Problem 71
Question
Write the area \(A\) of a circle as a function of its circumference \(C .\)
Step-by-Step Solution
Verified Answer
The area of a circle expressed as a function of its circumference is \(A = \frac{C^2}{4\pi}\)
1Step 1: Express Radius in Terms of Circumference
Take the formula for the circumference of a circle \(C = 2\pi r\) and solve it for \(r\). Dividing both sides by \(2\pi\) gives the radius in terms of the circumference: \(r = \frac{C}{2\pi}\)
2Step 2: Substitute Radius in Area Formula
Substitute \(r = \frac{C}{2\pi}\) into the area formula \(A = \pi r^2\). This yields \(A = \pi (\frac{C}{2\pi})^2\)
3Step 3: Simplify the Equation
Simplify the equation to obtain the area of the circle in terms of the circumference. This results in \(A = \frac{C^2}{4\pi}\)
Key Concepts
Circumference of a CircleArea of a CircleSolving for RadiusAlgebraic Manipulation
Circumference of a Circle
Understanding the circumference of a circle is crucial in various mathematical calculations. It is defined as the distance around the edge of the circle. To find this, we use the formula:
\[ C = 2\text\text\text{\text{pi}} r \]
where \( C \) represents the circumference, \( \text\text\text{\text{pi}} \) (approximately 3.14159) is a constant, and \( r \) is the radius of the circle. This formula is derived from the relationship between the radius and the circular path. It’s interesting to note that the circumference is always a little more than three times the diameter of the circle (which is twice the radius), a relationship that has fascinated mathematicians for centuries.
\[ C = 2\text\text\text{\text{pi}} r \]
where \( C \) represents the circumference, \( \text\text\text{\text{pi}} \) (approximately 3.14159) is a constant, and \( r \) is the radius of the circle. This formula is derived from the relationship between the radius and the circular path. It’s interesting to note that the circumference is always a little more than three times the diameter of the circle (which is twice the radius), a relationship that has fascinated mathematicians for centuries.
Area of a Circle
The area of a circle is the space contained within its circumference, often measured in square units. The formula for calculating the area is:
\[ A = \text\text\text{\text{pi}} r^2 \]
where \( A \) is the area and \( r \) is the radius. What's intriguing about this formula is that the area grows exponentially with the radius—doubling the radius results in a quadrupling of the area. This is because the area is directly proportional to the square of the radius, demonstrating a fundamental geometric principle of circles.
\[ A = \text\text\text{\text{pi}} r^2 \]
where \( A \) is the area and \( r \) is the radius. What's intriguing about this formula is that the area grows exponentially with the radius—doubling the radius results in a quadrupling of the area. This is because the area is directly proportional to the square of the radius, demonstrating a fundamental geometric principle of circles.
Solving for Radius
When you are given the circumference of a circle but need to find the radius, algebraic manipulation comes into play. You can rearrange the circumference formula to solve for radius:
\[ r = \frac{C}{2 \text\text\text{\text{pi}}} \]
The radius can be visualized as the distance from the center of the circle to any point on its edge. By using algebra to isolate \( r \) in the circumference formula, you effectively decode the dimensions of the circle, revealing its radius from a known circumference. This process is vital for learning to transition between different circle properties.
\[ r = \frac{C}{2 \text\text\text{\text{pi}}} \]
The radius can be visualized as the distance from the center of the circle to any point on its edge. By using algebra to isolate \( r \) in the circumference formula, you effectively decode the dimensions of the circle, revealing its radius from a known circumference. This process is vital for learning to transition between different circle properties.
Algebraic Manipulation
Algebraic manipulation is a foundational skill in mathematics, which allows you to reshape equations to reveal useful information. It involves operations such as adding, subtracting, multiplying, dividing, and factoring, to isolate one variable and solve for it. In the context of our circle problem, we used algebraic manipulation to express the area of the circle as a function of the circumference by substituting the expression for the radius obtained from the circumference formula into the area formula. Through simplifying, we arrived at:\[A = \frac{C^2}{4 \text\text\text{\text{pi}}} \]
This neat transformation is a perfect example of how algebraic manipulation helps us to interconnect different mathematical concepts, creating a more comprehensive understanding of the relationships between them.
This neat transformation is a perfect example of how algebraic manipulation helps us to interconnect different mathematical concepts, creating a more comprehensive understanding of the relationships between them.
Other exercises in this chapter
Problem 71
Find two functions \(f\) and \(g\) such that \((f \circ g)(x)=h(x)\) (There are many correct answers.) $$h(x)=(2 x+1)^{2}$$
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Find the inverse function of \(f\) algebraically. Use a graphing utility to graph both \(f\) and \(f^{-1}\) in the same viewing window. Describe the relationshi
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Use the fact that the graph of \(y=f(x)\) has \(x\) -intercepts at \(x=2\) and \(x=-3\) to find the \(x\) -intercepts of the given graph. If not possible, state
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Use a graphing utility to graph the function and determine whether it is even, odd, or neither. $$f(x)=-|x-5|$$
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