UPSC 2017 Maths Optional Paper 1 Q6a-ii — Step-by-Step Solution

8 marks · Section B

Variables separable · ODEs · asked 3× in 13 yrs · Read the full method →

Question

If the growth rate of the population of bacteria at any time $t$ is proportional to the amount present at that time and population doubles in one week, then how much bacterias can be expected after 4 weeks?

Technique

Solve $N'=kN$ , fix $k=\ln2$ from one-week doubling, evaluate at $t=4$ .

Solution

Step 1 — Model

“Growth rate proportional to amount present” is the standard Malthus law:

\frac{dN}{dt}=kN\quad\Longrightarrow\quad N(t)=N_0e^{kt},

where $N_0=N(0)$ is the initial population and $t$ is in weeks.

Step 2 — Use the doubling condition

Population doubles in one week: $N(1)=2N_0$ . Then

N_0e^{k}=2N_0\ \Rightarrow\ e^{k}=2\ \Rightarrow\ k=\ln 2.

So $N(t)=N_0e^{(\ln 2)t}=N_0\,2^{\,t}.$

Step 3 — Evaluate at $t=4$

N(4)=N_0\,2^{4}=16\,N_0.

Answer

\boxed{\;N(4)=16\,N_0\quad\text{(sixteen times the initial population).}\;}

UPSC 2017 Maths Optional Paper 1 Q6a-ii — Step-by-Step Solution

Question

Technique

Solution

Step 1 — Model

Step 2 — Use the doubling condition

Step 3 — Evaluate at t=4t=4t=4

Answer

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.

Step 3 — Evaluate at $t=4$