Students studying polar coordinates may be required to find the points of

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1 Finding Points of Intersection of Polar-Coordinate Graphs Warren W. Esty Mathematics Teacher, September 1991, Volume 84, Number 6, pp Mathematics Teacher is a publication of the National Council of Teachers of Mathematics (NCTM). More than 200 books, videos, software, posters, and research reports are available through NCTM S publication program. Individual members receive a 20% reduction off the list price. For more information on membership in the NCTM, please call or write: NCTM Headquarters Office 1906 Association Drive Reston, Virginia Phone: (703) Fax: (707) Internet: orders@nctm.org Article reprinted with permission from Mathematics Teacher, copyright September 1991 by the National Council of Teachers of Mathematics. All rights reserved. Students studying polar coordinates may be required to find the points of intersection of the polar-coordinate graphs of two functions, f and g. Their experiences with rectangular coordinates may lead them to expect that all the points of intersection can be found by solving the equation f g. They may be distressed to discover that points of intersection of the graphs can occur that do not correspond to solutions of that equation. Example 1. Consider the polar-coordinate graphs of and a circle, given in figure 1. Two points of intersection, B and C, obviously occur in the second and third quadrants. A natural approach to find these points is to set f equal to g and solve for. For the given functions this approach yields 1 3 cos 1, 3 cos 0, cos 0. f 1 3 cos g 1,

2 The solutions in the interval 0, 2 are given by 1 2 and 2 32, from which we obtain the points A 1, 2 and D 1, 32 in figure 1. The solutions of f g correspond to the two points of intersection on the y-axis, but none of the solutions corresponds to the points B and C in the second and third quadrants. Fig. 1. f 1 3 cos and g 1. Points B and C do not solve the equation f g. The points of intersection B and C occur when the two functions yield opposite functional values at angles that differ by, since the point r, is the same as the point r, in polar coordinates. Solving f g, we have 1 3 cos 1, 3 cos 2, cos 23, with solutions in 0, 2 given by 3 cos and , from which we obtain the points C 1, and B 1, What the graphics display shows so well is that the curves r f and r g pass through the points B and C at different times, that is, with different angles, and therefore these points of intersection will not be found by solving the equation that equates the two functions with the same angle,. (Example 1 is discussed further after theorem 2.) If any two functions are graphed in rectangular coordinates, all the points of intersection can be found by setting the functions equal to each other and solving. (See theorem 1 that follows.) The purpose of this article is to help students understand why the method of solving for points of intersection in polar coordinates is not precisely parallel to the rectangular-coordinate method. 2

3 Computer Graphics The computer program listed in the Appendix was created to plot two polar-coordinate graphs simultaneously and can plot individual graphs as well. Single plots could be used to familiarize students with polar-coordinate graphs before addressing the more complex problem of points of intersection. By plotting points with different angles at different times, this APPLEBASIC (or GW-BASIC) program vividly illustrates why points of intersection of the graphs of f and g can occur that do not correspond to solutions of the equation f g. It plots points from r f and r g simultaneously in increments of chosen by the user. A point corresponds to a solution of f g if and only if both plots go through the same point at the same time (that is, with equal angles). As the display develops, it is easy to see where the points of intersection are and which occur on both graphs at the same time. Points of intersection that are plotted at different times on the two graphs (such as points B and C in fig.1) cannot be found by solving f g. (See theorem 2 for the equations that yield the desired points.) Unfortunately, all the completed graphs in the figures lack the dynamic development over time that makes the computer display so appropriate. Input-output. The program listing includes, at lines 400 and 410, the pair of functions used to create figure 1, but the functions can easily be changed while running the program by rewriting lines 400 and 410 after entering zero in response to the first question. As long as the SAVE command is not used, changes in lines 400 and 410 will be only temporary. The independent variable must be T, and the form of the new definitions must be as follows: 400 F G... To plot one polar-coordinate graph instead of two, enter line 410 as 410 G F. Then only line 400 need be changed to create new graphs. The user is asked to supply four different inputs: (a) the number of points to plot as the variable goes through the angle (72 is recommended; the entry 0 is used to change functions), (b) the multiple of desired for the domain (usually 1), (c) a time-delay factor to permit slower development of the display (often the maximum speed, 1, works well), and (d) the maximum scale of the graph (which should be large enough to fit the graph on the screen). Figure 1 was created using the entries 72, 1, 1, and 3, respectively. Example 2. Consider the polar-coordinate graphs of f 3 cos 1 and g 1 sin given in figure 2. To obtain figure 2, load the program, type RUN, and then answer 0 to the first question. Then type two lines, 400 F 3 * COS T 1 and 410 G 1 SIN T, and restart the program by typing RUN. Then answer the four questions with 72, 1, 1, and 3, respectively. One of the intersections in the fourth quadrant, B, is found by solving the equation; and one, A, is not. 2 3

