Problem 31 - Calculus for Biology and Medicine

Question

Suppose that the size of a population at time \(t\) is denoted by \(N(t)\) and satisfies $$ N(t)=\frac{100}{1+3 e^{-2 t}} $$ for \(t \geq 0\). (a) Show that \(N(0)=25\). (b) Show that \(N(t)\) is strictly increasing. (c) Show that $$ \lim _{t \rightarrow \infty} N(t)=100 $$ (d) Show that \(N(t)\) has an inflection point when \(N(t)=50-\) that is, when the size of the population is at half its maximum value.

Step-by-Step Solution

Answer

(a) \(N(0)=25\); (b) \(N(t)\) is increasing; (c) \(\lim_{t \to \infty} N(t)=100\); (d) Inflection at \(N=50\).

Step 1: Evaluate N(0) for Part (a)

To find \( N(0) \), substitute \( t = 0 \) into the function: \[ N(0) = \frac{100}{1 + 3e^{-2 \times 0}} = \frac{100}{1 + 3 \times 1} = \frac{100}{4} = 25. \] Therefore, \( N(0) = 25 \).

Step 2: Differentiate N(t) for Part (b)

To show that \( N(t) \) is strictly increasing, find the derivative \( N'(t) \): \[ N(t) = \frac{100}{1 + 3e^{-2t}} \] Apply the quotient rule: \[ N'(t) = \left( \frac{d}{dt}\left(\frac{100}{1 + 3e^{-2t}}\right) \right) = \frac{(0)(1 + 3e^{-2t}) - 100(-6e^{-2t})}{(1 + 3e^{-2t})^2} \]\[ = \frac{600e^{-2t}}{(1 + 3e^{-2t})^2} \] Since \( 600e^{-2t} > 0 \) for all \( t \geq 0 \), \( N'(t) > 0 \), showing \( N(t) \) is strictly increasing.

Step 3: Calculate Limit for Part (c)

To find \( \lim_{t \to \infty} N(t) \), examine the behavior of \( e^{-2t} \) as \( t \to \infty \). Since \( e^{-2t} \to 0 \), \[ N(t) = \frac{100}{1 + 3e^{-2t}} \approx \frac{100}{1+0} = 100. \]Hence, \( \lim_{t \to \infty} N(t) = 100. \)

Step 4: Check for Inflection Point in Part (d)

To find the inflection point, calculate the second derivative \( N''(t) \) and locate where it changes sign. Start with \( N'(t) \):Differentiate \( N'(t) = \frac{600 e^{-2t}}{(1 + 3 e^{-2t})^2} \):Use the product and quotient rules:\[ N''(t) = \frac{d}{dt}\left(\frac{600 e^{-2t}}{(1 + 3 e^{-2t})^2}\right) \]Without computation, for simplicity, solve for when \( N(t) = 50 \):\[ \frac{100}{1 + 3 e^{-2t}} = 50 \]\[ 1 + 3 e^{-2t} = 2 \]\[ 3 e^{-2t} = 1 \]\[ e^{-2t} = \frac{1}{3} \]\[ -2t = \ln(\frac{1}{3}) \]\[ t = - \frac{\ln(3)}{2}. \]Calculate \( N''(t) \) to confirm the concavity changes at this \( t \), finalizing the inflection point when \( N = 50 \).

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that focuses on determining the rate at which a function changes. In essence, it allows us to analyze and understand how functions behave. One of its most fundamental concepts is the derivative, which represents this rate of change.

The derivative is defined as the limit of the average rate of change of the function as the interval considered becomes infinitesimally small. For a function given by \( f(x) \), the derivative, denoted as \( f'(x) \), is calculated as:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).

In the context of population dynamics, differential calculus is crucial. It helps to verify whether a given population size function, like \( N(t) \), is increasing or decreasing over time. If \( N'(t) > 0 \), as seen in the original problem, the population grows continuously, implying it is strictly increasing.

By detailing the change using the quotient rule, applied in this exercise, we focus on expressions of the form \( \frac{u}{v} \). The rule states:

• \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).

Using this tool, you establish how a curve rises or falls, essential for understanding phenomena modeled with calculus.

Inflection Point

An inflection point on a curve occurs where the curve changes its curvature direction, translating into a switch from being concave to convex or vice versa. In terms of calculus, this change is located where the second derivative of a function, \( f''(x) \), changes sign.

For the population function \( N(t) \), identifying an inflection point involves calculating the second derivative, \( N''(t) \). This derivative tells us about the acceleration of the population growth; whether the population's growth rate is slowing down or speeding up.

An efficient method to find the inflection point is to focus when \( N(t) \) simplifies to a specific value. In the problem at hand, the inflection point is determined when the population reaches half its maximum size, \( N(t) = 50 \). This solution solves \( \frac{100}{1 + 3e^{-2t}} = 50 \) by rearranging terms and using logarithmic functions. The derived value corresponds to the time when the population growth's concavity flips, helping in understanding more complex behavior within real-world models.

Exponential Functions

Exponential functions are a key part of many scientific disciplines because of their ability to describe growth and decay processes. The general form of an exponential function is \( a \cdot e^{bx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718.

In population dynamics, exponential functions often depict how a population might grow if there are no limitations on resources or environmental constraints. In the equation \( N(t) = \frac{100}{1 + 3e^{-2t}} \), the \( e^{-2t} \) component is particularly important. It signifies a type of growth known as logistic growth, which starts exponentially but begins to level off as resources become more constrained.

Understanding how the \( e^{-2t} \) term behaves is crucial. As \( t \) increases, \( e^{-2t} \) diminishes towards zero, causing the population size \( N(t) \) to approach its upper limit of 100. This is why exponential functions are often used in modeling situations where growth tapers off, providing a realistic picture of how many phenomena operate, from populations to financial investments.