Problem 59 - Calculus for Biology and Medicine

Question

Population Growth Suppose that the population size at time \(t\) is $$ N(t)=e^{2 t}, \quad t \geq 0 $$ (a) What is the population size at time 0 ? (b) Show that $$ \frac{d N}{d t}=2 N $$

Step-by-Step Solution

Answer

(a) Population size at time 0 is 1. (b) \(\frac{dN}{dt} = 2N\) is verified by differentiation.

Step 1: Evaluate Population Size at Time 0

To find the population size at time 0, substitute \(t = 0\) into the population function \(N(t) = e^{2t}\). This gives:\[N(0) = e^{2 imes 0} = e^0\]Since \(e^0 = 1\), the population size at time 0 is 1.

Step 2: Differentiate the Population Function

To find \(\frac{dN}{dt}\), differentiate the function \(N(t) = e^{2t}\) with respect to \(t\). Use the chain rule, where the derivative of \(e^{2t}\) is the original exponent derivative \(2\) times \(e^{2t}\):\[\frac{dN}{dt} = 2e^{2t}\]

Step 3: Verify the Differential Equation

We are asked to show \(\frac{dN}{dt} = 2N\). From Step 2, we found that \(\frac{dN}{dt} = 2e^{2t}\). Substituting \(N(t) = e^{2t}\) into \(2N\), we get:\[2N = 2(e^{2t}) = 2e^{2t}\]Since both sides are equal, \(\frac{dN}{dt} = 2N\) is true.

Key Concepts

Understanding Population Size

Population size in the context of calculus, especially within biological studies, refers to the number of individuals in a population at a given time. In this exercise, we are given a function for population size:

• Function: \(N(t) = e^{2t}\)
• Where \(t\) represents time in years.

To determine the initial population size, we substitute \(t = 0\) into the function:

• \(N(0) = e^{2 \times 0} = e^0\)
• Since \(e^0 = 1\), the population size at time 0 is 1.

This simple calculation reveals that our initial population, when time starts ticking (at \(t=0\)), consists of only one individual. This knowledge is essential for understanding how the population evolves over time, especially in predictive models of biological populations.

The Art of Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate at which a quantity changes. When studying populations, differentiation helps us understand the rate of change of the population size over time. To differentiate the function \(N(t) = e^{2t}\), we apply the chain rule, a method that deals with functions composed of other functions:

• First, find the derivative of the exponent, \(2t\), which is \(2\).
• Next, multiply this derivative by the original exponential function, \(e^{2t}\).

Thus, the derivative of \(N(t)\) is given by:

• \(\frac{dN}{dt} = 2e^{2t}\)

Differentiation provides insights into the dynamics of population growth, allowing biologists to explore how quickly a population is growing at any given point in time.

Exponential Function and Population Growth

An exponential function, like \(N(t) = e^{2t}\), is characterized by its constant relative growth rate. Exponential functions are crucial in modeling biological populations because they reflect rapid, uncontrolled growth under ideal conditions. Let's break down the key components:

• The base, \(e\), is an important mathematical constant approximately equal to 2.718. It's naturally occurring in continuous growth processes.
• The exponent \(2t\) indicates growth at a rate proportional to \(2\) multiple times per time unit \(t\).

Such a function implies that the population doubles approximately every unit time, illustrating an ideal scenario of population expansion. To verify the relationship \(\frac{dN}{dt} = 2N\),

• Substitute \(N(t) = e^{2t}\) back into \(2N\),
• Resulting in \(2 \cdot e^{2t} = 2e^{2t}\).

Understanding exponential functions equips students with the tools to assess real-world biological systems and predict future population trends.