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7sin1517sin452sin1205.18 square unitsA1801535130 293 The distance a, from the first station to the aircraft, is about 14.98 miles. Now that we know a, we can use right triangle relationships to solve for the height, h, of the aircraft. The aircraft is at an altitude of approximately 3.9 miles. Example 7.2.3. Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is 30° and at the second point is 45°. Assuming a straight coastline, find the distance from the second observation to the island. What point on the shore is closest to the island? How far is the island from this point? Solution. We sketch the problem below with the first observation point labeled as P and the second as Q. sin130sin3520sin13020sin3520sin35sin13014.98aaaasin15sin1514.98sin153.88hahahhh 15 14.98 milesa 294 In order to use the Law of Sines to find the distance d from Q to the island, we first need to find the measure of β which is the angle opposite the side of length 5 miles. To that end, we note that the angles γ and 45° are supplemental, so that. We can now find β. By the Law of Sines, we have Next, to find the point on the coast closest to the island, which we’ve labeled as C, we need to find the perpendicular distance from the island to the coast.1 Let x denote the distance from the second observation point Q to the point C and let y denote the distance from C to the island. Using the right triangle definition of sine, we
get 1 Do you see why C must lie to the right of Q? 180451351803018030135155sin30sin155sin30sin159.66 milesddd 295 Hence, the island is approximately 6.83 miles from the coast. To find the distance from Q to C, we note that so by symmetry, we get miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point. We close this section with the encouragement that, by working through the many problems in the Exercises, you will become proficient in applying the Law of Sines to real-world applications, and will be ready to move on to the Law of Cosines in Section 7.3. sin45sin4529.6626.83 milesydydyy1809045456.83xy 296 7.2 Exercises In Exercises 1 – 6, find the area of each triangle. Round each answer to the nearest tenth. 1. 3. 5. 2. 4. 6. 297 7. Find the area of the triangles. As in the text,, and are angle-side opposite pairs. (a) (b) (c),,,,, units units units, units 8. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in the figure. Determine the distance of the boat from station A and the distance of the boat from shore. Round your answers to the nearest whole foot. 9. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to
be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot. 10. A man and a woman standing 3.5 miles apart spot a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot. 11. Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile. 12. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1040 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. ,a,b,c13175a535328.01c5025a12.5b 298 13. Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be 75° and radios Sally immediately to find the angle of inclination from her position to the craft is 50°. How high off the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.) 14. A yield sign measures 30 inches on all three sides. What is the area of the sign? Grade: The grade of a road is much like the pitch of a roof in
that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 15 – 17, we first have you change road grades into angles and then use the Law of Sines in an application. 15. Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a 7% grade means that the road (hypotenuse) makes about a 4° angle with the horizontal. (It will not be exactly 4°, but it is pretty close. 16. What grade is given by a 9.65° angle made by the road and the horizontal? 17. Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road.1 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is 6°. Use the Law of Sines to find the height of the tree. (Hint: First show that the tree makes a 94° angle with the road.) 1 The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 299 Bearings (Another Classic Application): In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, N40°E (read “40° east of north”) is a bearing which is rotated clockwise 40° from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of. Similarly, S50°W would point into Quadrant III along the terminal side of because
we started out pointing due south (along ) and rotated clockwise 50° back to 220°. Counter- clockwise rotations would be found in the bearings N60°W (which is on the terminal side of ) and S27°E (which lies along the terminal side of ). These four bearings are drawn in the plane below. The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.) 18. Find the angle θ in standard position with which corresponds to each of the bearings given below. (a) due west (b) S83°E (c) N5.5°E (d) due south (e) N31.25°W (f) S72°41’12”W (g) N45°E (h) S45°W 502202701502970360 300 19. The Colonel spots a campfire at a bearing N42°E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the fire to be N20°W from his current position. Determine the distance from the campfire to each man, rounded to the nearest foot. 20. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N53°W which brings her to the Muffin Ridge Observatory. From there, she knows a bearing of S65°Ewill take her straight back to Sasquatch Point. How far will she have to walk to get from the Muffin Ridge Observatory to Sasquatch Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 21. The captain of the SS Bigfoot sees a signal flare at a bearing of N
15°E from her current location. From his position, the captain of the HMS Sasquatch finds the signal flare to be at a bearing of N75°W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50°E, find the distances from the flare to each vessel, rounded to the nearest tenth of a mile. 22. Carl spies a potential Sasquatch nest at a bearing of N10°E and radios Jeff, who is at a bearing of N50°E from Carl’s position. From Jeff’s position, the nest is at a bearing of S70°W. If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot. 23. A hiker determines the bearing to a lodge from her current position is S40°W. She proceeds to hike 2 miles at a bearing of S20°E at which point she determines the bearing to the lodge is S75°W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. 24. A watchtower spots a ship off shore at a bearing of N70°E. A second tower, which is 50 miles from the first at a bearing of S80°E from the first tower, determines the bearing to the ship to be N25°W. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile. 25. The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is 55°. From a point five stories below the original observer, the angle of inclination to the gargoyle is 20°. Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.) 301 26. Use the cases and diagrams in the proof of the Law of Sines (Theorem 7.1) to prove the area formulas given in Theorem 7.3. Why do those formulas yield square units when four quantities are being multiplied together? 302 7.3 The Law of Cosines Learning Objectives In this section you will:  Use The Law of Cosines to solve oblique triangles.  Solve SAS and SSS
