Compound interest, continuous growth, half-life, population models, and Newton's law of cooling — with worked examples for every formula.
General growth/decay model
A(t) = A₀ · e^(kt)
k > 0: growth | k < 0: decay | A₀ = initial amount | t = time
Compound interest (discrete)
A = P(1 + r/n)^(nt)
P = principal, r = annual rate (decimal), n = compounds/year, t = years
Continuous compounding
A = Pe^(rt)
Limit as n → ∞ of discrete formula. e ≈ 2.71828
Half-life
A(t) = A₀ · (1/2)^(t/h)
h = half-life period. Equivalent to A₀e^(kt) with k = −ln(2)/h
Doubling time
t_double = ln(2) / k ≈ 0.693 / k
k = continuous growth rate. Rule of 72: ≈ 72/(r%) years
Newton's Law of Cooling
T(t) = T_env + (T₀ − T_env) · e^(kt)
T_env = ambient temperature, T₀ = initial temperature, k < 0
| Compounding | n value | Formula |
|---|---|---|
| Annually | 1 | A = P(1 + r)^t |
| Semi-annually | 2 | A = P(1 + r/2)^(2t) |
| Quarterly | 4 | A = P(1 + r/4)^(4t) |
| Monthly | 12 | A = P(1 + r/12)^(12t) |
| Daily | 365 | A = P(1 + r/365)^(365t) |
| Continuously | ∞ | A = Pe^(rt) |
Example: $5,000 at 6% for 10 years — compare monthly vs. continuous
Monthly: A = 5000(1 + 0.06/12)^(12×10) = 5000(1.005)^120 ≈ $9,096.98
Continuous: A = 5000·e^(0.06×10) = 5000·e^0.6 ≈ $9,110.18
Continuous pays $13.20 more — more frequent compounding = slightly more earnings
After each half-life period h, the quantity is halved. After n half-lives: A = A₀ · (1/2)^n.
Example 1: Carbon-14 has a half-life of 5,730 years. A sample has 25% of its original C-14. How old is it?
A(t) = A₀ · (1/2)^(t/5730)
0.25 · A₀ = A₀ · (1/2)^(t/5730)
0.25 = (1/2)^(t/5730)
ln(0.25) = (t/5730) · ln(1/2)
t = 5730 · ln(0.25)/ln(0.5) = 5730 · (−1.386)/(−0.693)
t = 5730 × 2 = 11,460 years
Example 2: Find the decay constant k for C-14 (half-life 5,730 years)
At t = h: A₀/2 = A₀ · e^(kh)
1/2 = e^(k·5730)
ln(1/2) = 5730k
k = ln(0.5)/5730 ≈ −0.000121 per year
Exponential growth model: P(t) = P₀ · e^(rt), where r is the continuous growth rate.
Example: City has 50,000 people in 2000, growing at 3% per year. Population in 2025?
P(t) = 50,000 · e^(0.03t)
t = 2025 − 2000 = 25
P(25) = 50,000 · e^(0.75) = 50,000 × 2.117
≈ 105,850 people
Finding the rate k from two data points
Population was 100 in 1990, 350 in 2010 (t in years after 1990)
350 = 100 · e^(20k)
3.5 = e^(20k)
k = ln(3.5)/20 ≈ 0.0626 → about 6.26% per year
T(t) = T_env + (T₀ − T_env) · e^(kt)
k < 0 for cooling. T_env = surrounding temperature. T₀ = initial object temp.
Example: Coffee at 180°F cools to 150°F in 10 min in a 70°F room. What temp after 30 min?
Step 1: Find k using T(10) = 150
150 = 70 + (180 − 70)e^(10k) → 80 = 110e^(10k)
e^(10k) = 80/110 = 8/11 → k = ln(8/11)/10 ≈ −0.0318
Step 2: Find T(30)
T(30) = 70 + 110e^(−0.0318×30) = 70 + 110e^(−0.954)
T(30) ≈ 70 + 110 × 0.385 ≈ 112°F
General form: A(t) = A₀ · e^(kt), where A₀ is the initial amount, k is the growth rate (k > 0 for growth, k < 0 for decay), and t is time. For discrete compounding: A = P(1 + r/n)^(nt). For continuous compounding: A = Pe^(rt).
Half-life formula: A(t) = A₀ · (1/2)^(t/h), where h is the half-life period. Equivalently, A(t) = A₀ · e^(kt) where k = −ln(2)/h. To find when a quantity reaches a specific amount, substitute and solve for t using natural logarithms.
If a quantity grows at rate r (as a decimal), the doubling time is t = ln(2)/r ≈ 0.693/r. The Rule of 72 gives an approximation: doubling time ≈ 72/r% (where r% is the percentage rate). Example: 8% annual growth → doubling time ≈ 72/8 = 9 years.
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