Problem 55 - Algebra A Combined Function

Question

The formula \(y=y_{0} e^{k t}\) gives the population size y of a population that experiences a relative growth rate \(k(k\) is positive if growth is increasing and \(k\) is negative if growth is decreasing). In this formula, \(t\) is time in years and \(y_{0}\) is the initial population at time \(0 .\) Use this formula to solve Exercises 55 and \(56 .\) Round answers to the nearest year. (Source for data: U.S. Census Bureau and Federal Reserve Bank of Chicago) In \(2009,\) the population of Michigan was approximately 9,970,000 and decreasing according to the formula \(y=y_{0} e^{-0.003 t}\). Assume that the population continues to decrease according to the given formula and predict how many years after which the population of Michigan will be \(9,500,000 .\) (Hint: Let \(y_{0}=9,970,000 ; y=9,500,000\), and solve for \(t\).)

Step-by-Step Solution

Answer

The population will decrease to 9,500,000 after about 16 years.

Step 1: Identify variables

First, identify the variables in the given formula. We have \( y_0 = 9,970,000 \), \( y = 9,500,000 \), and the relative growth rate \( k = -0.003 \). We need to solve for \( t \), which will determine the time in years.

Step 2: Write down the equation

Substitute the given values into the formula: \[ 9,500,000 = 9,970,000 \, e^{-0.003 \, t} \] This equation represents the population decreasing over time.

Step 3: Isolate the exponential term

To isolate the exponential term, divide both sides of the equation by the initial population \( y_0 = 9,970,000 \):\[ \frac{9,500,000}{9,970,000} = e^{-0.003 \, t} \] Calculate the left side of the equation.

Step 4: Calculate the fraction

Perform the division: \[ \frac{9,500,000}{9,970,000} \approx 0.953 \]Now we have:\[ 0.953 = e^{-0.003 \, t} \]

Step 5: Apply the natural logarithm

To solve for \( t \), take the natural logarithm (ln) of both sides of the equation:\[ \ln(0.953) = -0.003 \, t \] This step will allow us to solve for \( t \).

Step 6: Solve for t

Simplify the equation:\[ t = \frac{\ln(0.953)}{-0.003} \] Use a calculator to compute the natural logarithm and divide by \(-0.003\).

Step 7: Compute the final answer

Calculate the value:\[ t = \frac{-0.04879}{-0.003} \approx 16.26 \]Rounding to the nearest year, \( t \approx 16 \).

Key Concepts

Population Modeling

Population modeling is a mathematical method used to estimate how a population grows or declines over time. It helps us make informed predictions and decisions based on available data. In this context, the model uses the formula \( y = y_0 e^{kt} \) to represent changes in population size.
Here’s how it works:

• \(y\) represents the future population that we want to predict.
• \(y_0\) is the initial population size at the starting point (time \(t = 0\)).
• \(k\) is the relative growth rate, indicating whether the population is increasing or decreasing over time.
• \(t\) stands for time in years, showing how far into the future or past you are predicting.

In our exercise, Michigan's population is modeled using a decreasing growth rate, where \( k = -0.003 \). This is shown by the negative sign, indicating a year-over-year decline. By setting the equation to the desired population number, students can solve for \( t \), which tells us how many years it will take to reach that population size.

Relative Growth Rate

The relative growth rate, denoted as \(k\) in our modeling formula, is a key factor in understanding how a population changes over time. It gives us a mathematical way to describe whether the population is growing or shrinking.

• If \(k\) is a positive number, the population is increasing.
• If \(k\) is negative, the population is decreasing as in our example with Michigan.

The relative growth rate can be thought of as the percentage change in the population per year. In our specific case, \(k = -0.003\) shows a slight annual decrease in the population size, meaning the population loses about 0.3% each year.
Understanding \(k\) is crucial because it allows us to predict future population sizes. By substituting different values, we can see how changes in \(k\) affect the whole model, giving us a versatile tool to plan and predict future scenarios.

Natural Logarithms

Natural logarithms (ln) are a fundamental part of solving exponential growth and decay problems. Taking the natural log is often necessary to isolate the variable we need, especially when solving for time \(t\).
In the exercise, once we have \( e^{-0.003t} = 0.953 \), taking \(\ln\) on both sides helps to solve for \(t\):

• The natural logarithm of \(e^x\) is simply \(x\). Therefore, \(\ln(e^{-0.003t}) = -0.003t\).
• This step simplifies the equation and removes the exponential, making it easier to solve for the unknown time \(t\).
• Applying the natural logarithm to \(0.953\) helps us compute the exact years needed for Michigan's population to reach the target size.

Using natural logarithms is an essential mathematical tool, especially in exponential functions. It provides a straightforward way to break down the complexity of these expressions, making the math simpler and the predictions clearer.