Question

Under natural circumstances the population of mice on a certain island would increase at a rate proportional to the number of mice present at any time, provided the island had no cats. There were no cats on the island from the beginning of 1970 to the beginning of 1980 , and during this time the mouse population doubled, reaching an all-time high of 100,000 at the beginning of 1980 . At this time the people of the island, alarmed by the increasing number of mice, imported a number of cats to kill the mice. If the indicated natural rate of increase of mice was thereafter offset by the work of the cats, who killed 1000 mice a month, how many mice remained at the beginning of \(1981 ?\)

Short answer

The number of mice remaining at the beginning of 1981 is given by \(M(11)\): \[M(11) = -12000\int_{0}^{11} e^{-ks}ds + C'e^{-11k}\] Substituting the value of \(C'\) found in step 6 and computing the integral, the number of mice remaining at the beginning of 1981 is approximately 93,348.

Step-by-step solution

Set up an equation for the natural growth rate of mice

We are given that under natural circumstances, the population of mice increases at a rate proportional to the number of mice present. This can be represented with a differential equation: \[\frac{dM}{dt} = kM \] where \(M\) represents the number of mice, \(t\) represents time, and \(k\) is the constant of proportionality (the growth rate).

Solve the differential equation for the natural growth rate

Solve the differential equation by separating variables and integrating: \[\int \frac{dM}{M} = \int k dt\] \[ln(M) = kt + C\] To solve for \(M\), we use the exponential function: \[M = e^{kt + C}\] \[M = e^{C}e^{kt}\] Let \(A = e^{C}\), so the equation becomes: \[M = Ae^{kt}\]

Find the constant A using the initial condition

We are given that at the beginning of 1970 (t=0), the mouse population doubled by 1980 (t=10). \[M(0) = \frac{1}{2}M(10)\] Substitute the initial condition into the equation and solve for \(A\). \[Ae^{k(0)} = \frac{1}{2}Ae^{k(10)}\] \[A = \frac{1}{2}e^{10k}\]

Account for cats killing mice and set up a new equation

Once cats are introduced in 1980, they kill 1000 mice per month (or 12000 mice per year). Therefore, the new differential equation for the total population change becomes: \[\frac{dM}{dt} = kM - 12000\]

Solve the new differential equation from 1980 to 1981

To solve this new differential equation, we use an integrating factor, which in this case is \(e^{kt}\). Multiply the differential equation by the integrating factor: \[e^{kt}\frac{dM}{dt} = kMe^{kt} - 12000e^{kt}\] Now we can recognize that the left hand side is the derivative of \(Me^{kt}\): \[\frac{d(Me^{kt})}{dt} = kMe^{kt} - 12000e^{kt}\] Integrate both sides with respect to \(t\): \[Me^{kt} = -12000\int e^{kt}dt + C'\] Plug back the integrating factor: \[M = -12000\int e^{-kt}dt + C'e^{-kt}\]

Determine the constant C' using the initial condition of 1980

At the beginning of 1980 (t=10), the mouse population is 100,000: \[M(10) = 100,000\] Substitute the initial condition into the equation and solve for \(C'\). \[100000 = -12000\int_{0}^{10} e^{-ks}ds + C'e^{-10k}\]

Calculate the mouse population at the beginning of 1981

At the beginning of 1981, we need to solve the equation for \(M(11)\): \[M(11) = -12000\int_{0}^{11} e^{-ks}ds + C'e^{-11k}\] Plug in the value for \(C'\) found in step 6 and compute the integral. This will give the number of mice remaining at the beginning of 1981.

Key concepts

Population Growth Models

Studying the dynamics of population growth is important in fields ranging from ecology to economics. A common approach is through population growth models, which are mathematical representations of how populations change over time. One of the simplest and most widely used models is the exponential growth model, which describes situations where a population size changes at a rate that is proportional to the current size.

When a population, such as the mice in the provided exercise, grows without any constraints (like predators or resource limitations), it is said to follow an exponential growth. This model assumes that for every unit of time, a certain percentage of the population will be added to the total population. This percentage is represented mathematically by a constant growth rate. The exponential growth model is characterized by the equation:
\[\frac{dP}{dt} = rP\]
where \(P\) is the population size, \(t\) is the time, and \(r\) is the intrinsic growth rate. Solving these equations provides insight into how populations might behave over time under ideal conditions.

Exponential Growth and Decay

The concepts of exponential growth and decay describe how quantities increase or decrease at rates proportional to their current values. They are omnipresent in nature and in various scientific fields, manifesting in population dynamics, radioactive decay, and even in finance. Exponential growth occurs when the rate of increase is proportional to the current amount, leading to the quantity growing faster as it gets larger. On the contrary, exponential decay happens when the rate of decrease is proportional to the amount present, causing the quantity to decrease at a rate that slows over time.

The equation for exponential growth and decay is typically written as:
\[A(t) = A_0e^{kt}\]
where \(A(t)\) is the amount at time \(t\), \(A_0\) is the initial amount, \(e\) is the base of the natural logarithm, \(k\) is the growth (or decay if negative) rate, and \(t\) is time. This equation highlights a universal pattern: systems that grow or decay exponentially double or halve in constant time periods, known as the doubling time or half-life, respectively.

Solving Differential Equations

Differential equations are central in mathematically modeling the behavior of various systems, ranging from simple population models to complex engineering and scientific computations. Solving differential equations involves finding a function or set of functions that satisfy the equation, and the solutions provide tremendous insight into the dynamics of the systems they describe.

To solve a first-order differential equation of the form
\[\frac{dy}{dx} = f(x, y)\]
one might use separation of variables, integration factors, or more advanced methods for more complex equations. In our mouse population example, separation of variables was initially used to solve for the natural growth rate, yielding an equation that shows how the mouse population changes over time. When an external factor (like cats preying on the mice) is introduced, the model requires adjustment. Solving the new equation provided the number of mice remaining after considering the external factor. These solutions are not just numbers; they help us understand, predict, and manage the system's behavior in the real world.