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Homework Help: Tangent of a circle question

  1. Sep 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A ball that is circling with x=cos(2t), y=sin(2t) flies off on a tangent at t=Pi/8. find its departure point and its position vector at a later time t (linear motion; compute its constant velocity v).


    2. Relevant equations
    v=dR/dt T=v/|v|= (dR/dt)/(ds/dt)=dR/ds.



    3. The attempt at a solution
    I have got to be the dumbest person on the planet. all I can think to do is this: R= cos(2t) + sin(2t). I thought after that it'd be a good idea to differentiate to find the velocity, but I'm not sure I'm doing it correctly. what I find is this: v=-2sin(2t)+2cos(2t). then I went to find |v|, but I'm don't know how to compute it when I've still got the t in there.

    whatever I do there, it's just not coming out right with the answer key. (thank goodness I have at least one odd-numbered problem tonight!!) what the author says is this:
    leaves at (sqrt(2)/2, sqrt(2)/2); v=(sqrt(2), sqrt(2)); R= (sqrt(2)/2, sqrt(2)/2) + v(t-Pi/8) which is the position at time t.

    helppppp! I promise this is my last post tonight! =)
     
  2. jcsd
  3. Sep 17, 2008 #2

    Dick

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    Differentiate the VECTOR (cos(2t),sin(2t)) and find its length! Don't differentiate cos(2t)+sin(2t). You aren't the dumbest person on the planet. I've been getting a bounty of dumb questions tonite and this is by no means the worst.
     
  4. Sep 18, 2008 #3
    this is a potential lightbulb moment!! I still have questions, though. are my differentiations correct other than the fact that I tried to add them? I'm not sure how to find the length since there are still t's involved.

    thanks for the morale boost, dick!
     
  5. Sep 18, 2008 #4

    Dick

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    Yes. The derivative vector is (-2*sin(2t),2*cos(2t)). The length of a vector (x,y) is sqrt(x^2+y^2). There's an identity that says cos^2(a)+sin^2(a)=??? Can you look that up for me? Uh, don't let my praise go to your head. :)
     
  6. Sep 18, 2008 #5
    well I don't know about that identity... I know cos^2(u)+sin^2(u)=1, but if we're talking about the same one, I'm not sure how that does me any good because what I'm getting under the square root is this:
    -4sin^2(4t)+4cos^2(4t).
     
  7. Sep 18, 2008 #6

    Dick

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    Noooo. You have 4*sin^2(2t)+4*cos^(2t). (-2*sin(2t))^2=4*sin^(2t). sin(2t)*sin(2t) is not sin^2(4t). Please rethink that and get back to me in the morning.
     
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