4 Fig. 2. f 3 cos 1 and g 1 sin. Point B corresponds to a solution of 3 cos 1 1 sin, but point A does not. Example 3. To produce the complicated picture in figure 3, in which some of the points of intersection correspond to solutions of the equation cos 1.8 cos and some do not, enter the lines 400 F COS T1.8 and 410 G COS 1 T1.8 and answer the questions with 72, 4.5, 1, and 1. Also, several of the intersections occur because each graph intersects with itself. Fig. 3. f cos /1.8, and g cos 1 /1.8, 0 9. Some of the points of intersection correspond to solutions of the equation cos /1.8 cos 1 /1.8, and some do not. Also, a number of the intersections occur because a graph intersects with itself. 4

5 The Mathematics The key to the difficulty of finding points of intersection of polar-coordinate graphs is that points do not have unique polar-coordinate representations. The lack of uniqueness is the reason that several equations (given in theorem 2) are required instead of just one, f g to guarantee finding all the points of intersection. For instructional purposes, it may be sufficient to illustrate visually how a point of intersection can appear with two distinct representations by using the computer program discussed previously. This section is devoted to the mathematics of why setting the functions equal and solving always works in rectangular coordinates but does not necessarily find all the points of intersection of two functions graphed in polar coordinates. Theorem 1A. In rectangular coordinates, if a solution to fx gx exists, then a corresponding point of intersection of the graphs of f and g exists. Theorem 1B. In rectangular coordinates, if a point of intersection of the graphs of f and g exists, then a corresponding solution to fx gx exists. Recall that the point x 0, y 0 is on the graph of f if fx 0 y 0. Proof of theorem 1A. Let x 0 be a solution to fx gx. Denote the common value of fx 0 and gx 0 by y 0. Then fx 0 y 0 and gx 0 y 0, and the point x 0, y 0 is on both graphs. Therefore, x 0, y 0 is a point of intersection of the graphs. Proof of theorem 1B. Suppose a point of intersection of the graphs of f and g exists. Since each point has a unique representation in rectangular coordinates, we can denote it uniquely by x 0, y 0. Then f x 0 y 0 because the point is on the graph of f and gx 0 y 0 because the point is on the graph of g. Thus fx 0 gx 0 by transitivity of equality. Therefore x 0 is a solution to the equation fx gx. Theorem 1A has a perfect polar-coordinate parallel that asserts that solutions to f g do correspond to points of intersection. The first proof can be paralleled exactly, replacing x by and y by r. No polar-coordinate parallel exists to theorem 1B, however, as proved by any one counterexample, such as the example in figure 1. Nevertheless, it is illuminating to determine why the proof of theorem 1B cannot be paralleled in polar coordinates. Recall that points do not have a unique representation in polar coordinates, for instance, 1, 3 1, 43 1, 73. The point r 0, 0 may be on the polar-coordinate graph of f even if f0 is not equal to r 0, since it may be on the graph with a different representation. Next consider a polar-coordinate parallel to the proof of theorem 1B. Suppose two polar-coordinate graphs intersect at a point. The point might be represented on the graph of f by r f, f and on the graph of g by r g, g. Then ff r f and gg r g, but transitivity of equality cannot be invoked to assert that ff gg, since the r coordinates may not be equal. Furthermore, the -coordinates may not be equal even if the r-coordinates are. Thus we cannot deduce that a exists such that f g. We cannot deduce the existence of a solution to f g corresponding to a point of intersection of the graphs. 5

