Question

A process creates a radioactive substance at the rate of \(2 \mathrm{~g} / \mathrm{hr}\) and the substance decays at a rate proportional to its mass, with constant of proportionality \(k=.1(\mathrm{hr})^{-1}\). If \(Q(t)\) is the mass of the substance at time \(t,\) find \(\lim _{t \rightarrow \infty} Q(t)\)

Short answer

Answer: The limit of the mass of the radioactive substance as time goes to infinity is 200 grams.

Step-by-step solution

Setting up the differential equation

Let's begin by setting up a differential equation to model the mass of the radioactive substance. Since the process creates the substance at a rate of 2 grams per hour and it decays at a rate proportional to its mass with a proportionality constant of k = 0.1 hr^(-1), the rate of change of Q(t) with respect to time, t, can be represented as: \(\frac{dQ}{dt} = 2 - kQ\) where k = 0.1 hr^(-1).

Solving the differential equation

Now, we'll solve the linear first-order differential equation: \(\frac{dQ}{dt} + kQ = 2\) Divide both sides by k: \(\frac{dQ}{dt} + 0.1Q = 2\) This is a linear first-order inhomogeneous differential equation. To solve, we will use an integrating factor, which is defined as: \(I(t) = e^{\int 0.1 dt}=e^{0.1t}\) Multiply both sides of the equation by \(I(t)\): \(e^{0.1t}\frac{dQ}{dt} +0.1e^{0.1t} Q = 2 e^{0.1t}\) Observe that the left side of our equation is the derivative of the product of our integrating factor \(I(t)\) and \(Q(t)\): \(\frac{d}{dt}\left( e^{0.1t}Q \right) = 2 e^{0.1t}\) Now, integrate both sides with respect to t: \(\int\frac{d}{dt}\left( e^{0.1t}Q \right) dt= \int 2 e^{0.1t} dt\) \(e^{0.1t}Q = \frac{20}{1} e^{0.1t} + C\) Now, solve for Q(t): \(Q(t) = \frac{20}{0.1} + Ce^{-0.1t}\) (1)

Finding the limit as t goes to infinity

Now, we need to find the limit of the mass of the radioactive substance, Q(t), as the time, t, goes to infinity: \(\lim_{t \rightarrow \infty} Q(t)\) Using equation (1), we have: \(\lim_{t \rightarrow \infty} \left( \frac{20}{0.1} + Ce^{-0.1t} \right)\) As \(t \rightarrow \infty\), the term with the exponential will go to zero, since \(-0.1t \rightarrow -\infty\): \(\lim_{t \rightarrow \infty} \left( \frac{20}{0.1} + Ce^{-0.1t} \right) = \frac{20}{0.1}\) Therefore, the limit of the mass of the radioactive substance as time goes to infinity is: \(\lim_{t \rightarrow \infty} Q(t)= 200 \;\mathrm{g}\)

Key concepts

First Order Differential Equations

First order differential equations are fundamental to understanding many natural processes, including the behavior of radioactive materials. Such an equation expresses the rate of change of a variable—in our case, the mass of a radioactive substance over time.
A first order differential equation has the general form
\[\frac{dQ}{dt} = f(t, Q)\]
where \( Q(t) \) represents our quantity of interest as a function of time, \( t \), and \( f(t, Q) \) is some function expressing how \( Q \) changes with \( t \). In our radioactive decay scenario, this equation accounts for two factors: the constant creation of the substance and its decay, which is proportional to its current mass.
Solutions to such equations are crucial as they allow us to predict future states of the system—at any given point, how much of the radioactive substance will there be? Understanding and solving first order differential equations like this one help scientists and engineers in a variety of fields, from pharmacokinetics to environmental science, manage and predict dynamic systems.

Integrating Factor Method

When tackling first order linear differential equations, the integrating factor method comes as a powerful tool. It transforms a non-exact differential equation into an exact one, which can then be integrated with respect to time.
In our equation, \(\frac{dQ}{dt} + 0.1Q = 2\), it seems challenging at first to separate and integrate the terms involving \(Q\). This is where the integrating factor, a function of time, steps in. Calculated as
\(I(t) = e^{\int k dt}\),
this factor, when multiplied with each term of our original differential equation, allows us to rewrite the left side of the equation as a product rule derivative. Thus, the complex looking term
\(e^{0.1t}\frac{dQ}{dt} +0.1e^{0.1t} Q\)
becomes the derivative of \(e^{0.1t}Q\), which is much simpler to integrate. Upon integration, we obtain a general solution that includes an arbitrary constant \(C\). This constant is typically determined by an initial condition, such as the initial mass of the substance, but in problems like the one at hand, it becomes irrelevant when considering the long-term behavior of the system.

Exponential Decay Limit

Understanding the exponential decay limit is vital when discussing the long-term fate of a radioactive substance. Exponential decay describes a situation where a quantity decreases at a rate proportional to its current value, which is precisely what happens with radioactive substances. The most intriguing aspect of exponential decay is its limit as time approaches infinity.
In our example, we found that as \(t\) grows large, the term \(Ce^{-0.1t}\) in our solution \(Q(t) = \frac{20}{0.1} + Ce^{-0.1t}\) shrinks to zero. Why is that? The exponential function \(e^{-kt}\) decreases very rapidly as \(t\) increases because \( -kt \) becomes increasingly negative, driving the value of the exponential closer to zero. This is why, despite any initial mass, the eventual mass of the substance is unaffected by the starting amount and stabilizes at a fixed value as time goes on.
With this in mind, students should realize the importance of the exponential decay limit concept in predicting the eventual behavior of many naturally decaying processes—whether that is radioactive decay, cooling of an object, or even depreciation of assets over time.