Problem 44 - Calculus for Biology and Medicine

Question

Assume that \(N(t)\) denotes the size of a population at time \(t\) and that \(N(t)\) satisfies the differential equation $$ \frac{d N}{d t}=r N $$ where \(r\) is a constant. (a) Find the per capita growth rate. (b) Assume that \(r<0\) and that \(N(0)=20\). Is the population size at time 1 greater than 20 or less than \(20 ?\) Explain your answer.

Step-by-Step Solution

Answer

(a) The per capita growth rate is \(r\). (b) If \(r < 0\), the population size at time 1 is less than 20.

Step 1: Understand the concept of per capita growth rate

The per capita growth rate is defined as the rate of change of the population size per individual in the population. Mathematically, it is expressed as \(\frac{1}{N}\frac{dN}{dt}\). Given the differential equation \(\frac{dN}{dt} = rN\), we divide both sides by \(N\) to find the per capita growth rate.

Step 2: Calculate the per capita growth rate

Based on the differential equation \(\frac{dN}{dt} = rN\), the per capita growth rate is \(\frac{1}{N} \cdot \frac{dN}{dt} = r\). Therefore, the per capita growth rate is the constant \(r\).

Step 3: Analyze the behavior of the population at time 1 for \(r < 0\)

Given \(N(0) = 20\) and \(r < 0\), we express the solution of the differential equation as \(N(t) = N(0)e^{rt}\). At \(t = 1\), this becomes \(N(1) = 20e^{r}\). Since \(r < 0\), \(e^{r} < 1\), which implies \(N(1) < 20\).

Step 4: Conclusion for part (b)

Since \(e^r < 1\) when \(r < 0\), the population size at time 1, \(N(1) = 20e^r\), is less than 20. This indicates the population size decreases when \(r\) is negative.

Key Concepts

Differential Equation

A differential equation is an equation that relates a function with its derivative. In other words, it connects the rate of change of a quantity to the quantity itself. This makes differential equations really important for understanding how systems change over time. For population dynamics, these equations can describe how the population size varies dependent on factors like birth and death rates.
In the provided problem, the differential equation is \( \frac{dN}{dt} = rN \). Here, \( N(t) \) is the function denoting population size, and \( \frac{dN}{dt} \) is its derivative representing the population growth rate. Meanwhile, \( r \) is a constant that determines the characteristics of the population growth—being positive indicates growth, while being negative indicates decay. This simple equation forms the foundation for modeling exponential population changes.

Per Capita Growth Rate

The per capita growth rate is a measure of how much each individual within a population contributes to the growth of the population. It provides a more comprehensive understanding of population changes by focusing on individual contributions rather than just the overall count.
To find the per capita growth rate in our example, we start with the differential equation \( \frac{dN}{dt} = rN \). By dividing both sides by \( N \), we derive the per capita growth rate as \( \frac{1}{N} \cdot \frac{dN}{dt} = r \). This statement tells us that the growth rate for each individual in the population is a constant \( r \). Thus, the per capita growth rate is directly tied to this constant. If \( r \) is positive, each individual increases the overall population size over time. Conversely, if \( r \) is negative, each individual effectively contributes to a decline in population size.

Exponential Growth

Exponential growth is a process where the rate of growth is proportional to the current size of the population. This leads to rapid increases or decreases depending on the sign of the growth rate. It can be described mathematically by the expression \( N(t) = N_0 e^{rt} \), where \( N_0 \) is the initial population size, \( e \) is the mathematical constant approximately equal to 2.718, and \( t \) is time.
In the given exercise, we have \( N(t) = 20 e^{rt} \) with \( N_0 = 20 \) and \( r<0 \). Since \( r \) is negative, the exponent \( rt \) becomes negative, leading \( e^{rt} \) to be less than 1. Therefore, the population \( N(t) \) decreases over time. For instance, at time \( t = 1 \), the population is \( N(1) = 20e^r \), demonstrating that the population will be less than the initial 20, confirming that negative \( r \) results in population decline.