Problem 36 - Precalculus Mathematics 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. $$\left\\{\begin{array}{l} x^{2}+y^{2}<9 \\ 2 x+y^{2} \geq 1 \end{array}\right.$$

Step-by-Step Solution

Answer

The vertices are at \((-3, \sqrt{7})\) and \((-3, -\sqrt{7})\). The solution set is bounded.

Step 1: Graph the First Inequality

The first inequality is \( x^2 + y^2 < 9 \). This is the inequality of a circle with center at the origin \((0,0)\) and radius 3. To graph it, draw a dashed circle with radius 3 since it does not include the boundary (indicated by strict \(<\) inequality symbol). Shade the inside of the circle to represent the region where \( x^2 + y^2 < 9 \).

Step 2: Graph the Second Inequality

The second inequality is \( 2x + y^2 \geq 1 \). Rearrange it to \( y^2 \geq 1 - 2x \). To graph it, first consider the equation \( y^2 = 1 - 2x \), which resembles a sideways parabola opening to the left. The boundary curve intersects the x-axis at \((0,1)\) and \((0,-1)\). Since \( y^2 \geq 1 - 2x \), the solution region includes the area to the right of, and along the parabola, so shade this area.

Step 3: Find the Intersection Points

To find the points of intersection (vertices) of the boundary curves of the inequalities, solve the system: \( x^2 + y^2 = 9 \) and \( y^2 = 1 - 2x \). Substituting the second equation into the first gives \( x^2 + (1-2x) = 9 \), simplifying to \( x = 2, -3 \). For \( x = 2 \), \( y^2 = 1 - 4 = -3 \) (not possible). For \( x = -3 \), \( y^2 = 1 + 6 = 7 \). Thus, points are \((-3, \pm \sqrt{7})\).

Step 4: Test the Boundedness of the Solution Set

Examine whether the intersection region is bounded. The circle \( x^2 + y^2 < 9 \) limits the region to a finite area inside it. Therefore, outside of the circle is not included in the solution set even if the second inequality permits it, so the overall solution set remains bounded.

Key Concepts

Graphing Inequalities

When it comes to graphing inequalities, it involves representing a set of solutions on a coordinate plane. For example, in this exercise, we have two inequalities: \(x^2 + y^2 < 9\) and \(2x + y^2 \geq 1\). The key is to understand the graphical representation of each inequality.
- **First Inequality**: \(x^2 + y^2 < 9\) represents a circle centered at the origin with a radius of 3. Since the inequality is "less than," the circle is dashed, indicating that points on the circle (i.e., the boundary) are not included in the solution set. So, you shade the interior of the circle to represent all points \((x, y)\) that satisfy the condition.
- **Second Inequality**: \(2x + y^2 \geq 1\) is a bit more complex. By manipulating it to \(y^2 \geq 1 - 2x\), it resembles a sideways parabola that opens to the left. The boundary of this parabola includes the boundary itself because of the "greater than or equal to" symbol. Therefore, you shade the region to the right of the parabola to depict the solutions.

Graphing these inequalities involves understanding each shape's role and how they represent different mathematical concepts visually. It's crucial to remember which boundary types (solid vs dashed lines) align with each inequality symbol.

Intersection Points

Finding intersection points is critical because it tells you where two regions overlap, providing the vertices of the solution set. For the system of inequalities given, these points are where the graphs of the inequalities share common solutions.
- To find intersection points, you simultaneously solve the boundary equations, \(x^2 + y^2 = 9\) and \(y^2 = 1 - 2x\). By substituting the second into the first, we arrive at \(x^2 + (1 - 2x) = 9\), which simplifies the process of finding solution point coordinates.
- Solving yields \(x = -3\), leading to \(y^2 = 7\). Therefore, the intersection points are at \((-3, \sqrt{7})\) and \((-3, -\sqrt{7})\). These points act as boundaries where both inequalities' conditions meet, representing critical vertices for the solution set.

This analysis of intersection points helps in understanding the limitations and relations between different inequalities in a system. Recognizing these points is vital to adequately describing the full nature of the solution region.

Bounded Solution Set

When we talk about a bounded solution set, we're asking if the solution region is contained within a finite area in the coordinate plane. In this context, the first inequality, \(x^2 + y^2 < 9\), naturally creates a bounded region because it represents an interior circle with a finite radius.
- Even though the second inequality, \(y^2 \geq 1 - 2x\), allows for a broad solution set spanning an infinite area to the right of the parabola, the first inequality restricts the overall solution set.
- The intersection of these two inequalities, therefore, results in a finite, circular region limited within and by the circle \(x^2 + y^2 = 9\). This exclusion of the external infinite space leads us to conclude that the solution set is bounded by the first inequality.

Understanding whether a solution set is bounded helps in visualizing the scope of possible solutions within system constraints. It's essential for determining how the region behaves with respect to space and boundaries in practical applications.