Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

4.6: Exponential Growth and Decay

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    When a plant or animal is alive it continually replenishes the carbon in its system. Some of this carbon is radioactive \(^{14}C\). When it dies the carbon it contains no longer replenishes, hence the \(^{14}C\) begins to decay. It is a chemical fact that the rate of decay is proportional to the amount of \(^{14}C\) in the body at that time. In equation form we have

    \[ \dfrac{dy}{dt}=ky.\]

    If we multiply both sides by \(dt\) and integrate, we get

    \[ \int \dfrac{dy}{y} = k dt \]


    \[ \ln y = kt + C_0. \]

    Exponentiating both sides to get rid of the \(\ln\) function gives

    \[ y = e^{kt + C_o} = e^{C_o} e^{kt}. \]

    Now let

    \[ C = e^{C_o}. \]


    \[ y = Ce^{kt} \]

    where \(C\) and \(k\) are constants.

    Example 1: Radiocarbon Dating a Skull

    You find a skull in a nearby Native American ancient burial site and with the help of a spectrometer, discover that the skull contains 9% of the \(^{14}C\) found in a modern skull. Assuming that the half life of \(^{14}C\) is 5730 years, how old is the skull?


    Since this is a radioactive decay question, we can say that

    \[ \dfrac{dy}{dt} = kt \]

    which has solution

    \[ y = Ce^{kt}. \]

    After 5730 years, there is

    \[ 1/2 C \]

    carbon 14 remaining. Hence:

    \[ \dfrac{1}{2} C = Ce^{k\, 5730}\]


    \[ 0.5 = e^{k\, 5730}. \]

    Taking \(\ln\) of both sides and dividing by 5730 gives

    \[ k = \dfrac{\ln 0.5}{5730} = -0.000121.\]

    Now we use the fact that there is 9% remaining today to give

    \[ 0.09 C = Ce^{kt}. \]

    To keep things compact we are still writing \(k\) instead of -0.000121.

    Now divide by \(C\):

    \[ 0.09 = e^{kt}.\]

    Take ln of both sides at divide by \(k\) to get

    \[ t =\dfrac{\ln 0.09}{k} = \dfrac{\ln 0.09}{-0.000121} = 19,905.\]

    So the skull is about 20,000 years old.


    1. Currently health care for senior citizens cost our government $400 per month. Assuming that the health care inflation rate will be at 8% for the next 40 years, write a differential equation that models the price of health care over this time. Solve this differential equation. How much will the government be spending on you when you are 65 years old?

    2. Suppose that there is a fruit fly infestation in the central valley. Being an environmentalist, you propose a plan to spread 50,000 infertile fruit flies in the area to control the situation. Presently, you have in your laboratory 1,000 fruit flies. In 1 week they will reproduce to a population of 3,000 fruit flies. The farmers want to know when you will be ready to drop your infertile fruit flies. What should you tell them?

    Larry Green (Lake Tahoe Community College)

    Integrated by Justin Marshall.