Problem 94 - Calculus and its applications

Question

The price of a ticket to the Super Bowl \(t\) years after 1967 can be estimated by $$ p(t)=0.696 t^{2}-13.290 t+61.857 $$. a) Use the function to predict the price of a Super Bowl ticket in 2014 b) Find the rate of change of the ticket price with respect to the year, \(d p / d t\). c) At what rate were ticket prices changing in \(2014 ?\)

Step-by-Step Solution

Answer

In 2014, the ticket price was approximately $975.29, increasing at a rate of $52.13 per year.

Step 1: Understanding the Problem

We have a quadratic function that predicts ticket prices for the Super Bowl based on the number of years since 1967. The goal is to calculate the price in 2014, find the rate of change of the price with respect to time, and specifically find this rate for the year 2014.

Step 1: Calculate the Number of Years Since 1967

First, we calculate the value of \(t\) for the year 2014. By subtracting 1967 from 2014, we find \(t = 2014 - 1967 = 47\).

Step 2: Predict the Ticket Price in 2014

Substitute \(t = 47\) into the price function: \[ p(t) = 0.696(47)^2 - 13.290(47) + 61.857 \]Calculate each part: - \(0.696 \times (47)^2 = 0.696 \times 2209 = 1538.064\)- \(13.290 \times 47 = 624.63\)Now calculate the overall price:\[ p(47) = 1538.064 - 624.63 + 61.857 = 975.291 \]

Step 3: Find the Rate of Change Formula

The rate of change of a function can be found by differentiating the function with respect to \(t\). Using the formula:\[ p'(t) = \frac{d}{dt}(0.696 t^2 - 13.290 t + 61.857) \]Differentiate each term:- \(\frac{d}{dt}(0.696 t^2) = 1.392 t\)- \(\frac{d}{dt}(-13.290 t) = -13.290\)Thus, the derivative is:\[ p'(t) = 1.392 t - 13.290 \]

Step 4: Calculate the Rate of Change in 2014

Substitute \(t = 47\) into the derivative to find the rate of change:\[ p'(47) = 1.392(47) - 13.290 \]Calculate each part:- \(1.392 \times 47 = 65.424\)Now find the rate:\[ p'(47) = 65.424 - 13.290 = 52.134 \]

Key Concepts

Quadratic Functions

Quadratic functions are an essential concept in mathematics, often appearing in various real-world situations. In essence, a quadratic function is a polynomial function of degree two, generally having the form \[ p(t) = at^2 + bt + c \] where \(a\), \(b\), and \(c\) are constants. A key characteristic of quadratic functions is their parabolic graph shape, which can either open upwards or downwards, depending on the sign of \(a\).

When it comes to predicting or modeling real-world phenomena, quadratic functions are valuable. For instance, the function provided in the problem, \[ p(t)=0.696t^{2}-13.290t+61.857 \], stands as a quadratic function estimating Super Bowl ticket prices.

The quadratic term, \(0.696 t^2\), is responsible for the parabola's curvature, where the sign and magnitude of \(a\) influence how steep or flat the parabola is. The linear term \(-13.290 t\) shifts the vertex horizontally, while the constant term \(61.857\) translates it vertically. Understanding these attributes allows you to anticipate how the function behaves across different domains of \(t\).

• The vertex of the parabola offers crucial information, indicating either a maximum or minimum value of the function.
• Intercepts on the graph help locate where the function crosses the axes, giving more context to the function's behavior.

The calculation of ticket price in 2014 involved substituting the corresponding year value into the function, ultimately providing an estimated price.

Rate of Change

The concept of the rate of change is fundamental in calculus and helps to understand how a quantity evolves over time. In simpler terms, the rate of change measures the difference in a function's output per unit change in the input. It is particularly useful when trying to determine how quickly something is increasing or decreasing, such as the price of Super Bowl tickets in this case.

To find the rate of change for a function such as the ticket price, you would calculate the derivative. The derivative measures the instantaneous rate of change of the function at any given point. For our quadratic function, the derivative is calculated as:\[ p'(t) = 1.392 t - 13.290 \]

• The term \(1.392 t\) indicates how the rate changes with time \(t\).
• The constant \(-13.290\) provides a baseline rate of change, applicable across all instances.

By evaluating this derivative at different points, you can determine how quickly the price increases or decreases with each passing year. For instance, in 2014, substituting \(t = 47\) shows how steep the price increase is, providing an immediate understanding of pricing dynamics that year.

Differentiation

Differentiation is one of the core operations in calculus, allowing us to determine the rate of change or the slope of a curve at any given point. It's an essential tool for analyzing functions and understanding dynamic systems. In differential calculus, you use differentiation to uncover the derivative of a function.

The derivative represents how a function's value changes as the input changes. For quadratic functions like \(p(t) = 0.696 t^2 - 13.290 t + 61.857\), differentiation reveals a linear function: \[ p'(t) = 1.392 t - 13.290 \].

• Each term in the derivative corresponds to a component of the original function, modified according to differentiation rules.
• The derivative shows us the rate of change at any particular moment, allowing predictions about future behavior or understanding past trends.

By differentiating our original ticket price function, we gain the means to calculate how pricing trends evolve, which is crucial for both financial forecasts and strategic planning in many fields. In 2014, using the derivative to determine the rate of price change offered valuable insights into how quickly prices were inflating.