Problem 29 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(A=A(S)\) be the area function for a square of side \(S .\) A takes as input the length \(S\) of the side of a square and gives as output the area of the square. (a) Find the following. \(\begin{array}{llll}\text { i. } A(4) & \text { ii. } A(W) & \text { iii. } A(\sqrt{2}+3) & \text { iv. } A(4+h) & \text { v. } A(x-1)\end{array}\) (b) Suppose that \(S\) is bigger than 1 . Which is larger, \(A(S-1)\) or \(A(S)-1 ?\) (Does the answer depend on the size of \(S ?\) If so, how?) (c) Explain in words the difference between \(A(S-1)\) and \(A(S)-1\). Which one of the expressions corresponds to the shaded area in the accompanying figure? What area corresponds to the other expression?

Step-by-Step Solution

Answer

The calculated areas are: A(4) = 16, A(W) = \( W^2 \), A(\( \sqrt{2}+3 \)) = 5 + 2\(\sqrt{2}\), A(4+h) = \( h^2 + 8h + 16 \), A(x-1) = \( x^2 - 2x + 1 \). When S > 1, A(S-1) is bigger than A(S)-1. The expression A(S-1) usually represents the shaded area in the figure, which illustrates a square with each side decreased by 1 unit.

Step 1: Calculate A(4)

The size of the side of a square is 4 units. By using the formula of the area \( A(S) = S * S \), we can substitute S with 4: \( A(4) = 4 * 4 = 16 \) square units.

Step 2: Calculate A(W)

In this case, \( W \) is the side length of the square. The area will be \( A(W) = W * W = W^2 \) square units.

Step 3: Calculate A(\( \sqrt{2}+3 \))

Substitute S with \( \sqrt{2}+3 \): \( A(\sqrt{2}+3) = (\sqrt{2}+3) * (\sqrt{2}+3) = (5+2\sqrt{2}) \) square units.

Step 4: Calculate A(4+h)

\ A(4+h) = (4+h) * (4+h) = 16 + 8h + h^2 \ square units.

Step 5: Calculate A(x-1)

A(x-1) = (x-1) * (x-1) = x^2 - 2x + 1 \ square units.

Step 6: Compare A(S-1) and A(S) - 1 for S > 1

A(S-1) = (S-1) * (S-1) = S^2 - 2S + 1 \ square units. On the other hand, A(S) - 1 = (S * S) - 1 = S^2 - 1 \ square units. When subtracted, we get an expression 2S - 2, which is positive for S > 1. Thus for S > 1, A(S-1) will always be greater than A(S) - 1.

Step 7: Explain and Visualize A(S-1) and A(S)-1

A(S-1) corresponds to the area of a square with each side decreased by 1 unit while A(S) - 1 corresponds to the area of the original square decreased by 1 square unit. Commonly, A(S-1) corresponds to the shaded area inside the square when a square of side length 1 is cut off from the edges.

Key Concepts

Squaring Functions

Squaring functions are mathematical expressions where a variable is multiplied by itself. In the language of algebra, if we have a variable like S, the squaring function for S is written as S². This function is central in geometry, particularly when dealing with areas of squares. To square a number or a variable is to raise it to the power of two, which is akin to finding the size of a square (a 2-dimensional object) whose sides are of length equal to that number or expression.

The area of a square is one of the simplest squaring functions; this is evident in the formula A(S) = S * S = S², where A represents the area, and S is the length of a square's side. Understanding squaring functions and their properties is fundamental in many areas of mathematics and is often one of the first applications of exponents that students learn.

Function Evaluation

Function evaluation is the process of finding the output of a function given an input. When you are given a function, such as the area function for a square mentioned in the exercise, evaluating it requires replacing the input variable with actual values or expressions and then simplifying.

For example, when you evaluate the area function A(S) at S = 4, you replace S in A(S) = S² with 4, giving you A(4) = 4² = 16 square units. By practicing the evaluation of functions with different inputs, you gain a deeper understanding of how the function behaves and how to manipulate it in various mathematical scenarios.

Mathematical Expressions

Mathematical expressions are combinations of numbers, variables, and operation symbols that represent a specific value or set of values. In calculus and other branches of mathematics, expressions are used to describe rules or relationships between variables, often forming functions.

Expressions can include operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction. Crucially, expressions must be well-formed, adhering to the rules of algebra, to ensure they are meaningful and calculable. For instance, the exercise shows different mathematical expressions representing the area of a square, such as A(W) = W² and A(x-1) = (x-1)². Each expression, when evaluated, yields a particular value representing the square's area for a certain side length.

Calculus Applications

Calculus, the mathematical study of change, has extensive applications in solving problems involving squaring functions and area calculations. In the context of these textbook problems, calculus is not explicitly used, but it provides essential tools like limits, derivatives, and integrals, which are crucial in more advanced area problems.

In calculus, you would learn how to find the rate of change of the area concerning the side length of a square using derivatives, or how to calculate the area under a curve using integrals. These techniques are powerful tools for analyzing and solving intricate problems in mathematics, physics, engineering, and many other fields. Thus, a solid understanding of the fundamentals, such as squaring functions and function evaluation, sets the stage for grasping the more complex concepts in calculus.