The size P of a certain insect population at time t in days

The size P of a certain insect population at time t (in days) obeys the function P(t) = 500 e^0 01t. Determine the number of insects at t = 0 days. What is the growth rate of the insect population? What is the population after 10 days? When will the insect population reach 700? When will the insect population double? What is the number of insects at t = 0 days? Insects

Solution

Solution:

Given That

P(t) = 500 e^0.01t

P = 500 e⁰ for initial population, time 0 days

P = 500

10 days,

P = 500 e^0.01*10

P = 500 e^0.1

P = 500*1.105

P = 552.5

Growth Rate looks like 0.01 percent, according to the equation model.

When reach 700 and when reach double of 500?

ln(P) = ln(500 e^0.01t)

P(t) = 600 e^(0.04t)

700 = 500 e^(0.01t)

700/500 = e^(0.01t)

ln(700/500) = ln(e^(0.01t))

ln(1.4) = 0.01t

t = ln(1.4)/0.01

t = 33.64 days

Solving for this answer is very similar to that of Part B, in that you plug in 1400 (or 700 x 2) for P(t)

and solve for the time t

P(t) = 500 e^(0.01t)

1400 = 500 e^(0.01t)

1400/500 = e^(0.01t)

ln(2.8) = ln(e^(0.01t))

ln(2.8) = 0.01t

t = ln(2.8)/0.01

t = 102.96 days