Growth, decay and cooling

If a quantity grows at a rate proportional to its size, dP/dt = kP with k > 0. Separating gives P = P₀e^(kt), where P₀ is the value at t = 0.

The time for P to double is found from 2 = e^(kT): T = (log 2)/k, the same from any starting value.

Worked example: money grows continuously at 10% a year, so k = 0.1. It doubles in 10 log 2 ≈ 6.93 years, and Rs 1000 becomes 1000e ≈ Rs 2718 after 10 years. At 7% a year it doubles in about 9.9 years.

Decay is the same equation with a negative constant: dN/dt = −λN gives N = N₀e^(−λt) and a half-life of (log 2)/λ.

Worked example: a culture that triples in 5 hours has e^(5k) = 3. After 10 hours it is e^(10k) = 3² = 9 times its starting size.

Worked example (cooling): a drink at 85 °C in a 25 °C room follows dT/dt = −k(T − 25), so T − 25 = 60e^(−kt). If it reaches 55 °C in 10 minutes, the gap 60 halves in 10 minutes, so after 20 minutes the gap is 15 and the drink is at 40 °C.