Triangles.  Use Heron’s Formula to find the area of a triangle.  Solve applied problems using the Law of Cosines. In Section 7.1, we developed the Law of Sines (Theorem 7.1) to enable us to solve triangles in the ‘Angle-Side-Angle’ (ASA), the ‘Angle-Angle-Side’ (AAS) and the ambiguous ‘Angle-Side-Side’ (ASS) cases. In this section, we develop the Law of Cosines which handles solving triangles in the ‘Side-AngleSide’ (SAS) and ‘Side-Side-Side’ (SSS) cases.1 The Law of Cosines We state and prove the Law of Cosines theorem below. Theorem 7.4. The Law of Cosines. Given a triangle with angle-side opposite pairs, and, the following equations hold or, solving for the cosine in each equation, we have To prove the theorem, we consider a generic triangle with the vertex of angle α at the origin, and with side b positioned along the positive x-axis. 1 Here, ‘Side-Angle-Side’ means that we are given two sides and the included angle – that is, the given angle adjacent to both of the given sides. ,a,b,c2222cosabcbc2222cosbacac2222coscabab222cos2bcabc222cos2acbac222cos2abcab 303 From this set-up, we immediately find that the coordinates of A and C are and. From Theorem 2.6, we know that since the point lies on a circle of radius c, the coordinates of B are. (This
would be true even if α were an obtuse or right angle so although we have drawn the case where α is acute, the following computations hold for any angle α drawn in standard position where.) We note that the distance between the points B and C is none other than the length of side a. Using the distance formula, we get The remaining formulas given in Theorem 7.4 can be shown by simply reorienting the triangle to place a different vertex at the origin. We leave these details to the reader. What’s important about a and α in the above proof is that is an angle-side opposite pair and b and c are the sides adjacent to α. The same can be said of any other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula, which has its roots in the Pythagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the 0,0A,0Cb,Bxy,cos,sinBxyBcc0180222222222222222222222222222222cossin0cossincossincos2cossincossin2cos12cosc2ossin1acbcacbcacbcacbcbcabcbcabcbcabcbc�
�� since cos,a 304 Pythagorean Theorem. If we have a triangle in which, then so we get the familiar relationship. What this means is that in the larger mathematical sense, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.2 Example 7.3.1. Solve the triangle in which, units, and units. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. Solution. We are given the lengths of two sides, and, and the measure of the included angle,. With no angle-side opposite pair to use, we apply the Law of Cosines to find b. In order to determine the measures of the remaining angles α and γ, we are forced to use the derived value for b. There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that, unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute angle is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles is positive, the sine of an angle alone is not enough to determine if the angle in question is acute or obtuse. We proceed with the Law of Cosines. When using the Law of Cosines, It’s always best to find the measure of the largest unknown angle first, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to find α first. To that end, we use the formula for from Theorem 7.4 and substitute, and. 2 This shouldn’t come as too much of a
shock. All of the theorems in Trigonometry can ultimately be traced back to the definition of the circular functions along with the distance formula and, hence, the Pythagorean Theorem. 90coscos900222cab507a2c7a2c502222222cos72272cos505328cos505.92 unitsbacacbbb,bcos7a5328cos50b2c 305 Since α is an angle in a triangle, we know the radian measure of α must lie between 0 and π radians. This matches the range of the arccosine function, so we have At this point, we could find γ using the Law of Cosines, to minimize the propagation of error. Or we can take the shortcut of using angle measures that have already been determined. We sketch the triangle below. As mentioned earlier in the preceding example, once we’d determined b it was possible to use the Law of Sines to determine the remaining angles. However, noting that this was the ambiguous ASS case, proceeding with the Law of Sines would require caution. It is advisable to first find the smallest of the unknown angles, since we are guaranteed it will be acute.3 In this case, we would find γ since the side 3 There can only be one obtuse angle in the triangle, and if there is one, it must be the largest. 
222222cos25328cos5027cos25328cos50227cos50cos5328cos50bcabc after simplifying27cos50arccos radians5328cos50114.99180180114.995015.01 306 opposite γ is smaller than the side opposite the other unknown angle, α. Using the Law of Sines with the angle-side opposite pair, we get The usual calculations produce and we find Example 7.3.2. Solve for the angles in the triangle with sides of length units, units and units. Solution. Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we find β first, since it is opposite the longest side, b. We then determine β through finding the arccosine of. As in the previous problem, now that we have obtained an angle-side opposite pair we could proceed using the Law of Sines. The Law of Cosines, however, offers us a rare opportunity to find the remaining angles using only the data given to us in the statement of the problem. Using the problem data, we get and. ,bsin50sin25328cos5015.011801805015.01114.994a7b5c�
