Problem 64 - Algebra 2

Question

In 2000 , there were about 300 million Internet users. That number is projected to grow to 1 billion in \(2005 .\) a. Let \(t\) represent the time, in years, since \(2000 .\) Write a function of the form \(y=a e^{c t}\) that models the expected growth in the population of Internet users. b. In what year might there be 500 million Internet users? c. In what year might there be 1.5 billion Internet users? d. Solve your equation for \(t .\) e. Writing Explain how you can use your equation from part (d) to verify your answers to parts \((\mathrm{b})\) and \((\mathrm{c}) .\)

Step-by-Step Solution

Answer

The function that models the expected growth in the population of Internet users is \(y = 300 e^{0.1192t}\), where \(t\) is the time since the year 2000. There might be 500 million Internet users in the year 2003 (rounding up from \(t=3.08\)) and there could be 1.5 billion Internet users in the year 2008 (rounding up from \(t=7.73\)). The equation for \(t\) to verify predictions is \(t = ln(\(y/300\))/0.1192.

Step 1: Find the values of a and c

First, we substitute \(t=0\) into the formula representing the year 2000 when there were 300 million users and get \(a=300\) million. Next, we substitute \(t=5\) representing the year 2005 and number of internet users \(y=1000\) million (or 1 billion) into the formula to get an equation 1000=300e\(^{5c}\) and solve for \(c\) to find \(c = ln(\(1000/300\))/5 = 0.1192\).

Step 2: Predict the year when there might be 500 million Internet users

Substitute \(y=500\) into the formula \(500=300e\(^{0.1192t}\)\) to solve for \(t\) to get: \(t = ln(\(500/300\))/0.1192 = 3.08\). Therefore, there might be 500 million Internet users in 2003.

Step 3: Predict the year when there might be 1.5 billion Internet users

Again, substitute \(y=1500\) into the formula \(1500=300e\(^{0.1192t}\)\) to solve for \(t\) to get: \(t = ln(\(1500/300\))/0.1192 = 7.73\). Therefore, there might be 1.5 billion Internet users in 2008.

Step 4: Solve the equation for \(t\)

Rearrange the equation \(y = 300 e^{0.1192t}\) to solve for \(t\): \(t = ln(\(y/300\))/0.1192.

Step 5: Use the equation from Step 4 to verify predictions

Substitute the number of Internet users into the equation from step 4, for example, for \(y=500\) millon users, we have \(t = ln(\(500/300\))/ 0.1192 = 3.08, which is the prediction made in step 2, and therefore, confirms our previous predictions.

Key Concepts

Function of the form y=ae^{ct}

Exponential growth modeling is a powerful tool to understand how populations or quantities grow over time. In the given problem, we use a function of the form \( y = ae^{ct} \) to model the growth of Internet users from 2000 onwards. Here:

• \( y \) represents the number of Internet users.
• \( a \) is the initial amount of users at \( t = 0 \) (the year 2000), which is 300 million in this case.
• \( e \) is the base of natural logarithms, approximately equal to 2.718.
• \( c \) is the growth rate constant.
• \( t \) refers to time in years since 2000.

By substituting known values into the equation at specific times, we can solve for \( c \) and make predictions about Internet user growth in future years. It’s essential to have accurate values for \( a \) and \( c \) to make reliable predictions.

Predicting Future Values

With our equation \( y = 300e^{0.1192t} \), we can predict the number of Internet users for various future dates. This equation helps in finding when a particular number of users will be reached. For instance:

• To find when there might be 500 million users, solve \( 500 = 300e^{0.1192t} \).
• By taking the natural logarithm of both sides, \( t \) can be solved, giving approximately 3.08 years from 2000, predicting 2003 as the year.

Such predictions revealed in the example allow us to make informed decisions about technological development and market strategies. By substituting different values of \( y \), it’s possible to extend these predictions further into the future.

Solving Exponential Equations for Variables

Solving equations like \( y = 300e^{0.1192t} \) for the variable \( t \) involves using logarithms. Here's a step-by-step explanation:

• First, divide both sides of the equation by the initial amount \( 300 \).
• Then take the natural logarithm (ln) of both sides to eliminate the base \( e \).
• Rearrange the equation to solve for \( t \): \( t = \frac{\ln(y/300)}{0.1192} \).

This equation lets you substitute any number of Internet users to determine \( t \), indicating how many years since 2000 that population will be reached. This approach is crucial for making accurate future forecasts using exponential models.

Verification of Mathematical Predictions

Verification is an integral part of mathematical modeling, ensuring that our predictions are reliable. After deriving the equation \( t = \frac{\ln(y/300)}{0.1192} \), it’s crucial to validate its accuracy.

• Plug back different \( y \) values, such as 500 million from Step 2 or 1.5 billion from Step 3, into our equation.
• Calculate \( t \) and see if it matches the previously predicted year (e.g., 2003 for 500 million users).
• Such confirmation strengthens the confidence in our model's credibility.

Repeated verification not only corroborates results but also exposes any potential calculation errors or model limitations, promoting more refined and robust predictions.