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:

$\frac{2*10^5}{1.5*10^5}=\frac{P_0e^{-3\lambda}}{P_0e^{-4\lambda}}=e^{-3\lambda+4\lambda}=e^{\lambda}$$e^{\lambda}=\frac{2}{1.5}\Rightarrow\lambda=ln(\frac{4}{3})$

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

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

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

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

Thus, our answer is P(10)=26697