Growth and Decay Problems

One model for growth and decay

Growth and decay problems use one model: $P=P_0\times b^t$P = P₀ × bᵗ. $P_0$P₀ is the amount at the start, when t = 0, because $b^0=1$b⁰ = 1 and so $P=P_0$P = P₀ there. The number b is the factor the amount is multiplied by in each unit of time. When b is more than 1 the amount grows; when b is between 0 and 1 it decays.

A colony of 100 bacteria that doubles every hour has $P_0=100$P₀ = 100 and b = 2, so after t hours there are $100\times 2^t$100 × 2ᵗ bacteria. After 3 hours that is $100\times 2^3=800$100 × 2³ = 800.

The colony after 0, 1, 2 and 3 hours: 100, 200, 400, 800.

From a sentence to a model

Most problems describe the change in words. Each wording fixes the factor b, and sometimes the exponent as well.

"Grows by 3% a year" keeps the whole amount and adds 3%, so b = 1.03 and t counts years. "Falls by 8% a year" keeps 92%, so b = 0.92. A 5% rise in a savings balance each year is × 1.05.

"Halves every 5 years" is a different kind of sentence. The factor $\frac{1}{2}$1/2 applies once every 5 years, not every year. In t years there are $\frac{t}{5}$t/5 halvings, so the exponent is $\frac{t}{5}$t/5: $P=P_0\times 0.5^{\frac{t}{5}}$P = P₀ × 0.5^(t/5). In the same way, "triples every 4 hours" gives $P=P_0\times 3^{\frac{t}{4}}$P = P₀ × 3^(t/4).

A half-life of 3 days gives $0.5^{\frac{t}{3}}$0.5^(t/3). From 80 g, 9 days is $\frac{9}{3}=3$9/3 = 3 halvings, so the mass left is $80\times 0.5^3=10\text{g}$80 × 0.5³ = 10 g. After n half-lives the fraction left is $\frac{1}{2^n}$1/2ⁿ.

A half-life of 3 days, from 80 g: 40 g after 3 days, 20 g after 6 and 10 g after 9.

Reading a model back

A model can also be read backward, to say what it describes. In $P=250\times 0.92^t$P = 250 × 0.92ᵗ the start is 250, and each step keeps 92% of the amount, so it falls by 8% a step. In $P=1500\times 1.12^t$P = 1500 × 1.12ᵗ the start is 1500 and it rises by 12% a step. In $A=40\times 2^{\frac{t}{6}}$A = 40 × 2^(t/6) the start is 40 and it doubles every 6 units of time.

Sometimes the quantity that decays is hidden. A cup of coffee cools toward the temperature of the room, and what shrinks by the same factor each minute is its extra heat, the difference between the coffee and the room. Coffee at 90°C in a room at 20°C has 70°C of extra heat. If that is multiplied by 0.9 every minute, then after 1 minute the extra heat is 63°C and the coffee is at 83°C. After 10 minutes the extra heat is $70\times 0.9^{10}=24.4^{\circ }C$70 × 0.9¹⁰ = 24.4°C, to 1 decimal place, so the coffee is at 44.4°C. It gets closer and closer to 20°C and never goes below it.

A growth problem, in full

A town has 12,000 people, and its population grows by 3% a year. Write a model for the population after t years, find the population after 5 years, and find the first whole number of years after which it is more than 15,000.

Build the model. The start is $P_0=12000$P₀ = 12000, and a 3% rise is a factor of 1.03, so $P=12000\times 1.03^t$P = 12000 × 1.03ᵗ.

After 5 years: $1.03^5=1.159274$1.03⁵ = 1.159274 to 6 decimal places, so P = 12000 × 1.159274 = 13911.3, to 1 decimal place. The population is about 13,911.

For 15,000, try whole years. After 7 years $P=12000\times 1.03^7=14758.5$P = 12000 × 1.03⁷ = 14758.5, to 1 decimal place, which is still under 15,000. After 8 years $P=12000\times 1.03^8=15201.2$P = 12000 × 1.03⁸ = 15201.2, which is over. So the population first passes 15,000 after 8 years.

