Word Problem - Integration

Mar 2016
170
0
Australia
Hi,

I don't think I know how to do all of this yet however:

The evolution of a population of rhinoceros, R(t), in hundreds, time in years, in anAfrican National Park is given by the equation,

dR/dt =1/5R(2 − R); R(0) = 0.5

(a) Solve the system exactly for R(t).
(b) What happens as the time t → ∞, i.e. what is the population a long time in the future?
(c) Write an Euler scheme and compute until the population levels off (using Excel, Matlab,Octave, LibreCalc or similar).
(d) Plot both of your solutions on the same set of axes and comment.

(a) dR/dt = 1/5 R(2-R)
Multiply R through brackets - dR/dt = 1/5 2R + R2 or dR/dt = 2R+R2/5
Divide both sides by dR - dt = (2R+R2)/5

If I am correct I can see dt = a number plus its derivative - does this mean I can use integration by substitution setting R2 as the number to be substituted? Or am I on the complete wrong planet?

Also for(b) am I looking for the absolute maxima of the equation?

Kind regards
Beetle
 
Mar 2016
170
0
Australia
For part (a) - I meant to say - is that just the product rule backwards?
 

Debsta

MHF Helper
Oct 2009
1,346
623
Brisbane
Just want to check the function. Is it \(\displaystyle \frac {dR}{dt} =\frac{1}{5}R(2-R)\) or is it \(\displaystyle \frac {dR}{dt} =\frac{1}{5}t(2-t)\)?
 
Mar 2016
170
0
Australia
Hi, it is the first one.
Cheers
Beetle
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
For part (a) - I meant to say - is that just the product rule backwards?
Integration by separation of variables, then partial fraction decomposition ...
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
dR/dt =1/5R(2 − R); R(0) = 0.5
$\dfrac{dR}{R(2-R)} = \dfrac{1}{5} \, dt$

partial fraction decomposition ...

$\dfrac{1}{R(2-R)} = \dfrac{A}{R} + \dfrac{B}{2-R}$

$1 = A(2-R) + BR$

$R = 0 \implies A = \dfrac{1}{2}$ ; $R=2 \implies B = \dfrac{1}{2}$

$\displaystyle \dfrac{1}{2} \int \dfrac{1}{R} + \dfrac{1}{2-R} \, dR = \dfrac{1}{5} \int \, dt$

$\displaystyle \int \dfrac{1}{R} - \dfrac{-1}{2-R} \, dR = \dfrac{2}{5} \int \, dt$

$\ln|R| - \ln|2-R| = \dfrac{2}{5} t + C$

$\ln\left|\dfrac{R}{2-R}\right| = \dfrac{2}{5} t + C$

$\dfrac{R}{2-R} = e^{2t/5} \cdot e^C$ ; let $e^C = A$

$\dfrac{R}{2-R} = Ae^{2t/5}$

$R = 2Ae^{2t/5} - RAe^{2t/5}$

$R + RAe^{2t/5} = 2Ae^{2t/5}$

$R(1 + Ae^{2t/5}) = 2Ae^{2t/5}$

$R = \dfrac{2Ae^{2t/5}}{1 + Ae^{2t/5}}$

$R(0) = \dfrac{1}{2} \implies A = \dfrac{1}{3}$

$R = \dfrac{2e^{2t/5}}{3 + e^{2t/5}}$

$R = \dfrac{2}{1 + 3e^{-2t/5}}$ ... a logistic curve


$\displaystyle \lim_{t \to \infty} \dfrac{2}{1 + 3e^{-2t/5}} = 2$
 

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skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
shortcut ...

general model for logistic growth is

$\dfrac{dy}{dt} = k\left(1 - \dfrac{y}{M}\right) \implies y = \dfrac{M}{1 + Be^{-kt}}$ where $M$ is the carrying capacity for $y$ and $B = \dfrac{M-y(0)}{y(0)}$

transforming $\dfrac{dR}{dt} = \dfrac{1}{5}R(2-R)$ to the form shown above ...

$\dfrac{dR}{dt} = \dfrac{2}{5}R\left(1-\dfrac{R}{2}\right)$

$k = \dfrac{2}{5}$, $M = 2$, $B = \dfrac{2-0.5}{0.5} = 3$

$y = \dfrac{M}{1 + Be^{-kt}} \implies R = \dfrac{2}{1+3e^{-2t/5}}$