Vector Valued Function Problem

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October 17th 2009, 05:07 PM
UCSociallyDead

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!
October 18th 2009, 10:49 AM
Media_Man

$r(t)=(t-\sin t,1-\cos t)$

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)
October 18th 2009, 01:00 PM
UCSociallyDead

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!
October 18th 2009, 02:45 PM
Media_Man

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. (Hi)