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Quick Differentiation Qu.

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data

    I have nearly finished a problem. I am at the stage where I have:

    [tex]B(s)\propto\frac{1}{s^{3}}[/tex]

    [tex]v=\frac{ds}{dt}[/tex]

    I want to find the rate of change of B(s), but expressed in terms of a velocity, rather than a displacement.

    2. Relevant equations

    So for displacement:

    [tex]B'(s) \propto\frac{-3}{s^{2}}[/tex]

    3. The attempt at a solution
    How can I express this in terms of v instead? My guess would be that:

    [tex]B'(s) \propto\frac{1}{v^{3}}[/tex]

    but I'm not sure thats correct. This does seem like quite a simple problem but its really got me stumped.

    Thanks in advance, Ewan
     
  2. jcsd
  3. Nov 25, 2007 #2
    Is the velocity constant?
    Do you mean by rate of change

    [tex]\frac{dB(s)}{dt}[/tex]?????
    And I think it would be better if you post whole question.
     
    Last edited: Nov 25, 2007
  4. Nov 25, 2007 #3
    Yes the velocity is constant. The rest of the question isn't relevant to my problem, but I can post if you want. And yes by rate of change I mean:

    [tex]\frac{dB(s)}{dt}[/tex]
     
    Last edited: Nov 26, 2007
  5. Nov 26, 2007 #4
    Can you post?
     
  6. Nov 28, 2007 #5
    Never mind I managed to solve it (i think!).

    [tex]B \propto \frac{-3}{s^{3}} \frac{dv}{dt}[/tex]

    using substitution of s for v.
     
    Last edited: Nov 28, 2007
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