# Calculus!!!

• Dec 11th 2008, 08:29 PM
Audriella
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!!!!
• Dec 11th 2008, 08:42 PM
Prove It
Quote:

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...
• Dec 11th 2008, 08:51 PM
Audriella
Quote:

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?
• Dec 11th 2008, 09:03 PM
Prove It
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 $0 \leq t \leq 2\pi$

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

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

$t = \{\frac{\pi}{3}, \frac{4\pi}{3}\}$.
• Dec 11th 2008, 09:27 PM
Audriella
thanks!!!
i think my issue was that i was trying to find the derivative to early...
thank you again!
• Dec 11th 2008, 09:28 PM
Prove It
Quote:

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...
• Dec 11th 2008, 09:49 PM
Audriella
Quote:

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!