\pi[/itex] to [itex]\pi[/itex] and so on I come up with (2a3 - 2b3)/(ab) + (2b3 - 2a3)/(ab). I'm guessing from the answer that I am justified in flipping that last fraction around to (2a^3 - 2b^3)/(ab). However, I'd like to know why, so I'll be able to do it in the future without knowing the answer. I think it's because the arc length must be positive and so I must flip (b3 - a3)/(ab) to (a3 - b3)/(ab) because b < a? So then do you take the absolute value of the integral as you move up the interval, not the integrand, when you find arc length?
     
  5. Nov 25, 2009 #4

    Dick

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    The reason is that you factored cos(t)^2*sin(t)^2 outside of the sqrt as cos(t)*sin(t). It really should be |cos(t)*sin(t)|. So you have to break the integral where cos(t)*sin(t) changes sign.
     
  6. Nov 25, 2009 #5
    Okay, I get it all now. I can't believe I missed that... Thanks both!
     
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