Problem 37 - Precalculus: Mathematical for Calculus

Question

Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned}x^{2}+y^{2} & \leq 8 \\\x & \geq 2 \\\y & \geq 0\end{aligned}\right.$$

Step-by-Step Solution

Answer

Vertices: (2, 0) and (2, 2). The solution set is bounded.

Step 1: Identify the Inequalities and Equations

The system of inequalities given is:1. \(x^2 + y^2 \leq 8\), which represents the region inside or on the circle centered at the origin with radius \(\sqrt{8}\).2. \(x \geq 2\), which is a half-plane to the right of the vertical line through \(x=2\).3. \(y \geq 0\), which is a half-plane above the x-axis.

Step 2: Graph the Circle Inequality

Graph the circle with radius \(\sqrt{8}\), centered at the origin \((0,0)\). The inequality \(x^2 + y^2 \leq 8\) means we include the interior and boundary of this circle.

Step 3: Graph the Vertical Line Inequality

Draw the vertical line \(x=2\). The solution \(x \geq 2\) includes all points to the right of this line. Use a solid line since our inequality is \(\geq\).

Step 4: Graph the Horizontal Line Inequality

Sketch the horizontal line \(y=0\) (the x-axis). The solution \(y \geq 0\) includes all points above the x-axis. As this equation uses \(\geq\), use a solid line.

Step 5: Determine the Intersection Region

The solution to the system of inequalities is the region where all three conditions overlap. This is the part of the circle that is to the right of the line \(x=2\) and above the x-axis.

Step 6: Find the Vertices of the Solution Region

The vertices of the solution region occur at the intersection of the boundaries:1. Intersection of \(x=2\) and the circle: Substituting \(x=2\) into \(x^2 + y^2 = 8\), we get \(y^2 = 4\), so \(y = \pm 2\). Since \(y \geq 0\), only \((2, 2)\) is valid.2. Intersection along the x-axis: Since \(y = 0\), solve \(x^2 = 8\) while respecting \(x \geq 2\), yielding the point \((2, 0)\).

Step 7: Evaluate Boundedness of the Solution Set

We observe that the solution set is bounded within the circle, and the constraints from the lines \(x=2\) and \(y=0\). The region is a segment of the circle, thus it is bounded.

Key Concepts

Graphing Inequalities

When graphing systems of inequalities, we are essentially plotting the regions of the coordinate plane that satisfy these inequalities. It's more than just drawing lines or curves. You are creating a visual representation of possible solutions.
First, let's look at each inequality individually and plot them:

• **Circle Inequality**: The inequality \(x^2 + y^2 \leq 8\) suggests we are dealing with a circle. The circle is centered at the origin with a radius of \(\sqrt{8}\), which is roughly 2.828. This means any point inside and on the boundary of this circle is included in the solution.
• **Vertical Line Inequality**: The inequality \(x \geq 2\) introduces a vertical line at \(x=2\). This line runs parallel to the y-axis. Everything to the right of this line (including points on the line itself) is part of the solution.
• **Horizontal Line Inequality**: The inequality \(y \geq 0\) corresponds to the x-axis. The region above (and including) the x-axis is part of the solution set.

Each inequality shapes a specific portion of the graph. Overlaying them helps illustrate where the solutions lie.

Bounded Region

A bounded region is a closed area within which all solutions exist. For the given system of inequalities, we get a specific area on the graph where all conditions overlap.
The circle created by \(x^2 + y^2 \leq 8\) provides a natural boundary. It restricts the area within a specific radius. The conditions \(x \geq 2\) and \(y \geq 0\) further define this region:

• The area is confined to one side of the line \(x = 2\). This line does not extend infinitely but is limited by the circle, creating a finite boundary.
• Similarly, the sine of \(y \geq 0\) confines this region to above the x-axis, again limited by the circle.

Consequently, the resultant shape is a wedge or sector-like region, showing all possible solutions while remaining confined within these distinct boundaries.

Intersection Points

Intersection points are critical for understanding where the boundaries of the inequalities meet. These points mark the vertices of the region where all conditions of the inequalities are satisfied simultaneously.
In this system, we find these intersection points by examining where the line boundaries meet the circle:

• **Intersection of \(x=2\) with the Circle**: By substituting \(x = 2\) into the circle equation \(x^2 + y^2 = 8\), we solve for \(y\). This results in \(y^2 = 4\), thus \(y = 2\). This gives us the first vertex at \((2, 2)\).
• **Intersection Along the x-axis**: Setting \(y = 0\) in the circle equation, then solving for \(x\), gives \(x^2 = 8\). Since \(x \ge 2\), we only take the positive result, leading to the point \((2, 0)\).

These vertices, \((2, 2)\) and \((2, 0)\), define the crucial points of the region which visualizes the solution set effectively.