Differential Equations: Real-World Scenarios (10)

easy 3 min read

Question

A bacterial colony grows at a rate proportional to its current population. If the population doubles in 3 hours, by what factor does it grow in 9 hours?

Solution — Step by Step

“Rate proportional to current population” translates to:

dNdt=kN\frac{dN}{dt} = kN

where NN is the population and k>0k > 0 is the growth constant.

dNN=kdt\frac{dN}{N} = k \, dt

Integrate: lnN=kt+C\ln N = kt + C, or N=N0ektN = N_0 e^{kt} where N0N_0 is the initial population.

At t=3t = 3, N=2N0N = 2N_0:

2N0=N0e3k2N_0 = N_0 e^{3k}

e3k=2e^{3k} = 2, so k=ln2/3k = \ln 2 / 3.

N(9)=N0e9k=N0e9ln2/3=N0e3ln2=N023=8N0N(9) = N_0 e^{9k} = N_0 e^{9 \ln 2 / 3} = N_0 e^{3 \ln 2} = N_0 \cdot 2^3 = 8N_0.

The population grows 8× in 9 hours.

Why This Works

Exponential growth is the unique solution to “rate proportional to amount.” Every doubling time multiplies the population by 2 — three doublings in 9 hours give 23=82^3 = 8.

This logic applies to compound interest, radioactive decay (with negative kk), Newton’s law of cooling, and population dynamics.

Speed shortcut: For exponential growth, “factor in time TT=2T/td= 2^{T/t_d}, where tdt_d is the doubling time. Skip the differential equation entirely for problems like this.

Alternative Method — Direct Ratio

Since growth is exponential, the factor in time tt is f(t)=ektf(t) = e^{kt}. The factor in 99 hours is (e3k)3=23=8(e^{3k})^3 = 2^3 = 8.

This avoids solving for kk explicitly — just exploit the exponential structure.

Common Mistake

Students often write linear growth: “If it doubles in 3 hours, it triples in 9 hours” — wrong. Linear growth is constant addition; exponential growth is constant multiplication.

Another trap: confusing doubling time with half-life formulas. For decay, N=N0(1/2)t/thN = N_0 (1/2)^{t/t_h} where tht_h is half-life. Same structure, just with 1/21/2 instead of 22.

CBSE Class 12 boards consistently include a “growth and decay” question for 4-6 marks. JEE Main rephrases this as Newton’s law of cooling or radioactive decay. The math is identical — it is just word problem disguise.

For a real-world twist: bacterial growth doesn’t continue forever (resources run out). The logistic equation dN/dt=kN(1N/K)dN/dt = kN(1 - N/K) models this saturation, but JEE/NEET stop at the simple exponential model.

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