2.8 Exponential Growth and Decay

Big idea

One sentence describes an enormous number of real processes: the rate of change is proportional to the amount present. A population with plenty of room grows faster when there are more individuals to reproduce. A lump of radioactive material decays faster when there are more atoms left to decay. Money in an account earning continuously compounded interest grows faster when there is more of it. In symbols all of these read $y' = ky$, and this section solves that equation once and then spends its time interpreting the answer.

The solution is the exponential function, and it could hardly be anything else. The exponential is defined by the property that it is its own derivative, so it is the one function whose rate of change is proportional to its value. What the argument below adds is uniqueness: not only does $y = y_0 e^{kt}$ satisfy the equation, nothing else does. That matters, because it means fitting two data points determines the process completely, which is exactly what a half-life measurement or a doubling-time measurement does.

The sign of $k$ splits the behaviour in two. Positive $k$ gives growth that eventually outruns any polynomial, which is why such a model can only be trusted where the “plenty of room” assumption holds. Negative $k$ gives decay toward zero that never quite arrives. In both cases the characteristic number is the time to change by a fixed factor, and that time does not depend on where you start: a sample takes just as long to fall from $80$ grams to $40$ as from $10$ grams to $5$.

Behind all of this is a general technique worth naming. An equation of the form $y' = f(x)g(y)$ can be attacked by separating the variables: gather everything with $y$ on one side, everything with $x$ on the other, and integrate both sides. Proportional growth is the simplest case, and the same method handles equations whose solutions are not exponential at all.

Decoder

If a quantity changes at a rate proportional to its current value, with proportionality constant $k$, then its value at time $t$ is its initial value multiplied by $e^{kt}$, and no other function has that property.

The uniqueness half is the part that needs an argument, and it is short. Suppose $y' = ky$. Consider the auxiliary function $u(t) = y(t)e^{-kt}$. By the product rule,

$$ u'(t) = y'(t)e^{-kt} - k y(t) e^{-kt} = \big( y'(t) - k y(t) \big) e^{-kt} = 0 $$

so $u$ is constant. Evaluating at $t = 0$ gives $u = y(0)$, hence $y(t) = y(0)e^{kt}$. The trick of multiplying by $e^{-kt}$ to force a derivative to collapse is worth remembering; it reappears whenever a linear equation has to be solved.

Read the constant $k$ carefully. It is a relative rate, with units of one over time, not an amount per unit time. A colony with $k = 0.5$ per hour grows by about $50$ percent of its current size each hour, so the actual increase per hour keeps rising even though $k$ never changes. Confusing a relative rate with an absolute one is the common modeling error here, and it shows up as a linear model where an exponential one belongs.

Definitions and results

The growth equation. The differential equation $y' = ky$ with initial value $y(0) = y_0$ has the unique solution

$$ y(t) = y_0 e^{kt} $$

Growth when $k > 0$, decay when $k < 0$, constant when $k = 0$.

Doubling time. If $k > 0$, the time for the quantity to double satisfies $e^{kT} = 2$, so

$$ T = \frac{\ln 2}{k} $$

It does not depend on the starting amount, so every doubling takes the same length of time.

Half-life. If $k < 0$, the time to fall to half satisfies $e^{k\tau} = \tfrac{1}{2}$, so $\tau = \dfrac{\ln 2}{|k|}$. Equivalently $k = -\dfrac{\ln 2}{\tau}$, which is how a stated half-life gets converted into a rate constant.

Powers of two form. Once the half-life $\tau$ is known, the solution can be written without $e$ at all, as $y(t) = y_0 \, 2^{-t/\tau}$. This is often the fastest route to a numerical answer, since three half-lives immediately gives one eighth of the original.

Compound interest. An account with continuous compounding at annual rate $r$ has balance $P e^{rt}$ after $t$ years, which is the growth equation with $k = r$.

Newton’s law of cooling. An object at temperature $T$ in surroundings held at $T_s$ obeys $T' = k(T - T_s)$ with $k < 0$. Writing $y = T - T_s$ turns this into the growth equation, so

$$ T(t) = T_s + \big( T_0 - T_s \big) e^{kt} $$

The temperature difference decays exponentially; the temperature itself approaches the ambient value.

Separable equations. An equation of the form $\dfrac{dy}{dx} = f(x)g(y)$ separates:

$$ \int \frac{dy}{g(y)} = \int f(x)\, dx $$

Each side is integrated in its own variable, one constant of integration suffices, and the initial condition determines it. Division by $g(y)$ is only legal where $g(y) \ne 0$, so check separately whether a constant solution $g(y) = 0$ has been lost.

