Problem 12 - Single Variable Calculus

Question

On the basis of data compiled by the World Health Organization, it is estimated that the number of people living with HIV worldwide from 1985 through 2006 is $$ N(t)=\frac{39.88}{1+18.94 e^{-0.2957 t}} \quad 0 \leq t \leq 21 $$ where \(N(t)\) is measured in millions and \(t\) in years with \(t=0\) corresponding to the beginning of 1985 . a. How many people were living with HIV worldwide at the beginning of \(1985 ?\) At the beginning of \(2005 ?\) b. Assuming that the trend continued, how many people were living with HIV worldwide at the beginning of \(2008 ?\)

Step-by-Step Solution

Answer

a. At the beginning of 1985, there were 2 million people living with HIV worldwide, and at the beginning of 2005, there were approximately 37 million people living with HIV worldwide. b. Assuming the trend continued, there were approximately 39 million people living with HIV worldwide at the beginning of 2008.

Step 1: Plug in t = 0 for 1985

First, plug in the value of \(t = 0\) for the beginning of 1985 into the given equation, and solve for \(N(t)\): \[N(0) = \frac{39.88}{1 + 18.94 e^{-0.2957 \times 0}}\]

Step 2: Solve for N(t) when t = 0

Simplify the equation and solve for \(N(0)\): \[N(0) = \frac{39.88}{1 + 18.94}\] \[N(0) = 2\] So, at the beginning of 1985, there were 2 million people living with HIV worldwide.

Step 3: Plug in t = 20 for 2005

Now, plug in the value of \(t = 20\) for the beginning of 2005 into the given equation, and solve for \(N(t)\): \[N(20) = \frac{39.88}{1 + 18.94 e^{-0.2957 \times 20}}\]

Step 4: Solve for N(t) when t = 20

Simplify the equation and solve for \(N(20)\): \[N(20) = \frac{39.88}{1 + 18.94 e^{-5.9140}}\] \[N(20) ≈ 37\] At the beginning of 2005, there were approximately 37 million people living with HIV worldwide. #b. Estimating the number of people living with HIV at the beginning of 2008#

Step 1: Plug in t = 23 for 2008

Assuming the trend continued, plug in the value of \(t = 23\) for the beginning of 2008 into the given equation, and solve for \(N(t)\): \[N(23) = \frac{39.88}{1 + 18.94 e^{-0.2957 \times 23}}\]

Step 2: Solve for N(t) when t = 23

Simplify the equation and solve for \(N(23)\): \[N(23) = \frac{39.88}{1 + 18.94 e^{-6.8009}}\] \[N(23) ≈ 39\] Assuming the trend continued, there were approximately 39 million people living with HIV worldwide at the beginning of 2008.

Key Concepts

Exponential functions

Exponential functions are mathematical functions that describe growth or decay processes where the rate of change is proportional to the current value. They can model various real-world phenomena, such as population growth, radioactive decay, and interest calculations.
In an exponential function, the variable appears in an exponent. For example, a simple exponential growth function can be described as: \[ f(t) = a \cdot e^{bt} \]where:

• \( a \) is the initial amount or size,
• \( e \) is the base of natural logarithms, approximately equal to 2.718,
• \( b \) is a constant that represents the growth rate,
• \( t \) is the time variable.

In the problem, the function for the number of people living with HIV includes an exponential decay component \( e^{-0.2957t} \) within a logistic growth equation. This indicates that initially, growth is rapid, then slows down, reflecting a more realistic approach to modeling real-world population data.

Population growth modeling

Population growth modeling helps us understand and predict how populations grow over time. One commonly used model is the logistic growth model. It is more realistic than the simple exponential model as it considers the carrying capacity, which is the maximum population size that an environment can support.
A logistic growth function is generally represented as:\[N(t) = \frac{{K}}{{1 + C e^{-rt}}}\]where:

• \( N(t) \) is the population size at time \( t \),
• \( K \) is the carrying capacity (the maximum sustainable population size),
• \( C \) is a constant derived from initial conditions,
• \( r \) is the growth rate,
• \( t \) is time.

In the exercise, the given function for HIV population number follows this logistic model. The constant 39.88 represents the asymptotic limit, implying that this might be the estimated maximum number of people living with HIV the model predicts. The equation constantly checks the balance between growth and limitations reflective of real-world dynamics.

Solving equations

Solving equations is a fundamental skill in calculus and involves finding the values of variables that satisfy a given mathematical statement. In single-variable calculus, solutions often involve performing algebraic manipulations and simplifications.
When tackling equations like \[\ N(t) = \frac{39.88}{1 + 18.94 e^{-0.2957t}} \],identifying initial conditions and evaluating the function at specific points in time (e.g., \( t=0, 20, 23 \)) is crucial. Here are the main steps:

• Plug in the desired variable value (e.g., \( t=0 \) for 1985) directly into the equation.
• Simplify the resulting expression by calculating the value of the exponential part.
• Perform arithmetic operations to find \( N(t) \).

This method allows you to effectively determine the population size at different times using known data. Understanding these steps in solving equations is key to deriving accurate population models and making informed predictions.