Problem 32
Question
Find an equation that expresses the area \(A\) of a circle as a function of its (a) radius \(r\) (b) diameter \(d\)
Step-by-Step Solution
Verified Answer
Question: Express the area of a circle as a function of its radius r and as a function of its diameter d.
Answer: The area of a circle can be expressed as (a) a function of radius r: \(A = \pi r^2\), and (b) a function of diameter d: \(A = \pi \frac{d^2}{4}\).
1Step 1: (a) Area of a circle as a function of radius r
The formula for the area of a circle with radius r is given by: \(A = \pi r^2\).
2Step 2: (b) Express the area A as a function of diameter d
The diameter of a circle is twice the radius, so we have the relation \(d = 2r\). To express area A as a function of diameter d, we substitute \(r = \frac{d}{2}\) in the area formula \(A = \pi r^2\).
So, we have: \(A = \pi \left(\frac{d}{2}\right)^2= \pi \frac{d^2}{4}\).
In conclusion, the area of a circle can be expressed as:
(a) a function of radius r: \(A = \pi r^2\)
(b) a function of diameter d: \(A = \pi \frac{d^2}{4}\).
Key Concepts
Circle GeometryMathematical FunctionsRadius and Diameter Relationships
Circle Geometry
Circle geometry is a fundamental aspect of mathematics that involves the study of the properties and the measures related to a circle. A key property is the area, which is the space enclosed by the circumference of the circle.
The formula to calculate the area of a circle is central in circle geometry and vital in various fields of science and engineering. The formula is derived from the circle's radius, which is the distance from the center of the circle to any point on its edge. Understanding the formula for the area of a circle, which is \( A = \pi r^2 \), where \( A \) represents the area and \( r \) is the radius, provides a solid foundation for comprehending other complex geometrical shapes and forms.
In addition to just knowing the formula, it is important to comprehend how the area can change proportionally with the square of the radius, meaning if the radius is doubled, the area increases by a factor of four. This quadratic relationship between radius and area is a key concept in circle geometry.
The formula to calculate the area of a circle is central in circle geometry and vital in various fields of science and engineering. The formula is derived from the circle's radius, which is the distance from the center of the circle to any point on its edge. Understanding the formula for the area of a circle, which is \( A = \pi r^2 \), where \( A \) represents the area and \( r \) is the radius, provides a solid foundation for comprehending other complex geometrical shapes and forms.
In addition to just knowing the formula, it is important to comprehend how the area can change proportionally with the square of the radius, meaning if the radius is doubled, the area increases by a factor of four. This quadratic relationship between radius and area is a key concept in circle geometry.
Mathematical Functions
Mathematical functions are core to understanding relationships between different quantities. When we say that the area of a circle is a function of its radius or diameter, we are describing a specific kind of relationship where the area depends on the radius or diameter values.
A function takes an input, processes it through a rule or formula, and then provides an output. In the context of circle area, when given the radius as the input, the function \( A(r) = \pi r^2 \) describes how every radius value is associated with one unique area. Similarly, for diameter, the function \( A(d) = \pi \frac{d^2}{4} \) reflects that there is also a unique area for each diameter value.
Understanding the concept of functions in mathematics also facilitates grasping how different geometrical formulas are interrelated and applied in real-world situations, such as designing circular objects or calculating the space in circular regions.
A function takes an input, processes it through a rule or formula, and then provides an output. In the context of circle area, when given the radius as the input, the function \( A(r) = \pi r^2 \) describes how every radius value is associated with one unique area. Similarly, for diameter, the function \( A(d) = \pi \frac{d^2}{4} \) reflects that there is also a unique area for each diameter value.
Understanding the concept of functions in mathematics also facilitates grasping how different geometrical formulas are interrelated and applied in real-world situations, such as designing circular objects or calculating the space in circular regions.
Radius and Diameter Relationships
The relationship between the radius and diameter of a circle is straightforward yet essential: the diameter is always twice the length of the radius. Expressed as an equation, this is \( d = 2r \).
Consequently, any formula that involves the radius can be converted to one that uses the diameter instead, by substituting \( r \) with \( \frac{d}{2} \) and vice versa. This substitution is useful in different mathematical and practical applications; for instance, when you only know the diameter of a circle but need to find the area, you can use the formula \( A = \pi \frac{d^2}{4} \) instead of the one based on radius.
Understanding this relationship also simplifies the calculation of other circle properties, such as circumference or volume if the circle is the base of a cylinder. It's a simple yet powerful concept that illustrates the interconnectivity of mathematical principles.
Consequently, any formula that involves the radius can be converted to one that uses the diameter instead, by substituting \( r \) with \( \frac{d}{2} \) and vice versa. This substitution is useful in different mathematical and practical applications; for instance, when you only know the diameter of a circle but need to find the area, you can use the formula \( A = \pi \frac{d^2}{4} \) instead of the one based on radius.
Understanding this relationship also simplifies the calculation of other circle properties, such as circumference or volume if the circle is the base of a cylinder. It's a simple yet powerful concept that illustrates the interconnectivity of mathematical principles.
Other exercises in this chapter
Problem 32
Verify that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\) for every \(x\) $$f(x)=2 x^{3}-5, \quad g(x)=\sqrt[3]{\frac{x+5}{2}}$$
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Assume \(h \neq 0 .\) Compute and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h} $$ $$f(x)=1 / x$$
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Find the approximate intervals on which the function is increasing, those on which it is decreasing, and those on which it is constant. $$f(x)=-x^{3}-8 x^{2}+8
View solution Problem 33
Assume \(h \neq 0 .\) Compute and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h} $$ $$f(x)=x^{2}+3$$
View solution