Fitting a model. Two data points determine both $y_0$ and $k$. Take the ratio of the two values to eliminate $y_0$, then take a logarithm to solve for $k$.

Worked examples

Growth from a rate constant

A culture grows according to $y' = 0.5y$, with $y$ measured in thousands of cells and $t$ in hours, and starts at $y(0) = 200$. Find the population after $4$ hours and the doubling time.

The solution is $y(t) = 200e^{0.5t}$, so

$$ y(4) = 200 e^{2} \approx 200(7.389) \approx 1478 $$

The doubling time is $T = \dfrac{\ln 2}{0.5} = 2\ln 2 \approx 1.386$ hours.

Check the two answers against each other. Four hours is a little under three doubling periods, since $3T \approx 4.16$ hours, so the population should be a little under $200 \cdot 2^3 = 1600$. The value $1478$ fits.

Decay from a half-life

A sample of a radioactive isotope with half-life $20$ years starts at $40$ grams. How much remains after $50$ years, and when does the sample fall to $5$ grams?

Using the powers-of-two form, $y(t) = 40 \cdot 2^{-t/20}$. At $t = 50$,

$$ y(50) = 40 \cdot 2^{-2.5} = \frac{40}{4\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \approx 7.07 \text{ grams} $$

For the second question, $5$ grams is one eighth of $40$, and one eighth is three halvings, so the answer is exactly $3 \times 20 = 60$ years. Solving algebraically confirms it: $2^{-t/20} = \tfrac{1}{8}$ forces $\tfrac{t}{20} = 3$.

Notice the consistency between the two parts. At $50$ years the amount is $7.07$ grams and at $60$ years it is $5$ grams, and $5\sqrt{2}$ is indeed larger than $5$ by the factor $\sqrt{2}$, which is what half of a half-life should cost.

Cooling toward room temperature

A metal block at $90$ degrees Celsius is left in a room held at $20$ degrees. After $10$ minutes it has cooled to $60$ degrees. What is its temperature after $20$ minutes?

Let $y = T - 20$ be the temperature difference. Then $y' = ky$ with $y(0) = 70$, so $y(t) = 70e^{kt}$. The measurement at $t = 10$ gives $y(10) = 40$, hence

$$ 70 e^{10k} = 40 \quad \Longrightarrow \quad e^{10k} = \frac{4}{7} $$

There is no need to solve for $k$ itself. At $t = 20$ the factor is $e^{20k} = \left(e^{10k}\right)^2 = \left(\tfrac{4}{7}\right)^2 = \tfrac{16}{49}$, so

$$ y(20) = 70 \cdot \frac{16}{49} = \frac{160}{7}, \qquad T(20) = 20 + \frac{160}{7} \approx 42.9 \text{ degrees} $$

The structure is worth seeing: the difference shrank by the factor $\tfrac{4}{7}$ in the first ten minutes and by the same factor again in the second ten, from $70$ to $40$ to about $22.9$. Equal time intervals multiply by equal factors, which is the defining feature of exponential decay.

A separable equation that is not exponential

Solve $\dfrac{dy}{dx} = x y^2$ with $y(0) = 1$.

Separate and integrate, assuming $y \ne 0$:

$$ \int \frac{dy}{y^2} = \int x\, dx \quad \Longrightarrow \quad -\frac{1}{y} = \frac{x^2}{2} + C $$

The initial condition gives $-1 = 0 + C$, so $C = -1$ and

$$ -\frac{1}{y} = \frac{x^2}{2} - 1 \quad \Longrightarrow \quad y = \frac{2}{2 - x^2} $$

Check by differentiating. With $y = 2(2 - x^2)^{-1}$, the chain rule gives $y' = -2(2-x^2)^{-2}(-2x) = \dfrac{4x}{(2-x^2)^2}$, while $xy^2 = x \cdot \dfrac{4}{(2-x^2)^2}$, the same expression. The initial value also checks: $y(0) = \tfrac{2}{2} = 1$.

This solution blows up as $x$ approaches $\sqrt{2}$, so it is only valid on the interval $-\sqrt{2} < x < \sqrt{2}$ containing the initial point. Also note the constant solution $y = 0$, which satisfies the equation but was divided away in the separation step and cannot be recovered by any choice of $C$.

Practice

Start with the proportional model. These problems give you a rate constant, a doubling time, a half-life or two data points, and ask for a value, a time or the constant itself. Convert whatever you are given into $k$ first.

Then separation of variables in general. Some of these produce exponentials and some do not; the method is the same either way, and the initial condition is applied after integrating, never before.

Quiz

Five items on solving the proportional growth equation, doubling time and half-life, cooling problems, and separable equations.