# The position of two particles on the $x$-axis are $x_1 = \sin t$ and $x_2 = \sin \left(t + \frac{\pi}{3}\right)$

The position of two particles on the $x$-axis are $x_1 = \sin t$ and $x_2 = \sin \left( t + \frac{\pi}{3}\right)$

(a) At what time(s) in the interval $[0,2\pi]$ do the particles meet?

(b) What is the farthest apart that the particles every get?

(c) When in the interval $[0,2\pi]$ is the distance between the particles changing the fastest?

I am studying for the AP Calculus BC exam and this is a problem out of the Calculus Problem book. I know the answers: (a) $\frac{\pi}{3}, \, \frac{4\pi}{3}$ (b) 1 (c) $\frac{\pi}{3}, \, \frac{4\pi}{3}$

I am have trouble with (a)

I rewrote $$\sin\left(t + \frac{\pi}{3}\right) \Rightarrow \frac{\sqrt{3}}{2}\sin t + \frac{1}{2}\cos t$$

I get this which I can't seem to solve for the answer given:

$$\left(2 - \sqrt{3}\right)\sin t = \cos t$$

For part (b), I would maximized the distance formula. $d(x)=\sqrt{x_1^2 + x_2^2}$

Set the 1st derivative to zero and use the 2nd derivative test to find the max.

for part (c) I would find the maximum of $d^{\prime}(x)$

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I think for a you could do sin t = sin (t + (pi/3)) and then solve for t. –  Link Nov 18 '12 at 14:58
@link, that is what I am trying to do, but its not working out. –  yiyi Nov 18 '12 at 14:59

(a)

For coincidence, $\sin(t+\frac{\pi}3)=\sin t$

So, $t+\frac{\pi}3=n\pi+(-1)^nt$ where $n$ is any intgere.

If $n=2m$(even), $t+\frac{\pi}3=2m\pi+t\implies 2m\pi=\frac{\pi}3$ which is impossible.

If $n=2m+1$(odd), $t+\frac{\pi}3=(2m+1)\pi-t\implies t=(6m+2)\frac{\pi}6=\frac{(3m+1)\pi}3$

Putting $m=0,t=\frac{\pi}3$

Putting $m=1,t=\frac{4\pi}3$

(b) We need to maximize $\mid\sin(t+\frac{\pi}3)-\sin t\mid$

Now, $\sin(t+\frac{\pi}3)-\sin t=\sin t(\cos \frac{\pi}3-1 )+\cos t \sin\frac{\pi}3=\sin(\frac{\pi}3-t)$

So, the distance will be maximum if $\sin(\frac{\pi}3-t)$ is minimum/maximum. But $-1\le \sin(\frac{\pi}3-t)\le 1$

So, the distance will be maximum $(=1)$ if $\sin(\frac{\pi}3-t)=\pm1\implies \cos(\frac{\pi}3-t)=0\implies \frac{\pi}3-t=(2r+1)\frac{\pi}2,t=\frac{\pi}3-(2r+1)\frac{\pi}2$ where $r$ is any integer.

(c)The change of distance=$$\mid\frac{d\{\sin t-\sin(t+\frac{\pi}3)\}}{dt}\mid=\mid\cos t-\cos(t+\frac{\pi}3)\mid$$

Now, $\cos t-\cos(t+\frac{\pi}3)=\cos t(1-\cos\frac{\pi}3)-\sin t\sin \frac{\pi}3=\cos(t+\frac{\pi}3)$

For the distance between the particles changing the fastest,

$\cos(t+\frac{\pi}3)=\pm 1$ as $-1\le \cos(\frac{\pi}3-t)\le 1$

$\implies \sin(t+\frac{\pi}3)=0$ $\implies t+\frac{\pi}3=s\pi$ where $s$ is any integer.

So, $t=s\pi-\frac{\pi}3$

Putting $s=0,t=-\frac{\pi}3$ which is not $[0,2\pi]$

Putting $s=1,t=\frac{2\pi}3$

Putting $s=2,t=\frac{5\pi}3$

Putting $s=3,t=\frac{8\pi}3>2\pi$ so is not acceptable.

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Awesome answer, but how did you find So, t+π3=nπ+(−1)nt where n is any intgere. –  yiyi Nov 18 '12 at 15:07
@MaoYiyi, this is standard formula of $\sin A=\sin B, A=n\pi+(-1)^B$, I'm trying to share a proof.But have you verified your the answers of $(c)$? –  lab bhattacharjee Nov 18 '12 at 15:10
could you also explain $\sin(t+\frac{\pi}3)-\sin t=\sin t(\cos \frac{\pi}3-1 )+\cos t \sin\frac{\pi}3=\sin(\frac{\pi}3-t)$ –  yiyi Nov 18 '12 at 15:14
@MaoYiyi, $\sin(t+\frac{\pi}3)-\sin t=\sin t\cos \frac{\pi}3+\cos t \sin \frac{\pi}3-\sin t=\sin t(1-\cos \frac{\pi}3)+\cos t \sin \frac{\pi}3$ –  lab bhattacharjee Nov 18 '12 at 15:17
@MaoYiyi, As $\sin A=\sin B\implies A=2m\pi+B$ or $A=2m\pi+\pi-B$ as $\sin(\pi-C)=\sin C$ –  lab bhattacharjee Nov 18 '12 at 16:23