1. ## Calculus!!!

Please help im goin crazy over here i have homework and a test due tomorrow and without help i know im not going to do so well...
so i am stuck on this problem...

There are two particles on the s-axis s1= sin t s2= sin (t + (pi/3))

at what time in the interval 0 < t < 2pi

I tried finding the derivatives of the two particles then having them = but that didn't seem to work i do not know if i should be graphing or what

thank you so much for your help!!!!

2. Originally Posted by Audriella

Please help im goin crazy over here i have homework and a test due tomorrow and without help i know im not going to do so well...
so i am stuck on this problem...

There are two particles on the s-axis s1= sin t s2= sin (t + (pi/3))

at what time in the interval 0 < t < 2pi

I tried finding the derivatives of the two particles then having them = but that didn't seem to work i do not know if i should be graphing or what

thank you so much for your help!!!!
At what time on the interval $0 \leq t \leq 2\pi$ what?

I don't know what you're asking...

3. Originally Posted by Prove It
At what time on the interval $0 \leq t \leq 2\pi$ what?

I don't know what you're asking...

ha ha oh wow sorry

At what time on the interval $0 \leq t \leq 2\pi$ do the particles meet?

4. Well you're right, just set them equal to each other...

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

$\sin{t} - \sin{(t + \frac{\pi}{3})} = 0$.

To simplify the left hand side, we use the identity

$\sin{x} - \sin{y} = 2\cos{\frac{1}{2}(x + y)}\sin{\frac{1}{2}(x - y)}$

So $2\cos{\frac{1}{2}(t + t + \frac{\pi}{3})}\sin{\frac{1}{2}[t - (t + \frac{\pi}{3})]} = 0$

$2\cos{\frac{1}{2}(2t + \frac{\pi}{3})}\sin{\frac{1}{2}(-\frac{\pi}{3})}=0$

$2\cos{(t + \frac{\pi}{6})}\sin{-\frac{\pi}{6}}=0$

$-\cos{(t + \frac{\pi}{6})}= 0$

$\cos{(t + \frac{\pi}{6})}= 0$

$t + \frac{\pi}{6} = \frac{\pi}{2} + \pi n, n \in \mathbf{Z}$

In the domain $0 \leq t \leq 2\pi$

$t + \frac{\pi}{6} = \{\frac{\pi}{2}, \frac{3\pi}{2}\}$

$t = \{\frac{\pi}{2}, \frac{3\pi}{2}\} - \frac{\pi}{6}$

$t = \{\frac{\pi}{3}, \frac{4\pi}{3}\}$.

5. thanks!!!
i think my issue was that i was trying to find the derivative to early...
thank you again!

6. Originally Posted by Audriella
thanks!!!
i think my issue was that i was trying to find the derivative to early...
thank you again!
Yeah, you don't even need the derivative...

7. Originally Posted by Prove It
Yeah, you don't even need the derivative...
that is so weird this is calculus that is supposed to be the one thing that is definite!