Final Assignment Cristina Aurrecoechea Fall 2005

# Multiple Solutions

Find as many solutions as possible for x,y,z that satisfy the equations:

First let's look at these surfaces in 3D. The first one is a hyperboloid shown in Figure 1. We observe the following about this surface:

• It does not intersect any of the three axes x=0, y=0, z=0, in any point.
• It only exists in 4 quadrants: For z > 0, the product xy must be positive, therefore the curve will exist in the first and third quadrant only.
• For z < 0, the product xy must be negative, therefore the curve will exist in the sixth and fourth eighth quadrants.

The right side of Figure 1 shows the projection in plane z = 0 of the intersection between the hyperboloid and z = 1 (in red, quadrants 1 and 3) and z = -1 (in blue, quadrants 2 and 4).

Figure 1

The second equation is a plane in 3D as shown in Figure 2. The right side shows the intersection of the plane with z=0 (a straight line). The plane will intersect the z axis (x=0, y=0) for z=-3 (unfortunately the 3D Figure 2 does not show the z axis for z < 0.)

Figure 2

The intersection of both surfaces is a curved line. Figure 3 shows the intersection in 3D. The right side of Figure 3 shows the projection on plane z = 0 of the intersection between the two surfaces with z = -1 (red), z = 1 (purple) and z=3 (black). In z = - 1 there are two intersecting points (for any z < 0), each in quadrants second (sixth) and fourth (eighth) ; for z = 1 there is no intersection between both surfaces (the resulting curve does not exist for z = 1). Finally for z = 3 there are again two intersection points, both in quadrant first.

Figure 3

So we have concluded that the resulting curve if projected into the xy plane, it exists in the first, second and fourth quadrants.

Given the two equations above, if we obtain z in the first equation and substitute z in the second equation, we obtain the following equation that is graphed in Figure 4:

Figure 4

To obtain multiple x,y,z solutions, we can give a value to z, for example z = -1; that gives us two equations with two unknowns x and y:

if we have x = -4/y, substituting in the second equation gives us a quadratic equation on y:

that gives us y = 3 and y = -2. For those values, x = -4/3 and x=2 respectively. Therefore two solutions are: (-4/3, 3, -1) and (2, -2, -1). We can repeat this process for any negative value of z and we will obtain 2 points for each z value. We can also repeat this process for any positive z value > 3 and we will have again two points for each z value.

An excel sheet would give us fast a set of values. Tables 1 and 2 are snapshots of the excel sheet (I dont know why the cell format --3 decimal digits-- did not stay).

 z x1 y1 x2 y2 -0.00005 231.440640589718 -345.660985884577 -230.440657256385 347.160960884577 -0.05 7.81116592853181 -10.2417488927977 -6.82783259519848 11.7167488927977 -1 2 -2 -1.33333333333333 3 -1.5 1.60656838300831 -1.65985257451246 -1.10656838300831 2.40985257451246 -2 1.33333333333333 -1.5 -1 2 -2.5 1.11948540282464 -1.42922810423696 -0.95281873615797 1.67922810423696 -3 0.942809041582063 -1.41421356237309 -0.942809041582063 1.41421356237309 -3.5 0.793507134681919 -1.44026070202288 -0.960173801348585 1.19026070202288 -4 0.666666666666667 -1.5 -1 1 -4.5 0.559377904685193 -1.58906685702779 -1.05937790468519 0.839066857027789 -5 0.469439638586153 -1.70415945787923 -1.13610630525282 0.70415945787923 -5.5 0.394788566824125 -1.84218285023619 -1.22812190015746 0.592182850236188 -6 0.333333333333333 -2 -1.33333333333333 0.5 -6.5 0.283000429159038 -2.17450064373856 -1.4496670958257 0.424500643738557 -7 0.241846858492329 -2.36277028773849 -1.57518019182566 0.362770287738494 -7.5 0.208152156786987 -2.56222823518048 -1.70815215678699 0.312228235180481 -8 0.18046042171637 -2.77069063257455 -1.84712708838304 0.270690632574555 -8.5 0.157578774549565 -2.98636816182435 -1.9909121078829 0.236368161824347 -9 0.138550085106622 -3.20782512765993 -2.13855008510662 0.207825127659933 -9.5 0.122615614797174 -3.43392342219576 -2.28928228146384 0.18392342219576 -10 0.109177280600309 -3.66376592090046 -2.44251061393364 0.163765920900464 -10.5 0.0977641685281049 -3.89664625279216 -2.59776416852811 0.146646252792157

Table 1: Negative z values

 z x1 y1 x2 y2 0 #DIV/0! #DIV/0! #DIV/0! #DIV/0! 0.5 #NUM! #NUM! #NUM! #NUM! 1 #NUM! #NUM! #NUM! #NUM! 1.5 #NUM! #NUM! #NUM! #NUM! 2 #NUM! #NUM! #NUM! #NUM! 2.5 #NUM! #NUM! #NUM! #NUM! 2.75 #NUM! #NUM! #NUM! #NUM! 2.827182 #NUM! #NUM! #NUM! #NUM! 2.827183 0.971132244801556 1.45689313279767 0.971262088531777 1.45669836720233 3 0.666666666666667 2 1.33333333333333 1 3.5 0.441689858180387 2.58746521272942 1.72497680848628 0.662534787270581 4 0.333333333333333 3 2 0.5 4.5 0.26516122770912 3.35225815843632 2.23483877229088 0.39774184156368 5 0.217786631287899 3.67332005306815 2.44888003537877 0.326679946931849 5.5 0.182934135039183 3.97559879744123 2.65039919829415 0.274401202558775 6 0.156290375283575 4.26556443707464 2.84370962471642 0.234435562925363 6.5 0.135338846400446 4.54699173039933 3.03132782026622 0.20300826960067 7 0.118498265999156 4.82225260100126 3.21483506733418 0.177747398998734

Table 2: Positive z values

Create another set of equations that also yield a useful exploration.

This is another set of equations that represent two surfaces intersecting.

Figure 5 shows the resulting curve in 3D.

Figure 5

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