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Math Help - Universal law of growth

  1. #1
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    Universal law of growth

    I need a serious help please

    Let Y be the metabolic rate of an organism, Yc the metabolic rate of a single cell, Nc(t) the total number of cells at time t, mc the mass of a cell, and Ec the energy required to create a new cell. The cell properties, Ec, mc, and Yc, are assumed to be constant and invariant with respect to the size of the organism. Thus

    Y = YcNc + Ec(dNc/dt).

    Let m be the total body mass of the organism at time t, and m = mcNc.(Note that Nc is the total number of cells in a body and is proportional to mass m, while the total number of capillaries Nn is proportional to 3/4 power of m.) Y=Y0(m)^(3/4)

    a.show that the above equation can be written as
    dm/dt = am^(3/4) - bm
    with a = Y0mc / Ec and b = Yc/Ec

    b.Let m = M be the mass of a matured organism, when it stops growing (dm/dt = 0). Find M, and show that the above equation can be rewritten as
    dm/dt = am^(3/4)[1-(m/M)^(1/4)].

    c.Let r = (m/M)^(1/4), and R = 1-r. Then the above equation becomes
    dR/dt = -(a/4M^(1/4))R.

    Solve this simple ordinary differential equation and show that a plot of ln(R(t)/R(0)) vs. the no-dimensinal time at/(4M^(1/4)) should yield a straight line with a slope -1 for any organism regardless of its size.

    d. Based on this scaling for time t, argue that, for a mammal, the interval between heartbeats should scale with its size as M^(1/4).

    I am completely lost,,,, please help ,,,, Thank you
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  2. #2
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    Quote Originally Posted by mathsohard View Post
    I need a serious help please

    Let Y be the metabolic rate of an organism, Yc the metabolic rate of a single cell, Nc(t) the total number of cells at time t, mc the mass of a cell, and Ec the energy required to create a new cell. The cell properties, Ec, mc, and Yc, are assumed to be constant and invariant with respect to the size of the organism. Thus

    Y = YcNc + Ec(dNc/dt).

    Let m be the total body mass of the organism at time t, and m = mcNc.(Note that Nc is the total number of cells in a body and is proportional to mass m, while the total number of capillaries Nn is proportional to 3/4 power of m.) Y=Y0(m)^(3/4)

    a.show that the above equation can be written as
    dm/dt = am^(3/4) - bm
    with a = Y0mc / Ec and b = Yc/Ec

    b.Let m = M be the mass of a matured organism, when it stops growing (dm/dt = 0). Find M, and show that the above equation can be rewritten as
    dm/dt = am^(3/4)[1-(m/M)^(1/4)].

    c.Let r = (m/M)^(1/4), and R = 1-r. Then the above equation becomes
    dR/dt = -(a/4M^(1/4))R.

    Solve this simple ordinary differential equation and show that a plot of ln(R(t)/R(0)) vs. the no-dimensinal time at/(4M^(1/4)) should yield a straight line with a slope -1 for any organism regardless of its size.

    d. Based on this scaling for time t, argue that, for a mammal, the interval between heartbeats should scale with its size as M^(1/4).

    I am completely lost,,,, please help ,,,, Thank you
    \displaystyle Y0m^{3/4}=YcNc+Ec\frac{dNc}{dt}\Rightarrow \frac{Y0m^{3/4}}{Ec}=\frac{YcNc+Ec\frac{dNc}{dt}}{Ec}\Rightarro  w \frac{Y0m^{3/4}}{Ec}=bNc+\frac{dNc}{dt}\Rightarrow\cdots

    Since every question is related to the above equation, finishing this should allow you to do the rest.
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    Is it possible to go little further for me ??? because I still have some difficulties dealing with derivatives
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  4. #4
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    Quote Originally Posted by mathsohard View Post
    Is it possible to go little further for me ??? because I still have some difficulties dealing with derivatives
    Multiply by mc

    \displaystyle\frac{Y0m^{3/4}mc}{Ec}=bNcmc+mc\frac{dNc}{dt}

    \displaystyle\Rightarrow am^{3/4}=b(Ncmc=m)+\frac{d(Ncmc=m)}{dt}\Rightarrow am^{3/4}-bm=\frac{dm}{dt}
    Last edited by dwsmith; January 13th 2011 at 12:16 PM.
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    for d, how can I interpret my equations??? I got up to C.
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  6. #6
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    Quote Originally Posted by mathsohard View Post
    for d, how can I interpret my equations??? I got up to C.
    What is your C?
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  7. #7
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    dR/dt = -(a/4M^(1/4))R.

    Solve this simple ordinary differential equation and show that a plot of ln(R(t)/R(0)) vs. the no-dimensinal time at/(4M^(1/4)) should yield a straight line with a slope -1 for any organism regardless of its size.

    I solved this................... ln(R(t)/R(0)) = -(a/4M^(1/4))t thus yield a straight line with a slope -1 vs the no-dimensinal time at/(4M^(1/4))
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  8. #8
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    Quote Originally Posted by mathsohard View Post
    dR/dt = -(a/4M^(1/4))R.

    Solve this simple ordinary differential equation and show that a plot of ln(R(t)/R(0)) vs. the no-dimensinal time at/(4M^(1/4)) should yield a straight line with a slope -1 for any organism regardless of its size.

    I solved this................... ln(R(t)/R(0)) = -(a/4M^(1/4))t thus yield a straight line with a slope -1 vs the no-dimensinal time at/(4M^(1/4))
    What is capital M?
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  9. #9
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    mass of a matured organism, oh so since it is all matured it should scale with its size as M^1/4 huh ??
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  10. #10
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    Quote Originally Posted by mathsohard View Post
    mass of a matured organism, oh so since it is all matured it should scale with its size as M^1/4 huh ??
    I am not a biology person so it beats me.
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