MATH 222 Problems on Exponential Models

1. Two ponds are stocked with fish, each pond with a different type. The first pond begins with 50 fish while the second begins with 200. One year later the first pond has 450 fish while the second has 1400. How long does it take the first pond to catch up with the second pond in population?

Solution:

Let P₁(t) and P₂(t) be the population functions for the two ponds. P₁(0) = 50 and P₂(0) = 200. P₁(1) = 450. Thus we have Thus we have r = ln(9). For the second pond we have We have s = ln(7). Set the two population functions equal to each other: Solving for t yields Taking natural logs of both sides gives

2. Using very careful measurements of a radioactive element, a researcher finds that 2% of his sample decays in 3 hours. Find the half-life of this element.

Solution:

Let M₀ be the initial mass. Then . Solving for k yields k = ln(.02)/3 = -1.304. The half-life T equals –ln(2)/k or .531 days.

3. A population obeying the logistic growth equation has an initial population of 40 and a carrying capacity of 200. After one month the population has grown to 60.

a) Find P(t) for this population.

b) Suppose that 10 individuals are removed from the population when it is at 120. How long does it take the population to return to the level of 120? Repeat the computation for 100 and for 80 .

Solution:

a) . Thus we have . Since N(1) = 60 we can compute -k which is So .

b) Solve N(t) = 120 and N(t) = 110 for t and subtract the earlier time from the later. Do the same for N(t) = 100, 90, 80, and 70. The results are:

N(3.324) = 120 and N(2.944) = 110, so the first time difference is 0.38.

N(2.572) = 100 and N(2.200) = 90, so the second time difference is 0.372.

N(1.820) = 80 and N(1.423) = 70, so the third time difference is 0.397.

4. A population obeying Logistic population growth has an initial population of 500 and a carrying capacity of 300. Suppose the growth constant is r = .06.

a) Find P(t).

b) Sketch the graph of P(t).

c) When does the population reach 400?

d) Find .

Solution:

a) All the parts are given except B. . Thus .

c) Solve P(t) = 400. t = 7.833.

d) . The population moves towards the carrying capacity.