Problem 59 - Calculus for Biology and Medicine

Question

Population Growth Suppose that the population size at time \(\underline{t}\) 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) The population at time 0 is 1; (b) \( \frac{dN}{dt} = 2N \) is shown by differentiation.

Step 1: Evaluate Population at Time 0

To find the population size at time 0, substitute \( t = 0 \) into the function \( N(t) = e^{2t} \). This gives us \( N(0) = e^{2 \times 0} = e^0 = 1 \). Therefore, the population size at time 0 is 1.

Step 2: Differentiate the Population Function

We need to find \( \frac{dN}{dt} \) by differentiating \( N(t) = e^{2t} \) with respect to \( t \). Using the chain rule, \( \frac{d}{dt}(e^{2t}) = 2e^{2t} \).

Step 3: Verify the Relationship

We are given to show that \( \frac{dN}{dt} = 2N \). From Step 2, we found \( \frac{dN}{dt} = 2e^{2t} \). Since \( N(t) = e^{2t} \), it follows that \( \frac{dN}{dt} = 2N \). Thus, the relationship is verified.

Key Concepts

Understanding Exponential Functions

Exponential functions are mathematical expressions where the variable is in the exponent. In simpler terms, these functions involve constant growth rates that multiply quickly. The general form of an exponential function is \( y = a \, e^{bx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718, and \( a \) and \( b \) are constants. One key feature of exponential functions is that as time progresses, changes occur at increasingly rapid rates. This is why such functions are particularly relevant in describing processes like population growth.
For example, in biology, exponential growth describes how populations of organisms grow under ideal conditions. The model reflects a continuous, increasing rate proportionate to the number of entities present. Hence, a population growing according to \( N(t) = e^{2t} \) means that each unit of time will see the population doubling, given that the natural growth rate doubles every time \( t \) increases. Hence, understanding exponential functions helps us predict long-term population behaviors that start small and increase rapidly.

Exploring Differentiation

Differentiation is the process of finding the derivative of a function. In simple terms, it measures how a function changes at any given point. This is crucial for understanding rates of change, which is why differentiation is a fundamental concept in calculus. The derivative of a function at a certain point gives us the slope of the tangent line to the curve at that point. When we differentiate \( N(t) = e^{2t} \), we're looking to find \( \, \frac{dN}{dt} \, \), which represents the rate of population growth at any time \( t \). Differentiation helps us translate the concept of continuous change over time into a manageable formula, which in this problem reads as \( 2e^{2t} \). This outcome shows us not just change, but how quickly the population grows at each instant, linking directly back to our understanding of exponential functions.

Applying the Chain Rule

The chain rule is a technique used in calculus for differentiating compositions of functions. It is applied when a function is "nested" within another function. In our problem, the chain rule is essential because \( N(t) = e^{2t} \) is an exponential function composed with a linear function \( 2t \). The chain rule formula is \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \). Here:

• The outer function \( f(x) = e^{x} \) has a derivative \( f'(x) = e^{x} \).
• The inner function \( g(t) = 2t \) has a derivative \( g'(t) = 2 \).

Therefore, for \( N(t) = e^{2t} \), applying the chain rule gives: \[ \frac{d}{dt}[e^{2t}] = e^{2t} \cdot 2 = 2e^{2t} \] This illustrates why the population growth rate as time progresses, \( \frac{dN}{dt} = 2N \), is true, confirming our earlier steps. The chain rule not only helps us differentiate complex functions but also connects different aspects of mathematical modeling in the context of real-world problems.