Assume at time t=0 there are 10^6 fish. find the population at time t ....?

Question

(according to the logistic equation: p ' = 0.003 p - 0.0001 p^2 - 0.002)

-& how long does it take the population to reach 1000?

--do i need to modify the equation first?

Answer 1

I'm assuming that the ' means that you have the derivative of the population function.

p' = -0.0001p² + 0.003p - 0.002

p' = -0.0001(p² - 30p + 20)

Now... you can factor this using the quadratic equation, split using partial fraction decomposition, and integrate. However, unfortunately, the factors of this equation are going to be complex.

Which makes me think that you mistyped the problem, maybe? Because if the original problem had been:

p' = 0.003 p - 0.0001 p^2 - 0.02

Then

p' = -0.0001(p² - 30p + 200) = -0.0001(p - 20)(p - 10)

Normally I don't change a problem to make it easier to solve, but I really think maybe you mistyped. If not, you can use this solution as a pattern to find the other one.

dp/dt = -0.0001(p - 20)(p - 10)

dp/[-0.0001(p - 20)(p - 10)] = dt

-10,000 ∫ 1/[(p - 20)(p - 10)] dp = ∫ dt

Now, use partial fractions to split the integral on the left:

1/[(p - 20)(p - 10)] = A/(p - 20) + B/(p - 10)

Multiply through by (p - 20)(p - 10):

1 = A(p - 10) + B(p - 20)

1 = Ap - 10A + Bp - 20B

0p + 1 = (A + B)p + (-10A -20B)

A + B = 0
-10A - 20B = 1
10A + 10B = 0 // Multiply first equation by 10
------------------
-10B = 1
B = -1/10
A + -1/10 = 0
A = 1/10

So back to the original equation:

-10,000 ∫ [1/10 1/(p - 20) - 1/10 1/(p - 10)] dp = ∫ dt

Split into two:

-1,000 ∫ (p - 20)ˉ¹ dp + 1,000 ∫ (p - 10)ˉ¹ dp = ∫ dt

Integrate:

-1,000 ln(p - 20) + 1,000 ln(p - 10) = t + C

Use log rules to isolate p:

-1,000 [ln(p - 20) - ln(p - 10)] = t + C

-1,000 (ln[(p - 20)/(p - 10)]) = t + C

ln[(p - 20)/(p - 10)] = -0.001t + C

(p - 20)/(p - 10) = e^(-0.001t + C) = e^(-0.001t)e^C = Ke^(-0.001t)

p - 20 = (p - 10)Ke^(-0.001t) = pKe^(-0.001t) - 10Ke^(-0.001t)

p - pKe^(-0.001t) = 20 - 10Ke^(-0.001t)

p(1 - Ke^(-0.001t)) = 10(2 - Ke^(-0.001t))

p(t) = 10[2 - Ke^(-0.001t)]/(1 - Ke^(-0.001t))

Whew!!!! Now you're given the initial population, so you can find K:

p(0) = 10^6 = 10(2 - K)/(1 - K)

10^5(1 - K) = 2 - K

10^5 - 10^5K = 2 - K

10^5 - 2 = 10^5K - K = (10^5 - 1)K

K = (10^5 - 2)/(10^5 - 1) = 0.99999

So, your final equation for p at any time t is:

p(t) = 10[2 - 0.99999e^(-0.001t)]/(1 - 0.99999e^(-0.001t))

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Then, finally, to answer your last question:

1,000 = 10[2 - 0.99999e^(-0.001t)]/(1 - 0.99999e^(-0.001t))

100 = [2 - 0.99999e^(-0.001t)]/(1 - 0.99999e^(-0.001t))

100(1 - 0.99999e^(-0.001t)) = 2 - 0.99999e^(-0.001t)

100 - 99.999e^(-0.001t) = 2 - 0.99999e^(-0.001t)

100 - 2 = 99.999e^(-0.001t) - 0.99999e^(-0.001t)

98 = (99.999 - 0.99999)e^(-0.001t)

e^(-0.001t) = 98/98.99001

-0.001t = ln 98/98.9901

t = -1000ln 98/98.9901 = 10.14237

Man....that took me over an hour... :c(