Growth and decay

What you will be able to do

• turn "the rate is proportional to the amount" into $\dfrac{dN}{dt} = kN$ and solve it;
• find the doubling time or the half-life, and use them to read a percentage rate;
• use the decay law to find the age of a sample, and meet the logistic equation;
• follow an equation step by step on the computer with Euler's method.

A rate that follows the amount

A colony of bacteria has more cells today than yesterday. Each cell divides at about the same pace, so a colony that is twice as big adds new cells twice as fast. The rate of growth is proportional to the number of cells. In symbols, with $N$ the number of cells at time $t$,

The constant $k$ is the rate constant. Its unit is "per hour", "per year" or whatever the time unit is. To solve, separate the variables:

1. Write $\dfrac{dN}{N} = k\,dt$.
2. Integrate both sides: $\ln N = kt + c$.
3. Write $N = e^{c}e^{kt}$. At $t = 0$ we have $N = N_0$, so $e^{c} = N_0$.

The solution is

If $k > 0$ the amount grows, and we call this exponential growth. If $k < 0$ the amount falls, and we call it exponential decay. The same equation covers a savings account with continuous interest, a population with plenty of food and a radioactive sample.

Doubling time and half-life

How long does a growing amount take to double? Put $N = 2N_0$ in the solution: $e^{kT} = 2$, so

For decay, put $N = \tfrac{1}{2}N_0$ and $k = -\lambda$ with $\lambda > 0$. The time for the amount to halve is the half-life:

Both times depend on $k$ alone. They do not depend on $N_0$. After $n$ doubling times the amount is $2^n N_0$, and after $n$ half-lives it is $N_0/2^n$. This lets you work without a calculator when the time is a whole number of half-lives.

Example. A sample of a radioactive isotope has a half-life of $12\ \text{h}$. Find $\lambda$, and the fraction left after $30\ \text{h}$.

1. $\lambda = \dfrac{\ln 2}{12\ \text{h}} = 0.05776\ \text{h}^{-1}$.
2. The fraction is $e^{-\lambda t} = e^{-0.05776 \times 30} = e^{-1.733} = 0.1768$. As a check, $30\ \text{h}$ is $2.5$ half-lives, and $2^{-2.5} = 0.1768$.

Percentage rates

A bank adds $5\%$ each year, so the amount is multiplied by $1.05$ every year. After $t$ years it is $N_0 \times 1.05^t$. Write this as $N_0 e^{kt}$ and we get $k = \ln 1.05 = 0.04879$ per year. In general, a yearly rate $r$ corresponds to $k = \ln(1 + r)$.

If the interest is added at every instant, we say it is compounded continuously. Then $k = r$ itself, and the factor after one year is $e^{r}$. For $r = 6\%$ and a deposit of $1000$ rupees over $10$ years:

• once a year: $1000 \times 1.06^{10} = 1790.85$;
• continuously: $1000 \times e^{0.6} = 1822.12$.

The continuous way pays slightly more, because the interest earns interest at every instant.

Radioactive decay and the age of a sample

A radioactive nucleus decays at random, but a large sample follows $\dfrac{dN}{dt} = -\lambda N$. The constant $\lambda$ is the decay constant. The activity $A = \lambda N$ is the number of decays each second, in becquerel (Bq). Activity falls by the same factor as $N$.

Living things take in carbon-14, and the amount of it in a living tree stays at a steady level. After death the carbon-14 decays, with a half-life of $5730$ years. We use this standard value. Older books print $5568$ years, the first value measured, so a textbook answer may differ slightly. If a sample has the fraction $f$ of the carbon-14 of living wood, then $f = e^{-\lambda t}$ and

Example. A beam has $40\%$ of the carbon-14 of living wood. Its age is $\dfrac{5730}{\ln 2}\ln 2.5 = 7574.6$ years, about $7600$ years.

