# Vector Valued Function Problem

• Oct 17th 2009, 05:07 PM
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!
• Oct 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)
• Oct 18th 2009, 01:00 PM
$speed=s(t)=|r'(t)|=\sqrt{2-2\cos t}$
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)