Problem 103 - College Algebra Essentials

Question

The formula \(A=37.3 e^{0.0095 t}\) models the population of California, \(A,\) in millions, \(t\) years after 2010 . a. What was the population of California in 2010 ? b. When will the population of California reach 40 million?

Step-by-Step Solution

Answer

a. The population of California in 2010 was 37.3 million. b. The population of California will reach 40 million in around the year 2028.

Step 1: Find the initial population

To find the population of California in 2010, we need to plug \(t=0\) into the given equation \(A=37.3 e^{0.0095 t}\). This is because \(t\) years represents the years after 2010, therefore, at 2010, \(t=0\).

Step 2: Calculate the initial population

If we substitute \(t=0\) into the equation, then the equation becomes \(A=37.3 e^{0.0095 * 0}\), which simplifies to \(A = 37.3\). Therefore, the population of California in 2010 was 37.3 million.

Step 3: Find the time when the population will reach 40 million

Now, we have to find the time \(t\) when the population will be 40 million. This means, we have to set \(A = 40\) in the given equation, and solve for \(t\). The equation now is \(40 = 37.3 e^{0.0095 t}\).

Step 4: Solve for t

To isolate \(t\), we first divide both sides of the equation by 37.3. This gives us \(\frac{40}{37.3} = e^{0.0095 t}\). Next, we apply the natural logarithm to each side, enabling us to remove the exponent on the right side. The equation becomes \(ln(\frac{40}{37.3}) = 0.0095 t\). Finally, to find \(t\), we divide both sides of the equation by 0.0095, solving for \(t\).

Step 5: Calculate the value of t

If we do the calculation in step 4, dividing \(ln(\frac{40}{37.3})\) by 0.0095, the result is approximately \(t = 18.26\) years. This indicates that the population of California will reach 40 million about 18.26 years after 2010, which corresponds to the year 2028 when rounded to the nearest whole year.

Key Concepts

Continuous Compound Growth

Continuous compound growth is a concept that describes how a quantity grows over time when it is being continuously compounded. This type of growth contrasts with simple interest growth, where interest is added at specific intervals, such as annually or monthly. In continuous compounding, the growth happens at an infinitesimally small rate, causing the exponential increase to be smoother and faster.

In mathematical terms, continuous compound growth is often modeled using the natural exponential function, represented by the equation:

• If a quantity grows continually at a rate of \(r\) over \(t\) time units, the future value \(A\) is given by \(A = A_0 e^{rt}\), where \(A_0\) is the initial amount and \(e\) is Euler's number (approximately 2.71828).

This equation forms the basis for many real-world calculations, including population growth, investment growth, and decay processes such as radioactive decay.

In our example, the formula \(A = 37.3 e^{0.0095 t}\) models the population growth of California with continuous compounding. Here, 0.0095 represents the continuous growth rate per year, and \(t\) represents the number of years since 2010.

Exponential Function

An exponential function is a mathematical expression wherein a constant base is raised to a variable exponent. The standard form of an exponential function is \(f(x) = a \, b^x\), where \(a\) is the initial amount and \(b\) is the base, corresponding to the growth or decay factor.

A key characteristic of exponential functions is that they exhibit rapid change. When the base \(b > 1\), the function represents exponential growth, while \(0 < b < 1\) indicates exponential decay.

In our example, the exponential function used is \(A = 37.3 e^{0.0095 t}\). The base here is Euler's number \(e\), which is utilized due to its natural properties in calculations involving continuous growth. Exponential functions like this are frequently applied in fields such as biology, finance, and demography to model real-life processes that grow or shrink at rates proportional to their current size.

Problem Solving with Exponential Models

Problem-solving with exponential models involves identifying the exponential relationship that accurately describes the situation at hand, and then using algebraic techniques to find unknown quantities.

When solving problems involving exponential growth and decay:

• Begin by identifying the variables and constants in the exponential function. In our problem, these include the initial population (37.3 million) and the growth rate (0.0095).
• Use the given information to set up equations. For instance, to find the population at a specific time, substitute the time value \(t\) into the equation.
• To find the time when the population reaches a certain amount, set the population \(A\) to the desired value, and solve for \(t\) using algebraic manipulation such as logarithms.

By following these steps, one can find solutions to real-world problems effectively, as demonstrated in determining when California's population will reach 40 million. The process involves calculations such as using the natural log function to unravel the exponent, ultimately solving for the time variable \(t\). This systematic approach ensures accurate predictions and insights based on exponential models.