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Math Help - Vector Valued Function Problem

  1. #1
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    Vector Valued Function Problem

    If someone could help me with this problem, I would greatly appreciate the help. I believe I have the solution to the first part, but I'm not sure about the second part.

    "Show that one arch of the cycloid r(t) = <t-sint, 1-cost> has length 8. Find the value of t in [0,2PI] where the speed is at a maximum"

    Thanks!
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  2. #2
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    r(t)=(t-\sin t,1-\cos t)

    Part I: Arc Length

    L=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left  (\frac{dy}{dt}\right)^2}dt

    =\int_0^{2\pi}\sqrt{(1-\cos t)^2+(\sin t)^2}dt

    =\int_0^{2\pi}\sqrt{2-2\cos t}dt

    =\sqrt2\int_0^{2\pi}\sqrt{1-\cos t}dt

    =\sqrt2\int_0^{2\pi}\frac{\sqrt{1-\cos^2 t}}{\sqrt{1+\cos t}}dt

    =\sqrt2\int_0^{2\pi}\frac{\sin t}{\sqrt{1+\cos t}}dt

    =2\sqrt2\int_0^{\pi}\frac{\sin t}{\sqrt{1+\cos t}}dt

    =2\sqrt2\int_2^0\frac{-1}{\sqrt{u}}dt=8

    Part II: Max Speed

    |r'(t)|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\  frac{dy}{dt}\right)^2}=0

    \sqrt{1-\cos t}=1; t=\pi (halfway between the start and end points)
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  3. #3
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    Could you explain why you have set the magnitude of r'(t) = 0? I thought that r'(t) was the velocity itself and by plugging in the solution you get setting it equal to zero to the original vector is the position.

    Thanks again!
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  4. #4
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    Same answer, corrected procedure

    Oops. Right you are, UCSociallyDead. Got ahead of myself.

    speed=s(t)=|r'(t)|=\sqrt{2-2\cos t}

    So acceleration=s'(t)=(2-2\cos t)^{-1/2}\sin t=0

    So t=0 or t=\pi, but if t=0, you have a zero in the denominator, which doesn't work. Therefore, the only stationary point occurs at t=\pi. I'll let you have the pleasure of taking another derivative to prove it's a max, not a min.
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