Compare unbounded growth with a carrying capacity · source reference

Printed question

A scientist and his student are investigating the growth of bacteria in a petri dish. At time $t$ minutes after the bacteria is first introduced into the dish, the number of bacteria in the dish is denoted by $x$, 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 $x$ as a function of $t$. [4]

(ii) The scientist claims that $\frac{dx}{dt}=m(100x-x^2)$, where $m$ is a constant. Show that $x=\frac H{1+Ae^{-Hmt}}$, where $H$ is a whole number to be determined. (You need not find the values of $A,m$.) [4]

(iii) Whose model do you think is better, the student’s or the scientist’s? Justify your answer. [1]

[2020/ACJC/Prelim/II/2]

Printed working

Student: $x'=k/\sqrt x$; $\int\sqrt x\,dx=\int kdt$; $2x^{3/2}/3=kt+c$.

At 0, $x=1$, $c=2/3$. At 10, $x=4$, $16/3=10k+2/3\Rightarrow k=7/15$.

$2x^{3/2}/3=7t/15+2/3\Rightarrow x=(7t/10+1)^{2/3}$.

Scientist: $x'=m(100x-x^2)=mx(100-x)$.

$\int dx/[x(100-x)]=\int mdt$; $\int[1/x+1/(100-x)]dx=\int100m\,dt$.

$\ln|x/(100-x)|=100mt+c$; $x/(100-x)=Ae^{100mt}$; $x=Ae^{100mt}(100-x)$.

$x(1+Ae^{100mt})=100Ae^{100mt}$; $x=100/(1+Ae^{-100mt})$ after renaming the reciprocal integration constant. Thus $H=100$.

Scientist’s model is better: long-term bacteria increases towards 10000, reflecting a maximum population in a petri dish.

Printed answer key

(i) $dx/dt=k/\sqrt x$; $x=(7t/10+1)^{2/3}$. (ii) $H=100$. (iii) Scientist.

Teacher note 1

Using partial fractions: $1/[x(100-x)]=A/x+B/(100-x)$; $1=A(100-x)+Bx$. Let $x=0$, $A=1/100$; let $x=100$, $B=1/100$. Thus $1/[x(100-x)]=(1/100)[1/x+1/(100-x)]$. Printed sidebar typo: its last fraction’s left side uses $x(15-x)$; context requires $x(100-x)$.

Mathematical review

(i) $\sqrt x\,dx=k\,dt$ gives $2x^{3/2}/3=kt+C$. x(0)=1 gives C=2/3; x(10)=4 gives k=7/15. Hence $x=(1+7t/10)^{2/3}$ for t≥0.

(ii) $1/[x(100-x)]=(1/100)(1/x+1/(100-x))$. Integration gives $\ln[x/(100-x)]=100mt+C$, so $x=100/(1+Ae^{-100mt})$ and H=100. The initial data force positive growth and a positive A; part (ii) does not request their values.

(iii) The scientist’s model tends to x=100, or 10000 bacteria, whereas the student’s model grows without bound. A finite petri dish supports a limited population, making the logistic model more plausible on this criterion; observations alone do not prove the specific carrying capacity.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 6. Working page(s): 22, 23. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.