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5.9: Growth and Decay

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    We’ve seen a differential equation pop up several times already, and it is the most common and simplest of all differential equations:

    \(G'(t) = k G(t)\)

    When \(k\) is positive, this is saying that \(G\) is growing at a rate proportional to the value of the function at any given point. As we’ve seen, population tends to follow this rule, but several other things do as well. When \(k\) is negative, this is saying that \(G'\) is decreasing at a rate proportional to its value, and this is true for several things as well. These are called the growth and decay equations respectively.

    And there is a simple solution to the differential equation \(G'(t) = kG(t)\). It is \(G(t) = Ae^{k t}\). Let’s see some examples


    A radioactive isotope decays at a rate of \(0.003\) times its mass in grams per day. Initially, a sample contains \(40\) grams of the isotope at \(t = 0\).
    1. How much of the isotope will there be left at \(t = 365\)?
    2. At what time will there be \(1\) gram left?

    Let \(I(t)\) be the mass of isotope in grams at time \(t\). Thus, \(I(0) = 40\), since we start with \(40\) grams. Since the isotope decays at a rate of \(0.003\) times its current mass, we see that \(I'(t) = -0.003 \cdot I(t)\). The negative is in there because it is a decay rate — the amount of isotope is going down. We know the solution to a differential equation like this is

    \(I(t) = A e^{-0.003 t}\)

    Since \(I(0) = 40\), we also have \(I(0) = A e^{-0.003(0)} = A e^0 = A\), and hence \(A = 40\). Our equation for the mass of the isotope is now \(I(t) = 40 e^{-0.003 t}\)

    From here, we can now tell how much isotope will be left at \(t = 365\). We plug in \(t = 365\) and have

    \(I(365) = 40 e^{-0.003 (365)} \approx \boxed{13.38 \text{ grams}}.\)

    This solves part (1).

    To find out when there will be \(1\) gram left, we solve

    \[\begin{align*} 1 & = 40 e^{-0.003t} \\ \frac{1}{40} & = e^{-0.003 t} \\ \ln \left( \frac{1}{40} \right) & = -0.003t \\ \frac{1}{-0.003} \ln \left( \frac{1}{40} \right) & = t \end{align*}\]

    Simplifying this, we see \(t \approx \boxed{1229.62\text{days}}\) or a little over \(3\) years. This solves part (2).

    This page titled 5.9: Growth and Decay is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Tyler Seacrest via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.