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Taking the log of an exponential function and finding the slope

  1. Mar 13, 2014 #1
    This is part of a differential equations group project problem where I solve a set of differential equations to obtain the solution to a function. The part that I am stuck at involves taking the log of an exponential function, though there may be a mistake on the book's part, but I'm not sure.

    1. The problem statement, all variables and given/known data
    2. Relevant equations
    [itex]\frac{d}{dt}T(t) = λ - δT(t) - βV(t)T(t)[/itex]
    [itex]\frac{d}{dt}I(t) = βV(t)T(t) - μI(t)[/itex]
    [itex]\frac{d}{dt}V(t) = NμI(t) - γV(t)[/itex]
    T represents population of uninfected T cells in units of cells.
    I represents population of infected T cells in units of cells.
    V represents population of virus in units of virions.
    λ is the rate of T cell production by the human body per day in units of 1/days
    δ is the rate constant of T cells naturally dying off per day in units of 1/days
    μ is the rate constant of Infected T cells dying off (bursting) per day resulting in the spread of virus in units of 1/days
    γ is the rate constant of virus decaying per day in units of 1/days
    N is the number of virus per cell in units of virions/cell
    β is the infection rate constant in units of 1/(days*virions)

    The problem states to set β = 0, assuming there is a drug that completely removes infection of T cells, allowing me to solve by 1st order linear differential equation methods to get V(t):
    [itex]V(t) = \frac{NμI_0}{γ-μ}e^{-μt} + (V_0 - \frac{NμI_0}{γ-μ})e^{-γt}[/itex]
    with initial conditions
    [itex]V(0) = V_0 \text{ and } I(0) = I_0[/itex]

    3. The attempt at a solution
    Taking the log of V(t) and then taking the derivative, I get:
    [itex]\frac{d}{dt}log[V(t)] = \frac{d}{dt}\frac{ln[V(t)]}{ln(10)} = \frac{1}{ln(10)V(t)}\frac{d}{dt}V(t)[/itex]
    and
    [itex]\frac{d}{dt}V(t) = \frac{-Nμ^2I_0}{γ-μ}e^{-μt} - γ(V_0 - \frac{NμI_0}{γ-μ})e^{-γt}[/itex]

    Assuming μ is smaller than γ, the term containing [itex]e^{-γt}[/itex] should go to 0 when t is large, leaving us with:
    [itex]\frac{d}{dt}log[V(t)] = \frac{V'(t)}{ln(10)V(t)} = \frac{\frac{-Nμ^2I_0}{γ-μ}e^{-μt}}{ln(10)\frac{NμI_0}{γ-μ}e^{-μt}} = \frac{-μ}{ln(10)} \text{for large t}[/itex]

    The problem is the factor of [itex]\frac{1}{ln(10)}[/itex] causing the slope of the log graph to be inconsistent with what the problem statement claims it should be: -μ.
     
    Last edited: Mar 13, 2014
  2. jcsd
  3. Mar 13, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    When doing calculus, "log" means "natural logarithm".
     
  4. Mar 13, 2014 #3
    While that would get rid of the 1/ln10 factor, I doubt I could make that assumption in this case because for one this book does use ln notation, and there are graphs for this problem that depicts virus amount in a log base 10 scale (where it goes from 1 to 10 to 100 to... 10^n... every major gridline). The problem even states to determine μ based on the slope of those graphs.
     
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