��222222cos245724515acbac151arccos radians5101.54,b5arccos radians 44.42729arccos radians 34.0535 307 We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may differ slightly from those posted in the Examples and the Exercises. Example 7.3.2 is a great example of this, in that the approximate values we record for the measures of the angles sum to 180.01°, which is geometrically impossible. Solving Applied Problems Using the Law of Cosines Next, we have an application of the Law of Cosines. Example 7.3.3. A researcher wishes to determine the width of a vernal pond as drawn below. From a point P, he finds the distance to the western-most point of the pond to be 950 feet, while the distance to the northern-most point of the pond from P is 1000 feet. If the angle between the two lines of sight is 60°, find the width of the pond. 308 Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to find the length of the missing side opposite the given angle. Calling this length w (for width), we get We next take the square root to get Heron’s Formula In Section 7.2, we used the proof of the Law of Sines to develop Theorem 7.3 as an alternate formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive another such formula – Heron’s Formula. Theorem 7.5. Heron’s Formula. Suppose a
, b and c denote the lengths of the three sides of a triangle. Let s be the semiperimeter of the triangle, that is, let. Then the area A, enclosed by the triangle, is given by We prove Theorem 7.5 using Theorem 7.3. Using the convention that the angle γ is opposite the side c, we have from Theorem 7.3. In order to simplify computations, we start by manipulating the expression for A2. The Law of Cosines tells us, so substituting this into our equation for A2 gives 222950100029501000cos60952,500w952500976 feetw12sabcAssasbsc1sin2Aab22222222221sin21sin41cos4cossin1Aababab since 222cos2abcab 309 At this stage, we recognize the last factor as the semiperimeter,. To complete the proof, we note that 222222222222222222222222222222222221cos41421444444162abAababcababcababababcababababcaba�
�� Law of Cosines2222222222222222222222162216221616bcababccaabbaabbccaabbaabbccababccabca difference of squares perfect square trinomials16162222babcabcbcaacbabcabcbcaacbabcabc�
�� difference of squares122abcsabc2222abcsaaabcabca 310 Similarly, we find and. Hence, we get so that as required. We close with an example of Heron’s Formula. Example 7.3.4. Find the area enclosed by the triangle in Example 7.3.2. Solution. In Example 7.3.2, we are given a triangle with sides of length units, units and units. Using these values, we find. Using Heron’s Formula, we get 2acbsb2abcsc22222bcaacbabcabcAsasbscsAssasbsc4a7b5c145782s8848785841396469.80 square unitsAssasbsc 311 7.3 Exercises In Exercises 1 –
4, solve for the length of the unknown side. Round to the nearest tenth. 1. 3. 2. 4. In Exercises 5 – 8, find the measure of angle A. Round to the nearest tenth. 5. 6. 7. 312 8. 9. Find the measure of each angle in the triangle. Round to the nearest tenth. In Exercises 10 – 19, use the Law of Cosines to find the remaining side(s) and angles(s) if possible. 10. 12. 14. 16. 18. 11. 13. 15. 17. 19. In Exercises 20 – 25, solve for the remaining side(s) and angle(s), if possible, using any appropriate technique. 20. 22. 24. 21. 23. 25. 7, 12, 59.3ab104, 25, 37bc153, 8.2, 153ac3, 4, 90ab120, 3, 4bc7, 10, 13abc1, 2, 5abc300, 302, 48abc5, 5, 5abc5, 12, 13abc18, 63, 20ab37, 45, 26abc16, 63, 20ab22, 63, 20ab42, 117, 88bc7, 170, 98.6c 313 In Exercises 26 – 29, find the area of the triangle. Round to the nearest hundredth. 26. 27. 28. 29. 30. Find the area of the triangles. (a) (b) (c) 7, 10, 13abc300, 302,
48abc5, 12, 13abc 314 31. A rectangular octagon is inscribed in a circle with a radius of 8 inches. Find the perimeter of the octagon. 32. A rectangular pentagon is inscribed in a circle of radius 12 cm. Find the perimeter of the pentagon. 33. The hour hand on my antique Seth Thomas schoolhouse clock is 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o’clock. Round your answer to the nearest hundredth of an inch. 34. A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern- most point of the crater and 2 miles to the southern-most point. If the angle between the two lines of sight is 117°, what is the diameter of the crater? Round your answer to the nearest hundredth of a mile. 35. From the Pedimaxus International Airport a tour helicopter can fly to Cliffs of Insanity Point by following a bearing of N8.2°E for 192 miles and it can fly to Bigfoot Falls by following a bearing of S68.5°E for 207 miles.1 Find the distance between Cliffs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile. 1 Please refer to the 7.2 Exercises for an introduction to bearings. 315 36. Cliffs of Insanity Point and Bigfoot Falls from Exercise 35 both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliffs of Insanity Point? Round your angle to the nearest tenth of a degree. 37. A naturalist sets off on a hike from a lodge on a bearing of S80°W. After 1.5 miles, she changes her bearing to S17°W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. 38. The HMS Sasquatch leaves port on a bearing of N23°E and travels for 5 miles. It then changes course and follows a heading of S41°E for 2 miles. How far is it from port? Round your answer to the nearest hundredth
of a mile. What is its bearing to port? Round your angle to the nearest degree. 39. The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N32°E. A storm moves in and after 100 miles, the captain of the Bigfoot finds he has drifted off course. If his bearing to the harbor is now S70°W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the nearest tenth of a degree. 40. From a point 300 feet above level ground in a firetower, a ranger spots two fires in the Yeti National Forest. The angle of depression made by the line of sight from the ranger to the first fire is 2.5° and the angle of depression made by line of sight from the ranger to the second fire is 1.3°. The angle formed by the two lines of sight is 117°. Find the distance between the two fires. Round your answer to the nearest foot. (Hint: In order to use the 117° angle between the lines of sight, you will first need to use right angle Trigonometry to find the lengths of the lines of sight. This will give you a SAS case in which to apply the Law of Cosines.) 316 41. If you apply the Law of Cosines to the ambiguous ASS case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to each of the following in order to demonstrate this result. (a) (b) (c) 42. Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are being multiplied together. 18, 63, 20ab16, 63, 20ab22, 63, 20ab 317 CHAPTER 8 POLAR COORDINATES AND APPLICATIONS Chapter Outline 8.1 Polar Coordinates 8.2 Polar Equations 8.3 Grap
