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Math Help - [SOLVED] problem solving in differential equations

  1. #1
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    [SOLVED] problem solving in differential equations

    This one has been bugging me. I cant seem to figure it out. can anyone help..??



    A tumor may be regarded as a population of multiplying cells. It is found empirically that the birth rate of the cells in a tumor decreases exponentially with time so that β( t) = β0e-αt ( where α and β0 are positive constants).
    And hence
    dP/dt = β0e-αt P
    P(0) = P0

    1.Solve this initial value problem for P(t) = P0exp ( 0/α(1- e-αt)

    Observe that P(t) approaches the finite limiting population P0exp(0/α) as t approaches infinity.
    2.Suppose that at time t=0 there are P0 = 106 cells and that P(t) is then increasing at the rate of 3 x105 cells per month. After 6 months the tumor has doubled (in size and in number of cells). Solve numerically for α , and then find the limiting population of the tumor.
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  2. #2
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    Differential equation

    Hello mathprincess24
    Quote Originally Posted by mathprincess24 View Post
    This one has been bugging me. I cant seem to figure it out. can anyone help..??



    A tumor may be regarded as a population of multiplying cells. It is found empirically that the birth rate of the cells in a tumor decreases exponentially with time so that β( t) = β0e-αt ( where α and β0 are positive constants).
    And hence
    dP/dt = β0e-αt P
    P(0) = P0

    1.Solve this initial value problem for P(t) = P0exp ( 0/α(1- e-αt)

    Observe that P(t) approaches the finite limiting population P0exp(0/α) as t approaches infinity.
    2.Suppose that at time t=0 there are P0 = 106 cells and that P(t) is then increasing at the rate of 3 x105 cells per month. After 6 months the tumor has doubled (in size and in number of cells). Solve numerically for α , and then find the limiting population of the tumor.
    I have a solution based on your figures, but it doesn't seem to make much sense, in that the limiting value is reached almost immediately. Anyway, here it is...

    \frac{dP}{dt} = \beta_0e^{-\alpha t}P

    \Rightarrow \int\frac{dP}{P} = \int\beta_0e^{-\alpha t}dt

    \Rightarrow ln(P) = -\frac{\beta_0}{\alpha}e^{-\alpha t}+c

    \Rightarrow P = e^{-\frac{\beta_0}{\alpha}e^{-\alpha t}}\times e^c

    P(0)=P_0 =e^{-\frac{\beta_0}{\alpha}}\times e^c

    \Rightarrow e^c = P_0e^{\frac{\beta_0}{\alpha}}

    \Rightarrow P=P_0e^{\frac{\beta_0}{\alpha}(1-e^{-\alpha t})}

    So as t \rightarrow \infty, P \rightarrow P_0e^{\frac{\beta_0}{\alpha}}

    So far, so good.

    Now P_0 = 106 and \frac{dP}{dt}= 3 \times 10^5 (This is what I assume you mean by 3 x 105). I think there may be a problem here: if the initial rate of growth is 300,000 cells per month, the terminal value will be reached very soon. So to double from 106 to 212 in 6 months doesn't seem very sensible. But based on this value...

    \beta_0P_0 = 3\times 10^5

    \Rightarrow \beta_0 = 2830

    When t = 6, P = 212 = 106e^{2830(1-e^{-6\alpha})/\alpha}

    \Rightarrow e^{2830(1-e^{-6\alpha})/\alpha}=2

    \Rightarrow \alpha = 4083 (using a spreadsheet to do the number-crunching)

    This gives the limiting value as 212, which is reached almost immediately. So, have we got the numbers wrong?

    Grandad

    PS. Of course! Perhaps you mean P_0 = 10^6. That makes a lot more sense. This gives \beta_0 = 0.3
    , \alpha = 0.3915, and the limiting value as 2.15 \times 10^6.
    Last edited by Grandad; February 19th 2009 at 07:02 AM. Reason: Add PS
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  3. #3
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    Thanks Grandad! I got the first part on my own after staring for hours and a simple arithmetic mistake. the second part was still tricky for me and the was a typo in the numbers. P0 = 106
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    okay so the program doesnt want to paste it correctly, but it is 10 to the 6th power.
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  5. #5
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    LaTex

    Hello mathprincess24
    Quote Originally Posted by mathprincess24 View Post
    okay so the program doesnt want to paste it correctly, but it is 10 to the 6th power.
    I'm glad we sorted it out. For future reference, to write 10 to the 6th power,

    • Write 10^6
    • Select the text you have just written (drag the mouse over it)
    • Press the \Sigma button on the toolbar. This will put [ math] before and [ /math] after.
    • Press the 'Preview Post' button to check that it looks how you want it to.

    Tip: if you want to see how I've created any of the other text, like \frac{dP}{dt}, or \int\beta_0e^{-\alpha t}dt, just click on it, and a window will pop up showing the code to you. Go on, try it now!

    Grandad
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