We’ve seen a differential equation pop up several times already, and it is the m
We’ve seen a differential equation pop up several times already, and it is the most common and simplest of all differential equations:
[latex]G'(t) = k G(t)[/latex]
When [latex]k[/latex] is positive, this is saying that [latex]G[/latex] 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 [latex]k[/latex] is negative, this is saying that [latex]G'[/latex] 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 [latex]G'(t) = kG(t)[/latex]. It is [latex]G(t) = Ae^{k t}[/latex]. Let’s see some examples
- How much of the isotope will there be left at [latex]t = 365[/latex]?
- At what time will there be [latex]1[/latex] gram left?
Let [latex]I(t)[/latex] be the mass of isotope in grams at time [latex]t[/latex]. Thus, [latex]I(0) = 40[/latex], since we start with [latex]40[/latex] grams. Since the isotope decays at a rate of [latex]0.003[/latex] times its current mass, we see that [latex]I'(t) = -0.003 \cdot I(t)[/latex]. 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
[latex]I(t) = A e^{-0.003 t}[/latex]
Since [latex]I(0) = 40[/latex], we also have [latex]I(0) = A e^{-0.003(0)} = A e^0 = A[/latex], and hence [latex]A = 40[/latex]. Our equation for the mass of the isotope is now [latex]I(t) = 40 e^{-0.003 t}[/latex]
From here, we can now tell how much isotope will be left at [latex]t = 365[/latex]. We plug in [latex]t = 365[/latex] and have
[latex]I(365) = 40 e^{-0.003 (365)} \approx \boxed{13.38 \text{ grams}}.[/latex]
This solves part (1).
To find out when there will be [latex]1[/latex] 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 [latex]t \approx \boxed{1229.62\text{days}}[/latex] or a little over [latex]3[/latex] years. This solves part (2).