When growth meets a limit

No colony grows for ever. Food, space or disease set a limit. The logistic equation keeps the idea "growth is proportional to the amount" but slows it down near the limit $M$, called the carrying capacity:

When $P$ is small the bracket is close to $1$, and the growth is almost exponential. When $P$ is near $M$ the bracket is close to $0$, and the growth stops. The solution, which the next course shows how to find, is

You can check it without any method. Put $u = Ae^{-kt}$, so that $\dfrac{du}{dt} = -ku$ and $P = \dfrac{M}{1 + u}$.

1. Differentiate: $\dfrac{dP}{dt} = -\dfrac{M}{(1 + u)^2}\dfrac{du}{dt} = \dfrac{Mku}{(1 + u)^2}$.
2. Now $1 - \dfrac{P}{M} = 1 - \dfrac{1}{1 + u} = \dfrac{u}{1 + u}$. So $kP\left(1 - \dfrac{P}{M}\right) = k\dfrac{M}{1 + u}\dfrac{u}{1 + u} = \dfrac{Mku}{(1 + u)^2}$.
3. The two sides agree. At $t = 0$, $P = \dfrac{M}{1 + A} = P_0$ for the stated $A$.

The growth rate $kP(1 - P/M)$ is largest when $P = \dfrac{M}{2}$, and then it is $\dfrac{kM}{4}$. The curve has its steepest point there.

Euler's method

Many equations have no neat formula. A computer can still follow the solution. At a point we know the slope $\dfrac{dN}{dt}$ from the equation. Take a small step $h$ along that slope, then find the slope at the new point and step again. This is Euler's method:

For $\dfrac{dN}{dt} = 0.4N$ with $N_0 = 100$ and $h = 0.5$, the first step gives $100 + 0.5 \times 40 = 120$, and the second gives $120 + 0.5 \times 48 = 144$. The picture shows the four steps to $t = 2$. They stay below the true curve, because the curve bends upwards and each step uses the slope at its left end. A smaller step gives a smaller error.

Practise with new numbers

Try it in Python

The program uses Euler's method for $\dfrac{dN}{dt} = 0.4N$ and compares the result at $t = 2$ with the exact answer. It tries three step sizes.

Common mistakes

• Using the wrong unit for $k$. If time is in hours, $k$ is per hour. A half-life in days gives $\lambda$ per day.
• Writing $e^{kt}$ for a yearly percentage. A rate of $5\%$ a year gives $k = \ln 1.05$, not $0.05$. The two are close for small rates, but not equal.
• Forgetting the minus sign in decay. The law is $N = N_0 e^{-\lambda t}$ with $\lambda > 0$.
• Adding instead of multiplying. Doubling every $3$ hours means multiplying by $2$ each time. It does not add $N_0$.
• Trusting one big Euler step. The estimate improves as the step shrinks.

Remember these

• $\dfrac{dN}{dt} = kN$ gives $N = N_0 e^{kt}$. A negative $k$ is decay.
• Doubling time $\dfrac{\ln 2}{k}$ and half-life $\dfrac{\ln 2}{\lambda}$ do not depend on the starting amount.
• A yearly rate $r$ means $k = \ln(1 + r)$. Continuous interest at rate $r$ means $k = r$.
• The logistic solution is $P = \dfrac{M}{1 + Ae^{-kt}}$. Its steepest growth, $\dfrac{kM}{4}$, is at $P = \dfrac{M}{2}$.
• Euler's method: new value = old value + $h \times$ slope.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The number of cells in a colony is multiplied by $3$ every $4$ hours. Find $k$ per hour, to four decimal places.

2. A sample has a half-life of $8$ days. What fraction of it is left after $20$ days?

3. A deposit earns $7\%$ a year, compounded continuously. How long does it take to double?

4. A piece of wood has $30\%$ of the carbon-14 of living wood. Find its age, using a half-life of $5730$ years.

5. In the logistic equation with $k = 0.4$ per year and $M = 800$ animals, what is the largest growth rate?

References

• OpenStax, Calculus Volume 2, Chapter 4, section 4.2 Direction Fields and Numerical Methods, and section 4.4 The Logistic Equation.
• OpenStax, Calculus Volume 1, Chapter 3, section 3.8 Exponential Growth and Decay.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the sections on differential equations and their applications.