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Tricky equation to solve

  1. Sep 25, 2012 #1
    Hi,

    I need to solve this equation that stems from a differential equation, though that isn't really important. Here's the equation that needs to be solved for t:

    3t+100=4((9t^3+900t^2+30000t)/(3t+100)…

    I've tried myself and always seem to hit a dead end. Using a calculator, I know the answer is 100/3(2^2/3-1) or ~19.58, but can't find a way to get to this answer.

    I should mention, I would like this in a fully worked solution if possible. The subject does not allow calculators.
     
  2. jcsd
  3. Sep 26, 2012 #2

    Simon Bridge

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    The three dots at the end of an equation usually means "and so on" - meaning to continue some sort of pattern. Is that what you meant here?
    (there's also an extra bracket in there.)

    Guessing "no" ... rewriting - you need t to satisfy the relation: $$(3t+100)^2 = 4(9t^3+900t^2+30000t)$$your first task (brute force) is to put the equation into standard form.

    From the look of it, it is a cubic, so as many as three values of t may make this equation true. You should already know how to solve for the roots of a cubic.
     
    Last edited: Sep 26, 2012
  4. Sep 26, 2012 #3
    Sorry, wrote the question wrong:
    3t+100=4((9t^3+900t^2+30000t)/(3t+100)^2)
     
    Last edited: Sep 26, 2012
  5. Sep 26, 2012 #4
    Here's some more info on the problem that may help:

    x(t) is the amount of an element in a container

    The original d.e:
    $$dx/dt+6x/(100+3t)=3$$

    100+3t=v, the total volume of liquid in a container over time. The max of this is 300.

    Solving the d.e gives:
    $$x(t)=(9t^3+900t^2+30000t)/(3t+100)^2)$$

    Now, and this is where I am stuck, I have to find when the percentage of the element within the total liquid is ≥ 25%. This occurs when $$x/v=1/4$$, so $$3t+100=4x$$, hence:
    $$3t+100=4((9t^3+900t^2+30000t)/(3t+100)^2)$$
     
    Last edited: Sep 26, 2012
  6. Sep 26, 2012 #5

    Simon Bridge

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    Hmmm I get ##x(t)= c(3t-100)^2-3t+100## where c is the constant of integration.

    But if you are still happy with yours - you still have a cubic.
    Put it in standard form.
     
  7. Sep 26, 2012 #6
    Oh sorry, I forgot to mention the intial vlaue is x(0)=0
     
  8. Sep 26, 2012 #7
    So I guess this is what you mean by standard form:

    $$9t^3+900t^2+30000t-1000000=0$$
     
  9. Sep 26, 2012 #8

    Ray Vickson

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    It is easier (and more insightful) to re-scale this: set t = 100s in the equation, to get
    [tex] (3s+1)^3 = 12(3s^3 + 3s^2 + s). [/tex] It is easier to get high accuracy on a calculator when you do this, plus it is easier to manipulate, etc.

    RGV
     
  10. Sep 26, 2012 #9

    SammyS

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    Following Ray's suggestion, let u = 3s+1. so that [itex]\displaystyle s=\frac{u-1}{3}\ .[/itex]

    That should make the algebra even cleaner.
     
  11. Sep 26, 2012 #10

    Simon Bridge

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    In which case, c=-1/100 ... have you checked your DE solution?

    Sammy and Ray have good suggestions for making things clearer.
    You can look up how to solve a cubic equation.... and that was "standard form" yes.
     
    Last edited: Sep 26, 2012
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