Problem 55
Question
Solve for \(y\) and use a graphing utility to graph each of the resulting equations in the same viewing window. (Adjust the viewing window so that the circle appears circular.) \(x^{2}+y^{2}=16\)
Step-by-Step Solution
Verified Answer
The equations for the circle are \(y=\sqrt{16-x^{2}}\) and \(y=-\sqrt{16-x^{2}}\). Upon graphing, they give a circle with a radius of 4 units centered at the origin.
1Step 1: Rewrite the Equation for y
Start by isolating \(y\) on one side of the equation. Subtract \(x^{2}\) from both sides to get \(y^{2}=16-x^{2}\). Then take the square root of both sides to yield two equations: \(y=\sqrt{16-x^{2}}\), representing the upper half of the circle, and \(y=-\sqrt{16-x^{2}}\), representing the lower half.
2Step 2: Graph the Equations
Use a graphing utility to plot both functions \(y=\sqrt{16-x^{2}}\) and \(y=-\sqrt{16-x^{2}}\). Make sure to adjust the viewing window to correctly display the circular shape of the plotted equations. The x-values should range from -4 to 4, given that the radius of the circle is 4 units.
Key Concepts
Equation ManipulationGraphing UtilitySquare RootsCircle Equations
Equation Manipulation
The process of equation manipulation involves transforming the given equation to solve for a particular variable, in this case, \( y \). To isolate \( y \), the first step is to get \( y^2 \) on one side of the equation by subtracting \( x^2 \) from both sides. This gives us the equation \( y^2 = 16 - x^2 \).
The next step is to solve for \( y \) by taking the square root of both sides of the equation. Remember that taking the square root introduces two solutions, a positive and a negative root. Hence, from \( y^2 = 16 - x^2 \), we derive the equations \( y = \sqrt{16 - x^2} \) and \( y = -\sqrt{16 - x^2} \).
These two equations represent different parts of the circle. The positive square root corresponds to the upper half, and the negative to the lower half.
The next step is to solve for \( y \) by taking the square root of both sides of the equation. Remember that taking the square root introduces two solutions, a positive and a negative root. Hence, from \( y^2 = 16 - x^2 \), we derive the equations \( y = \sqrt{16 - x^2} \) and \( y = -\sqrt{16 - x^2} \).
These two equations represent different parts of the circle. The positive square root corresponds to the upper half, and the negative to the lower half.
Graphing Utility
Graphing utilities are valuable tools for visualizing mathematical equations. They allow you to plot equations on a graph, revealing geometrical relationships that might not be immediately apparent from the equations themselves.
When graphing the circle derived from \( x^{2} + y^{2} = 16 \), we use the equations \( y = \sqrt{16 - x^2} \) and \( y = -\sqrt{16 - x^2} \). Plotting these reveals the top and bottom halves of the circle, respectively.
When graphing the circle derived from \( x^{2} + y^{2} = 16 \), we use the equations \( y = \sqrt{16 - x^2} \) and \( y = -\sqrt{16 - x^2} \). Plotting these reveals the top and bottom halves of the circle, respectively.
- Ensure the graphing calculator or software is set to handle both functions at the same time.
- Adjust the viewing window to ensure the circle appears circular, rather than an ellipse due to axis scaling differences. A suitable range is from -4 to 4 for both \( x \) and \( y \) axes.
Square Roots
The concept of square roots emerges when solving the equation \( y^2 = 16 - x^2 \). Taking the square root of both sides helps isolate \( y \).
Remember, every positive number has two square roots: one positive and one negative. Thus, the solution \( y = \sqrt{16 - x^2} \) represents the positive root, while \( y = -\sqrt{16 - x^2} \) captures the negative counterpart. These roots graphically depict the top and bottom halves of the circle.
Calculating the square roots directly involves understanding the behavior and limits of \( x \). Here, the expression \( 16 - x^2 \) must remain non-negative, meaning \( x \) values are confined to \([-4, 4]\). This range confirms the circle's radius, being equidistant from its center at the origin.
Remember, every positive number has two square roots: one positive and one negative. Thus, the solution \( y = \sqrt{16 - x^2} \) represents the positive root, while \( y = -\sqrt{16 - x^2} \) captures the negative counterpart. These roots graphically depict the top and bottom halves of the circle.
Calculating the square roots directly involves understanding the behavior and limits of \( x \). Here, the expression \( 16 - x^2 \) must remain non-negative, meaning \( x \) values are confined to \([-4, 4]\). This range confirms the circle's radius, being equidistant from its center at the origin.
Circle Equations
Circle equations in the coordinate plane are expressed as \( x^{2} + y^{2} = r^{2} \), where \( r \) is the radius. For this exercise, \( r^2 = 16 \), so \( r = 4 \).
The center of this circle is \((0, 0)\), which is also the origin of the coordinate plane. The radius extends out from this point, defining the circle's boundary.
When manipulating and graphing the equation for the circle, the values of \( x \) and \( y \) are confined to the range of the circle's radius. Thus, appropriately setting your graphing window to encompass \([-4, 4] \) ensures that both portions of the circle are included. This allows a complete and accurate graph of the circle to be visualized.
The center of this circle is \((0, 0)\), which is also the origin of the coordinate plane. The radius extends out from this point, defining the circle's boundary.
When manipulating and graphing the equation for the circle, the values of \( x \) and \( y \) are confined to the range of the circle's radius. Thus, appropriately setting your graphing window to encompass \([-4, 4] \) ensures that both portions of the circle are included. This allows a complete and accurate graph of the circle to be visualized.
Other exercises in this chapter
Problem 55
Use a graphing utility to graph the function and use the Horizontal Line Test to determine whether the function is one-to-one and so has an inverse function. $$
View solution Problem 55
Find all values of \(x\) such that \(f(x)=0.\) $$f(x)=\frac{9 x-4}{5}$$
View solution Problem 55
Write an equation of the line that passes through the points. Use the slope- intercept form (if possible). If not possible, explain why and use the general form
View solution Problem 56
Sketch the graph of the piecewise-defined function by hand. $$f(x)=\left\\{\begin{array}{ll} x+6, & x \leq-4 \\ 3 x-4, & x>-4 \end{array}\right.$$
View solution