Problem 43 - Precalculus with Limits

Question

In Exercises 33-46, sketch the graph (and label the vertices) of the solution set of the system of inequalities. $$\left\\{\begin{array}{l}{x^{2}+y^{2} \leq 36} \\ {x^{2}+y^{2} \geq 9}\end{array}\right.$$

Step-by-Step Solution

Answer

The graph of the system of inequalities will look like a ring centered at the origin extending from a circle of radius 3 to a circle of radius 6.

Step 1: Draw Circle 1

Begin by drawing the first circle corresponding to the inequality \(x^2 + y^2 \leq 36\). This circle has a radius of 6 units and is centered at the origin (0,0). Remember to include all points inside the circle and on the boundary.

Step 2: Draw Circle 2

Now draw the second circle corresponding to the inequality \(x^2 + y^2 \geq 9\). This circle has a radius of 3 units and is centered at the origin as well. Include all points outside of this circle and on its boundary.

Step 3: Mark the Solution Set

Mark or shade the solution set. Since we are considering points which satisfy both inequalities, we shade points that are within the larger circle and outside the smaller circle. It forms a 'ring' or 'donut' shape around the origin between the two circles.

Key Concepts

Understanding Inequalities

Inequalities are mathematical expressions that show the relative sizes or order of two values. They tell us how numbers compare to each other, whether they are less than (<), greater than (>), less than or equal to (\(leq\)), or greater than or equal to (\(geq\)). Unlike equations, which have one solution, inequalities typically have a range of solutions that satisfy the conditions set by the inequality.

For example, in the system of inequalities from the provided exercise, each inequality defines a set of points in the coordinate plane that fulfills its condition. The first inequality, \(x^2 + y^2 \leq 36\), tells us that the solution includes all points within and on the circle with radius 6 centered at the origin. Similarly, the second inequality, \(x^2 + y^2 \geq 9\), designates an area that includes all points outside of and on a circle with radius 3, again centered at the origin. These graphical representations turn abstract inequalities into visual regions where solutions exist.

Identifying the Solution Set

The solution set of a system of inequalities comprises all the ordered pairs that satisfy all of the inequalities in the system at once. In the context of our exercise, it means finding the set of points that lie in both of the shaded regions described by the two inequalities.

In practice, to mark the solution set, we simultaneously consider the regions dictated by each inequality. We then overlap these regions to find where they intersect. In our exercise, this intersection is the annular region (the 'ring' or 'donut shape') between two circles. This shared area is where the points satisfy both the condition of being within or on the larger circle (\(x^2 + y^2 \leq 36\)) and outside or on the smaller circle (\(x^2 + y^2 \geq 9\)). To convey understanding, it's helpful to visualize or sketch the individual regions first before overlapping them, making it easier to identify the solution set.

Sketching Graphs of Inequalities

Sketching the graph of an inequality on the coordinate plane translates the inequality from a symbolic expression to a visual representation, allowing students to better understand and solve the problem. Here are simplified steps to sketch the systems of inequalities like those in the exercise:

• Sketch the boundaries of each inequality, typically using solid lines for '\(leq\)' or '\(geq\)' and dashed lines for '<' or '>'.
• Decide which side of the boundary line defines the region that satisfies the inequality. This can often be determined by testing a point not on the line (the origin (0,0) is a convenient choice if it's not on the line).
• Shade the region of the plane that satisfies the inequality.
• For a system of inequalities, find the overlapping shaded regions from all inequalities. This common area represents the solution set for the system.

It's crucial to sketch accurately because the graph provides a visual proof of the solution set, and any error in sketching can misrepresent the actual solution.