Exponential Growth & Decay
- Recognize dy/dt = ky as the exponential growth/decay model
- Write its solution y = y₀·e^(kt) and interpret the constants
- Distinguish growth (k > 0) from decay (k < 0)
The signature equation of proportional change
Whenever a quantity changes at a rate proportional to its current amount, it obeys dy/dt = ky. Populations, radioactive samples, and continuously compounded money all follow this law. The constant k is the proportionality (relative growth) rate. This single separable equation is so common it is worth memorizing its solution rather than re-deriving it each time.
The solution and its constants
Solving dy/dt = ky by separation gives y = y₀·e^(kt), where y₀ = y(0) is the initial amount and k is the same rate constant. The exponent kt controls the shape: if k > 0 the quantity grows without bound (exponential growth); if k < 0 it decays toward zero (exponential decay). The amount y₀ simply scales the curve to match the starting value.
A bacterial culture grows so that dP/dt = 0.4P. If P(0) = 200, how many bacteria are there at t = 5 (time in hours)?
- 1.The equation dP/dt = kP with k = 0.4 has solution P = P₀·e^(kt).
- 2.Insert the initial amount P₀ = 200: P(t) = 200·e^(0.4t).
- 3.Evaluate at t = 5: P(5) = 200·e^(0.4·5) = 200·e².
- 4.Since e² ≈ 7.389, P(5) ≈ 200·7.389 ≈ 1478.
Recognize the model by its wording: "rate proportional to the amount present" is the tell for dy/dt = ky. You can jump straight to y = y₀·e^(kt) and spend your effort finding y₀ and k from the given data.
The solution to dy/dt = ky with y(0) = y₀ is:
A radioactive sample decays according to y = y₀·e^(kt). Which value of k represents decay?
For half-life or doubling-time questions, set y = y₀/2 (or 2y₀) in y = y₀·e^(kt); the y₀ cancels, leaving an equation you solve with a natural log. The half-life depends only on k, never on the starting amount.
Answer the 2 checkpoints as you read.
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