# One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with f

Mathematics

## Question

One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. let p represent the number of female insects in a population and s the number of sterile males introduced each generation. let r be the per capita rate of production of females by females, provided their chosen mate is not sterile. then the female population is related to time t by t = p + s p[(r − 1)p − s] dp. suppose an insect population with 10,000 females grows at a rate of r = 1.2 and 400 sterile males are added. evaluate the integral to give an equation relating the female population to time. (note that the resulting equation can't be solved explicitly for p. remember to use absolute values where appropriate.)

• ### 1. User Answers Mindaka

To be clear, the given relation between time and female population is an integral:
$$t = \int { \frac{P+S}{P[(r - 1)P - S]} } \, dP$$

The problem says that r = 1.2 and S = 400, therefore substituting:
$$t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]} } \, dP$$

$$\int { \frac{P+400}{P(0.2P - 400)} } \, dP$$

In order to evaluate this integral, we need to write this rational function in a simpler way:
$$\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} + \frac{B}{(0.2P - 400)}$$

where we need to evaluate A and B. In order to do so, let's calculate the LCD:
$$\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P - 400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)}$$

the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
= P(0.2A + B) - 400A

The two sides must be equal to each other, bringing the system:
$$\left \{ {{0.2A + B = 1} \atop {-400A = 400}} \right.$$

Which can be easily solved:
$$\left \{ {{B=1.2} \atop {A=-1}} \right.$$

Therefore, our integral can be written as:
$$t = \int { \frac{P+400}{P(0.2P - 400)} } \, dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP$$
= $$- \int { \frac{1}{P} \, dP + 1.2\int { \frac{1}{0.2P-400} } \, dP$$
= $$- \int { \frac{1}{P} \, dP + 6\int { \frac{0.2}{0.2P-400} } \, dP$$
= - ln |P| + 6 ln |0.2P - 400| + C

Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000
| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5)
Therefore, the equation relating female population with time requested is:
t =  - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)