Population Growth Model Development

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Maya
Maya on 2 Feb 2023
Answered: Denish on 18 May 2024
Research in cell and tissue engineering often involves growing cells in the lab in a dish. Imagine
having a single 120 cm2 dish that has been seeded with 1500 cells/cm2. The dish can only
sustain 9x107 cells. With this seeding of the dish, the maximum that the dish can sustain
(carrying capacity) is reached in about 20-25 days. The growth rate, r, of this cell type is known
to be 0.75 cells per day (๐‘๐‘–๐‘Ÿ๐‘กโ„Ž ๐‘Ÿ๐‘Ž๐‘ก๐‘’ โ€• ๐‘‘๐‘’๐‘Ž๐‘กโ„Ž ๐‘Ÿ๐‘Ž๐‘ก๐‘’).
This cell population follows a logistic population growth model:
๐‘‘๐‘ƒ(๐‘ก)
๐‘‘๐‘ก = ๐‘Ÿ๐‘ƒ(๐‘ก)(1 โ€• ๐‘ƒ(๐‘ก)/๐พ ),
where P(t) is the size of the population at time, t, K is a constant corresponding to the
saturation level (carrying capacity) and r > 0 is the birth rate.
1. Write a MATLAB script for the numerical solution of this cell population problem
utilizing the Euler differential equation solver as demonstrated in class.
  5 Comments
Maya
Maya on 2 Feb 2023
I keep trying to input your code, but it consistently gives me a Parse error at '=' and at ')' for line 29 (in my script, line 29 is dy/dt = f(t,y);)
Torsten
Torsten on 2 Feb 2023
The part you could use is only the lower part of what I posted, not the upper line which is the mathematical problem description.

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Answers (1)

Denish
Denish on 18 May 2024
% Years from 1950 to 2020 years = 1950:2020;
% World population data (in billions) population = [2.557 2.768 2.981 3.165 3.333 3.552 3.692 3.842 4.035 4.442 ... 4.853 5.308 5.726 6.114 6.496 6.866 7.346 7.814 8.301 8.827 ... 9.357 9.847 10.325 10.771 11.173 11.515 11.837 12.123 12.374 12.606 ... 12.816 13.021 13.227 13.426 13.621 13.815 14.008 14.197 14.384 14.570 ... 14.756 14.942 15.128 15.315 15.503 15.691 15.879 16.069];
% Plotting plot(years, population, 'LineWidth', 2); title('World Population Over Time'); xlabel('Year'); ylabel('Population (in billions)'); grid on;

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