Solving For Doubling Time In Exponential Regression Equations A Water Lily Example

As a student delving into the fascinating world of mathematical modeling, you've likely encountered scenarios where quantities grow exponentially. One such scenario involves the growth of a water lily population, which can be modeled using an exponential regression equation. Let's explore how to determine the doubling time of this population using the given equation and the appropriate algebraic techniques.

The Regression Equation: A Window into Exponential Growth

The regression equation provided, y = 3.915(1.106)^x, is a powerful tool for understanding the growth dynamics of the water lily population. In this equation:

• y represents the population size at a given time.
• x represents the time elapsed, typically measured in days.
• 3.915 is the initial population size, the population at time x = 0.
• 1.106 is the growth factor, indicating that the population increases by 10.6% each day.

This equation tells us that the water lily population starts at 3.915 and grows exponentially over time. The base of the exponential term, 1.106, is crucial because it determines the rate of growth. A base greater than 1 signifies exponential growth, while a base between 0 and 1 indicates exponential decay.

To find the doubling time, we need to determine how long it takes for the population to double its initial size. This means we want to find the value of x (which we'll call D for doubling time) when the population y is twice the initial population. Since the initial population is 3.915, we are looking for the time D when the population reaches 2 * 3.915 = 7.830.

Identifying the Equations to Solve for Doubling Time

The core question we aim to address is: Which two equations can we solve to find D, the number of days it takes for the water lily population to double? To answer this, we need to translate the concept of doubling time into a mathematical equation.

As discussed earlier, doubling time is the time it takes for the population to become twice its initial size. The initial population in our model is 3.915. Therefore, we are looking for the time D when the population y reaches 2 * 3.915 = 7.830. We can set up an equation to represent this situation:

• 7. 830 = 3.915(1.106)^D

This equation directly represents the scenario where the population has doubled. It states that the population size (7.830) is equal to the initial population (3.915) multiplied by the growth factor (1.106) raised to the power of the doubling time (D). Solving this equation for D will give us the number of days it takes for the population to double.

Another way to think about doubling is to consider the population at time D being twice the initial population. We can express this as:

• 2 * 3.915 = 3.915(1.106)^D
• 7. 830 = 3.915(1.106)^D

This equation is mathematically equivalent to the first one. It emphasizes the concept of doubling by explicitly stating that the population at time D is twice the initial population.

Now, let's consider the options provided in the question:

• 2 = 3.915(1.106)^D

This equation is incorrect because it sets the doubling factor (2) equal to the entire population model. It doesn't account for the initial population size of 3.915.

• 7. 830 = 3.915(1.106)^D

This equation is correct as it directly represents the doubling condition. It sets the doubled population size (7.830) equal to the population model at time D.

Therefore, the correct equation to solve for the doubling time D is:

• 7. 830 = 3.915(1.106)^D

Solving for Doubling Time: A Step-by-Step Approach

Now that we've identified the correct equation, let's delve into the process of solving for the doubling time, D. The equation we need to solve is:

• 7. 830 = 3.915(1.106)^D

To isolate the exponential term, we need to divide both sides of the equation by 3.915:

• 7. 830 / 3.915 = (1.106)^D
• 2 = (1.106)^D

This simplified equation tells us that the population has doubled (represented by the factor of 2) after D days. Now, to solve for D, we need to use logarithms. Logarithms are the inverse operation of exponentiation, and they allow us to bring the exponent down as a coefficient.

We can take the logarithm of both sides of the equation. It doesn't matter which base of logarithm we use (natural logarithm, common logarithm, etc.), as long as we use the same base on both sides. For simplicity, let's use the natural logarithm (ln):

• ln(2) = ln((1.106)^D)

Using the logarithmic property that ln(a^b) = b * ln(a), we can rewrite the right side of the equation:

• ln(2) = D * ln(1.106)

Now, to isolate D, we divide both sides of the equation by ln(1.106):

• D = ln(2) / ln(1.106)

Using a calculator, we can find the approximate values of ln(2) and ln(1.106):

• ln(2) ≈ 0.6931
• ln(1.106) ≈ 0.1004

Therefore,

• D ≈ 0.6931 / 0.1004
• D ≈ 6.90

So, it takes approximately 6.90 days for the water lily population to double.

Implications and Applications of Doubling Time

Understanding doubling time is crucial in various real-world applications, particularly in fields like biology, ecology, and finance. In the context of water lily populations, knowing the doubling time can help us:

• Predict population growth: By knowing how long it takes for the population to double, we can estimate the population size at future times.
• Manage aquatic ecosystems: Rapid growth of water lilies can sometimes lead to ecological imbalances. Understanding the growth rate can help in implementing effective management strategies.
• Model resource consumption: If water lilies are used as a resource, knowing the doubling time can help in sustainable harvesting practices.

In a broader sense, the concept of doubling time applies to any exponentially growing quantity. For instance, in finance, it helps estimate how long it takes for an investment to double at a given interest rate. In epidemiology, it's used to understand the spread of infectious diseases.

Conclusion: Mastering Exponential Growth

In conclusion, understanding exponential growth and the concept of doubling time is a valuable skill in mathematics and its applications. By correctly identifying the equations that represent the doubling condition and employing logarithmic techniques, we can accurately calculate the time it takes for a population to double. This knowledge empowers us to make predictions, manage resources, and understand the dynamics of various real-world phenomena that exhibit exponential growth.

The regression equation y = 3.915(1.106)^x provides a powerful model for understanding the growth of the water lily population. By setting up the equation 7.830 = 3.915(1.106)^D and solving for D, we can determine the doubling time. This process involves applying logarithmic properties and careful algebraic manipulation. The result, approximately 6.90 days, gives us a crucial insight into the growth rate of the water lily population and its implications for the ecosystem. Through this exercise, we not only learn how to solve mathematical problems but also gain a deeper appreciation for the power of mathematical modeling in understanding the world around us.