Question

In 1990 the Department of Natural Resources released 1000 splake (a crossbreed of fish) into a lake. In 1997 the population of splake in the lake was estimated to be 3000. Using the Malthusian law for population growth, estimate the population of splake in the lake in the year 2020.

Short answer

By using Malthusian law for population growth,theestimated value of the population of splake in the lake in the year 2020 is110868.

Step-by-step solution

Analyzing the given statement 

Given, that in1990, thepopulation of splake in the lake was1000 and it was estimated to be 3000 in 1997. One has to find the estimated population of splake in the year 2020 by using Malthusian law for population growth and the formula for this is,

$\mathit{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{p}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{kt}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

where p(t) is the population at time t, p₀ is the initial population and k is a constant.

Initial condition

If one is set to be the year1990,then by formula(1),

$\mathit{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{1000}\mathbf{\right)}{\mathit{e}}^{\mathbf{kt}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

where p(t) is the population of splake at a time t.

Find the value of k

The population of splake in the lake was estimated to be 3000 in 1997 and the difference between the years 1990 and 1997 is 7years. Therefore,

$\mathit{p}\mathbf{\left(}\mathbf{7}\mathbf{\right)}\mathbf{=}\mathbf{3000}$

Now in equation (2), if we put t=7, then

$$\begin{gathered} p\left(7\right)=\left(1000\right)e^{7k} \\ 3000=\left(1000\right)e^{7k} \\ \frac{3000}{1000}=e^{7k} \\ {e}^{7k}=3 \\ 7k=\ln 3 \\ k=\frac{\ln 3}{7} \\ k=0.156945 \end{gathered}$$

One will use this value of k, to find the estimated value of the population of splake in the lake in the year 2020.

Find the estimated value of the population of splake in the lake in the year 2020

Now as the difference between the years 1997 and 2020 is 23years, and (from step 3), here one will take 1997 as the initial year i.e., we will substitute ${\mathit{p}}_{\mathbf{0}}\mathbf{=}\mathbf{3000}$in (1). Therefore,

$$\begin{gathered} \mathit{p}\mathbf{\left(}\mathbf{23}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{3000}\mathbf{\right)}{\mathit{e}}^{\mathbf{\left(}\mathbf{23}\mathbf{\right)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{156945}\mathbf{)}} \\ \mathit{p}\mathbf{\left(}\mathbf{23}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{3000}\mathbf{\right)}{\mathit{e}}^{\mathbf{3}\mathbf{.}\mathbf{60973}} \\ \mathit{p}\mathbf{\left(}\mathbf{23}\mathbf{\right)}\mathbf{=}\mathbf{110868} \end{gathered}$$

Hence, theestimated value of the population of splake in the lake in the year 2020 is 110868.