3. The population P (in thousands) of a colony of rabbits at time (t) years is given by
P(t) = 5 + ae−bt, t ≥ 0
where a and b are positive constants. If it is given that the initial population is 10, 000 rabbits and
one year later this population has reduced by 1, 000, find the following;
(a) The values of a and b. (Do not approximate your answer).
(b) The time (to 2 decimal places) when the population becomes two-thirds of the initial population.
(c) The time when there is no change in the rabbit population as time changes. Explain your answer.
since P(1) = 1000
and P(0) = 10000
5+a = 10000
5+9995e^-b = 1000
e^-b = 995/9995 = 0.0995
b = -ln 0.0995 = 2.31
so P(t) = 5 + 9995 e^(-2.31 t)
now you can answer the questions.
To find the values of a and b in the given function, we can use the information provided about the initial population and its reduction after one year.
(a) The initial population is given as 10,000 rabbits, which means when t = 0, P(0) = 10,000. Substituting this value into the population function, we have:
10,000 = 5 + ae^(-b*0)
10,000 = 5 + a
a = 10,000 - 5
a = 9,995
After one year, the population has reduced by 1,000 rabbits. Using this information, we can find the value of b. When t = 1, P(1) = 10,000 - 1,000 = 9,000. Substituting these values into the population function, we have:
9,000 = 5 + 9,995e^(-b*1)
Solving this equation:
9,000 = 5 + 9,995e^(-b)
8,995 = 9,995e^(-b)
e^(-b) = 8,995/9,995
e^(-b) = 0.900950475
Taking the natural logarithm (ln) of both sides to solve for b:
ln(e^(-b)) = ln(0.900950475)
-b = ln(0.900950475)
b = -ln(0.900950475)
Therefore, the values of a and b are a = 9,995 and b = -ln(0.900950475).
(b) To find the time when the population becomes two-thirds of the initial population (2/3 * 10,000 = 6,666.67), we can use the population function and solve for t:
6,666.67 = 5 + 9,995e^(-b*t)
Substituting the value of b found earlier:
6,666.67 = 5 + 9,995e^(ln(0.900950475)*t)
Simplifying the equation:
6,666.67 - 5 = 9,995e^(ln(0.900950475)*t)
6,661.67 = 9,995e^(ln(0.900950475)*t)
Dividing both sides by 9,995:
(6,661.67/9,995) = e^(ln(0.900950475)*t)
Taking the natural logarithm (ln) of both sides:
ln(6,661.67/9,995) = ln(e^(ln(0.900950475)*t))
ln(6,661.67/9,995) = ln(0.900950475)*t
Solving for t:
t = ln(6,661.67/9,995) / ln(0.900950475)
Therefore, the time when the population becomes two-thirds of the initial population is t = ln(6,661.67/9,995) / ln(0.900950475).
(c) To find the time when there is no change in the rabbit population, we set P(t) = 0 and solve for t:
0 = 5 + 9,995e^(-b*t)
Substituting the value of b found earlier:
0 = 5 + 9,995e^(ln(0.900950475)*t)
Simplifying the equation:
9,995e^(ln(0.900950475)*t) = -5
Since e raised to any power will always be positive, it is not possible for this equation to hold true. Therefore, there is no time when there is no change in the rabbit population.