hing Polar Equations 8.4 Polar Representations for Complex Numbers 8.5 Complex Products, Powers, Quotients and Roots Introduction Chapter 8 takes us from graphing on the Cartesian coordinate plane to graphing on a polar grid, using the pole and polar axis for reference. We begin in Section 8.1 by plotting points defined by polar coordinates. The geometry and connection between the pole and origin, polar axis and positive x-axis, lead to the conversion of points between polar and rectangular coordinates. Section 8.2 continues this theme by converting equations back and forth between polar form and rectangular form. Graphing is the focus of Section 8.3, beginning with circles and lines in the coordinate plane, then moving on to more complicated polar graphs. Throughout Section 8.3, techniques are introduced and emphasized that enable the student to complete polar graphs by hand, without the aid of technology. The theme of Section 8.4 is complex numbers, as represented on the complex plane. A polar definition for complex numbers is introduced, and practice is provided for converting between rectangular and polar forms. This chapter ends with Section 8.5 and the introduction of ‘arithmetic‘ with complex numbers. DeMoivre’s Theorem is included as a means for determining powers and roots of complex numbers. This chapter introduces new concepts that rely on the trigonometric tools and skills developed up to this point in the course. It provides valuable insight into polar graphs and complex numbers. 318 8.1 Polar Coordinates Learning Objectives In this section you will:  Graph points in polar coordinates.  Convert points in polar coordinates to rectangular coordinates and vice versa. Up to this point, we have graphed points in the Cartesian coordinate plane by assigning ordered pairs of numbers to points in the plane. We defined the Cartesian coordinate plane using two number lines, one horizontal and one vertical, which intersect at right angles at a point called the origin. To plot a point, say, we start at the origin, travel horizontally to the left 3 units, then up 4 units. Alternatively, we could start at the origin, travel up 4 units, then to the left 3 units and arrive at the same location. For the most part, the motions of the Cartesian system (over and up) describe a rectangle, and most points can be thought of as the corner diagonally across the rectangle from the origin.1 For this reason, the Cartesian coordinates of a point are often called rectangular coordinates
. In this section, we introduce polar coordinates, a new system for assigning coordinates to points in the plane. Plotting Polar Coordinates We start with an origin point, called the pole, and a ray called the polar axis. We then locate a point P using two coordinates,, where r represents a directed distance from the pole2 and θ is a measure of 1 Excluding, of course, the points in which one or both coordinates are 0. 2 We will explain more about this momentarily. 3,4P,rxyP(-3,4) 319 rotation from the polar axis. Roughly speaking, the polar coordinates of a point measure how far out the point is from the pole (that’s r) and how far to rotate from the polar axis (that’s θ). Example 8.1.1. Plot the point P with polar coordinates. Solution. We start at the pole, move out along the polar axis 4 units, then rotate radians counter- clockwise. First Second The Resulting Point We may also visualize this process by thinking of the rotation first.3 To plot this way, we rotate radians counter-clockwise from the polar axis, then move outwards from the pole 4 units. Essentially, we are locating a point on the terminal side of which is 4 units away from the pole. 3 As with anything in Mathematics, the more ways you have to look at something, the better. Take some time to think about both approaches to plotting points given in polar coordinates. ,r54,65654,6P5656Polar AxisPole r r  ,Pr�
�Pole 4rPole 56Pole 54,6P 320 First Second The Resulting Point If, we begin by moving, from the pole, in the opposite direction of the polar axis. Example 8.1.2. Plot. Solution. We start at the pole, moving 3.5 units in the opposite direction of the polar axis. We then rotate units counter-clockwise. First Second The Resulting Point If we interpret the angle first, we rotate radians, then move back through the pole 3.5 units. Here we are locating a point 3.5 units away from the pole on the terminal side of, not. 0r3.5,4Q44544Pole 56Pole 56Pole 54,6PPole 3.5rPole 4Pole 3.5,4Q 321 First Second The Resulting Point As you may have guessed, means the rotation away from the polar axis is clockwise instead of counter-clockwise. Example 8.1.3. Plot. Solution. To plot, we have the following. First Second The Resulting Point From an ‘angles first’ approach, we rotate then move out 3.5 units from the pole. We see that R is the point on the terminal side of which is 3.5 units from the pole. 033.5,4R33.5,4R3434
Pole 4Pole 4Pole 3.5,4QPole 3.5rPole 34Pole 33.5,4R 322 First Second The Resulting Point Multiple Representations for Polar Coordinates The points Q and R in the above examples are, in fact, the same point despite the fact that their polar coordinate representations are different. Unlike Cartesian coordinates where and represent the same point if and only if and, a point can be represented by infinitely many polar coordinate pairs. We explore this notion in the following examples. Example 8.1.4. Plot the point, given in polar coordinates, and then give two additional expressions for the point, one of which has and the other with. Solution. Whether we move 2 units along the polar axis and then rotate 240° or rotate 240° then move out 2 units from the pole, we plot below. We now set about finding alternate descriptions for the point P. Since P is 2 units from the pole,. Next, we choose angles θ for each of the r values. The given representation for P is so the angle θ we choose for the case must be coterminal with 240°. (Can you see why?) One ,ab,cdacbd2,240P0r0r2,240P,r2r2,2402rPole 34Pole 34Pole 33.5,4RPole 240Pole 2,240P 323 such angle is so one answer for this case is. For
