Population Models

• Oct 19th 2010, 10:36 PM
Population Models
Okay I have been doing this for a while now and I got t = 3, but clearly it was wrong because it actually equals 1.6 days.

I set up the model like this

dp/dt = k(p)^1/2

This is the problem

Let t denote time (in days) and let P denote the population of some species at time t. Suppose that the birth rate β (number of births per day per unit of population) is proportional to P^1/2 and that the death rate δ (number of deaths per day per unit of population) equals 0. Suppose the initial population is 4, and after one day the population is 16. When will the population equal 100?
• Oct 19th 2010, 11:30 PM
mr fantastic
Quote:

Okay I have been doing this for a while now and I got t = 3, but clearly it was wrong because it actually equals 1.6 days.

I set up the model like this

dp/dt = k(p)^1/2

This is the problem

Let t denote time (in days) and let P denote the population of some species at time t. Suppose that the birth rate β (number of births per day per unit of population) is proportional to P^1/2 and that the death rate δ (number of deaths per day per unit of population) equals 0. Suppose the initial population is 4, and after one day the population is 16. When will the population equal 100?

• Oct 20th 2010, 12:16 AM
okay I'm guessing i setup the model correctly
So, I took the integrals of (1/(p)^(1/2)) dP = k dt

and I got 2p^(1/2) = kt + c

p = ((kt/2) + c)^2

c = Po^1/2 (initial population)

so i plugged in the initial values to find the constant k value

16 = ( k/2 + (4)^(1/2))^2 ... Took the square root of both sides

4 = (k/2 + 2)

2 = k/2

4 = k

then for the new value of P (100) i plugged in the values

100 = ( 4t/2 + (4)^1/2)^2

square root

10 = 2t +2

8 = 2t

4 = t

I actually got 4, but the Professor said it was equal to 1.6 days.
• Oct 20th 2010, 12:25 AM
mr fantastic
Quote:

okay I'm guessing i setup the model correctly
So, I took the integrals of (1/(p)^(1/2)) dP = k dt

and I got 2p^(1/2) = kt + c

p = ((kt/2) + c)^2

c = Po^1/2 (initial population)

so i plugged in the initial values to find the constant k value

16 = ( k/2 + (4)^(1/2))^2 ... Took the square root of both sides

4 = (k/2 + 2)

2 = k/2

4 = k

then for the new value of P (100) i plugged in the values

100 = ( 4t/2 + (4)^1/2)^2

square root

10 = 2t +2

8 = 2t

4 = t

I actually got 4, but the Professor said it was equal to 1.6 days.

p = ((kt/2) + c)^2

Substitute t = 0, p = 4: c = 2.

Therefore p = ((kt/2) + 2)^2.

Substitute t = 1, p = 16. Get k = 4.

p = (2t + 2)^2.

Substitute p = 100: 100 = (2t + 2)^2 => t = 4.

Looks like you're correct.
• Oct 20th 2010, 12:29 AM