Question

The rodent population of a certain isolated island increases at a rate proportional to the number of rodents present at any time \(t\). If there are \(x_{0}\) rodents on the island at time \(t=0\) and twice that many at time \(T>0\), how many rodents will there be at (a) time \(2 T\), (b) time \(3 T\), (c) time \(n T\), where \(n\) is a positive integer.

Short answer

(a) There will be \(4x_0\) rodents at time \(2T\). (b) There will be \(8x_0\) rodents at time \(3T\). (c) There will be \(2^nx_0\) rodents at time \(nT\), where \(n\) is a positive integer.

Step-by-step solution

(Step 1: Set up the differential equation)

Since the rate of increase of the rodent population is proportional to the number of rodents, we can represent this as a first-order differential equation: \[\frac{dx}{dt} = kx \] where \(x\) is the number of rodents at any time \(t\), and \(k\) is the constant of proportionality.

(Step 2: Solve the differential equation)

To solve the differential equation, we can first rewrite it using the separation of variables method: \[\frac{dx}{x} = k dt\] Now, integrate both sides: \[\int\frac{dx}{x} = \int k dt\] \[\ln|x| = kt + C_1\] Next, solve for \(x\) by taking the exponent of both sides: \[x(t) = Ce^{kt}\] where \(C = e^{C_1}\) is an integration constant. We now have the general solution of the differential equation.

(Step 3: Find initial conditions from given information)

We are given that at \(t=0\), there are \(x_{0}\) rodents, and at \(t=T\), there are \(2x_{0}\) rodents. We will use these initial conditions to find the constant \(C\) and \(k\). At \(t=0\), we have: \[x(0) = Ce^{k\cdot0} \Rightarrow x_0 = C\] At \(t=T\), we have: \[x(T) = x_0 e^{kT} \Rightarrow 2x_0 = x_0 e^{kT}\] Divide by \(x_0\): \[2 = e^{kT}\] Now we can find \(k\): \[k = \frac{1}{T}\ln(2)\] So our specific solution is: \[x(t) = x_0 e^{\frac{1}{T}\ln(2)t}\]

(Step 4: Find rodent populations at given times)

(a) At \(2T\), we want to find \(x(2T)\): \[x(2T) = x_0 e^{\frac{1}{T}\ln(2)(2T)} = x_0e^{2\ln(2)} = x_0\cdot 2^2 = 4x_0\] There will be \(4x_0\) rodents at time \(2T\). (b) At \(3T\), we want to find \(x(3T)\): \[x(3T) = x_0 e^{\frac{1}{T}\ln(2)(3T)} = x_0e^{3\ln(2)} = x_0\cdot 2^3 = 8x_0\] There will be \(8x_0\) rodents at time \(3T\). (c) At \(nT\), we want to find \(x(nT)\): \[x(nT) = x_0 e^{\frac{1}{T}\ln(2)(nT)} = x_0e^{n\ln(2)} = x_0\cdot 2^n\] There will be \(2^nx_0\) rodents at time \(nT\), where \(n\) is a positive integer.

Key concepts

Exponential Growth

Exponential growth occurs in situations where a quantity increases at a rate proportionate to its current value, such as in population dynamics and compound interest scenarios. In the context of the rodent population problem, we witness exponential growth since the population doubles after a constant time interval, here labeled as T.

This pattern, characterized by the repeated doubling at regular intervals, illustrates the 'explosive' nature of exponential growth. Mathematically, exponential growth is modeled by functions of the form x(t) = x_0e^{kt}, where x_0 is the initial quantity at time t = 0, k is the growth constant, and e is the base of the natural logarithm. In our rodent population example, the larger the value of n in nT, the more pronounced the exponential growth becomes, resulting in increasingly large populations of rodents over time.

Separation of Variables

Separation of variables is a fundamental technique for solving ordinary differential equations (ODEs), which involves separating the variables (often x and t in our case) on opposite sides of the equation. After separation, each side of the equation is integrated independently.

In the rodent population example, we start with the ODE dx/dt = kx and then rearrange it to dx/x = k dt. The integration of both sides yields ln|x| = kt + C_1, guiding us to the general solution of x(t) = Ce^{kt}. This step is crucial since it converts the ODE into a form that can be solved explicitly for the unknown function x(t), reflecting the population at any time t.

Initial Conditions

Initial conditions in the context of differential equations are the values that the unknown function of the differential equation, x(t), takes at a specific starting time, usually t = 0. These conditions are essential for determining the particular solution to a differential equation from its family of general solutions.

For the rodent population, the initial condition tells us that there were x_0 rodents at t = 0. With this, we established that C = x_0. Another set of conditions defined by the problem allowed us to calculate the constant k by observing the population at time T being twice x_0. These initial conditions, when plugged in the general solution, tailor the equation specifically for this rodent population scenario and allow us to predict future populations at any given time t.

Population Dynamics

Population dynamics involve the study of how and why the number of individuals in a population changes over time. An understanding of factors like birth rates, death rates, immigration, and emigration is essential for creating models that describe these phenomena.

In our exercise, the rodent population growth on the island is an isolated system - no immigration or emigration - and the change is only due to birth rates that are proportional to the current population. The doubling of the population at time T, and then the subsequent quadrupling at time 2T indicates a consistent growth pattern unaffected by external variables. This unrealistic assumption of constant growth without limits is typical in simpler models and helps students understand the underlying mechanics of population changes before introducing more complex factors such as carrying capacity and environmental constraints.