Question

The population, P, of a species of fish is decreasing at a rate that is proportional to the population itself. If P=200000 when t=3 and P=150000 when t=4, what is the population when t=10?Round your answer to the nearest integer.

Answer 1

As we know, the type of function that can be used to modelate populations is the exponential function, which, in our case will look like the following:

`P(t)=P_0e^(-\lambda t)`

Where P0 is the initial population, and lambda is a parameter which tells us how fast the population decays. By replacing the values we have, we get:

`2*10^5=P_0e^(-3\lambda),\text{ }1.5*10^5=P_0e^(-4\lambda)`

In order to find out the first parameter, let us divide both equations:

`(2*10^5)/(1.5*10^5)=(P_0e^(-3\lambda))/(P_0e^(-4\lambda))=e^(-3\lambda+4\lambda)=e^(\lambda)`
`e^(\lambda)=(2)/(1.5)\Rightarrow\lambda=ln((4)/(3))`

With this, we can find out the initial population by replacing lambda on any equation:

`2*10^5=P_0e^{-3*ln((4)/(3))}\Rightarrow P_0=\frac{2*10^5}{e^{-3ln((4)/(3))}}\approx474074.0741`

Thus, the population when t=10 will be approximately:

`P(10)\approx474074.0741e^{-10*ln((4)/(3))}=26696.77734`

Thus, our answer is P(10)=26697