Problem 83 - College Algebra with Modeling and Visualization

Question

Let the distance in feet that a car travels in \(t\) seconds be given by \(d(t)=8 t^{2}\) for \(0 \leq t \leq 6\) (a) Find \(d(t+h)\) (b) Find the difference quotient for \(d\) and simplify. (c) Evaluate the difference quotient when \(t=4\) and \(h=0.05 .\) Interpret your result.

Step-by-Step Solution

Answer

The average rate of change is 64.4 feet per second at \(t=4\) to \(t=4.05\).

Step 1: Substitute to Find \(d(t+h)\)

Given the function \(d(t) = 8t^2\), substitute \(t+h\) into the function to find \(d(t+h)\). This results in:\[d(t+h) = 8(t+h)^2\]Expanding the square, we get:\[d(t+h) = 8(t^2 + 2th + h^2) = 8t^2 + 16th + 8h^2\]

Step 2: Compute the Difference Quotient

The difference quotient is given by:\[\frac{d(t+h) - d(t)}{h}\]Substitute \(d(t+h) = 8t^2 + 16th + 8h^2\) and \(d(t) = 8t^2\) into the expression:\[\frac{(8t^2 + 16th + 8h^2) - 8t^2}{h} = \frac{16th + 8h^2}{h}\]Simplify by factoring out \(h\):\[\frac{16th + 8h^2}{h} = 16t + 8h\]

Step 3: Evaluate the Difference Quotient at \(t=4\) and \(h=0.05\)

Substitute \(t = 4\) and \(h = 0.05\) into the simplified difference quotient:\[16t + 8h = 16(4) + 8(0.05)\]Calculate the result:\[= 64 + 0.4 = 64.4\]

Step 4: Interpret the Result

The difference quotient calculated, 64.4, represents the average rate of change of distance with respect to time over the interval from \(t = 4\) to \(t = 4.05\) seconds. This is the approximate speed of the car in feet per second during this interval.

Key Concepts

Average Rate of Change

The average rate of change is a fundamental concept in calculus that helps us understand how a quantity changes over time. It tells us how much a function's output changes per unit increase in the input.
For the car travel problem, the distance the car travels over time can be represented by a function, specifically a polynomial function. By finding the average rate of change, we gain insights into the car's speed over a specific time period.

• The formula used to calculate the average rate of change for a function \(f\) over an interval \([a, b]\) is given by: \[ \frac{f(b) - f(a)}{b-a} \]
• In our exercise, the difference quotient represents the average rate of change for the distance function \(d(t)\). When evaluated, it tells us the car's average speed over a tiny time interval.
• For instance, when \(t = 4\) and \(h = 0.05\), this quotient gives us the rate of change, interpreting the car's speed between 4 seconds and 4.05 seconds.

The key takeaway is that as the interval \(h\) becomes smaller, the average rate of change approaches the instantaneous rate of change, or the derivative.

Polynomial Functions

Polynomial functions are important in mathematics, especially in algebra and calculus. They are expressions involving variables raised to whole number powers, added or subtracted, and possibly multiplied by coefficients.
In our scenario, the function \(d(t) = 8t^2\) is a polynomial function.

• The function is specifically a quadratic polynomial because the highest degree of the variable \(t\) is 2.
• Polynomial functions can describe various physical and real-world phenomena, such as distances in physics problems or financial growth in economics.
• They are often chosen for modeling because they are simple yet flexible enough to provide accurate estimates and predictions.

Understanding polynomial functions allows us to utilize their properties in mathematical operations, such as differentiation and integration, broadening our ability to analyze more complex systems.

Algebraic Expressions

Algebraic expressions play a crucial role in forming and simplifying mathematical models like polynomial functions. They consist of numbers, variables, and operators (like +, −, ×, ÷) organized into well-defined mathematical terms.
For instance, the expression \(8(t+h)^2\) is expanded and simplified using algebraic techniques.

• This process involves distributing the terms and combining like terms to produce a simpler expression: \[ 8(t^2 + 2th + h^2) = 8t^2 + 16th + 8h^2 \]
• Simplifying such algebraic expressions is a foundational skill in math, as it prepares expressions for further operations like finding the difference quotient discussed earlier.
• Algebraic manipulation also helps reveal underlying patterns and relationships between quantities, crucial for both solving equations and communicating complex mathematical ideas clearly.

Mastering algebraic expressions enables students to handle various mathematical challenges and lays the groundwork for advanced topics in calculus and beyond.