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Q3.2-9E

Expert-verified
Found in: Page 100

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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.

By using Malthusian law for population growth, the estimated value of the population of splake in the lake in the year 2020 is 110868.

See the step by step solution

Step 1: Analyzing the given statement

Given, that in 1990, the population of splake in the lake was 1000 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,

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

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

Step 2: Initial condition

If one is set to be the year 1990, then by formula (1),

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

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

Step 3: 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,

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

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

$p\left(7\right)=\left(1000\right){e}^{7k}\phantom{\rule{0ex}{0ex}}3000=\left(1000\right){e}^{7k}\phantom{\rule{0ex}{0ex}}\frac{3000}{1000}={e}^{7k}\phantom{\rule{0ex}{0ex}}{e}^{7k}=3\phantom{\rule{0ex}{0ex}}7k=\mathrm{ln}3\phantom{\rule{0ex}{0ex}}k=\frac{\mathrm{ln}3}{7}\phantom{\rule{0ex}{0ex}}k=0.156945\phantom{\rule{0ex}{0ex}}$

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

Step 4: 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 ${{\mathbit{p}}}_{{\mathbf{0}}}{\mathbf{=}}{\mathbf{3000}}$ in (1). Therefore,

${\mathbit{p}}\mathbf{\left(}\mathbf{23}\mathbf{\right)}{\mathbf{=}}\mathbf{\left(}\mathbf{3000}\mathbf{\right)}{{\mathbit{e}}}^{\mathbf{\left(}\mathbf{23}\mathbf{\right)}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{156945}\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}{\mathbit{p}}\mathbf{\left(}\mathbf{23}\mathbf{\right)}{\mathbf{=}}\mathbf{\left(}\mathbf{3000}\mathbf{\right)}{{\mathbit{e}}}^{\mathbf{3}\mathbf{.}\mathbf{60973}}\phantom{\rule{0ex}{0ex}}{\mathbit{p}}\mathbf{\left(}\mathbf{23}\mathbf{\right)}{\mathbf{=}}{\mathbf{110868}}\phantom{\rule{0ex}{0ex}}$

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