Compare unbounded growth with a carrying capacity

A scientist and his student are investigating the growth of bacteria in a petri dish. At time tt minutes after the bacteria is first introduced into the dish, the number of bacteria in the dish is denoted by xx, in hundreds. Initially, 100 bacteria were introduced into the dish. In 10 minutes, the number of bacteria became 400.

(i) The student claims that the growth rate of the bacteria in the dish is inversely proportional to the square root of the number of bacteria in the dish. Write down a differential equation for this situation and solve it to get xx as a function of tt. marks

(ii) The scientist claims that dxdt=m(100x−x2)\frac{dx}{dt}=m(100x-x^2), where mm is a constant. Show that x=H1+Ae−Hmtx=\frac H{1+Ae^{-Hmt}}, where HH is a whole number to be determined. (You need not find the values of A,mA,m.) marks

(iii) Whose model do you think is better, the student’s or the scientist’s? Justify your answer. mark

[2020/ACJC/Prelim/II/2]

Total marks: 9

Solution

(i)

Power-law growth

Initially 100 bacteria (x=1x=1 hundred); after 10 minutes 400 (x=4x=4). If dxdt\frac{dx}{dt} is inversely proportional to x\sqrt x, find x(t)x(t).

Explore this idea

Hint

  1. dxdt=kx\frac{dx}{dt}=\frac k{\sqrt x}.

  2. Separate: x dx=k dt\sqrt x\,dx=k\,dt.

Answer · (i)

x′=715xx\prime=\frac7{15\sqrt x}; x=(1+7t/10)2/3x=(1+7t/10)^{2/3} (hundreds).

Explore the idea

The student's model

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of x at t = 01
Read the solution at t10
Try an example
x at t = 10
4
Rate dx/dt there
0.233
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

The solution keeps growing: there is no upper limit in this model.

Ideas to try
  1. Read xx at t=10t=10: it should be 4 (400 bacteria).
  2. Does the curve ever level off?
(ii)

Logistic model

The scientist claims dxdt=m(100x−x2)\frac{dx}{dt}=m(100x-x^2). Show that x=H1+Ae−Hmtx=\dfrac H{1+Ae^{-Hmt}} with HH a whole number.

Explore this idea

Hint

  1. Partial fractions: 1x(100−x)=1100(1x+1100−x)\frac1{x(100-x)}=\frac1{100}\left(\frac1x+\frac1{100-x}\right).

Answer · (ii)

x=100/(1+Ae−100mt)x=100/(1+Ae^{-100mt}); H=100H=100.

Explore the idea

The logistic model (m = 0.001)

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of x at t = 01
Read the solution at t50
Try an example
x at t = 50
59.986
Rate dx/dt there
2.4
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

The curve levels off at the carrying capacity x=100x=100 hundred: 10 000 bacteria.

Ideas to try
Start above 100: the curve falls to 100.
(iii)

Choosing a model

Whose model is better?

Hint

  1. Compare long-term behaviour with a finite petri dish.

Answer · (iii)

Logistic model is more plausible because its finite capacity is 10000 bacteria.