With few exceptions, we have spent our time in this course graphing equations relating two variables. In this section, we explore graphing inequalities relating two variables which usually a two-dimensional region in the plane instead of a one dimensional line or curve. In our first example, we restrict our attention to looking at regions in the \(xy\)-plane bounded by equations which describe \(y\) as a function of \(x\).

A few remarks about Example ineqgraphex01 are in order. First note that each of the regions presented here can be views as graphs of relations as described in Section Relations. Moreover, there are many ways to describe a region so our answers to number describeregionex above are by no means unique.

In particular, our solution \(R = \{ (x,y) \, | \, -1 \leq x \leq 2, \, x^2 \leq y \leq x+2 \}\) to number parabolalineregionex can be visualized as ‘filling up’ the region \(R\) from the bottom curve, \(y = x^2\) to the top curve, \(y = x+2\) as \(x\) runs from the leftmost extent of the region at \(x = -1\) to the rightmost extent at \(x = 2\). The GeoGebra inteactive below captures this motion.

The notion that each \(x\) determines where to start and stop filling the region is a consequence of us viewing the bounding curves as functions of \(x\). That is, for each \(x\), we can determine the lower boundary of the region, \(y = x^2\) and the upper boundary of the region, \(y = x+2\).

There are times in Calculus where it may be convenient to describe the region \(R\) as filling up left-to-right as \(y\) varies from the bottom most extent, \(y = 0\) to the top most extent, \(y = 4\). In this case, we need to describe the bounding curves as functions of \(y\). To that end, we solve \(y = x+2\) and \(y = x^2\) for \(x\) and get three functions of \(y\): \(x = y-2\), \(x = \sqrt {y}\) and \(x = -\sqrt {y}\). The GeoGebra interactive below shows this ‘left to right’ filling action along with the bounding curves, reframed as functions of \(y\).

Based on the diagram, we see we need to describe \(R\) as two pieces. The first piece being bounded on the left by \(x = -\sqrt {y}\) and on the right \(x = \sqrt {y}\) from \(y = 0\) to \(y=1\), and the second piece bounded on the left by \(x = y-2\) and on the right by \(x= \sqrt {y}\) from \(y = 1\) to \(y = 4\). That is,

\[ R = \{ (x,y) \, | \, 0 \leq y \leq 1, \, -\sqrt {y} \leq x \leq \sqrt {y} \} \cup \{ (x,y) \, | \, 1 \leq y \leq 4, \, y-2 \leq x \leq \sqrt {y} \}. \]

Not all regions in the plane are best described using functions of \(x\) or \(y\). Suppose, for instance, we wish to sketch the region \(\{ (x,y) \, | \, x^2 < 4 - y^2. \}\) Algebraically, we wish to plot all points \((x,y)\) for which the inequality \(x^2<4 - y^2\) is true. One way to proceed is to mimic the ‘sign diagram’ routine we use for solving nonlinear inequalities in one variable: rewrite the inequality so as to obtain \(0\) on one side of the inequality, find the zeros of the non-zero side, choose test values determined by the zeros, and record our solution.

First, we gather all of the terms on one side and leave a \(0\) on the other: \(x^2 + y^2 -4 < 0\). Next, we find the zeros of the left hand side, that is, where is \(x^2 + y^2 - 4 = 0\). Rewriting, we get \(x^2+y^2 = 4\) which describes the circle of radius \(2\) centered at the origin. In other words, instead of obtaining a few numbers which divide the real number line into intervals, we get an equation of a curve, in this case, a circle, which divides the plane into two regions - the ‘inside’ and ‘outside’ of the circle.

Just like we used test values to determine whether or not an interval belongs to the solution of the inequality, we use test points in the each of the regions to see which of these belong to our solution set. We choose \((0,0)\) to represent the region inside the circle and \((0,3)\) to represent the points outside of the circle. When we substitute \((0,0)\) into \(x^2 + y^2 -4 < 0\), we get \(-4 < 4\) which is true. This means \((0,0)\) and all the other points inside the circle are part of the solution. On the other hand, when we substitute \((0,3)\) into the same inequality, we get \(5 < 0\) which is false. This means \((0,3)\) along with all other points outside the circle are not part of the solution. What about points on the circle itself? Choosing a point on the circle, say \((0,2)\), we get \(0 < 0\), which means the circle itself does not satisfy the inequality. As a result, we leave the circle dashed in the final diagram.

We put this technique to good use in the following example.

We close this section with a follow-up to Example circuitex in Section MatMethods. Recall in the circuit diagrammed below, we have two batteries with source voltages \(V\!B_1\) and \(V\!B_2\), measured in volts \(V\), along with six resistors with resistances \(R_1\) through \(R_6\), measured in kiloohms, \(k\Omega \).Recall if we think of electrons flowing through the circuit, we can think of the voltage sources as providing the ‘push’ which makes the electrons move, the resistors as obstacles for the electrons to overcome, and the mesh current as a net rate of flow of electrons around the indicated loops.

Using Ohm’s Law and Kirchhoff’s Voltage Law, we can relate the voltage supplied to the circuit by the two batteries to the voltage drops across the six resistors in order to find the four ‘mesh’ currents: \(i_1\), \(i_2\), \(i_3\) and \(i_4\), measured in milliamps, \(mA\). This gives rise to the following system of linear equations:

\[ \left \{ \begin{array}{rcl} \left (R_1 + R_3\right )i_1 - R_3i_2 - R_1i_4 & = & V\!B_1 \\ -R_3i_1 + \left (R_2 + R_3 + R_4\right )i_2 - R_4i_3 - R_2i_4 & = & 0 \\ -R_4i_2 + \left (R_4 + R_6\right )i_3 - R_6i_4 & = & -V\!B_2 \\ -R_1i_1 - R_2i_2 - R_6i_3 + \left (R_1 + R_2 + R_5 + R_6\right )i_4 & = & 0 \\ \end{array} \right .\]

In Example circuitex, we found that under the assumptions \(V\!B_1 = 10 V\), \(V\!B_2 = 5 V\), and all the resistances are all \(1 k\Omega \), the mesh currents worked out to be \(i_1 = 10.625 \, \, mA\), \(i_2 = 6.25 \, \, mA\), \(i_3 = 3.125 \, \, mA\), and \(i_4 = 5 \, \, mA\).

In our final example, we assume \(V\!B_1 = 10 V\) and \(V\!B_2 = 5 V\) and work to find what combination of resistances would combine to produce these mesh currents.