Question

The field mouse population in Example 1 satisfies the differential equation $$ d p / d t=0.5 p-450 \text { . } $$ (a) Find the time at which the population becomes extinct if \(p(0)=850\). (b) Find the time of extinction if \(p(0)=p_{0}\), where \(0

Short answer

Answer: The extinction time for an initial population of 850 field mice is approximately 4.36 years. If the extinction is to occur in 1 year, the initial population should be approximately 579.60 field mice.

Step-by-step solution

Identify the type of differential equation

The differential equation is given by: $$ \frac{dp}{dt} = 0.5p - 450 $$ This is a first-order linear differential equation.

Solve the differential equation for a general solution

To solve the given differential equation, we can use separation of variables. Divide both sides of the equation by \(p - 900\): $$ \frac{dp}{p - 900} = 0.5 dt $$ Integrate both sides with respect to their respective variables: $$ \int \frac{dp}{p - 900} = \int 0.5 dt $$ $$ \ln|p - 900| = 0.5t + C $$ Solve for \(p\): $$ p(t) = 900 + Ce^{0.5t} $$ This is the general solution to the given differential equation.

Solve for extinction time with \(p(0) = 850\)

Plug the initial condition \(t=0\) and \(p(0) = 850\) into the general solution: $$ 850 = 900 + Ce^0 $$ Solve for \(C\): $$ C = -50 $$ The particular solution with this initial condition is: $$ p(t) = 900 - 50e^{0.5t} $$ Find the extinction time by setting \(p(t) = 0\): $$ 0 = 900 - 50e^{0.5t} $$ $$ t = \frac{2\ln(18)}{1} \approx 4.36 \text{ years} $$ Thus, for \(p(0) = 850\), the time of extinction is approximately 4.36 years.

Solve for extinction time with \(p(0) = p_{0}\)

Plug the initial condition \(t=0\) and \(p(0) = p_{0}\) into the general solution: $$ p_{0} = 900 + Ce^0 $$ Solve for \(C\): $$ C = p_{0} - 900 $$ The particular solution with this initial condition is: $$ p(t) = 900 + (p_{0} - 900)e^{0.5t} $$ To find the extinction time, set \(p(t) = 0\): $$ 0 = 900 + (p_{0} - 900)e^{0.5t} $$ $$ t = \frac{2\ln(\frac{900}{900-p_{0}})}{1} $$ Thus, for \(p(0) = p_{0}\), the time of extinction is given by: $$ t = 2\ln\left(\frac{900}{900-p_{0}}\right) $$

Solve for initial population if extinction occurs in 1 year

We are given that the extinction time is 1 year, so we can plug \(t = 1\) into the extinction time equation from Step 4: $$ 1 = 2\ln\left(\frac{900}{900-p_{0}}\right) $$ Solve for \(p_{0}\): $$ \frac{1}{2} = \ln\left(\frac{900}{900-p_{0}}\right) $$ $$ e^{1/2} = \frac{900}{900-p_{0}} $$ $$ p_{0} = 900 - \frac{900}{e^{1/2}} \approx 579.60 $$ Thus, if the population is to become extinct in 1 year, the initial population \(p_{0}\) should be approximately 579.60 field mice.

Key concepts

First-order linear differential equations

First-order linear differential equations are an important class in the study of calculus and differential equations. They generally take the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). In a first-order linear equation, the highest derivative is first degree—or exponent one. This means their solutions often involve exponential functions.

• Output depends linearly on both its derivative and function.
• They are generally solved using methods like separation of variables or integrating factors.

In our problem, the equation \( \frac{dp}{dt} = 0.5p - 450 \) fits this form, with \( P(x) = 0.5 \) and \( Q(x) = -450 \). To solve, we separate variables or identify integrating factors to simplify and solve the equation. The final solutions allow us to understand the behavior or change in values over time, which is very useful for many applications, including population dynamics.

Population dynamics

Population dynamics studies how populations (like animals or bacteria) change over time. It’s a critical area in biology and ecology that involves understanding factors influencing population sizes and growth patterns. The differential equation in our example models population growth and determines critical points, like extinction.

• Models account for birth rates, death rates, and environmental carrying capacity.
• The term \( 0.5p - 450 \) shows how growth rate decreases as population nears limits.

In the field mouse example, we find that when \( p \) hits 900 mice, the growth rate balances death and new births. If the population starts below this point, extinction is possible as shown in part (a) and (b) of the exercise. Population studies utilize such equations to predict changes and plan conservation or management strategies. They are essential for sustainable development and maintaining biodiversity.

Mathematical modeling

Mathematical modeling is the process of using mathematical equations and concepts to represent real-world phenomena. It’s a bridge between mathematics and practical applications, providing predictions and insights into complex systems. In the exercise, we used a differential equation to model population changes.

• Models simplify reality to give practical solutions and predictions.
• Through models, complex behaviors like extinction times are calculable.

Such models are widely used across fields—from engineering to economics to biology. In our case, we used a mathematical model to predict when the population of field mice could go extinct based on initial conditions. Adjusting parameters (like changing initial populations) allows us to see various outcomes and make informed decisions. Mathematical models hence support fields in decision-making, policy formulation, and understanding nature's interactions and behaviors.