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.)
Accepted Solution
A:
The relation between population and time is: [tex]t = \int { \frac{P+S}{P[(r - 1)P - S]} } \, dP [/tex]
Since r = 1.2 and S = 400, we get: [tex]t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]} } \, dP [/tex]
This is a rational function that we can write as: [tex]\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} + \frac{B}{(0.2P - 400)} [/tex]
In order to evaluate A and B, let's calculate the LCD: [tex]\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P - 400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)} [/tex]
which brings to: P + 400 = 0.2AP - 400A + BP = P(0.2A + B) - 400A
Since, the left side must be equal to the right side, we get the following system of equations: [tex] \left \{ {{0.2A + B = 1} \atop {-400A = 400}} \right. [/tex]
Which gives: [tex] \left \{ {{B=1.2} \atop {A=-1}} \right. [/tex]
We now have to evaluate the constant C. In order to do so, we consider 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 is: t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)