the case, we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate to arrive at a location coterminal with 240°. Hence, our answer here is. We check our answers by plotting them. Example 8.1.5. Plot the point and give two additional expressions for the point, one with and the other with. Solution. We plot by first moving 4 units to the left of the pole and then rotating radians. We find our point lies 4 units from the pole on the terminal side of. To find alternate descriptions for P, we note that the distance from P to the pole is 4 units, so any representation for P must have. As we noted above, P lies on the terminal side of, so this, coupled with, gives us as one of our answers. To find a different representation for P 1202,1202r602,6074,6P0r0r74,6766,r4r64r4,6Pole 120 2,120PPole 60 2,60PPole 76Pole 74,6P 324 with, we may choose any angle coterminal with the angle in the original representation of. We pick and get as our second answer. Example 8.1.6. Plot the point and give two additional expressions for the point, one with and the other with. Solution. To plot, we move along the polar axis 117
units from the pole and rotate clockwise radians as illustrated below. Since P is 117 units from the pole, any representation for P satisfies. For the case, we can take θ to be any angle coterminal with. In this case, we choose and get as one answer. For the case, we visualize moving left 117 units from the pole and 4r74,6P5654,65117,2P0r0r5117,2P52,r117r117r52323117,2117rPole 6 4,6PPole 54,6P 56 52 117rPole 5117,2P 325 then rotating through an angle θ to reach P. We find that satisfies this requirement, so our second answer is. Example 8.1.7. Plot the point and give two additional expressions for the point, one with and the other with. Solution. We move three units to the left of the pole and follow up with a clockwise rotation of radians to plot. We see that P lies on the terminal side of. 2
117,23,4P0r0r43,4P34Pole 3117,2P 32Pole 2 117,2PPole 4Pole 3,4P 326 Since P lies on the terminal side of, one alternative representation for P is. To find a different representation for P with, we may choose any angle coterminal with. We choose for our final answer of. Now that we have had some practice with plotting points in polar coordinates, it should come as no surprise that any given point expressed in polar coordinates has infinitely many other representations in polar coordinates. The following property of the polar coordinate system summarizes characteristics of different polar coordinates that determine the same point in the plane. Equivalent Representations of Points in Polar Coordinates Suppose and are polar coordinates where, and the angles are measured in radians. Then and determine the same point P if and only if one of the following is true:   and and for some integer k for some integer k All polar coordinates of the form represent the pole regardless of the value of θ. The key to understanding this result, and indeed the whole polar coordinate system, is to keep in mind that means (directed distance from pole, angle of rotation). If, then no matter how much rotation is performed, the point never leaves the pole. Thus, is the pole for all values of θ. 3433,4
3r47473,4,r','r0r'0r,r','r'rr'2k'rr'21k0,,r0r0,Pole 34 33,4PPole 73,4P 74 327 Now let’s assume that neither r nor r' is zero. If and determine the same point P then the (non-zero) distance from P to the pole in each case must be the same. Since this distance is controlled by the first coordinate, we have either or. 1. If then, when plotting and, the angles and have the same initial side. Hence, if and determine the same point, we must have that is coterminal with. We know that this means for some integer k, as required. 2. If, on the other hand,, then when plotting and the initial side of is rotated radians away from the initial side of. In this case, must be coterminal with. Hence, which we rewrite as for some integer k. Conversely, 1. If and for some integer k, then the points and lie the same (directed) distance from the pole on the terminal sides of coterminal angles, and hence are the same point. 2. Suppose that and for some integer k. To plot P, we first move a directed distance r from the pole; to
plot, our first step is to move the same distance from the pole as P, but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated exactly π radians apart. Since for some integer k, we see that is coterminal to and it is this extra π radians of rotation which aligns the points and. Converting Between Rectangular and Polar Coordinates To move points from the polar coordinate system to the Cartesian (rectangular) coordinate system, or vice versa, we identify the pole and polar axis in the polar system to the origin and positive x-axis, respectively, in the rectangular system. ,r','r'rr'rr'rr,r','r',r','r''2k'rr,r','r'''2k'21k'rr'2k,Pr'','Pr'rr'21k'P'212kk'
P'P 328 If we have a polar point in Quadrant I, we can form a right triangle by first dropping a perpendicular line segment from the point to the point, on the positive x-axis, to form a vertical leg. To form a horizontal leg, we sketch the line segment from the origin to the point. Finally, the hypotenuse is the line segment from the origin to the polar point. The lengths of the legs of this triangle, x and y, are the corresponding rectangular coordinates for the polar point. Using right triangle trigonometry, we can express x and y in terms of r and θ: Translating a polar point to its rectangular representation is fairly straightforward using these two formulas for x and y. Suppose, on the other hand, we want to translate a rectangular point to its polar representation. We can find r using the Pythagorean Theorem. ,r,r,0r,0r,r,xy,rcoscosxrxrsinsinyryr,xy,r222xyrpoleoriginpolar axispositivex-axis ,rpoleoriginpolar axispositivex-axisxy ,r  Then, to determine θ, we use the tangent. 329 For Quadrant II, III or IV, we can use a reference angle,, and include the appropriate signs on x and y, as determined by the quadrant in which they lie. Recall that, since for, to find an angle θ in Quadrant II or III it will be necessary to add π to obtain the correct angle. Also keep in mind that a point in polar coordinates can be expressed in many ways since where k is an integer. We get the following result. Theorem 8.1. Conversion Between Rectangular and Polar Coordinates: Suppose P
is represented and in polar coordinates as in rectangular coordinates as. Then   and and (provided To verify this result, we check out the three cases: ), and. 1. In the case, the theorem is an immediate consequence of Theorem 2.6. Recall that We apply the quotient identity to verify that. tanyxarctantan22,,2rrk,xy,rcosxrsinyr222xyrtanyx0x0r0r0r0rcoscossinsinxxrryyrrsintancostanyxxy ,r   330 2. If, then we know an alternate representation for is. Using and, we apply Theorem 2.6 as follows. Moreover, case too. and, so the theorem is true in this 3. The remaining case is, in which case is the pole. Since the pole is identified with the origin in rectangular coordinates, the theorem in this case amounts to checking ‘ ’. The following example puts Theorem 8.1 to good use. Example 8.1.8. Convert each point in rectangular coordinates given below into polar coordinates with and. Use exact values if possible and
