How Does the Gompertz Equation Model Tumor Growth?

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In summary, a growth of cancerous tumors can be modeled by the Gompertz equation as long as N isn't too small. The equation is dN/dt = -rN*ln(N/K), where N(t) is proportional to the number of cells in the tumor, and r,K>0 are parameters. After solving the equation, the simplified solution is N = K e^{e^{-rt}}.
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Jamin2112
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Homework Statement



As long as N isn't too small, the growth of cancerous tumors can be modeled by the Gompertz equation,

dN/dt = -rN*ln(N/K),

where N(t) is proportional to the number of cells in the tumor, and r,K>0 are parameters.

Homework Equations





The Attempt at a Solution



Separate out.

dN/[N*ln(N/K)]= -rdt

After much toil, I figured out that the left side is simply d/dN (ln(ln(N/K))

==> ln(ln(N/K)) = -rt + C
==> ln(N/K) = e-rt+C
==> N/K = ee-rt+C
==> N = Kee-rt+C

Seems overly complicated. Is there some property of ln(ln(f)) that would help simplify? Suggestions, please.
 
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  • #2
Well the solution works out so it is correct. If you get a solution to a differential equation and are unsure of it, plug it back into your equation to find if the two sides equal each other.

One thing you can do to simplify it is to get rid of that +C in your second exponent.

[tex]N = K e^{e^{-rt +C}}[/tex]

This can be rewritten:

[tex]N = K e^{e^{-rt}e^{C}}[/tex]

[tex]N = K e^{e^{-rt}}e^{e^{C}}[/tex]

The [tex]e^{e^{C}}[/tex] is a constant we can absorb into K. Giving a simplified solution:

[tex]N = K e^{e^{-rt}}[/tex]
 
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