Problem 11 - Calculus and its applications

Question

Bottled water sales. Since \(2000,\) sales of bottled water have increased at the rate of approximately \(9.3 \%\) per year. That is, the volume of bottled water sold, \(G,\) in billions of gallons, \(t\) years after 2000 is growing at the rate given by \(\frac{d G}{d t}=0.093 G\). a) Find the function that satisfies the equation, given that approximately 4.7 billion gallons of bottled water were sold in 2000 . b) Predict the number of gallons of water sold in 2025 . c) What is the doubling time for \(G(t)\) ?

Step-by-Step Solution

Answer

a) \( G(t) = 4.7e^{0.093t} \), b) Approximately 48.08 billion gallons, c) Doubling time is about 7.45 years.

Step 1: Find the General Solution to the Differential Equation

Given the equation \( \frac{dG}{dt} = 0.093G \), this is a first-order differential equation which can be solved by separation of variables.First, separate the variables:\[ \frac{1}{G}dG = 0.093 dt \]Next, integrate both sides:\[ \int \frac{1}{G}dG = \int 0.093 dt \]\[ \ln|G| = 0.093t + C \] Solving for \(G(t)\), we take the exponential of both sides:\[ G(t) = e^{0.093t + C} = Ce^{0.093t} \] Where \(C = e^C\) is the new constant of integration.

Step 2: Apply Initial Condition to Find Particular Solution

From the problem, we know that in the year 2000 (i.e., \(t = 0\)), the volume of bottled water sold is \(4.7\) billion gallons. Substitute these values into the particular solution equation:\[ G(0) = Ce^{0.093 \times 0} = C = 4.7 \]Therefore, the particular solution is:\[ G(t) = 4.7e^{0.093t} \]

Step 3: Predict the Gallons of Water Sold in 2025

To predict sales for 2025, substitute \(t = 25\) years (since 2025 is 25 years after 2000) into the equation \(G(t) = 4.7e^{0.093t}\):\[ G(25) = 4.7e^{0.093 \times 25} \]Calculate the exponent:\[ 0.093 \times 25 = 2.325 \]Now, compute:\[ G(25) = 4.7e^{2.325} \]Using the exponential approximation:\[ G(25) \approx 4.7 \times 10.23 \approx 48.08 \]So, approximately 48.08 billion gallons of bottled water will be sold in 2025.

Step 4: Calculate the Doubling Time for the Function

The doubling time formula using exponential growth is:\[ T = \frac{\ln 2}{r} \]Where \(r = 0.093\) is the growth rate. Calculate the doubling time:\[ T = \frac{\ln 2}{0.093} \]\(\ln 2 \approx 0.693\), thus:\[ T = \frac{0.693}{0.093} \approx 7.45 \]Therefore, the doubling time for \(G(t)\) is approximately 7.45 years.

Key Concepts

Differential Equations

Differential equations represent mathematical equations that involve functions and their derivatives. In this case, we are looking at a differential equation that models exponential growth. The rate of growth of sales of bottled water is directly proportional to the current volume of sales. This is encapsulated in the equation \( \frac{dG}{dt} = 0.093G \). It means that the change in the volume of bottled water sold over time \( t \), denoted as \( \frac{dG}{dt} \), is 9.3% of the current volume \( G \). By tackling such problems, we can predict future growth and comprehend complex relationships in real-world scenarios like sales growth over time.
To solve the differential equation given, we use a technique called separation of variables, which allows us to isolate \( G \) on one side of the equation and \( t \) on the other. This method helps us in finding a general solution, which is a crucial first step before applying specific conditions given in a problem.

Integration Techniques

Integration techniques are essential when dealing with differential equations. After separating the variables in our equation, we integrate both sides to find \( G(t) \). For this problem, you integrate \( \int \frac{1}{G} \, dG = \int 0.093 \, dt \). On the left side, this results in \( \ln |G| \) and on the right, it translates to \( 0.093t + C \), where \( C \) is the integration constant.
By solving for \( G(t) \), we take the exponential of both sides which gives us \( G(t) = Ce^{0.093t} \). This solution represents the exponential growth equation in its general form. The constant \( C \) can be determined through initial conditions, such as knowing the amount of sales at a particular time, in this case, the year 2000.

• Application of initial conditions: Sets a particular solution relevant to the context
• Exponentiating to solve: Ensures the solution represents the original growth model

This method of integration is a versatile technique you will encounter frequently in calculus, especially when solving real-world growth models.

Doubling Time Calculation

Doubling time is a concept used to determine how long it will take for a quantity growing exponentially to double in size. It's especially useful when analyzing continuous growth problems like this one. The formula for calculating doubling time is \( T = \frac{\ln 2}{r} \), where \( r \) is the growth rate.
For the bottled water sales problem, substituting the growth rate \( 0.093 \) into the formula gives:\[ T = \frac{0.693}{0.093} \approx 7.45 \text{ years} \]
This result means that every 7.45 years, the sales of bottled water will double if the growth rate remains constant. Understanding this concept is crucial for businesses and economists when planning for future demand and resources. It helps in making informed decisions regarding investments, resource allocation, and understanding market dynamics.