Exponential growth and decay

For A > 0, a positive k gives growth and a negative k gives decay. If t is measured in hours, k has units per hour and N′ has quantity units per hour.

N(t) = 120e^0.18t
N′(t) = 21.6e^0.18t = 0.18N(t)

The 21.6 multiplies the exponential alone. The coefficient multiplying the entire quantity N is 0.18. Keeping those two forms separate prevents a common rate error.

For N = Ae^kt, N(0) = A. For N = C + Ae^kt, the initial value is C + A instead. The chosen time origin matters: t = 0 means the start of the model’s clock, not necessarily the beginning of the real process.

T(t) = 18 + 72e^−t/4
T(0) = 18 + 72 = 90°C

If the model starts when measurements begin, use t ≥ 0. Extending it to earlier times requires a separate justification.

If k = 0.12 per year, the increase over a whole year is about 12.75%, not exactly 12%. If k = −0.20, the one-year decrease is about 18.13%, not 20%.

A fixed 5% increase per unit:
N = A(1.05)ᵗ = Ae^{(ln 1.05)t}
k = ln(1.05)

The approximation eᵏ ≈ 1 + k can be useful when |k| is small, but it is an approximation. For a time interval Δt, use multiplier e^kΔt.

A population is modelled by N = 240e^0.08t, with t in hours. At t = 5:

N(5) = 240e^0.4 ≈ 358.04
N′(5) = 19.2e^0.4 ≈ 28.64 per hour

The model predicts about 358 individuals and an instantaneous growth rate of about 28.6 individuals per hour. It is a smooth approximation to a count; round a requested count at the end, not every intermediate step.

A tracer mass is modelled by Q = 90e^−0.15t micrograms, with t in hours. At t = 4, Q ≈ 49.39 micrograms and Q′ ≈ −7.409 micrograms per hour.

To find when Q is below 30:

90e^−0.15t < 30
−0.15t < ln(1/3)
t > ln(3)/0.15 ≈ 7.324 hours

Division by the negative coefficient reverses the inequality. At the exact boundary time the mass equals 30; it is strictly below afterwards. The logarithm equations lessons explain the inverse step.

Set N(t + T)/N(t) to 2 or 1/2. The initial multiplier A cancels. For k = −0.15 per hour, the half-life is about 4.621 hours.

For the cooling model T = 18 + 72e^−t/4, the temperature starts at 90°C and approaches 18°C. For every finite t ≥ 0, the exponential term is positive, so T > 18.

T′(t) = −18e^−t/4 = −(T − 18)/4
T = 42 ⇒ e^−t/4 = 1/3
t = 4ln(3) ≈ 4.394 minutes

The rate is proportional to T − 18, not to T. Halving the excess above 18°C is different from halving the whole temperature. Reaching 18°C exactly would require infinite time in this idealised model.

A decreasing reading follows y = C + Ae^kt. Its horizontal asymptote is y = 5, and the readings are y(0) = 29 and y(6) = 11:

C = 5
A = 29 − 5 = 24
11 − 5 = 24e^6k
e^6k = 1/4
k = −ln(2)/3

The ratio to use is (11 − 5)/(29 − 5), not 11/29. The resulting model is y = 5 + 24e^{−(ln 2)t/3}.

Unrestricted exponential growth eventually exceeds finite resources or available space. A good fit over a short interval does not make a far-future prediction reliable. Changes in conditions can alter the fitted rate.

A cooling model assumes the surrounding temperature stays fixed. A decay model may need a nonzero background term if measurements never approach zero. State which quantity approaches the limit, the units of time, and the observed interval supporting the prediction.