Question

You should solve the following problems without using a graphing calculator. True or False For small values of \(t\) the solution to logistic differential equation \(d P / d t=k P(100-P)\) that passes through the point \((0,10)\) resembles the solution to the differential equa- tion \(d P / d t=k P\) that passes through the point \((0,10) .\) Justify your answer.

Short answer

True, for small values of \(t\), the solution to the logistic differential equation that passes through the point (0,10) appears to resemble the solution to the exponential differential equation that passes through the point (0,10).

Step-by-step solution

Understand the Dynamics

The logisitc differential equation is given by \(d P / d t=k P(100-P)\), whereas the exponential differential equation is given by \(d P / d t=k P\). In the logistic differential equation, the population growth rate decreases as \(P\) approaches its carrying capacity \(M=100\) whereas in the exponential differential equation, population growth rate is proportional to the population size.

Solve the First Differential Equation

Separation of variables gives \(\int \frac{d P}{P(100-P)}=\int k dt\). Solving this integral, finds \( P(t) = \frac{100}{1 + 9e^{-kt}}\).

Solve the Second Differential Equation

Find the solution to the exponential differential equation \(d P / d t=k P\) to be \( P(t) = 10e^{k t}\) by simple integration.

Comparison of Both Equations at Small 't'

For small values of \(t\), \(e^{-kt}\) is very close to \(1\), thus the logistic equation solution becomes \(P(t) = \frac{100}{1 + 9}\) which is roughly \(10\), and the solution to the second differential equation at \(t=0\) is \(P(0) = 10e^{k . 0} = 10\).

Key concepts

Separation of Variables

To solve a differential equation like the logistic differential equation, a method called separation of variables can be used. This technique is pivotal to understanding and solving ordinary differential equations, especially when we can't just simply integrate both sides as a single entity. The process involves rearranging the equation so that all the terms with the variable we are interested in (in this case, population, denoted by P) are on one side, and all the terms with the independent variable (usually time, t) are on the other.

The logistic differential equation is typically written as \( \frac{dP}{dt} = kP(M - P) \), where M is the carrying capacity. By using separation of variables, we rewrite it to isolate P and t, resulting in \( \frac{dP}{P(M - P)} = k dt \). Integrating both sides separately takes us a step closer to finding the population as a function of time, P(t). The beauty of this method is its broad applicability to a range of problems beyond population dynamics, making it an essential tool in any mathematical toolbox.

Exponential Growth

The concept of exponential growth is central to understanding how populations increase when resources are unlimited. It's characterized by the rate of growth being proportional to the current size of the population. This can be expressed by the differential equation \(\frac{dP}{dt} = kP \), where P is the population size, t is time, and k is a constant that represents the growth rate.

Exponential growth is initially very similar to logistic growth, as can be observed by looking at solutions to both equations for small values of t. For a very short time after the start, the limiting effects of resources on population growth aren't yet significant, and the population grows approximately exponentially. Mathematically, the solution to the exponential growth differential equation is \( P(t) = P_0e^{kt} \), where \( P_0 \) is the initial population size. The graph of exponential growth skyrockets upwards as time increases, which is unrealistic in nature due to resource limitations, leading us to the concept of carrying capacity.

Carrying Capacity

In population dynamics, carrying capacity refers to the maximum number of individuals of a species that an environment can sustain indefinitely given the available resources such as food, water, and living space. This concept is what differentiates logistic growth from exponential growth. In the real world, exponential growth cannot continue indefinitely because resources are limited.

In the logistic differential equation, \(\frac{dP}{dt} = kP(M - P) \), M represents the carrying capacity, set at 100 in the given exercise. As the population P approaches M, the growth rate slows down and eventually stops when the population reaches the carrying capacity. This creates an S-shaped curve, unlike the J-shaped curve of exponential growth. The interesting aspect of carrying capacity is that it provides a realistic model for population growth and its inherent constraints. The concept shows us that the real-world growth of populations is moderated by the environment and resources, eventually leading to a stable equilibrium.