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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. If P represents the number of female insects in a population, S the number of sterile males introduced each generation, and r the population's natural growth rate, 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 = 0.10 and 900 sterile males are added in each generation. Evaluate the integral to give an equation relating the female population to time. (Note that the resulting equation cannot be solved explicitly for P.)

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 class=

Respuesta :

tehorsesnameisfriday
t = integral [(P+S)/P((r-1)P-S)] dP

now, to do partial fraction decomposition, consider

(P+S)/P((r-1)P-S) = A/P + B/((r-1)P-S)

P + S = A((r-1)P -S) + PB

if P = 0, then:

A = -1

If (r-1)P - S = 0

=> p = S/(r-1)

B = 1 + r - 1

B = r

Therefore

t = integral(-1/P + (r-1+1)/((r-1)P - S))dP

t = -ln|P| + ln|(r-1)P-S| + 1/r-1 * integral((r-1)/[(r-1)P-S])

t = -ln|P| + ln|(r-1)P-S| + ln|(r-1)P-S|/(r-1)

t = -ln|P| + (r/r-1)ln|(r-1)P-S|

substituting known constants, we get:

t = -ln|P|-1/9(ln|-0.9P - 900|)

t = -ln|P| - 1/9(ln|-0.9P -900|)

t = -ln|P| - 1/9(ln(|(-9/10)(P + 1000)|))

t = -ln(P) - 1/9ln(P + 1000)- 1/9ln(9/10)

as P >= 0, ln(P) and ln(P + 1000) doesn’t need abs value

Note that our expression for t doesn’t account for P=0 as ln(P) is undefined at P = 0.

The expression for t is also undefined at P = -1000, but that falls outside our domain P >= 0, so we don’t really need to worry about that.

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