Bacteria growth after a product substitution · source reference

Printed question

It is known that a particular type of bacteria grows very well under certain controlled conditions in a specially prepared Petri dish. The researcher believes that the growth rate of such bacteria can be modelled by $t^2\frac{dx}{dt}-2xt=-x^2$, where $x$ milligrams is the amount of bacteria grown in the dish after $t$ hours.

(i) Using the substitution $x=ut^2$, show that the differential equation can be reduced to $\frac{du}{dt}=-u^2$. [2]

(ii) Find $x$ in terms of $t$, given that there was 0.2 milligrams of bacteria after 15 minutes. Hence find the amount of this particular type of bacteria after 4 hours. [4]

(iii) Explain if this mathematical model is a realistic one. [2]

[2012/NJC/Prelim/I/4]

Printed working

$x=ut^2$ (1); $dx/dt=2tu+t^2du/dt$ (2).

$t^2(2tu+t^2du/dt)-2t(ut^2)+(ut^2)^2=0\Rightarrow t^4du/dt=-u^2t^4\Rightarrow du/dt=-u^2$.

$\int u^{-2}du=\int-1dt\Rightarrow-1/u=-t+C'$; $t^2/x=t-C'$; $x=t^2/(t+C)$ where $C=-C'$.

At 15 minutes, $t=15/60=1/4$ and $x=0.2$: $0.2=(0.25)^2/(0.25+C)\Rightarrow0.05+0.2C=0.0625\Rightarrow C=1/16$.

$x=16t^2/(16t+1)$. At $t=4$, $x=16(4)^2/(16(4)+1)=256/65$ or 3.94.

As $t\to\infty$, $x\to\infty$. The particular solution suggests bacteria grow indefinitely as time passes. The model is not realistic: a petri dish should have a limit to the maximum amount of bacteria.

Printed answer key

(ii) $x=16t^2/(16t+1)$; $256/65$ or 3.94. (iii) No, as there should be a limit to maximum amount of bacteria in a petri dish.

Teacher note 1

Since the given DE and DE to be shown involve $dx/dt,du/dt$, differentiate the given substitution with respect to $t$. Given substitution (1), substitution after differentiation (2): substitute both into the given DE to get a DE of form $du/dt=f(u)$.

Teacher note 2

Given $du/dt=f(u)$, bring all “$u$” terms to the left where $du/dt$ is before integrating: $(1/f(u))du/dt=1$, $\int(1/f(u))(du/dt)dt=\int1dt$, $\int du/f(u)=\int1dt$. The intermediate integral can be skipped once familiar.

Mathematical review

(i) $x'=t^2u'+2tu$, so substitution gives $t^4u'=-u^2t^4$, hence $u'=-u^2$ for $t\ne0$.

(ii) Integrate: $1/u=t+C$, so $x=t^2/(t+C)$. At $t=1/4$, $1/5=(1/16)/(1/4+C)$ gives $C=1/16$. Thus $x=16t^2/(16t+1)$ for $t\ge0$. At 4 hours, $x=256/65\approx3.94$ mg. Direct differentiation gives $x'=32t(8t+1)/(16t+1)^2$ and verifies the original equation.

(iii) As $t\to\infty$, x grows without bound, approximately linearly. A finite dish cannot support unlimited mass, so the model is limited to the observed growth period.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 1. Working page(s): 3. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.