The gold curve is $P=12000\times 1.03^t$P = 12000 × 1.03ᵗ, and the white line is the level 15,000. The marked points are the start, 12,000, the population after 5 years, about 13,911, and after 8 years, about 15,201, the first whole year above the line.

A decay problem, in full

A sample of a radioactive substance weighs 60 mg, and its half-life is 5 years. Write a model for the mass after t years, and find the mass after 15 years and after 8 years.

The mass halves once every 5 years, so the exponent is $\frac{t}{5}$t/5: $M=60\times 0.5^{\frac{t}{5}}$M = 60 × 0.5^(t/5).

After 15 years there have been $\frac{15}{5}=3$15/5 = 3 half-lives, so $M=60\times 0.5^3=60\times \frac{1}{8}=7.5$M = 60 × 0.5³ = 60 × 1/8 = 7.5 mg.

After 8 years the exponent is $\frac{8}{5}=1.6$8/5 = 1.6, which is not a whole number of half-lives. $0.5^{1.6}=0.329877$0.5^1.6 = 0.329877 to 6 decimal places, so M = 60 × 0.329877 = 19.8 mg, to 1 decimal place. Check: after 5 years the mass is 30 mg and after 10 years it is 15 mg, and 19.8 mg lies between them, as 8 years lies between 5 and 10.

The same model can be read year by year. $0.5^{\frac{1}{5}}=0.871$0.5^(1/5) = 0.871 to 3 decimal places, so the mass is multiplied by about 0.871 each year: halving every 5 years is the same as losing about 12.9% a year.

$M=60\times 0.5^{\frac{t}{5}}$M = 60 × 0.5^(t/5). The mass is 30 mg at 5 years, 15 mg at 10 and 7.5 mg at 15. The point at 8 years, 19.8 mg, sits on the curve between the halvings.

The usual mistakes

Leaving the 5 out of the exponent $\frac{t}{5}$t/5. $60\times 0.5^t$60 × 0.5ᵗ halves every year, not every 5 years: after 15 years it would leave $60/2^{15}$60 / 2¹⁵, far too little.

Writing a percentage as the wrong factor. 3% growth is × 1.03, not × 0.03 and not × 1.3.

Adding the same percentage of the start each year. 3% of 12,000 is 360, but after 5 years the town is not 12,000 + 5 × 360 = 13,800. Each year the 3% is taken from the population that year, so the rise grows, and the model gives 13,911.

Doubling once and then multiplying by the hours. A colony of 50 that doubles every hour has $50\times 2^4=800$50 × 2⁴ = 800 after 4 hours, not 50 × 2 × 4 = 400.

Finding when it passes a million

The next problem asks when a colony of 500 bacteria, doubling every hour, first passes a million. That means solving $500\times 2^n=1000000$500 × 2ⁿ = 1000000, and dividing both sides by 500 gives $2^n=2000$2ⁿ = 2000. The whole hours can be found by trying powers of 2: $2^{10}=1024$2¹⁰ = 1024 is too small and $2^{11}=2048$2¹¹ = 2048 is enough. The exact value of n uses common logarithms: the common logarithm of a number is the power of 10 that makes it.

A calculator gives the common logarithm of 2 as log 2 = 0.3010, to 4 decimal places, so $2=10^{0.3010}$2 = 10^0.3010. Raising a power to a power multiplies the indices, so $2^n=10^{0.3010n}$2ⁿ = 10^(0.3010n): the logarithm of $2^n$2ⁿ is n times the logarithm of 2. And $2000=2\times 10^3=10^{0.3010}\times 10^3=10^{3.3010}$2000 = 2 × 10³ = 10^0.3010 × 10³ = 10^3.3010, so the logarithm of 2000 is 0.3010 + 3 = 3.3010.

So taking the logarithm of both sides of $2^n=2000$2ⁿ = 2000 gives n × 0.3010 = 3.3010, and $n=3.3010/0.3010=10.97$n = 3.3010 / 0.3010 = 10.97, which lies between 10 and 11, as the trial found.