Problem 75
Question
A rectangle is bounded by the \(x\) -axis and the semicircle \(y=\sqrt{36-x^{2}},\) as shown in the figure. Write the area \(A\) of the rectangle as a function of \(x\) and determine the domain of the function.
Step-by-Step Solution
Verified Answer
The area A of the rectangle as a function of x is \(A(x) = 2x\sqrt{36-x^2}\) and the domain of this function is [-6, 6].
1Step 1: Express the area of the rectangle as a function of x
Based on the geometry, the width of the rectangle is \(2x\) (from -x to +x) and the height is given by the equation of the semi-circle which is \(y=\sqrt{36-x^2}\). Thus, the area \(A\) of the rectangle can be written as the product of its width and height: \(A(x) = 2x * \sqrt{36-x^2}\).
2Step 2: Simplify the function
Using the properties of multiplication, the function can be rewritten as: \(A(x) = 2x\sqrt{36-x^2}\)
3Step 3: Compute the domain of the function
The domain of \(A(x)\) are the values of \(x\) for which the function is defined. Since the height \(y\) of the rectangle cannot be negative and \(x\) is the horizontal distance from the center of the semicircle, \(x\) can be any real number such that \(-6 \leq x \leq 6\). Therefore the domain of \(A(x)\) is \([-6,6]\)
Key Concepts
Area calculationFunction of a variableDomain of a function
Area calculation
Understanding the area of a rectangle involves a straightforward calculation. A rectangle's area is determined by multiplying its width by its height.
In the context of this exercise, we are examining a rectangle formed using the x-axis and a semicircle defined by the equation \(y = \sqrt{36-x^2}\).
It’s important to grasp that this setup allows us to view the rectangle’s width as twice the distance from a point on the x-axis to the center of the semicircle, or \(2x\).
Meanwhile, the height of the rectangle corresponds to the y-value from the semicircle's equation.
Putting these together, the area \(A\) can be expressed as a function of \(x\) by combining width and height:
\[A(x) = 2x \times \sqrt{36-x^2}\]
This expression represents how the area varies based on the variable \(x\).
In the context of this exercise, we are examining a rectangle formed using the x-axis and a semicircle defined by the equation \(y = \sqrt{36-x^2}\).
It’s important to grasp that this setup allows us to view the rectangle’s width as twice the distance from a point on the x-axis to the center of the semicircle, or \(2x\).
Meanwhile, the height of the rectangle corresponds to the y-value from the semicircle's equation.
Putting these together, the area \(A\) can be expressed as a function of \(x\) by combining width and height:
- Width: \(2x\)
- Height: \(\sqrt{36-x^2}\)
\[A(x) = 2x \times \sqrt{36-x^2}\]
This expression represents how the area varies based on the variable \(x\).
Function of a variable
A function describes a relationship between one set of values and another. In this context, we need to analyze how the area \(A\) of the rectangle changes as \(x\) changes.
In geometry, expressing a quantity like area as a function of a variable helps to understand how changes in the variable influence the overall measurement.
In our problem, we have the function \(A(x) = 2x \sqrt{36 - x^2}\). This indicates that the area is dependent on \(x\), a value representing half the width of the rectangle. This makes \(x\) a critical aspect of how the area is determined.
The concept serves to show that by increasing or decreasing the value of \(x\), we have a direct impact on the area of the rectangle.
Thus, if \(x\) increases toward its maximum value, the rectangle’s potential area will also change accordingly. This functional relationship is essential for comprehensively understanding how modifications in base dimensions directly alter the area.
In geometry, expressing a quantity like area as a function of a variable helps to understand how changes in the variable influence the overall measurement.
In our problem, we have the function \(A(x) = 2x \sqrt{36 - x^2}\). This indicates that the area is dependent on \(x\), a value representing half the width of the rectangle. This makes \(x\) a critical aspect of how the area is determined.
The concept serves to show that by increasing or decreasing the value of \(x\), we have a direct impact on the area of the rectangle.
Thus, if \(x\) increases toward its maximum value, the rectangle’s potential area will also change accordingly. This functional relationship is essential for comprehensively understanding how modifications in base dimensions directly alter the area.
Domain of a function
The domain of a function consists of all possible values that the variable can take so the function remains defined. It is crucial to consider physical and mathematical constraints.
In this exercise, we are considering \(A(x) = 2x \sqrt{36-x^2}\).
The function \(\sqrt{36-x^2}\) introduces a constraint since square roots are only defined for non-negative values.
This means that \(36-x^2 \geq 0\), with the solution being \(-6 \leq x \leq 6\).
This defines the interval where \(x\) can be situated, ensuring the rectangle maintains its proper geometric form.
Additionally, since \(x\) represents a spatial dimension from the center outward, it fits within the confines of the semicircle.
This mathematical boundary ensures the function only provides realistic, usable values for the dimensions of the rectangle.
In this exercise, we are considering \(A(x) = 2x \sqrt{36-x^2}\).
The function \(\sqrt{36-x^2}\) introduces a constraint since square roots are only defined for non-negative values.
This means that \(36-x^2 \geq 0\), with the solution being \(-6 \leq x \leq 6\).
This defines the interval where \(x\) can be situated, ensuring the rectangle maintains its proper geometric form.
Additionally, since \(x\) represents a spatial dimension from the center outward, it fits within the confines of the semicircle.
This mathematical boundary ensures the function only provides realistic, usable values for the dimensions of the rectangle.
- Mathematical Condition: \(36-x^2 \geq 0\)
- Domain of \(A(x)\): \([-6, 6]\)
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