Question

Gorilla Population A certain wild animal preserve can support no more than 250 lowland gorillas. Twenty-eight gorillas were known to be in the preserve in \(1970 .\) Assume that the rate of growth of the population is \(\frac{d P}{d t}=0.0004 P(250-P)\) where time \(t\) is in years. (a) Find a formula for the gorilla population in terms of \(t\) . (b) How long will it take for the gorilla population to reach the carrying capacity of the preserve?

Short answer

The exact functions will depend on the outcome of the integrations and the constant of integration, but the end result should yield the gorilla population as a function of time, \(P(t)\), and the time \(t\) for the population to reach the preserve's carrying capacity.

Step-by-step solution

Separation of Variables

To solve the differential equation, we need to separate variables. Firstly, rewrite the differential equation into the form where all terms of \(P\) are on one side and those with \(t\) on the other side: \(\frac{dP}{P(250-P)} = 0.0004 dt\)

Integrate both sides

Next, integrate both sides of the equation to solve for \(P\):\(\int{\frac{1}{P(250-P)}} dP = \int{0.0004} dt\)

Use Partial Fraction Decomposition

The left hand side requires partial fraction decomposition. We rewrite \(\frac{1}{P(250-P)}\) as \(\frac{A}{P} + \frac{B}{250-P}\). Solve for A and B to get \(\frac{1}{P(250-P)} = \frac{-1}{P} + \frac{1}{250-P}\)

Integrate using the partial fractions

Integrate both sides using these new fractions:\(-\int{\frac{1}{P}} dP + \int{\frac{1}{250-P}} dP = 0.0004\int{dt}\)

Find an equation for the gorilla population (P)

The integrals will yield the logarithmic form, solve them to get:\( -ln|P| + ln|250-P| = 0.0004 t + C\)This equation can be solved for \(P\) to give the gorilla population as a function of time.

Solve for constant C

Initial condition is given as \(P(0) = 28\), use this to solve for C.

Calculate time for population to reach carrying capacity

Set \(P(t) = 250\) and solve for \(t\) to find out when the gorilla population will reach the carrying capacity of the preserve.