Question

Consider a population \(p\) of field mice that grows at a rate proportional to the current population, so that \(d p / d t=r p\). (a) Find the rate constant \(r\) if the population doubles in 30 days. (b) Find \(r\) if the population doubles in \(N\) days.

Short answer

Answer: The rate constant for population doubling in 30 days is \(r = \frac{\ln(2)}{30}\) and for doubling in N days, it is \(r = \frac{\ln(2)}{N}\).

Step-by-step solution

Solve the differential equation

To solve the differential equation \(d p / d t = r p\), we can use the separation of variables method. Separate the variables: \(\frac{d p}{p} = r dt\) Now, integrate both sides: \(\int\frac{d p}{p} = \int r dt\) After integrating, we have: \(\ln(p) = rt + C\)

Obtain the population function \(p(t)\)

We want to get the population function \(p(t)\). To do this, exponentiate both sides of the equation: \(p(t) = e^{rt + C} = e^C e^{rt}\) Let's use a constant \(P_0\) as the initial population at time \(t = 0\). Hence, we have: \(p(t) = P_0 e^{rt}\). Now we will use the given conditions to find the unknown parameter \(r\).

(a) Find the rate constant \(r\) for doubling in 30 days

When the population doubles in 30 days, the population function becomes: \(p(30) = 2P_0\) Now, substitute the value of \(p(30)\) into the population function: \(2P_0 = P_0 e^{30r}\) Solve for \(r\): \(e^{30r} = 2\) Take the natural logarithm of both sides: \(30r = \ln(2)\) Then, solve for \(r\): \(r = \frac{\ln(2)}{30}\)

(b) Find the rate constant \(r\) for doubling in N days

When the population doubles in N days, the population function becomes: \(p(N) = 2P_0\) Now, substitute the value of \(p(N)\) into the population function: \(2P_0 = P_0 e^{Nr}\) Solve for \(r\): \(e^{Nr} = 2\) Take the natural logarithm of both sides: \(Nr = \ln(2)\) Then, solve for \(r\): \(r = \frac{\ln(2)}{N}\) In summary, we found the rate constant \(r\) when the population doubles in 30 days to be \(r = \frac{\ln(2)}{30}\) and when the population doubles in N days, the rate constant is \(r = \frac{\ln(2)}{N}\).

Key concepts

Separation of Variables

Separation of variables is a powerful technique for solving differential equations. It involves manipulating the equation so that all terms containing one variable are on one side, and all terms containing the other variable are on the opposite side. This allows us to turn a complex problem into two simpler problems that can be solved separately.

For the differential equation given by \(\frac{dp}{dt} = rp\), separation of variables looks like this:

• Move all terms involving \(p\) to one side: \(\frac{dp}{p} = r\,dt\).
• Now, each side only has one variable, so we can integrate both sides separately.

After integrating, we find: \(\ln(p) = rt + C\), where \(C\) is the constant of integration. This equation shows the relationship between population \(p\) and time \(t\) by exhibiting the exponential growth nature of the solution.

Exponential functions indicate growth that increases at a consistent rate, a concept important in biological and physical sciences. With the separation of variables, solving for \(p(t)\) becomes manageable, leading to solutions that help us understand how populations and other phenomena change over time.

Exponential Growth

Exponential growth is a fascinating concept often described using real-world scenarios, like population dynamics. It characterizes a process where the rate of change is proportional to the quantity present. This means the larger the population, the faster it grows.

The exponential function, \(p(t) = P_0 e^{rt}\), models this growth effectively. Here:

• \(p(t)\) is the population at time \(t\).
• \(P_0\) is the initial population when \(t = 0\).
• \(r\) is the growth rate, determining how quickly the population grows.

They key feature of exponential growth is the doubling time. In our context, if a population doubles every 30 days, this time period reflects how rapid \(r\) is in influencing the population size.

Mathematically, the doubling time can be visually inferred by setting \(p(t) = 2P_0\). By isolating \(r\), we find the rate at which this doubling occurs, enlightening us on the pace and sustainability of biological growth patterns.

Initial Value Problem

An initial value problem is a form of differential equation that not only involves a function and its derivatives but also includes specific values at a starting point. Initial conditions are given to determine a unique solution to the differential equation.

In our example, \(p(0) = P_0\) is the initial population at time \(t = 0\). This initial condition helps in pinpointing the exact solution among a family of possible solutions to the differential equation \(\frac{dp}{dt} = rp\).

Initial value problems usually require:

• The differential equation, which describes the relationship between the variables.
• An initial condition, which provides a starting point for the function.

By applying the initial set condition, solutions like \(p(t) = P_0 e^{rt}\) are not just general but tailored specifically to the given problem. This allows for precise predictions about the state of the system at any future time \(t\).