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Calculus Differential equations (introductory)

Exponential growth and decay models

20 practice questions 0 video lessons Theory + worked examples

Exponential Growth and Decay

California Calculus • Standard 27.0 • Differential Equations

Exponential Growth and Decay is a topic in Differential Equations in the California Calculus Standards. It is aligned to Standard 27.0, which requires students to solve elementary differential equations, including growth-and-decay problems.

Exponential growth and decay follow \(\dfrac{dy}{dt}=ky\), which solves to \(y=y_0 e^{kt}\) — growth when \(k>0\) and decay when \(k<0\), governed by a doubling time or half-life.

California Calculus › Differential Equations › Exponential Growth and Decay  —  Standard 27.0

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Theory

Many quantities change at a rate proportional to their size, \(\dfrac{dy}{dt}=ky\). Solving this gives the exponential model \(y=y_0 e^{kt}\): growth when \(k>0\) and decay when \(k<0\), governed by a doubling time or a half-life.

When the rate of change of a quantity is proportional to the amount present, the model is

\[\dfrac{dy}{dt}=ky\ \Rightarrow\ y=y_0 e^{kt},\]

where \(y_0\) is the initial amount (the value at \(t=0\)) and \(k\) sets the rate.

  • \(k>0\): exponential growth (populations, compound interest);
  • \(k<0\): exponential decay (radioactive material, cooling).

The doubling time (growth) and half-life (decay) are the times for the amount to double or halve; both are \(\dfrac{\ln 2}{|k|}\).

Key idea: \(y_0\) is always the amount at \(t=0\), since \(e^{0}=1\). The sign of \(k\) alone decides growth versus decay.
An exponential growth curve rising ever faster When k is positive the model grows without bound, rising faster and faster as time increases. t y growth: k > 0
Growth (\(k>0\)): rises ever faster.
An exponential decay curve falling toward zero When k is negative the model decays toward zero; the half-life is the time for the amount to halve. t y half-life decay: k < 0
Decay (\(k<0\)): falls toward \(0\), halving each half-life.

The model and the doubling/half-life:

\[y=y_0 e^{kt},\qquad\dfrac{dy}{dt}=ky\]
y equals y naught e to the kt; its rate is k times y
\[\text{doubling time}=\dfrac{\ln 2}{k},\qquad\text{half-life}=\dfrac{\ln 2}{-k}\]
doubling time and half-life are natural log of 2 over the magnitude of k
Halving repeatedly is often quickest: after \(n\) half-lives the amount is \(y_0\left(\dfrac{1}{2}\right)^{n}\).

How to work with a growth/decay model

  1. Identify \(y_0\) and \(k\) (or a doubling time / half-life).
  2. Write \(y=y_0 e^{kt}\), or the halving form \(y_0\left(\dfrac{1}{2}\right)^{t/H}\).
  3. Substitute the time to find the amount, or set \(y\) to a target and solve for \(t\) with logarithms.
Example 1 — Doubling population
A population of \(500\) doubles every hour. Find it after \(3\) hours.
Solution

Doubling each hour multiplies by \(2\) three times.

\(y\)\(=\)\(500\cdot 2^{3}\)
\(=\)\(500\cdot 8=4000\)
population equals 4000
Example 2 — Radioactive half-life
A \(100\)-unit sample has a half-life of \(5\) years. How much remains after \(15\) years?
Solution

\(15\) years is \(\dfrac{15}{5}=3\) half-lives, so halve three times (\(2^3=8\)).

\(y\)\(=\)\(100\cdot\left(\dfrac{1}{2}\right)^{3}\)
\(=\)\(100\cdot\dfrac{1}{8}\)
\(=\)\(12.5\)

\(12.5\) units remain.

amount remaining is 12.5 units
Example 3 — Find k from doubling time
A quantity following \(y=y_0 e^{kt}\) doubles after time \(T\). Find \(k\).
Solution

Set \(y=2y_0\) and solve for \(k\).

\(2\)\(=\)\(e^{kT}\)
\(k\)\(=\)\(\dfrac{\ln 2}{T}\)
k equals natural log of 2 over T
Example 4 — Write the model
A population satisfies \(\dfrac{dy}{dt}=0.03y\), \(y(0)=5000\). Write \(y(t)\).
Solution

The solution of \(\dfrac{dy}{dt}=ky\) is \(y=y_0 e^{kt}\).

\(y(t)\)\(=\)\(5000\,e^{0.03t}\)

Since \(k>0\), it is growth.

model is 5000 e to the 0.03 t

Common pitfalls

Sign of \(k\). Positive \(k\) is growth, negative \(k\) is decay — a decay problem must use \(k<0\).
\(y_0\) is the value at \(t=0\). Do not confuse it with the amount at some later time.
Count half-lives carefully. After \(n\) half-lives the amount is \(y_0(1/2)^{n}\); match \(n\) to the elapsed time divided by the half-life.

Frequently asked questions

What is the exponential growth model?

\(y=y_0 e^{kt}\), the solution of \(\dfrac{dy}{dt}=ky\). With \(k>0\) the quantity grows, with \(k<0\) it decays.

What does y0 represent?

The initial amount, the value of \(y\) at \(t=0\). Because \(e^{0}=1\), \(y(0)=y_0\).

What is half-life?

The time it takes a decaying quantity to fall to half its amount. It equals \(\dfrac{\ln 2}{-k}\) and is constant for exponential decay.

How do you find k from a doubling time?

Set \(2=e^{kT}\) and solve: \(k=\dfrac{\ln 2}{T}\), where \(T\) is the doubling time.

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