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Math Help - Where are the limits of this question?

  1. #1
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    Where are the limits of this question?

    Question 7


    At first I thought it was polar co-ordinates but it was parametric so I used the arc length formula to try and find the length of the curve in ONE of the quadrants. I would later times this by 4 to get the length of the curve all together. So arc length = \int{\sqrt{\frac{dx}{dt}^{2}+\frac{dy}{dt}^{2}}}dt

    so after lots of cancelling and using everybody's favourite trig identity sin^2(t) + cos^2(t) = 1 I get it down to

    3a\int{sint cos t}dt

    Okay now use a substitution of u = sint

    to get \frac{3a}{2} sin^{2}t but where are the limits I need to use to get this done? Can someone show me what the limits I need to use are?

    Thanks

    NOTE using t instead of theta
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  2. #2
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    If you're going to calculate the length of the arc in the first quadrant then simply use 0 \leq \theta \leq \frac{\pi}{2}
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  3. #3
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    The problem says " 0\le \theta \le 2\pi"! But because you will have signs that cancel out, integrate from 0 to \pi/2 and multiply by 4.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    The problem says " 0\le \theta \le 2\pi"! But because you will have signs that cancel out, integrate from 0 to \pi/2 and multiply by 4.
    Im not quite sure what you mean? Have you simply down 2pi/4 but you mention about signs cancelling? Could you please elaborate

    Thanks
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  5. #5
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    If you put in 0 and 2\pi as the limits then \sin^2 x|_0^{2\pi}=0

    A length is obviously not negative, but if you integrate around the entire circle you will have four lengths that cancel each other out because two of them will end up negative.
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