Example
One of the most straightforward applications is the decay of radioactive isotopes – elements that emit radiation due to unstable nuclei.
In the scientific field, we would be interested in determining how much of a given isotope will decay at a particular time. As a basis, scientists will refer to its half-life – a measure of time that will tell us when half of the material will decay.
For example, if the half-life of Zirconium-89 is 78.41 hours, then Zr-89 would have decayed by half after 78.41 hours.
Identify
Let's say we want to find the amount of Zr-89 after 48 hours if its mass was \(m_1 =100g\) initially.
In this type of problem, it's best to list the details of the problem:
- Initial condition: \(m_1 = 100\) when \(t_1 = 0\)
- Unknown condition: \(m_2\) when \(t_2 = 48\)
- Half-life condition: \(m_3 = 50\) when \(t_3 = 78.41\)
Find the Model
Since the amount is directly proportional to its rate of change \(m \propto m^{\prime}\), it observes the decay application of DE; hence, the model for such a phenomenon, as derived earlier, is: \(m = Ce^{kt}\).
After identifying this model (general solution), we need to find the constants \(C\) and \(k\) based on the conditions. Another way of saying this is we need to find a particular solution based on the given criteria:
For \(C\), let's use the initial criteria:
- \(100=Ce^{k(0)}\)
- \(C=100\)
For \(k\), consider the half-life condition:
- \(50=100e^{k(78.41)}\)
- \(k=-8.840×10^{-3}\)
The constant of proportionality is negative; hence, we can confirm it is a decay model.
Now that we know these constants, we can now form the model
\(m = 100e^{(-8.840×10^{-3})(t)}\).
This expression is the function we use to determine the amount of Zr-89 at any given point in time. We can see a graphical view of the function below.
Evaluate
At this point, we only need to substitute \(t=48\) hours to determine the answer to the problem:
- \(m = 100e^{(-8.840×10^{-3})(t)}\)
- \(m = 100e^{(-8.840×10^{-3})(48)}\)
- \(m = 65.42\)
The result means that after 48 hours, there would be 65.42 grams of Zr-89 left.
Summary
Suppose a quantity \(y\) is a function of time \(t\) and is directly proportional to its rate of change \(y^{\prime}\). In that case, we can model the event as a differential equation: \(y^{\prime} = ky\), where \(k\) is the constant of proportionality.
The general solution, \(y=Ce^{kt}\), has the following variables: \(y\) is the output, \(C\) is the initial value, \(k\) is the constant of proportionality, and \(t\) is time
Such events that follow this model are growth and decay situations. It is a growth model if \(k \gt 0\). Otherwise, it is a decay model if \(k \lt 0\).
