# Exponential Growth and Decay Models

> Exponential growth and decay in JEE Mathematics: dN/dt = kN solved as N0 e^(kt), doubling time, half-life, Newton cooling and carbon dating logic.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/exponential-growth-decay-math
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Exponential Growth and Decay Models", PrepElephant, https://prepelephant.com/topics/jee/mathematics/exponential-growth-decay-math

## Direct answer

Populations, bank balances under continuous compounding, radioactive nuclei and cooling bodies all obey one differential equation: dN/dt = kN, solved by N = N0 e^(kt), where N0 is the value at t = 0 and k is the growth (positive) or decay (negative) constant. Doubling time and half-life both equal ln2/|k| — about 0.693/|k| — independent of the current amount, which is the defining signature of exponential change. Newton's law of cooling, dT/dt = -k(T - Ts), is the same equation in disguise for the excess temperature T - Ts, giving T = Ts + (T0 - Ts)e^(-kt). Carbon dating exploits the decay form with the C-14 half-life of 5730 years: a sample retaining one quarter of the original C-14 has passed through two half-lives, an age of about 11460 years.

## What you must remember

- **The model:** dN/dt = kN ⟹ N = N0 e^(kt); separate variables, integrate, apply the initial condition — three lines every time.
- **Doubling time / half-life:** t_double = t_half = ln2/|k| ≈ 0.693/|k|, the same for every starting value; k = ln2/t_half when a half-life is given.
- **Newton's law of cooling:** dT/dt = -k(T - Ts) ⟹ T = Ts + (T0 - Ts)e^(-kt); the excess over the surroundings decays exponentially, so excess temperature halves in a fixed time.
- **Continuous compounding:** dA/dt = rA ⟹ A = Pe^(rt); the effective annual rate is e^r - 1, always above the nominal r.
- **Two-point calibration:** N(t1)/N0 given fixes k through k = (1/t1) ln(N(t1)/N0) — most word problems resolve this first.
- **Carbon dating anchor:** C-14 half-life 5730 years, so k = ln2/5730 ≈ 1.21 × 10^(-4) per year; fraction remaining = (1/2)^(t/5730).
- **Linearisation trick:** ln N against t is a straight line with slope k — recognising exponential data and reading k off a log-linear picture is an examined skill.

## One population question, fully worked

A culture doubles in 25 hours. How long to triple? First calibrate k: 2N0 = N0 e^(25k) gives k = ln2/25 per hour. Now ask when N = 3N0: 3 = e^(kt), so t = ln3/k = 25 ln3/ln2 = 25 × 1.585 ≈ 39.6 hours. Notice the structure — tripling time is (ln3/ln2) times the doubling time, a dimensionless multiplier that never touches the units. The same skeleton runs the decay direction: a substance decays to 25 percent; the ratio is (1/2)^2, so exactly two half-lives have elapsed, and with a half-life of 5730 years the age is 11460 years. And the cooling variant: coffee at 90°C in a 20°C room cools to 60°C; the excess fell from 70 to 40, a factor 4/7, so kt = ln(7/4); when the excess halves to 35 (coffee at 55°C), another factor 1/2 requires t' with kt' = ln2 — the two times compare as ln(7/4)/ln2 ≈ 0.807 of each other. Every such problem is ratio-taking followed by one logarithm.

## How the exam frames it

JEE Main phrases these as short word problems with clean numbers: doubling in 3 hours (find k as ln2/3), population growth from a census pair, a residue percentage after a stated half-life — answers expressed in terms of ln2 and ln3 are routine. Advanced leans on the cooling law and its modelling assumptions (temperature excess, not temperature itself, decays; the body must be small compared to the surroundings), and occasionally couples growth with a threshold condition — find when the population crosses a barrier, which inverts the exponential into a logarithm. The systematic errors: writing N = N0 e^(kt) with k positive for a decay problem (the formula then grows); using log base 10 against base e inconsistently inside one problem; and in cooling questions, applying the exponential to T rather than the excess T - Ts, which produces elegant nonsense. The differential equation itself is separable-variables syllabus (NCERT Class 12 application chapter), examined in both papers with reliable frequency.

## Frequently asked questions

### What is the solution of the growth equation dN/dt = kN?

N = N0 e^(kt) with N0 the initial value; k > 0 gives growth and k < 0 decay, and the derivation is direct separation of variables.

### How are doubling time and half-life related to the rate constant?

Both equal ln2/|k| ≈ 0.693/|k| — independent of the amount present, which is what makes exponential change self-similar.

### What does Newton's law of cooling state?

dT/dt = -k(T - Ts): the rate of cooling is proportional to the excess over the surroundings, so T = Ts + (T0 - Ts)e^(-kt) and the excess decays exponentially.

### How is the age of a sample found from carbon dating?

From the fraction of C-14 remaining: t = 5730 × log2(N0/N) years, using the 5730-year half-life; one quarter remaining means 11460 years.

### Why does the effective rate of continuous compounding exceed the nominal rate?

Because A = Pe^(rt) compounds instantly, giving an annual multiplier e^r, so the effective rate is e^r - 1, which exceeds r for r > 0.