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Minimum speed question

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    See Q4 https://www.physicsforums.com/attachment.php?attachmentid=39736&d=1318081996"

    2. Relevant equations

    3. The attempt at a solution
    I am not sure whether my direction of thinking is correct.
    I have a bad feeling about my work.
    Is my work correct?

    http://a4.sphotos.ak.fbcdn.net/hphotos-ak-ash4/317084_2287980633629_1072324659_2450377_542887368_n.jpg" [Broken]
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 12, 2011 #2
    Hi there!
    Personally, I would have done this slightly differently:
    First, recall that in polar coordinates:
    |v| = \sqrt{\dot{r}^2+r^2\dot{\theta}^2}
    Differentiating that would yield the result, directly.
    Also, since you'll have to find the acceleration anyway, another derivative could also help with establishing the extrema, in other words, whether they are minima or maxima.
    I hope that works for you,
  4. Oct 12, 2011 #3
    Also, by this method, your efforts seem to contain an error there, so you may need to look into it again, in any case(I get the extremum at t=c).
  5. Oct 12, 2011 #4
    I also get extremum at t=c, but it turns out to be maximum, since dv/dt is increasing for 0<t<c and decreasing for c<t<2c
  6. Oct 13, 2011 #5
    Well, there may be an arithmatical discrepancy there, that one might want to follow;
    Let's derive it together:
    \frac{d|v|}{dt} = -\frac{2b}{c^3}(c-t)
    By using your approach, if we plug in, a value to the "left" of t=c, say, 0.5c, we would get: -b/c^2
    And, by inserting 1.5c(literally, to the "right" of t=c), it merely reduces to: +b/c^2.
    If you want to take it a step further, in order to verify your conclusions, one could always use the second derivative, i.e d^2v/dt^2;
    Remembering that:
    if \frac{d^2v}{dt^2}\vert_{t=c} > 0 \Rightarrow \min
    In our case, the second derivative, as is easily verifiable, is constant, and equal to 2*b/c^3,clearly, minimal, for all events and purposes.
    Was that more useful?
  7. Oct 13, 2011 #6
    Thanks a lot!
  8. Oct 16, 2011 #7
    I suddenly think of a question.
    Is [itex]\frac{d|\vec{s}|}{dt}[/itex]=[itex]\vec{|v|}[/itex]?
    Because they look different. If they are not the same, what is the physical meaning of [itex]\frac{d|\vec{s}|}{dt}[/itex]? Does it related to speed?
  9. Oct 18, 2011 #8
    You're very astute in pointing out a deficiency in the definition we may occasionally give to velocity as it pertains to trajectory.
    S-as a curve, or path, would lead to the total course taken by a particle. Differentiating it, leads to the rate of change of it, its curvature, parameters, so forth, overtime, a sort of average 1/k(s), where "k" is the osculation of the journey, rather than the velocity; It's "speed", literally speaking.
    By velocity, or dr/dt, we take the discrete, local, momentary rate of change of the path, that we are so desirous of in calculus.
    You're very accurate in pointing out the finer nuances of the trade, and kudos for that!
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