round any approximate values to two decimal places. Check your answer by converting back to rectangular coordinates. 1. 2. 3. 4. Solution. 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking the time to plot the point before we do any calculations. Plotting shows that it lies in Quadrant IV. With and, we get 0r,r,rcoscossinsincoscoscosxrrrsinsinsinyrrr2222xyrrtantanyx0r,0,r0,0000r022,23P3,3Q0,3R3,4S2,23P
2,23P2x23y2222222341216rxyxy 2,23P 53 331 So and, since we are asked for, we choose. To find θ, we have that This tells us θ has a reference angle of, and since P lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have, so we choose. Hence, our answer is. To check, we convert back to rectangular coordinates and we find The result is the point in rectangular coordinates, as required. 2. The point is in Quadrant III. Using, we get so We find, which means θ has a reference angle of. Since Q lies in Quadrant III, we choose, which satisfies the requirement that. Our final answer is. 4r0r4rtan2323yx3025354,35,4,3rcos54cos31422xrsin54sin334223yr2,23
3,3Q3xy2223318r183232r since we are asked for r03tan13454025,32,4rxy 3,3Q 54 332 To check, we find The resulting point verifies our solution. 3. The point lies along the negative y-axis. While we could go through the usual computation4 to find the polar form of R, in this case we can find the polar coordinates of R using the definition. Since the pole is identified with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation of R we know. Since we require, we choose. Concerning θ, the angle satisfies with its terminal side along the negative y-axis, so our answer is. To check, we note 4 Since x = 0, we would have to determine θ geometrically. cos532cos423223xrsin532sin423223yr3,3
0,3R,r3r0r3r320233,2cos33cos2300xrsin33sin2313yrxy 0,3R 32 333 4. The point lies in Quadrant II. With and, we get so. As usual, we choose and proceed to determine θ. We have Since this isn’t the tangent of one of the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisfy, we choose radians. Hence, our answer is. To check our answers requires a bit of tenacity since we need to simplify expressions of the form and. These are good review exercises and are hence left to the reader. We find and, so that Now that we’ve had practice converting representations of points between the rectangular and polar coordinate systems, we move on to the next section where we will convert equations from one system to another. 3,4S3x4y2223425r5r50rtan4343yx024arctan3�
��4,5,arctan5,2.213r4cosarctan34sinarctan343cosarctan3544sinarctan35cos3553xrsin4554yrxy 3,4S 4arctan3 334 8.1 Exercises In Exercises 1 – 16, plot the point given in polar coordinates and then give three different expressions for the point such that (a) and (b) and (c) and 1. 5. 9. 13. 2. 6. 10. 14. 3. 7. 11. 15. 4. 8. 12. 16. In Exercises 17 – 36, convert the point from polar coordinates into rectangular coordinates. 20. 24. 28. 18. 22. 26. 17. 21. 25. 29. 31. 33. 19. 23. 27. 30. 32. 34. 0
r020r00r22,375,413,3255,26712,653,422,713,2620,354,421,33,2113,62.5,445,3,75,42,3711,6�
�20,33,5254,679,295,41342,6117,1176,arctan210,arctan343,arctan345,arctan312,arctan21,arctan5231,arctan42,arctan223 35. 335 36. In Exercises 37 – 56, convert the point from rectangular coordinates into polar coordinates with and. 37. 41. 45. 49. 53. 38. 42. 46. 50. 54.
39. 43. 47. 51. 55. 40. 44. 48. 52. 56. ,arctan1213,arctan50r020,53,37,73,33,02,24,4331,44333,10105,56,85,258,1210,6105,12525,151524,712,926,4465265,55 8.2 Polar Equations In this section you will: 336 Learning Objectives  Convert an equation from rectangular coordinates into polar coordinates.  Convert an equation from polar coordinates into rectangular coordinates. Just as we’ve used equations in x and y to represent relations in rectangular coordinates, equations in the variables r and θ represent relations in
polar coordinates. We use Theorem 8.1 to convert equations between the two systems. Converting from Rectangular to Polar Coordinates One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with and every occurrence of y with, and use identities to simplify. This is the technique we employ in the following three examples. Example 8.2.1. Convert from an equation in rectangular coordinates into an equation in polar coordinates. Solution. We start by substituting and into and then simplify. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us.1 1 Study this example and see what techniques are employed, then try your best to apply these techniques in the Exercises. cosrsinr2239xycosxrsinyr2239xy222222222222222239cos3sin9cos6cos9sin9cossin6cos0cossin6cos06cos06cos0xyrrrrrrrrrrrrrr after subtracting 9 from both sides since after
factoring22cos+sin=1θθ 337 We get or. Recognizing the equation as describing a circle, we exclude the first since describes only a point (namely the pole/origin). We choose for our final answer. Note that when we substitute into, we recover the point, so we aren’t losing anything by disregarding. Example 8.2.2. Convert from an equation in rectangular coordinates into an equation in polar coordinates. Solution. We substitute and into. This gives or. Solving the latter for θ, we get for integers k. As we did in the previous example, we take a step back and think geometrically. We know describes a line through the origin. As before, describes the origin but nothing else. Consider the equation. In this equation, the variable r is free, meaning it can assume any and all values including. If we imagine plotting points for all conceivable values of r (positive, negative and zero), we are essentially drawing the line containing the terminal side of which is none other than. 0r6cosr2239xy0r6cosr26cosr0r0ryxcosxrsinyryxsincoscossin0cossin0yxrrrrr after rearranging after factoring0rcossin04k
yx0r40r,4r4yxxy 2239xyxy yx 338 Hence, we can take as our final answer.2 Example 8.2.3. Convert from an equation in rectangular coordinates into an equation in polar coordinates. Solution. We substitute and into. Either or. We can solve the latter equation for r by dividing both sides of the equation by. As a general rule we never divide through by a quantity that may be equal to 0. In this particular case, we are safe since if then then both and, for the equation to hold, would also have to be 0. Since there are no angles with and, we are not losing any information by dividing both sides of by. Doing so, we get 2 We could take it to be θ = –π ∕ 4 + πk for any integer k. 42yxcosxrsinyr2yx2222222sincossincos0cossin0cossinyxrrrrrrrr0r2cossinr2cos2cos0cos0�
