Applications Growth Decay and Motion: Concepts and Examples

Growth, decay, and motion problems share one mathematical heartbeat: the rate of change is proportional to the current amount. Populations grow, radioactive atoms decay, hot objects cool, and falling bodies accelerate, all following differential equations of the form dN/dt = kN or close relatives.

These models matter because they predict the future from the present: how many bacteria at midnight, how old a fossil is, when a forensic sample cooled. The mathematics is always exponential, and the same equation reappears in biology, physics, chemistry, and finance.

Key Concepts

• Proportional rate: dN/dt = kN says the change per unit time is a fixed fraction k of the current size. Positive k means growth, negative k means decay.
• Exponential solution: separating variables gives N(t) = N₀e^(kt), where N₀ is the initial amount at t = 0.
• Doubling time: the time to double is T = (ln 2)/k, independent of N₀; every doubling takes the same time.
• Half-life: for decay with rate constant λ > 0, the half-life is (ln 2)/λ, the time for half the material to decay.
• Newton's law of cooling: dT/dt = −k(T − Tₛ), where Tₛ is the surrounding temperature; the solution T(t) = Tₛ + (T₀ − Tₛ)e^(−kt) approaches Ts exponentially.
• Motion with constant acceleration: dv/dt = g integrates to v = gt + v₀, the simplest rate equation; with air resistance the equation becomes dv/dt = g − kv, giving exponential approach to terminal velocity.

Formulas

• Growth/decay: dN/dt = kN gives N(t) = N₀e^(kt).
• Doubling time: doubling time = (ln 2)/k for k > 0.
• Half-life: (ln 2)/λ for decay dN/dt = −λN.
• Cooling: T(t) = Tₛ + (T₀ − Tₛ)e^(−kt).

Worked Example

A bacterial culture has 500 cells at t = 0 and doubles every 2 hours. How many cells are present after 6 hours?

1. Write the model: N(t) = 500e^(kt) with unknown growth constant k per hour.
2. Use the doubling data: N(2) = 1000, so 1000 = 500e^(2k), giving e^(2k) = 2 and 2k = ln 2, so k = (ln 2)/2 ≈ 0.3466 per hour.
3. Evaluate at t = 6: N(6) = 500e^(6 × 0.3466) = 500e^(2.0796). Since e^(2.0796) = e^(3 ln 2) = 2³ = 8, we get N(6) = 500 × 8 = 4000 cells.
4. Cross-check: 6 hours contain exactly 3 doubling periods of 2 hours, and 500 → 1000 → 2000 → 4000. Both routes agree.

Common Misconceptions

• Adding instead of multiplying per period. Exponential growth multiplies by a fixed factor each period; adding a fixed number each period is linear growth, a completely different model.
• Half-life depends on the amount. It does not: (ln 2)/λ involves only the decay constant, so 1 kg and 1 g of the same isotope share the same half-life.
• Cooling reaches room temperature in finite time. The exponential approaches Ts asymptotically; it gets arbitrarily close but the model never quite arrives.

Real-World Applications

• Biology: bacterial cultures, unconstrained animal populations, and early epidemic spread follow exponential growth.
• Archaeology: carbon-14 dating uses the known half-life of about 5730 years to date organic remains.
• Forensics: Newton's law of cooling estimates time of death from body temperature.
• Finance: continuously compounded interest is exponential growth with k as the interest rate.
• Engineering: capacitor discharge and terminal velocity of falling objects follow the same exponential mathematics.

Whenever a rate is proportional to the amount present, the answer is exponential: N₀ekt. Find k from doubling or half-life data, watch the signs for growth versus decay, and one formula predicts populations, ages of fossils, cooling times, and investment growth alike.