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Finding a differential equation of a growing object

  1. Sep 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Mass , r, of a round object grows in proportions with its SA. The density of the object is constant. Find the differential equation for the change in 'r' of the object over time. Arbitrary constants can be combined but final answer must depend on only r and constants.

    2. Relevant equations

    Mass is proportional to volume.

    3. The attempt at a solution

    I have,

    density (D) = r/V (r= mass)
    r = (D)*V This shows mass is proportional to volume.

    SA = 4πR^2
    V = (4/3)πR^3

    Isolate R in SA equation.
    Lets call SA = A

    A = 4πR^2
    r = (A/4π)^0.5

    Sub into V

    V = (4/3)π(A/4π)^1.5

    dR/dt = r*V = r*(4/3)π(A/4π)^1.5

    However I still have A in the differential equation.

    Am I on the right track?
     
  2. jcsd
  3. Sep 12, 2014 #2
    I am a bit confused. I did not get the last equation.. Isn't it dimensionally inconsistent? On the left you have length over time and on the right you have mass times volume. Are you not supposed to find dr/dt? Why do you write dR/dt? I thought of starting like:
    [tex]\frac{dr}{dt}=cA[/tex]
    Since it says "proportional" I guess we should include a constant of proportionality, c.
     
  4. Sep 12, 2014 #3

    mfb

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    r is a really bad choice for a mass of a sphere, that looks like the radius. Okay, you cannot change that.

    Where does that come from? It is not true.
    Edit: Ah, you mean R here. Well, there goes the r confusion...

    You can express that A in the last equation in terms of R or r.

    Why?

    Hint: This problem gets much easier if you express everything in terms of the radius.
     
  5. Sep 12, 2014 #4
    You're right, my mistake. It's suppose to be dr/dt. If the equation is dr/dt = cA , that'll still leave us with a variable R(radius) without any r (mass) terms.
     
  6. Sep 12, 2014 #5
    Yes, but what if you isolate R in the equation of volume? Then you also have r = D*V, so I think this is the solution.
     
  7. Sep 12, 2014 #6
    So, V = (4/3)πR^3, then R = (3V/(4π))^(1/3). But what can I do with R? How can I incorporate it into the DE with only r terms and constants?

    I wrote dr/dt = r*V but I think I meant to write dr/dt = r/V because that equals to density which is the only info given + not knowing if the eqn made sense.
     
  8. Sep 12, 2014 #7
    You actually need to first use [itex]A=4πR^2[/itex]. Then:
    [itex]\frac{dr}{dt}=cA[/itex]
    [itex]\frac{dr}{dt}=c4πR^2[/itex]
    [itex]\frac{dr}{dt}=c4π(\frac{3V}{4π})^{2/3}[/itex]
    [itex]\frac{dr}{dt}=c4π(\frac{3r}{4πD})^{2/3}[/itex]
     
  9. Sep 12, 2014 #8
    That makes sense. But one question, density (D) is still in the equation. Is there any way to turn that into terms of r? or do we just combine D with a constant?
     
  10. Sep 12, 2014 #9
    Since it is a constant the expression in terms r should also be a constant, and since r is variable, another variable will appear to make the expression constant. Besides, since D is a constant, is it not allowed to be in the equation?
     
  11. Sep 12, 2014 #10

    mfb

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    Staff: Mentor

    D is constant, that's fine. The final answer depends on D, there is no way to get rid of it.
     
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