Problem 119 - Precalculus : Building Concepts and Connections

Question

The surface area of a sphere is given by \(A(r)=4 \pi r^{2},\) where \(r\) is in inches and \(A(r)\) is in square inches. The function \(C(x)=6.4516 x\) takes \(x\) square inches as input and outputs the equivalent result in square centimeters. Find \((C \circ A)(r)\) and explain what it represents.

Step-by-Step Solution

Answer

The function \((C \circ A)(r)\) represents the surface area of a sphere with radius \(r\), measured in square centimeters. The result of this composite function is \(25.70688\pi r^{2}\).

Step 1: Understand the Function Composition

The composition \((C \circ A)(r)\) means that you first apply function \(A\) to \(r\), and then apply function \(C\) to the result of \(A(r)\). This gives the surface area of the sphere in square centimeters.

Step 2: Apply Function A to r

First, you calculate the surface area of sphere with radius \(r\) using function \(A\). \(A(r)=4\pi r^{2}\).

Step 3: Apply Function C to A(r)

Then, you apply function \(C\) to the result of \(A(r)\) to convert the surface area from square inches to square centimeters. \(C(A(r)) = C(4\pi r^{2}) = 6.4516 \cdot 4\pi r^{2}\).

Step 4: Simplify the Result

Simplify \(C(A(r))=C(4\pi r^{2}) = 6.4516 \cdot 4\pi r^{2}\) to get the final result of: \((C \circ A)(r) = 25.70688\pi r^{2}\).

Key Concepts

Surface Area of a Sphere

Understanding the concept of surface area is crucial when studying geometric shapes. The surface area can be seen as the total amount of space that the surface of a three-dimensional object covers. For a sphere, which is a perfectly round object, like a basketball, every point on the surface is an equal distance (radius, r) from the center.

The formula to calculate the surface area of a sphere, as mentioned in the problem, is given by the expression \(A(r) = 4 \pi r^2\). This formula is derived from integral calculus, but it’s important to understand what it represents: \(4\) is a constant that appears as part of the formula, \(\pi\) approximates the ratio of the circumference of a circle to its diameter, and \(r^2\) emphasizes that surface area is two-dimensional, hence the square of the radius. When you input a sphere's radius into this formula, you obtain its surface area in square units, allowing you to quantify the space it would take up on a flat surface.

Unit Conversion

Unit conversion plays an essential role in mathematics and science, as it allows us to compare and work with different measurements effectively. In the context of the exercise, we're looking at converting the surface area of a sphere from square inches to square centimeters.

The function \(C(x) = 6.4516 x\) is used for this conversion because there are exactly 6.4516 square centimeters in one square inch. That's the conversion factor needed. This conversion is crucial when we need to communicate measurements in a different standard or apply formulas that require uniform measurement units. Without proper conversion, comparisons or calculations can lead to incorrect results, making this concept fundamental in real-world applications as well.

Simplifying Mathematical Expressions

Simplification of mathematical expressions is a process that makes expressions easier to interpret or work with. In the provided solution, the expression \(C(A(r)) = 6.4516 \cdot 4\pi r^2\) gets simplified to \(25.70688\pi r^2\). This is done through multiplication, combining the constants \(6.4516\) and \(4\) into a single constant, and keeping the variables and their exponents intact.

Why Simplify?

Simplification allows us to more easily see the relationship between variables and understand the impact of changing one variable on the outcome. In computational terms, a simplified expression can often be calculated more efficiently.

Simplifying is not about changing the value of the expression; it's about expressing that value in a clear, concise, and preferably elegant form which helps in various branches of mathematics, including algebra and calculus, due to its facilitation of problem-solving and equation manipulation.