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Magazine Chapter 12: Q41E (page 518)
Population Model The differential equation \(\frac{{dP}}{{dt}} = (kcost)\), where \(k\) is a positive constant, is a model of human population \(P(t)\) of a certain community. Discuss an interpretation for the solution of this equation. In other words, what kind of population do you think the differential equation describes?
Short Answer
The differential equation that describes the population is \({\rm{p = }}{{\rm{e}}^{{\rm{ksint}}}}{{\rm{c}}_{\rm{1}}}\).
Step by step solution
Define a derivative of the function.
The derivative of a function of a real variable in mathematics describes the sensitivity of the function value (output value) to changes in its argument (input value).
Calculus uses derivatives as a fundamental tool. When a derivative of a single-variable function exists at a given input value, it is the slope of the tangent line to the function's graph at that point.
Determine the differential equation.
Let the differential equation be,
\(\begin{aligned}{c}\frac{{{\rm{dp}}}}{{\rm{p}}}{\rm{ = kcost\;dt}}\\\int {\frac{{{\rm{dp}}}}{{\rm{p}}}} {\rm{ = k}}\int {{\rm{cos}}} {\rm{t\;dt}}\end{aligned}\)
\({\rm{ln|p| = ksint + c}}\)
Determine an interpretation.
The population increases in the first cycle from \({\rm{t = 0}}\) to \({\rm{t = 1}}{\rm{.5}}\), and decreases in the next cycle from \({\rm{t = 1}}{\rm{.5}}\) to \({\rm{t = 4}}{\rm{.5}}\) in a sinusoidal way. It increases and decreases with the \({\rm{6}}{\rm{.5}}\) unit period of time. During each cycle, the population increases and decreases below the average level.
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