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Thread: Finding length of path over an interval

  1. #1
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    Finding length of path over an interval

    (2cost - cos2t, 2sint - sin2t) 0 <= t <= pi/2
    Here's what I have so far:
    x'(t)^2 = 4sin^2(t) - 8(sint)(sin2t) + 4sin^2(2t)
    y'(t)^2 = 4cos^2(t) - 8(cost)(cos2t) + 4cos^2(2t)
    x'(t)^2 + y'(t)^2 = 4 - 8(sint)(sin2t) + 4sin^2(2t) - 8(cost)(cos2t) +
    4cos^2(2t)

    What do I do next? And/or am I even doing it right?
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  2. #2
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    You will want to use the formula
    $\displaystyle s = \int_{t_0}^{t_1} |v(t)|\, dt$

    Where v(t) is the speed function.
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  3. #3
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    Well, I'm wondering how I can integrate it as the resultant integrand will be:
    integral( SQRT(4 - 8(sint)(sin2t) + 4sin^2(2t) - 8(cost)(cos2t) + 4cos^2(2t)), t, 0, pi/2 )

    I'm not sure how I can simplify it when this part is in the square root: 8(sint)(sin2t) + 4sin^2(2t) - 8(cost)(cos2t) + 4cos^2(2t)
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  4. #4
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    Well
    $\displaystyle 4\sin^2 2t + 4\cos^2 2t = 4$
    And
    $\displaystyle -8\sin t \sin 2t - 8 \cos t \cos 2t = -8\cos t$

    EDIT: Fixed typos
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  5. #5
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    Hello, ZeroVector!

    You're on the right track . . .


    $\displaystyle \begin{Bmatrix}x &=& 2\cos t - \cos2t \\ y &=& 2\sin t - \sin2t\end{Bmatrix}\;\;0 \leq t \leq \frac{\pi}{2} $

    Here's what I have so far:

    . . $\displaystyle x'(t)^2 \:=\: 4\sin^2\!t - 8\sin t\sin2t + 4\sin^2\!2t$
    . . $\displaystyle y'(t)^2 \:=\: 4\cos^2\!t - 8\cos t\cos2t + 4\cos^2\!2t$

    $\displaystyle x'(t)^2 + y'(t)^2 \:=\: 4\underbrace{\left[\sin^2\!t + \cos^2\!t\right]}_{\text{This is 1}}$

    . . . . . . . . . . . . . . . . . . $\displaystyle - 8\underbrace{\left[\cos t\cos2t + \sin t\sin2t\right]}_{\text{This is }\cos t} $

    . . . . . . . . . . . . . . . . . . . . . . $\displaystyle + 4\underbrace{\left[\sin^2\!2t + \cos^2\!2t\right]}_{\text{This is 1}}$

    . . . . . . . . . . $\displaystyle =\;8 - 8\cos t \;=\;8(1-\cos t)$

    . . . . . . . . . . $\displaystyle =\;16\left(\dfrac{1-\cos t}{2}\right) \;=\;16\sin^2\!\frac{t}{2}$


    Hence: .$\displaystyle \sqrt{x'(t)^2 + y'(t)^2} \;=\;4\sin\frac{t}{2}$


    Therefore: .$\displaystyle \displaystyle L \;=\;\int^{\frac{\pi}{2}}_0 \sin\frac{t}{2}\,dt$


    Got it?
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  6. #6
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    Thanks! By the way, shouldn't the integral you wrote at the end have a 4 in front of it?
    I think the main part was that I didn't think that I was supposed to use things like angle sum/double angle/half angle (which, unfortunately, my teacher pre-calculus never taught) formulas.

    Also, I was wondering, how did you get the text to look "nice"? Is there some sort of web app that does it for you?
    Last edited by ZeroVector; Oct 17th 2010 at 09:08 PM.
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