Problem 77 - Algebra and Trigonometry

Question

Find a function whose graph is the given curve. The top half of the circle \(x^{2}+y^{2}=9\)

Step-by-Step Solution

Answer

The function is \(y = \sqrt{9 - x^2}\) with domain \(-3 \leq x \leq 3\).

Step 1: Understand the Equation of a Circle

The given equation is \(x^2 + y^2 = 9\), which represents a circle centered at the origin (0,0) with a radius of 3. To confirm, the general form of a circle equation is \((x-h)^2 + (y-k)^2 = r^2\), where \(h\) and \(k\) represent the center and \(r\) the radius.

Step 2: Isolate the Variable y

To find the function representing the top half of the circle, we must express \(y\) in terms of \(x\). Starting with the equation \(x^2 + y^2 = 9\), solve for \(y^2\) by subtracting \(x^2\) from both sides: \(y^2 = 9 - x^2\).

Step 3: Solve for y

Take the square root of both sides to solve for \(y\): \(y = \pm \sqrt{9 - x^2}\). Here, \(+\sqrt{9 - x^2}\) represents the top half of the circle because it yields positive \(y\)-values.

Step 4: Express as a Function

The function representing the top half of the circle is \(y = \sqrt{9 - x^2}\). This function gives the upper half because it includes only the positive square root.

Step 5: Note Domain Restrictions

Recognize that the domain of the function \(y = \sqrt{9 - x^2}\) is restricted to \(-3 \leq x \leq 3\) because these are the horizontal limits of the circle with radius 3.

Key Concepts

Function of a Circle

A circle's equation is a beautiful way to represent both the shape and position of a circle on a graph. The general circle equation is given by \((x-h)^2 + (y-k)^2 = r^2\). To break this down:

• \(h\) and \(k\) are the coordinates of the circle's center.
• \(r\) is the radius of the circle.

In our exercise, the circle is described by the equation \(x^2 + y^2 = 9\). Here, you can see that both \(h\) and \(k\) are zero, meaning the circle is centered at the origin \(0,0\). The number 9 represents \(r^2\), so the radius \(r\) is 3, as \(3^2=9\).
To extract a function from this equation, focus on one half of the circle. The top half can be represented by the function \(y = \sqrt{9 - x^2}\). This form discards the negative root, capturing just the positive part, which draws the upper portion of the circle.

Domain Restrictions

Domain restrictions are crucial for defining a function accurately. They specify the permissible range of values for \(x\), ensuring the function produces valid results.
For the circle equation \(x^2 + y^2 = 9\), the function \(y = \sqrt{9 - x^2}\) serves the top half. However, we can only calculate a real number for \(y\) if the expression inside the square root, \(9 - x^2\), is non-negative.

• The values of \(x\) must satisfy \(9 - x^2 \geq 0\).
• Simplifying gives \(-3 \leq x \leq 3\).

This restriction aligns with the horizontal limits of the circle, given its radius of 3. Thus, the domain of \(y = \sqrt{9 - x^2}\) spans from -3 to 3, allowing the drawing of the top half of the circle within these constraints.

Graph of a Circle

Graphing a circle is like sketching a perfect symmetrical shape on a plane. When graphing the equation \(x^2 + y^2 = 9\), we start by plotting the center at the origin \(0,0\) and use the radius of 3 to draw the circle.
The full circle stretches from x-coordinates -3 to 3 horizontally and encompasses y-coordinates -3 to 3 vertically. The circle is symmetric about both x and y axes due to its center at the origin.

• The function \(y = \sqrt{9 - x^2}\) traces only the top half.

This is because it yields positive y-values, representing the upper portion above the x-axis. Understanding this split ensures that when you plot, you map only the top semi-circle while respecting the established domain and range.

Square Root Function

The square root function is a fascinating mathematical tool that helps in creating semi-circular graphs, like those representing the equation of a circle.
In the context of \(y = \sqrt{9 - x^2}\), the square root function focuses on determining the top half of the circle.

• Taking a square root gives non-negative results.
• This is why \(y = \sqrt{9 - x^2}\) reflects only positive y-values.

The symmetry arising from the square root is harnessed by considering only \(\sqrt{9 - x^2}\) and not \(-\sqrt{9 - x^2}\).
This decision simplifies the function into one that draws only the arc above the x-axis, leaving behind its lower symmetrical part, thus representing the top half of the circle effectively.