��2cossinrsincos0sin02cossinr2cos2sincos1sincoscossectanrxy 2yx 339 As before, the case is recovered in the solution when. So we state our final solution as. Converting from Polar to Rectangular Coordinates As a general rule, converting equations from polar to rectangular coordinates isn’t as straight forward as the reverse process. We will begin with the strategy of rearranging the given polar equations so that the expressions,, and/or present themselves. Example 8.2.4. Convert from an equation in polar coordinates into an equation in rectangular coordinates. Solution. Starting with, we can square both sides. We may now substitute to get the equation. As we have seen, squaring an equation does not, in general, produce an equivalent equation. The concern here is that the equation might be satisfied by more points than. On the surface, this appears to be the case since is equivalent to, not just. However, any point with polar coordinates can be represented as, which means any point whose polar coordinates satisfy the relation has an equivalent3 representation which satisfies. 3 Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs, (3,0) and (–3,π) are different, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically speaking, relations are sets of ordered pairs, so the equations r2 = 9 and r = –3 represent different relations since they correspond to different sets of ordered pairs. Since polar coordinates were defined geometrically to describe the location of points in the plane, however, we concern ourselves
only with ensuring that the sets of points in the plane generated by two equations are the same. This was not an issue, by the way, in algebra when we first defined relations as sets of points in the plane. Back then, a point in the plane was identified with a unique ordered pair given by its Cartesian coordinates. 0rsectanr0sectanr222rxycosrxsinrytanyx3r3r222339rrr222rxy229xy29r3r29r3r3r3,3,,r3r3r 3r 340 Thus, we state our final solution as. Example 8.2.5. Convert from an equation in polar coordinates into an equation in rectangular coordinates. Solution. We begin by taking the tangent of both sides of the equation. Since, we get the following. Of course, we pause a moment to wonder if, geometrically, the equations and generate the same set of points.4 The same argument presented in Example 8.2.4 applies equally well here. We conclude that our answer of is correct. Example 8.2.6. Convert from an equation in polar coordinates into an equation in rectangular coordinates. Solution. Once again, we need to manipulate a bit before using the conversion formulas given in Theorem 8.1. We could square both sides of this equation like we did in Example 8.2.4 to 4 There are infinitely
many solutions to, and is only one of them. Additionally, we went from, in which x cannot be 0, to in which we assume x can be 0. 229xy43434tantan3tan3tanyx33yxyx433yx3yx1cosr1cosrtan3433yx3yx 43 341 obtain an on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r and continue manipulating the equation so that we can apply the conversion formulas from Theorem 8.1. In the last step, we applied Theorem 8.1 and we now have the equation as a solution. It can be shown that this is a legitimate solution by confirming the results when. We will forego this verification for now, as well as the verification that points with coordinates which satisfy will also satisfy. To the right is a graph of the polar equation, from this example. This curve is referred to as a cardioid. In the next section, we will graph cardioids, along with other polar equations. In practice, much of the pedantic verification of the equivalence of equations is left unsaid. Indeed, in most textbooks, squaring equations like to arrive at happens without a second thought. Your instructor will ultimately decide how much, if any, justification is warranted. 2r222222222222cos1coscoscoscosrrxrrrrrrrrrrxyxxy�
�� multiplying through by r adding to both sides squaring both sides substituting =2cosryx and 22222xyxxy0r,r222cosrrr2cosrrr1cosr3r29r 1cosr 342 8.2 Exercises In Exercises 1 – 20, convert the equation from rectangular coordinates into polar coordinates. Solve for r in all but #4 through #7. In Exercises 4 – 7, you need to solve for θ. 1. 5. 9. 13. 17. 19. 2. 6. 10. 14. 3. 7. 11. 4. 8. 12. 15. 16. 18. 20. In Exercises 21 – 40, convert the equation from polar coordinates into rectangular coordinates. 21. 25. 29. 33. 37. 22. 26. 30. 34. 38. 23. 27. 31. 35. 39. 24. 28. 32. 36. 40. 41. Convert the origin into polar coordinates in four different ways. 42. With the help of your classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates. 6x3x7y0yyx3yx2yx2225xy22117xy419yx31xy23yx24xy2220xyy2240xxy�
�22xyx227yyx2224xy2239xy2214412xy7r3r2r423324cosr5cosr3sinr2sinr7secr12cscr2secr5cscr2sectanr22sinr12cosr1sinrcsccotr0,0 343 8.3 Graphing Polar Equations Learning Objectives In this section you will:  Learn techniques for graphing polar equations.  Graph polar equations. In this section, we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. Since any given point in the plane has infinitely many different representations in polar coordinates, practice with graphing polar equations will be an essential part of the learning process. We begin with the Fundamental Graphing Principle for polar equations. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points
which satisfy the equation. That is, a, point is on the graph of an equation if and only if there is a representation of P, say such that and satisfy the equation. Graphing a Simple Polar Equation – Constant Radius or Constant Angle Our first example focuses on some of the more structurally simple polar equations. Example 8.3.1. Graph the following polar equations. 1. 2. 3. 4. Solution. In each of these equations, only one of the variables r and θ is present, resulting in the missing variable taking on all values without restriction. This makes these graphs easier to visualize that others. 1. In the equation, θ is missing. The graph of this equation is, therefore, all points which have a polar coordinate representation, for any choice of θ. Graphically, this translates into tracing out all of the points 4 units away from the origin. This is exactly the definition of circle, centered at the origin, with a radius of 4. ,Pr','r'r'4r32r54324r4,xy 4r 344 2. Once again, we have θ missing in the equation. Plotting all of the points of the form gives us a circle of radius centered at the origin. 3. In the equation, r is missing, so we plot all of the points with polar representations. What we find is that we are tracing out the line which contains the terminal side of when plotted in standard position. 4. As in the previous problem, the variable r is missing in the equation,. Plotting for various values of r shows us that we are tracing out the y-axis. Hopefully, our experience in Example 8.3.1 makes the following result clear. 32r32,32545,4r54323,2r�
