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Homework Help: Particles and Derivatives

  1. Sep 7, 2008 #1
    The problem states that a particle's position at time t is given by the equation:
    [tex]x(t)=2\pi t+cos(2\pi t)[/tex]
    Therefore, the velocity of the particle at time t would be the first derivative of the above equation:
    [tex]x'(t)=v(t)=2\pi-2\pi(sin(2\pi t))[/tex]
    I was asked what the maximum velocity of the particle was, and was able to determine that it is 2, but that is only because I used a graphing aid. I am not quite sure where to start on how to find the maximum, other than know what the graph looks like in one's mind's eye, but that seems a little too extreme. I have also calculated the second derivative (acceleration of the particle) if it is needed:
    [tex]x''(t)=a(t)=4\pi^2cos(2\pi t)[/tex]
    I would show other attempt at work, but we really weren't shown how to do this.
    But yes, we still have to do it.
    Thank you for your help.
  2. jcsd
  3. Sep 7, 2008 #2


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    The max of v isn't 2. Working this in your minds eye is a perfectly good way to do it. sin(2*pi*t) starts at 0 goes up to 1, goes down to -1, goes back to zero and then just keeps repeating. For which of those values is v a maximum?
  4. Sep 7, 2008 #3
    hmmm...that does make sense. Thank you very much for your reply, i think the [tex]2\pi[/tex] outside the sine section of the function was confusing me.

    Now that I have double checked my graphing aid again, i find that it gives me a completely different answer, something to the tune of 12....
    now i am quite sure that you are correct! thank you again for the help, i appreciate it a lot.
    Last edited: Sep 7, 2008
  5. Sep 7, 2008 #4


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    You have [itex]x'= 2\pi - 2\pi sin(2\pi t)[/itex] and want to find its maximum value. Okay, forget about the fact that it is a first derivative and just think of maximizing [itex]y= 2\pi - 2\pi sin(2\pi t)[/itex]. [itex]y'= 4\pi^2 cos(2\pi t)= 0[/itex] when [itex]2\pi t= \pi/2[itex]. or when [itex]2\pi t= 3\pi/2[/itex]. When [itex]2\pi t= \pi/2[/itex] [itex]x'= y= 2\pi- 2\pi sin(\pi/2)= 0. When [itex]2\pi t= 3\pi/2[/itexs] [itex]x'= y= 2\pi- \2pi sin(3\pi/2)= 2\pi+ 2\pi= 4\pi[/itex]. The maximum value is [itex]4\pi[/itex].
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