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Math Help - Calculus!!!

  1. #1
    Newbie Audriella's Avatar
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    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!!!!
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  2. #2
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    Quote Originally Posted by Audriella View Post

    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...
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  3. #3
    Newbie Audriella's Avatar
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    Quote Originally Posted by Prove It View Post
    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?
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  4. #4
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    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}\}.
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  5. #5
    Newbie Audriella's Avatar
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    thanks!!!
    i think my issue was that i was trying to find the derivative to early...
    thank you again!
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  6. #6
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    Quote Originally Posted by Audriella View Post
    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...
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  7. #7
    Newbie Audriella's Avatar
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    Quote Originally Posted by Prove It View Post
    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!
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