Vector Valued Function Problem

**UCSociallyDead** · October 17th 2009, 05:07 PM

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!

**Media_Man** · October 18th 2009, 10:49 AM

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

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)

**UCSociallyDead** · October 18th 2009, 01:00 PM

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!

**Media_Man** · October 18th 2009, 02:45 PM

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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