Problem 15 - Calculus Concepts

Question

Airline Fuel The amount of airline fuel consumed by Southwest Airlines each year between 2004 and 2008 can be modeled as $$ f(t)=-0.009 t^{2}+0.12 t+1.19 \text { billion gallons } $$ where \(t\) is the number of years since 2004 (Source: Based on data from Bureau of Transportation Statistics) a. Calculate the amount of fuel consumed in \(2007 .\) b. Use the algebraic method to develop a formula for the derivative of \(f\) c. How quickly was the amount of fuel used by Southwest Airlines changing in \(2007 ?\) Interpret the result.

Step-by-Step Solution

Answer

a. 1.469 billion gallons. b. \( f'(t) = -0.018t + 0.12 \). c. Increasing by 0.066 billion gallons/year in 2007.

Step 1: Identify the Year for Calculation

Given that \( t \) is the number of years since 2004, we first calculate \( t \) for the year 2007. Since 2007 is three years after 2004, we have \( t = 3 \).

Step 2: Substitute the Value of t into f(t)

The function for fuel consumption is given by \( f(t) = -0.009t^2 + 0.12t + 1.19 \). Substitute \( t = 3 \) into this equation to find the fuel consumed in 2007:\[f(3) = -0.009(3)^2 + 0.12(3) + 1.19\]Simplify to get:\[f(3) = -0.009(9) + 0.36 + 1.19 = -0.081 + 0.36 + 1.19 = 1.469 \text{ billion gallons}\]

Step 3: Differentiate the Function to Find f'(t)

To find the rate of change of the fuel consumption, we differentiate \( f(t) \) with respect to \( t \). If \( f(t) = -0.009t^2 + 0.12t + 1.19 \), then:\[f'(t) = \frac{d}{dt}(-0.009t^2) + \frac{d}{dt}(0.12t) + \frac{d}{dt}(1.19)\]This simplifies to:\[f'(t) = -0.018t + 0.12\]

Step 4: Evaluate f'(t) at t = 3

Now, substitute \( t = 3 \) into the derivative \( f'(t) \) to find the rate of change of fuel consumption in 2007:\[f'(3) = -0.018(3) + 0.12 = -0.054 + 0.12 = 0.066\]This result indicates that the amount of fuel consumed was increasing by \( 0.066 \) billion gallons per year in 2007.

Key Concepts

Derivatives

When we talk about derivatives in calculus, we refer to the tool that helps us find the rate at which something changes. Imagine you're watching a car speed up. The rate at which it speeds up is akin to what a derivative does, but with mathematical functions instead of cars.

For a function like our airline fuel consumption model, which is \( f(t) = -0.009t^2 + 0.12t + 1.19 \), the derivative, denoted \( f'(t) \), tells us how the fuel consumption is changing over time. It's like capturing the speed of change. When we differentiated the function, the result \( f'(t) = -0.018t + 0.12 \) gives us a new function that we can use to calculate the exact rate of change at any year \( t \).

To differentiate each term:

• The derivative of \(-0.009t^2\) becomes \(-2 \cdot 0.009t = -0.018t\).
• The derivative of \(0.12t\) is constant \(0.12\).
• The derivative of the constant \(1.19\) is zero, because constants don't change.

This process shows why derivatives are powerful. They reveal the underlying "movement" or "trend" within functions.

Functions

Functions are like machines, but for numbers. You put a number in, follow the function, and you get a number out. In math, a function is typically expressed as \( f(t) \), where \( t \) is your input. For our specific example of airline fuel consumption, the function \( f(t) = -0.009t^2 + 0.12t + 1.19 \) helped us model the fuel used each year.

Every term in the function plays a role:

• The \(-0.009t^2\) part represents how fuel usage might be increasing or decreasing over time in a non-linear fashion.
• The \(0.12t\) indicates the linear trend, showing either a steady rise or fall in fuel usage as years go by.
• The constant \(1.19\) represents a baseline fuel consumption, unaffected by the number of years passed.

Using functions, we can model real-world scenarios like our airline's annual fuel consumption over different years. By substituting specific values of \( t \) into the function, like \( t = 3 \) for the year 2007, we calculated the total fuel consumed that year.

Rate of Change

The rate of change is all about how fast something is increasing or decreasing over time. In our scenario, we are interested in how the fuel consumption of Southwest Airlines changes from one year to the next. It's similar to how you might wonder how quickly your garden grows or how fast your savings add up.

By evaluating the derivative \( f'(t) \) at \( t = 3 \), we found \( f'(3) = 0.066 \). This tells us that in the year 2007, the fuel consumption by Southwest Airlines was increasing by 0.066 billion gallons per year. This positive figure means the fuel usage was rising during that specific year.

Understanding the rate of change is crucial. It helps airlines and other businesses strategize, forecast, and plan for future needs, ensuring they're prepared for yearly changes in resources, costs, and logistics. The derivative function and the rate it provides make complex data digestible and actionable for decision-makers. This clarity is powerful in many fields, not just mathematics.