Problem 1 - The Calculus with Analytic Geometry

Question

If \(A\) in. \(^{2}\) is the area of a square and \(s\) in. is the length of a side of the square, find the average rate of change of \(A\) with respect to \(s\) as \(s\) changes from (a) \(4.00\) to \(4.60\); (b) \(4.00\) to 4.30; (c) \(4.00\) to 4.10. (d) What is the instantaneous rate of change of \(A\) with respect to \(s\) when \(s\) is \(4.00\) ?

Step-by-Step Solution

Answer

The average rates are 8.60, 8.30, and 8.10 square inches per inch. The instantaneous rate is 8 square inches per inch.

Step 1 - Understand the problem

The area of a square depends on the length of its side. The area can be written as a function of the side length: \[ A(s) = s^2 \]We need to find the average rate of change of the area with respect to the side length over the given intervals and the instantaneous rate of change at a specific side length.

Step 2 - Formula for Average Rate of Change

The average rate of change of a function \( f(x) \) over an interval \([a, b]\) is given by:\[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \]Here, \( f(s) = A(s) \), so we will use this formula to find the averages.

Step 3 - Calculate Average Rate of Change for (a)

For the interval \(4.00\) to \(4.60\),\[ \text{Average Rate of Change} = \frac{A(4.60) - A(4.00)}{4.60 - 4.00} \]We know:\[ A(4.60) = (4.60)^2 = 21.16 \]\[ A(4.00) = (4.00)^2 = 16.00 \]Thus,\[ \text{Average Rate of Change} = \frac{21.16 - 16.00}{4.60 - 4.00} = \frac{5.16}{0.60} \ \text{Average Rate of Change} = 8.60 \text{ square inches per inch} \]

Step 4 - Calculate Average Rate of Change for (b)

For the interval \(4.00\) to \(4.30\),\[ \text{Average Rate of Change} = \frac{A(4.30) - A(4.00)}{4.30 - 4.00} \]We know:\[ A(4.30) = (4.30)^2 = 18.49 \]\[ A(4.00) = (4.00)^2 = 16.00 \]Thus,\[ \text{Average Rate of Change} = \frac{18.49 - 16.00}{4.30 - 4.00} = \frac{2.49}{0.30} \ \text{Average Rate of Change} = 8.30 \text{ square inches per inch} \]

Step 5 - Calculate Average Rate of Change for (c)

For the interval \(4.00\) to \(4.10\),\[ \text{Average Rate of Change} = \frac{A(4.10) - A(4.00)}{4.10 - 4.00} \]We know:\[ A(4.10) = (4.10)^2 = 16.81 \]\[ A(4.00) = (4.00)^2 = 16.00 \]Thus,\[ \text{Average Rate of Change} = \frac{16.81 - 16.00}{4.10 - 4.00} = \frac{0.81}{0.10} \ \text{Average Rate of Change} = 8.10 \text{ square inches per inch} \]

Step 6 - Formula for Instantaneous Rate of Change

The instantaneous rate of change of the function (which is also the derivative) at a point \( s \) is given by:\[ \frac{dA}{ds} = 2s \]

Step 7 - Calculate Instantaneous Rate of Change

When \( s = 4.00 \), the instantaneous rate of change is:\[ \frac{dA}{ds}\bigg|_{s=4.00} = 2 \times 4 = 8 \text{ square inches per inch} \]

Key Concepts

Derivatives

In calculus, a derivative is a measure of how a function changes as its input changes. It's a fundamental tool for understanding the behavior of functions. The derivative of a function at a point gives the slope of the tangent line to the function's graph at that point. For a function represented as \( f(x) \), its derivative is often denoted as \( f'(x) \) or \( \frac{df}{dx} \). This tells us how quickly \( f(x) \) is changing at any given value of \( x \).

Instantaneous Rate of Change

The instantaneous rate of change of a function is closely linked to the concept of a derivative. It can be thought of as the function's rate of change at a single point, rather than over an interval. To find the instantaneous rate of change, you essentially find the derivative of the function. For example, in the given exercise, the function representing the area \( A \) of a square with side length \( s \) is \( A(s) = s^2 \). The instantaneous rate of change of \( A \) with respect to \( s \) is the derivative of \( s^2 \), which is \( 2s \). Therefore, at \( s = 4.00 \), the instantaneous rate of change is \( 2 \times 4 = 8 \) square inches per inch.

Functions and Their Rates of Change

Functions are mathematical entities that relate an input to an output. Understanding how functions change is crucial for analyzing any phenomenon described by them. The rate of change of a function can be either average or instantaneous. The average rate of change provides a general rate over an interval, while the instantaneous rate gives us the exact rate at a particular point. For instance, in the exercise, the average rate of change is calculated using the formula: \[ \frac{f(b) - f(a)}{b - a} \] where \( f(s) = A(s) \) and specified intervals of \( s \). To get the instantaneous change, differentiate the function \( A(s) = s^2 \) to get \( A'(s) = 2s \). By understanding these concepts, you can better analyze and predict the behavior of the functions you study.