- Mathematical Modelling

LESSON 1

(a) A pond is initially empty and is then filled gradually with water. After \(t\) minutes, the depth of the water, \(x\) metres, satisfies the differential equation
\(\displaystyle {dx\over dt}={\sqrt{4+5x}\over 5(1+t)^2}\)
Solve this differential equation to find \(x\) in terms of \(t\).
(b) Another pond is gradually filling with water, after \(t\) minutes, the surface of the water forms a circle of radius \(r\) metres. The rate of change of the radius is inversely proportional to the area of the surface of the water.
(i) Write down a differential equation, in the variables \(r\) and \(t\) and a constant of proportionality, which represents how the radius of the surface of the water is changing with time.
(You are not required to solve your differential equation.)
(ii) When the radius of the pond is 1 metre, the radius is increasing at a rate of 4.5 metres per second. Find the radius of the pond when the radius is increasing at a rate of 0.5 metres per second.
SOLUTION
(a) Solve as a separable differential equation
\(\displaystyle {dx\over dt}={\sqrt{4+5x}\over 5(1+t)^2}\)
\(\displaystyle {1\over \sqrt{4+5x}}dx={1\over 5(1+t)^2} dt\)
\(\displaystyle ∫(4+5x)^{-{1\over2}} dx={1\over5} ∫(1+t)^{-2} dt\)
\(\displaystyle {(4+5x)^{{1\over2}}\over 5\left({1\over2}\right)} ={1\over5} [-(1+t)^{-1}]+c\)
\(\displaystyle {2\over5} \sqrt{4+5x}=-{1\over5} \left({1\over 1+t}\right)+c\)
Since the pond is initially empty, this means that at \(t=0, x=0\). We can use these values to determine \(c\)
When \(x=0, t=0\)
\(\displaystyle {2\over5} \sqrt{4+5(0)}=-{1\over5} \left(1\over 1+0\right)+c\)
\(1=c\)
  • Substitute and solve for \(x\)
\(\displaystyle {2\over5} \sqrt{4+5x}=1-{1\over5} \left({1\over 1+t}\right)\)
\(\displaystyle \sqrt{4+5x}={5\over2}-{1\over2} \left(1\over 1+t\right)\)
\(\displaystyle 4+5x=\left[{5\over2}-{1\over 2(1+t)} \right]^2\)
\(\displaystyle 5x=\left[{5\over2}-{1\over 2(1+t)}\right]^2-4\)
\(\displaystyle x={1\over5} \left[{5\over2}-{1\over 2(1+t)} \right]^2-{4\over5}\)

(b) (i) Since \(\displaystyle {dr\over dt}\) represents the rate of change of the radius and there is inverse
proportionality
\(\displaystyle {dr\over dt}={k\over πr^2}\)
(ii) Use the fact that
\(\displaystyle r=1, {dr\over dt}=4.5\)
\(\displaystyle ⟹{9\over2}={k\over π(1)^2}\)
\(\displaystyle ⟹{9\over2}={k\over π}\)
\(\displaystyle ⟹k={9π\over2}\)
  • Rewrite the differential equation with the determined value of \(k\).
\(\displaystyle {dr\over dt}={9\over 2r^2}\)
  • Use equation to determine required value
When \(\displaystyle {dr\over dt}={1\over 2}\)
\(\displaystyle {1\over2}={9\over 2r^2}\)
\(r^2=9\)
\(r=3\)

VIDEO LESSON


LESSON 2
The number of bacteria in a liquid culture is observed to grow at a rate proportional to the number of cells present. At the beginning of the experiment there are 10,000 cells and after three hours there are 500,000. How many will there be after one day of growth if this unlimited growth continues? What is the doubling time of the bacteria?
SOLUTION
  • Let \(y(t)\) represent the number of bacteria present at time \(t\). Then the rate of change is
\(\displaystyle {dy\over dt}=ky\)
where \(k\) is the constant of proportionality.
  • Solve as a separable differential equation
\(\displaystyle {1\over y}\:dy=k\:dt\)
\(\displaystyle ∫{1\over y}\:dy=∫k\: dt\)
\(\ln ⁡y=kt+c\)
\(e^{\ln ⁡y} =e^{kt}+c\)
\(y=e^{kt+c}\)
\(y=e^c e^{kt}\)
\(y=Ae^{kt}\) where \(A=e^c\)
\(y(t)=Ae^{kt}\)
  • Use the fact that the initial population occurs when \(t=0\) (initial population) to determine the value of \(A\)
\(y(0)=Ae^{k(0)} =A\)
Therefore \(A\) is the initial population.
\(⟹A=10000\)
  • Use the information from the problem to determine \(k\)
\(500 000=10 000e^{3k}\)
\(50=e^{3k}\)
\(\ln ⁡50=3k\)
\(\displaystyle k={1\over 3} \ln ⁡50\approx 1.304\)
\(y=10 000e^{1.304t}\)
Doubling time refers to the amount of time for the bacteria to double, in number, from its original number.
The original number is 10,000 so we need to determine how long it takes to get to 20,000.
\(20000=10000e^{1.304t}\)
\(2=e^{1.304t}\)
\(\ln ⁡2=1.304t\)
\(\displaystyle t={\ln ⁡2\over 1.304}\approx 0.532\) hours