6 Of course, all the points of intersection of the polar-coordinate graphs of f and g do correspond to solutions of equations just not necessarily the obvious equation f g. To discover all the equations that need to be checked we need to consider the various alternative representations of points. Points of intersection that do not satisfy f g can occur in three distinct ways: (a) the two r-coordinates may be equal but the angles differ by an integral multiple of 2; (b) the two r-coordinates may be opposite and the angles differ by plus an integral multiple of 2; and (c) both curves may pass through the pole, but with angles that are not the same. Theorem 2. In polar coordinates, the points of intersection of the graphs of f and g correspond to (1) the solutions of the equations f g 2n for all integral n; (2) the solutions of the equations f g 2n for all integral n; and (3) the pole, if a solution exists to f 0 and to g 0, regardless of whether the two solutions are the same. Proof. A point of intersection is either at the pole or not. First consider the case when the point is not at the pole. If one representation of the point is r f, f, then all others are given by r or r f, f f, f 2n 2n for integral values of n. Precisely two r-coordinate values are possible. If the two representations have the same r-coordinate, then r and g g r f f 2n, that is, the point will be found by solving the equations in (1). If the two r-coordinates are negatives of one another, then r and g f g r f 2n, that is, the point will be found by solving the equations in (2). Next consider the case when the point of intersection is the pole. The pole can be represented by 0, for any. Thus, r f 0 and r g 0, but the -coordinates may be any number. That fact justifies (3). Example 1, revisited. Reconsider the functions graphed in figure 1. The points of intersection at 1, 2 and 1, 32 are found by solving f g. Point B in the second quadrant occurs on the graph of g with r 1 when is in the second quadrant but with r 1 on the graph of f when is in the fourth quadrant. Thus point B can be found by solving f g 2, as in theorem 2, part 2. The computer graph can easily illustrate this point. Note that regardless of n. Enter the line 410 G 1 g 2n 1, in addition to the original line 400. The developing graphs pass through the points B and C in figure 1 at the same time, thus indicating that they can be found by solving f g 2n, as stated in part 2 of theorem 2. With this change in g the final plot is identical to that shown in figure 1; but the points B and C are the solutions of f g. And the graphs pass through B and C simultaneously but through A and D at different times. 6

7 With the original two functions, the computer graph of g can be made to stop at point C by entering in answer to the question about the multiple of to use. It is then easy to see that f went through point C with a different angle radian and that the current angle yields a point from f further out in the third quadrant. Example 4. The graphs of f sin and g cos yield a particularly simple example in which the equation f g does not yield all the points of intersection (fig. 4). The solution set in 0, 2 is 2, 32. At 2, r 22; and at 32, r 22. This result yields the point of intersection in the first quadrant twice. But the graphs of r sin and r cos cross at the pole, where r 0, which is a point of intersection not found by solving sin cos. Since sin 0 0, the point 0, 0 is on the graph of the sine function; and since cos 2 0, the point 0, 2 is on the graph of the cosine function. But 0, 0 and 0, 2 represent the same point, the pole, which is therefore a point of intersection of the two graphs, by theorem 2, part 3. 2 Fig. 4. f sin, and g cos. Point A corresponds to a solution of sin cos, but the pole, P, does not. Conclusion Solving for the points of intersection of two graphs in polar coordinates is not the simple problem it is in rectangular coordinates. Whether the proofs are analyzed in conjunction with computer graphics, the reason for the difference will be remembered by students who watch the computer simultaneously plot two graphs in polar coordinates. 7

8 APPENDIX The following computer program written in APPLEBASIC (DOS 3.1 and higher) will simultaneously plot two polar-coordinate graphs and can also plot individual graphs. For a GW-BASIC version of the program use the modifications following the program listing. 10 TEXT 20 PRINT " SUGGESTED ANSWERS TO THE INPUT QUESTIONS" 30 PRINT " ARE MARKED WITH AN ASTERISK(*). " 40 PRINT " " 50 PRINT " TO CHANGE FUNCTIONS ENTER ' 0' NEXT. " 60 PRINT " " 70 PRINT " HOW MANY POINTS IN ANGLE PI? (72*)" 80 INPUT P 90 IF P > 0 THEN PRINT " " 110 PRINT " " 120 PRINT " ENTER NEW LINES 400 AND 410 NOW. " 130 PRINT " DEFINE FUNCTIONS F AND G WITH THE" 140 PRINT " VARIABLE NAMED ' T '. FOR EXAMPLE, TYPE: " 150 PRINT " " 160 PRINT " 400 F 3* COST 1" 170 PRINT " 410 G 1 SINT" 180 PRINT " RUN" 181 PRINT " " 182 PRINT " TO CREATE SINGLE GRAPHS, ENTER" 183 PRINT " LINE 400 AS ABOVE AND ' 410 G F'". 184 PRINT " THEREAFTER, CHANGE ONLY LINE 400. " 190 GOTO PRINT " MULTIPLE OF 2*PI FOR THE DOMAIN? (1*)" 210 INPUT N 220 PRINT " ENTER THE TIME DELAY FACTOR (1*)." 225 PRINT " LARGER NUMBERS SLOW THE PLOTTING. " 230 INPUT D 240 PRINT " MAXIMUM SCALE VALUE? (3*)" 8

