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Find time t at which the speed of the particle is minimized

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data

    If r(t) = −3t2i+2tj+(t2 −3t)k gives the position
    of a particle at time t, find the time t at which the speed of the
    particle is minimized.


    2. Relevant equations



    3. The attempt at a solution

    i know for minimization and maximization, you find the derivative and solve for the variable and see which values produces a largest or smallest value when plugged into the original equation right?

    so, since this is asking for speed, would it be right to find the deriv of this (v(t)) and find the absolute value of it (|(v(t)| = s(t)) then find the derivative of s(t) and solve for t?
    and just a side question but, s(t) is just the magnitude of the velocity vector right?
     
  2. jcsd
  3. Sep 30, 2009 #2
    How can you take the absolute value of a vector? Do you mean that you want to dot it with itself? That would give you a scalar valued function of t that would have the value of the length of the velocity vector squared at a particular t.

    You could then minimize that.
     
  4. Sep 30, 2009 #3

    Dick

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    Yes, I think that's all pretty right.
     
  5. Sep 30, 2009 #4
    so basically minimize s(t) not s'(t)?
     
  6. Sep 30, 2009 #5

    Dick

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    Well, yeah. Minimize s(t)=|v(t)|. Which as aPhilosopher pointed out is the same as sqrt(v(t).v(t)). Which in turn is minimized at the same time as v(t).v(t).
     
    Last edited: Oct 1, 2009
  7. Sep 30, 2009 #6
    i didn't know that s(t) was the same as sqrt( v(t) dot v(t) ), is this a theorem...or just common sense? X|
     
  8. Sep 30, 2009 #7

    Dick

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    What's your expression for the magnitude of a vector v(t)? Isn't it the same as sqrt(v(t) dot v(t))?
     
  9. Oct 1, 2009 #8
    It's sooooort of common sense once you've been shown how to think about it the right way.

    In one definition, you have the magnitude of the vector squared times the cosine of zero.

    In the other definition, you have (x1, x2, x3)2 = x12 + x22 + x32. This is just one side of the Pythagorean theorem in three dimensions. The other side is the length squared.
     
  10. Oct 1, 2009 #9

    HallsofIvy

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    All mathematics is "common sense once you think about it the right way"!
     
  11. Oct 1, 2009 #10
    Ahhhhhh, but is thinking about it the right way "common sense"? Perhaps you hold common sense in higher esteem than I do ;)
     
  12. Oct 1, 2009 #11

    HallsofIvy

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    Look carefully at what I said, "Philospher"! I did not say that "looking at it the right way" is common sense. I said that, once you have looked at it the right way, the rest is "common sense".
     
  13. Oct 1, 2009 #12
    If by common sense, you mean easy, or at least doable, then I agree whole heartedly! I was just taking a dig at common sense. My apologies for the confusion.
     
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