Assuming P(0)=2 and t ® ¥ P(t) ® 152, find A and B; find k after 5 months; then how many years the population exceeds 100 wolves.As part of a wildlife study a pair of wolves, one male and one...

Assuming P(0)=2 and t ® ¥ P(t) ® 152, find A and B; find k after 5 months; then how many years the population exceeds 100 wolves.

As part of a wildlife study a pair of wolves, one male and one female, is placed in a protected zone of Alaska large enough to sustain a population of 152 wolves.  The pair breeds and the wolf population in the zone grows larger.  A possible mathematical model for the population growth is the function P(t) = A(1 – Be-kt) + 2, where P(t) represents the number of wolves after t months while A, B and k are constants.

Assuming P(0) = 2 and, as t ® ¥ P(t) ® 152, find the values of
A and B.  After five months there are 12 wolves in the zone.  Use this information to find the value of k correct to three significant figures.  Then in how many years will the population exceed 100 wolves?

Asked on by adolphina

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neela | High School Teacher | (Level 3) Valedictorian

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The given model for the population p(t) of wolves at time t.

P(t) = A{1-Be^(-kt)}+2.............(1),  where t is in months.

The initial population of wolves is P(0) .

P(0) =2, when t = 0 by data.

=> P(0) = A{1-B*e^(-k*0)}+2 = 2

=> A(1-B) = 0. Or B = 1.

The population of wolves could grow to a maximum of 152 only. We pressume the population of wolves becomes stationary is for a very large value of t.

=> For t = infinity, P(infinity) = A(1-Be^(-k*infinity)}+2 = 152.

=> A(1-B*0}+2 = 152, as e^-k*(infinity) = 0.

=> A +2 = 152.

=> A = 152-2 = 150. Or A = 150 ....(3).

We put A = 150  and B = 1 in the given equation (1):

=> P(t) = 150{1-e^(-kt)}+2..... (4).

Now we use the fact that in 5 months the population of wolves is 12.

So P(5) = 12. We put t= 5 in (4):

=> P(5) = 150(1-e^(-k*5)}+2 = 12.

=> 150{1-e^(-5k)} = 12-2 = 10.

=> 1-e^(-5k) = 10/150 = 1/15.

=> -e^(-5k) = 1/15-1 = -14/15.

=>e^(5k) = 15/14

We take logarithms:

5k = ln15/14.

k = (1/5)ln(15/14).

k= ln(15/14)^(1/5).

-k =  ln(15/14)^(-1/5).

-k =  ln(14/15)^(1/5).

=> P(t) = 150{1-(14/15)^(t/5)}+2.

Therefore  the required population model is P(t) = 150{1-(14/15)^(t/5)}+2.

We now answer when P(t) = 100, what is t?

=> P(t) = 150{1-(14/15)^(t/5)}+2 = 100.

=> {1-(14/15)^(t/5)} = (100-2)/150 = 98/150 = 49/75

=> {1-(14/15)^(t/5)} = 49/75.

=>  -(14/15)^(t/5) = 49/75-1 = -26/75.

=> (14/15)^(t/5) = 26/75.

=> (t/5)ln(14/15) = ln(26/75).

=> t = 5{ln(26/75)}/ln((14/15)

=>  t = 76.78 months.

Therefore it requires 76.78 months for the population of wolves with a model equation P(t) = 150{1-(14/15)^(t/5)}+2 to grow from the initial population of 2 to 100.

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