9 250 INPUT S 260 W 90S 270 D D*D*D 280 HOME: HGR: HCOLOR HPLOT 125, 1 TO 125, HPLOT 1,90 TO 250, HPLOT W,89 TO W, HPLOT 125W,89 TO 125W, HPLOT 124,90W TO 126,90W 340 HPLOT 124,90+W TO 126,90+W 370 IT 2*N*P 380 FOR 1 0 TO IT 390 T # *I/P 400 F 1 3*COS(T) 410 G X1 F*COS(T) 430 Y1 F*SIN(T) 440 X2 G*COS(T) 450 Y2 G*SIN(T) 460 X1 125W*X1 470 Y1 90W*Y1 480 X2 125W*X2 490 Y2 90W*Y2 500 IF I 0 THEN HPLOT U1, V1 TO X1, Y1 520 HPLOT U2, V2 TO X2, Y2 530 U1 X1: V1 Y1 540 U2 X2: V2 Y2 550 FOR K 1 TO D 560 NEXT K 570 NEXT END. 9

10 APPLEBASIC modifications. The program is complete as listed, but the APPLEBASIC program may halt unexpectedly if a scale factor is entered that is too small to fit the whole graph on the screen. The insertion of the following lines will protect against that possibility by automatically using a larger scale factor if necessary. 265 SX 1.388*S 421 IF ABSX1 > SX THEN IF ABSY1 > S THEN IF ABSX2 > SX THEN IF ABSY2 < S THEN S S1 453 GOTO 265 GW-BASIC modifications. The aforementioned program uses the APPLEBASIC HPLOT command. A similar program will run using IBM compatible versions of BASIC (assuming they permit graphics), such as GW-BASIC 2.14, where the LINE command replaces the HPLOT command. Lines , , and remained unchanged. Replace line 10 with 10 CLS, replace lines with 280 SCREEN 1: CLS 290 LINE 0,90250, LINE 125,0125, LINE 125W,89125W, LINE 125W,89125W, LINE 124,90W126,90W 340 LINE 124,90W126,90W, and replace lines 510 and 520 with 510 LINE U1,V1X1,Y1 and 520 LINE U2,V2X2,Y2. No need arises for the APPLEBASIC modifications mentioned earlier. 10

11 Program notes. The coordinate systems of the HPLOT and LINE commands are such that 0, 0 is the upper-left corner of the screen. This program treats the screen as 250 horizontal units by 180 vertical units, and references to 125 and 90 in the program refer to the center of the screen. Line 270 defines W, which is the screen equivalent of one unit in the original system. Line 280 clears the screen and puts it in the graphics mode. Lines draw the axes and mark one unit from the pole in each direction. Line 460 shifts and rescales the original x-coordinate value. Line 470 subtracts because the HPLOT command coordinate system is upside down with respect to the usual x-y coordinate system. Line 240 merely creates a large number from the time-delay factor D so that the development of the picture can be slowed by the loop If the program is used to plot two graphs, I recommend omitting lines , since they clutter the screen and merely inform the user how to plot a single graph. Line 225 can be omitted also. If your computer expects a different screen width, graphs that should be circles may appear as ellipses. To remedy that problem, the user can implement a width-changing command described in the manual; or simply insert a line reading 275 WX 100S and change W in lines 310, 320, 460, and 480 to WX. If, in line 275, the number 100 which would expand the horizontal scale by a factor of is not correct, another number should be tried. The author thanks Lyle Anderson, mathematics professor at Montana State University, and the referees for suggestions that substantially improved this article. 11