Problem 61 - College Algebra with Modeling and Visualization

Question

The distance \(D\) in feet that an object has fallen after \(t\) seconds is given by \(D(t)=16 t^{2}\) (a) Bvaluate \(D(2)\) and \(D(4)\) (b) Calculate the average rate of change of \(D\) from 2 to \(4 .\) Interpret the result.

Step-by-Step Solution

Answer

D(2) is 64 feet, D(4) is 256 feet, and the average rate of change is 96 feet per second.

Step 1: Evaluate D(2)

To find how far the object falls in 2 seconds, substitute \(t = 2\) into the function: \[D(2) = 16(2)^2 = 16 \times 4 = 64 \; \text{feet}.\]

Step 2: Evaluate D(4)

To find how far the object falls in 4 seconds, substitute \(t = 4\) into the function: \[D(4) = 16(4)^2 = 16 \times 16 = 256 \; \text{feet}.\]

Step 3: Calculate the Average Rate of Change

The average rate of change from \(t = 2\) to \(t = 4\) is given by the formula: \(\text{Average Rate of Change} = \frac{D(4) - D(2)}{4 - 2}\). Calculate it: \[\text{Average Rate of Change} = \frac{256 - 64}{4 - 2} = \frac{192}{2} = 96 \; \text{feet per second}.\]

Step 4: Interpret the Result

The average rate of change of 96 feet per second represents the average speed at which the object falls between 2 seconds and 4 seconds. This means that, on average, the object is falling at a speed of 96 feet per second during this time interval.

Key Concepts

Quadratic Functions

Quadratic functions are an essential part of algebra and many applications in the real world. A quadratic function is usually written in the form of \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and the highest power of \( x \) is a square (hence, "quadratic"). In our specific function, \( D(t) = 16t^2 \), the term \( a = 16 \), and there are no \( b \) or \( c \) values, making it a simple, pure quadratic function.
Quadratic functions often describe parabolic shapes when graphed, which can either open upwards or downwards. In the context of physics, such as the problem we're exploring, they are often used to portray the motion of objects under the influence of gravity (like falling objects).
Key features of quadratic functions include:

• **Vertex**: The highest or lowest point on the graph, depending on whether it opens upwards or downwards.
• **Axis of Symmetry**: A vertical line through the vertex, around which the parabola is symmetric.
• **Graph Shape**: A parabola that opens upwards if \( a > 0 \) and downwards if \( a < 0 \).

Understanding these functions helps in predicting the path of objects in motion and analyzing their properties.

Rate of Change

The rate of change is a fundamental concept in algebra that tells us how a quantity changes relative to another quantity. In essence, it's a way to capture the difference in a function's value over a particular interval. When dealing with quadratic functions like our example \( D(t) = 16t^2 \), the rate of change can inform us about aspects such as speed and acceleration.
For any function \( f(x) \), the average rate of change between two points \( x_1 \) and \( x_2 \) is calculated as:
\[ \text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]
This formula essentially gives us the slope of the line connecting these two points on the function's graph. It's particularly useful for quadratic functions, as their rates of change are not constant, they increase or decrease depending on the position along the curve. In our problem, the average rate of change is 96 feet per second between 2 and 4 seconds.
This tells us how fast, on average, the object is falling during this time. It's crucial to interpret this number correctly to understand the dynamics of the situation.

Evaluation of Functions

Evaluating functions is a simple yet powerful tool that holds importance in numerous mathematical problems. To evaluate a function means to find its value for a given input. This process helps determine the effect of certain parameters by substituting values into the function's formula.
In the exercise, we evaluate \( D(t) \) for specific times \( t = 2 \) and \( t = 4 \). This involves substituting these values directly into \( D(t) = 16t^2 \) to find how far the object has fallen:

• For \( t = 2 \), \( D(2) = 16 \times 2^2 = 64 \) feet.
• For \( t = 4 \), \( D(4) = 16 \times 4^2 = 256 \) feet.

This step-by-step substitution allows us to accurately calculate outcomes and, in this case, understand how far an object has fallen under gravity.
Evaluating functions is central to solving real-world problems where mathematical models describe phenomena, helping us make predictions or verify known results.