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The following question consider the Gompertz equation, a modification for logistic growth, which is often used for modeling cancer growth, specifically the number of tumor cells.

Exponential models
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asked 2021-08-16
The following question consider the Gompertz equation, a modification for logistic growth, which is often used for modeling cancer growth, specifically the number of tumor cells. When does population increase the fastest in the threshold logistic equation \(\displaystyle{P}'{\left({t}\right)}={r}{P}{\left({1}-{\frac{{{P}}}{{{K}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}?\)

Expert Answers (1)

2021-08-17
The maximum population can be found by solving \(\displaystyle{P}'={0}\) for P while the fastest growth can be reached by equating the differentiation of the population rate by zero, then solving
(i.e) by solving \(\displaystyle{P}\text{}{0}\) for P as follows
\(\displaystyle{P}\text{}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({r}{P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}\right)}={0}\)
\(\displaystyle\Rightarrow{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}\right)}={0}\)
\(\displaystyle\Rightarrow{\left({P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}\right)}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({\ln}{\left|{P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}\right|}\right)}={0}\) (logarithmic diff.)
\(\displaystyle\Rightarrow{\left({P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}\right)}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({\ln{{P}}}+{\ln{{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}}}+{\ln{{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}}}\right)}={0}\)
\(\displaystyle\Rightarrow{\left({P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}\right)}{\left[{\frac{{{P}'}}{{{P}}}}-{\frac{{{\frac{{{P}'}}{{{k}}}}}}{{{1}-{\frac{{{P}}}{{{k}}}}}}}+{\frac{{{T}{\frac{{{P}'}}{{{P}{2}}}}}}{{{1}-{\frac{{{T}}}{{{P}}}}}}}\right]}={0}\) (chain rule)
\(\displaystyle\Rightarrow{\left({P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}\right)}{\left[{\frac{{{1}}}{{{P}}}}-{\frac{{{\frac{{{1}}}{{{k}}}}}}{{{1}-{\frac{{{P}}}{{{k}}}}}}}+{\frac{{{\frac{{{T}}}{{{P}{2}}}}}}{{{1}-{\frac{{{T}}}{{{P}}}}}}}\right]}={0}\) (dividing by P')
\(\displaystyle\Rightarrow{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}-{\frac{{{P}}}{{{k}}}}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}+{\frac{{{T}}}{{{P}{2}}}}{P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}={0}\)
\(\displaystyle\Rightarrow{\left({1}-{\frac{{{2}{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}+{\frac{{{T}}}{{{p}{2}}}}{P}{\left({1}-{\frac{{{P}}}{{{k}}}}\right)}={0}\) (simplify)
\(\displaystyle\Rightarrow{\left({1}-{\frac{{{2}{P}}}{{{k}}}}\right)}{\left({1}-{\frac{{{T}}}{{{P}}}}\right)}+{\left({\frac{{{T}}}{{{P}}}}-{\frac{{{T}}}{{{k}}}}\right)}={0}\) (simplify)
\(\displaystyle\Rightarrow{1}-{\frac{{{2}{P}}}{{{k}}}}-\neg{\left\lbrace{\frac{{{T}}}{{{P}}}}\right\rbrace}+{\frac{{{2}{T}}}{{{k}}}}+\neg{\left\lbrace{\frac{{{T}}}{{{P}}}}\right\rbrace}-{\frac{{{T}}}{{{k}}}}={0}\) (simplify)
\(\displaystyle\Rightarrow{1}-{\frac{{{2}{P}}}{{{k}}}}+{\frac{{{T}}}{{{k}}}}={0}\) (simplify)
\(\displaystyle\Rightarrow{k}-{2}{P}+{T}={0}\) (multiply by k)
\(\displaystyle\Rightarrow{P}={\frac{{{T}+{k}}}{{{2}}}}\) (simplify)
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