Problem 82
Question
Find a function whose graph is the given curve. The bottom half of the circle \(x^{2}+y^{2}=9\)
Step-by-Step Solution
Verified Answer
The function is \( y = -\sqrt{9 - x^2} \), valid for \( -3 \leq x \leq 3 \).
1Step 1: Understand the Equation of the Circle
The given equation of the circle is \( x^2 + y^2 = 9 \). This represents a circle centered at the origin with a radius of 3, since \( r^2 = 9 \), and thus \( r = 3 \).
2Step 2: Solve for y
To find the function corresponding to the bottom half of the circle, solve for \( y \) in terms of \( x \). Start with the equation \( x^2 + y^2 = 9 \) and solve for \( y^2 \): \( y^2 = 9 - x^2 \).
3Step 3: Determine the Negative Square Root
Take the square root of both sides to solve for \( y \). For the bottom half of the circle, we need the negative square root: \( y = -\sqrt{9 - x^2} \).
4Step 4: Domain of the Function
The function \( y = -\sqrt{9 - x^2} \) is defined only where the expression under the square root is non-negative. Hence, \( -3 \leq x \leq 3 \), which matches the circle's domain.
Key Concepts
Domain of a FunctionSolving for yCircle GraphSquare Root Function
Domain of a Function
Understanding the domain of a function is crucial, especially when dealing with equations like circles. The domain of a function refers to all possible input values, usually defined by the variable \( x \). For the function from the circle's equation \( y = -\sqrt{9 - x^2} \), the domain is limited by the necessity for values under the square root to be zero or positive.
Therefore, the expression \( 9 - x^2 \) must be greater than or equal to zero. Solving \( 9 - x^2 \geq 0 \) gives us \(-3 \leq x \leq 3 \). This range represents all possible \( x \) values that keep the function defined and real. In essence, the domain is the horizontal span of the part of the circle represented by our equation, covering all \( x \) values for the bottom half of the circle.
Therefore, the expression \( 9 - x^2 \) must be greater than or equal to zero. Solving \( 9 - x^2 \geq 0 \) gives us \(-3 \leq x \leq 3 \). This range represents all possible \( x \) values that keep the function defined and real. In essence, the domain is the horizontal span of the part of the circle represented by our equation, covering all \( x \) values for the bottom half of the circle.
Solving for y
Solving for \( y \) in an equation like a circle's is essential for expressing it as a function. Starting with the original equation of a circle \( x^2 + y^2 = 9 \), our goal is to isolate \( y \) on one side.
First, rearrange to express \( y^2 \) as \( y^2 = 9 - x^2 \). This isolates \( y^2 \) by moving \( x^2 \) to the other side of the equation. The challenge with circles is deciding between the positive or negative square root when solving \( y^2 \).
For the bottom half of the circle, take the negative square root: \( y = -\sqrt{9 - x^2} \). This solution will yield \( y \) values below the x-axis, representing the lower part of the circle. Solving for \( y \) converts the implicit relation into an explicit function.
First, rearrange to express \( y^2 \) as \( y^2 = 9 - x^2 \). This isolates \( y^2 \) by moving \( x^2 \) to the other side of the equation. The challenge with circles is deciding between the positive or negative square root when solving \( y^2 \).
For the bottom half of the circle, take the negative square root: \( y = -\sqrt{9 - x^2} \). This solution will yield \( y \) values below the x-axis, representing the lower part of the circle. Solving for \( y \) converts the implicit relation into an explicit function.
Circle Graph
A circle graph visually represents all points equidistant from a central point. When given the equation \( x^2 + y^2 = 9 \), we know this describes a circle centered at the origin with a radius of 3, calculated using \( r = \sqrt{9} \).
Each point \((x, y)\) on the circle satisfies the given equation, maintaining a constant distance from the circle's center at \((0,0)\). The graph is symmetric, meaning it looks the same if folded along the x or y-axis.
In focusing on only the bottom half of this circle, the circle graph is sliced horizontally along the x-axis, graphing only the points where \( y \leq 0 \). This approach highlights how functions can represent specific parts of geometric shapes.
Each point \((x, y)\) on the circle satisfies the given equation, maintaining a constant distance from the circle's center at \((0,0)\). The graph is symmetric, meaning it looks the same if folded along the x or y-axis.
In focusing on only the bottom half of this circle, the circle graph is sliced horizontally along the x-axis, graphing only the points where \( y \leq 0 \). This approach highlights how functions can represent specific parts of geometric shapes.
Square Root Function
A square root function involves taking the square root of an expression, often resulting in two possible outputs: positive or negative. It's vital to determine the correct root when finding functions.
In the equation \( y = -\sqrt{9 - x^2} \), we are looking for the bottom half of a circle, so we use the negative square root from solving \( y^2 = 9 - x^2 \). The square root function restricts the range due to the square root's non-negative domain properties, which is handled by the substitution \( q \geq 0 \).
This affects both the shape and limit of our function. The derived square root constraint \(-3 \leq x \leq 3\) ensures that the function remains real and fully represents the semicircle's bottom half.
In the equation \( y = -\sqrt{9 - x^2} \), we are looking for the bottom half of a circle, so we use the negative square root from solving \( y^2 = 9 - x^2 \). The square root function restricts the range due to the square root's non-negative domain properties, which is handled by the substitution \( q \geq 0 \).
This affects both the shape and limit of our function. The derived square root constraint \(-3 \leq x \leq 3\) ensures that the function remains real and fully represents the semicircle's bottom half.
Other exercises in this chapter
Problem 81
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