Exponential Growth and Decay Models

Model exponential growth and decay for AP Calculus AB. Solve dy/dt = ky and apply to real-world contexts.

The differential equation dydt=ky\frac{dy}{dt} = ky models exponential growth (k>0k > 0) and decay (k<0k < 0). This is one of the most commonly tested DEs on AP Calculus AB.

The Solution

y=y0ekty = y_0 e^{kt}

where y0=y(0)y_0 = y(0) is the initial value.

Growth ($k > 0$)

Doubling time: td=ln2kt_d = \frac{\ln 2}{k}.

Decay ($k < 0$)

Half-life: th=ln2kt_h = \frac{\ln 2}{|k|}.

Worked Example: Population

Problem

Population doubles every 5 years. P(0)=1000P(0) = 1000.

k=ln250.1386k = \frac{\ln 2}{5} \approx 0.1386. P(t)=1000e0.1386tP(t) = 1000e^{0.1386t}.

After 12 years: P(12)=1000e1.6635278P(12) = 1000e^{1.663} \approx 5278.

Solution

Worked Example: Radioactive Decay

Problem

Half-life 8 hours. 200g initially.

k=ln28k = \frac{-\ln 2}{8}. N(t)=200e0.0866tN(t) = 200e^{-0.0866t}.

After 24 hours: N(24)=200e2.07925N(24) = 200e^{-2.079} \approx 25g.

Solution

Worked Example: Newton's Cooling

Problem

dTdt=k(TTs)\frac{dT}{dt} = k(T - T_s)T(t)=Ts+(T0Ts)ektT(t) = T_s + (T_0 - T_s)e^{kt}.

Solution

Practice Problems

    1. dydt=0.03y\frac{dy}{dt} = 0.03y, y(0)=500y(0) = 500. Find y(10)y(10).
    1. A substance decays: 100g → 60g in 4 hours. Find the half-life.
    1. Coffee at 90°C cools in a 20°C room. After 5 min it's 70°C. Find kk.

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Key Takeaways

  • dydt=ky\frac{dy}{dt} = kyy=y0ekty = y_0 e^{kt}.

  • k>0k > 0: growth. k<0k < 0: decay.

  • Doubling time = ln2k\frac{\ln 2}{k}. Half-life = ln2k\frac{\ln 2}{|k|}.

  • Newton's cooling: dTdt=k(TTs)\frac{dT}{dt} = k(T - T_s).

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