# Thread: Vector Valued Function Problem

1. ## 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!

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

Part I: Arc Length

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

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

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

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

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

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

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

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

Part II: Max Speed

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

$\displaystyle \sqrt{1-\cos t}=1; t=\pi$ (halfway between the start and end points)

3. 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!

4. ## Same answer, corrected procedure

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

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

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

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