Question

Problem #10: A model for a certain population P() is given by the initial value problem P(10-1-10-9 P), P(0) - 1000000 dt where t is measured in months (a) What is the limiting value of the population? (b) At what time (i.e., after how many months) will the populaton be equal to one fifth of the limiting value in (a)? (Do not round any numbers for this part. You work should be all symbolic.) Problem #10(a): 100000000 Enter your answer symbolically, as in these examples -10*In(7/99) Problem #10(b): -101n() Just Save Submit Problem #10 for Grading Problem #10 | Attempt #1 Your Answer:10(a) 100000000 10 10(a) 10(b) 10(b)-10 In(, | 10(b) 10(a) 10(b) Your Mark: 10(a) 2/2v 10(b) 0/2x 10(a) 10(b)

Answer 1

A certain population p(t) is given by a initial value problem delta p/dt = p (10^-1 - 10^-9 p) p(a) = 1000000 where 't' is measured in months where: a = 10^-1 b = 10^-9 Now, we determine the limiting value = a/b = 10^-1/10^-9 = 10^+8 limiting value = 100,00,000 Now, let us calculate at what time will the population be equal to 1/5^th of limiting value 1/5^th of limiting value = 100000000/5 = 200000000 Then put it in logarithmic equation 20000000 = 10^-1 times 10000000/(10^-9 + (10^-1 - 10 times 10000000) e^-t 10^-1) 20000000 = 1000000/0.0009 + 0.99 e^-t/10 = 0.005 0.099 e^-t/10 = 0.0041 /t/10 = ln [0.0041/0.099] - t/10 = ln [0.0414] -t/10 = -3.1844 t = 3.1844 times 10 t = 31.844 sec At t = 31.844 sec the population will be equal to one fifth of limiting values