��xy 32rxy 54 54xy 32 32 345 Theorem 8.2. Graphs of Constant r and θ: Suppose a and α are constants,.  The graph of the polar equation.  The graph of the polar equation of radius on the Cartesian plane is a circle centered at the origin on the Cartesian plane is the line containing the terminal side of α when plotted in standard position. Graphing Polar Equations Containing Variables r and θ Suppose we wish to graph. A reasonable way to start is to treat θ as the independent variable, r as the dependent variable, evaluate at some ‘friendly’ values of θ and plot the resulting points. We generate the table below, followed by a graph of the resulting points in the xy-plane. π 0araa6cosrrf6cosr,r6cosr,r066,05432532,443232,432030,2200,27432732,43432332,4�
�266,266,xy 346 Despite having nine ordered pairs, we only get four distinct points on the graph. For this reason, we employ a slightly different strategy. We graph on the θr-plane1 and use it as a guide for graphing the equation on the xy-plane. We first see that as θ ranges from 0 to, r ranges from 6 to 0. In the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis, when, and gradually returns to the origin, at. in the θr-plane in the xy-plane The arrows drawn in the above figures are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve. In the xy-plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise way to generate the graph than computing the actual function values, but is markedly faster. Next, we repeat the process as θ ranges from to. Here, the r-values are all negative. This means that in the xy-plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative x-axis. 1 The graph looks exactly like in the xy-plane, and for good reason. At this stage, we are just graphing the relationship between r and θ before we interpret them as polar coordinates on the xy-plane. 6cosr2026cosr6cosr6cosr26cosyx,rxy
COORDINATESθ θ 6,0 32,4 0,2 ,r 347 in the θr-plane in the xy-plane As θ ranges from to, the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the for these values of θ match the r values for θ in, we have that the curve begins to retrace itself at this point. Proceeding further, we find that when, we retrace the part of the curve in Quadrant IV that we first traced out as. The reader is invited to verify that plotting any range of θ outside the interval results in retracing some portion of the curve.2 We present the final graph below. 2 The graph of r = 6cos(θ) looks suspiciously like a circle, for good reason. See Example 8.2.1. 6cosr6cosr32r,232220,xyCOORDINATESθ θ 0,2 ,r 332,4 6, 348 in the θr-plane in the xy-plane Example 8.3.2. Graph the polar equation. Solution. We first plot the fundamental cycle of on the θr-axes. To help us visualize what is going on graphically,
we divide up into the usual four subintervals,, and, and proceed as we did above. 1. As θ ranges from 0 to, r decreases from 4 to 2. This means that the curve in the xy-plane starts 4 units from the origin on the positive x-axis and gradually pulls in toward a point 2 units from the origin on the positive y-axis. in the θr-plane in the xy-plane 6cosr6cosr42sinr42sinr0,20,2,23,23,22242sinr42sinrrθ θ xyθ θ 3 3 6rθ θ xyθ θ 349 2. Next, as θ runs from to, we see that r increases from 2 to 4. In the xy-plane, picking up where we left off, we gradually pull the graph toward the point 4 units away from the origin
on the negative x-axis. in the θr-plane in the xy-plane 3. Over the interval, we see that r increases from 4 to 6. On the xy-plane, the curve sweeps out away from the negative x-axis toward the negative y-axis. in the θr-plane in the xy-plane 4. Finally, as θ takes on values from to, r decreases from 6 back to 4. The graph on the xy-plane pulls in from the negative y-axis to finish where we started. 242sinr42sinr3,242sinr42sinr322rθ θ xyθ θ xyθ θ rθ θ  350 in the θr-plane in the xy-plane We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval results in retracing portions of the curve, so we are finished. in the θr-plane in the xy-plane Example 8.3.3. Graph the polar equation. Solution. The first thing to note when graphing on the θr-plane over the interval is that the graph crosses through the θ-axis. This corresponds to the graph of the curve passing through the origin in the xy-plane, and our first task is to determine when this happens by determining the values of θ for which. 42sinr
42sinr0,242sinr42sinr24cosr24cosr0,20rrθ θ xyθ θ rθ θ xyθ θ 4 4 2 6 351 Solving for θ in gives and. Since these values of θ are important geometrically, we break the interval into six subintervals:,,,, and. 1. As θ ranges from 0 to, r decreases from 6 to 2. Plotting this on the xy-plane, we start 6 units out from the origin on the positive x-axis and slowly pull in towards the positive y-axis. in the θr-plane in the xy-plane 2. On the interval, r decreases from 2 to 0, which means the graph is heading into (and will eventually cross through) the origin. Not only do we reach the origin when, the curve hugs the line as it approaches the origin. 024cos01cos2r0,223430,20,22,23�
�2,34,343,323,22224cosr24cosr2,232323rθ θ xyθ θ 352 in the θr-plane in the xy-plane 3. On the interval, r ranges from 0 to. Since, the curve passes through the origin in the xy-plane, following the line and continues upwards through Quadrant IV toward the positive x-axis. With increasing from 0 to 2, the curve pulls away from the origin to finish at a point on the positive x-axis. in the θr-plane in the xy-plane 4. Next, as θ progresses from to, r ranges from to 0. Since, we continue our graph in the first quadrant, heading into the origin along the line. 24cosr24cosr2,320r23r24cosr�
��24cosr4320r43rθ θ  23xyθ θ 23rθ θ  23xyθ θ 23 353 in the θr-plane in the xy-plane 5. On the interval, r returns to positive values and increases from 0 to 2. We hug the line as we move through the origin and head toward the negative y-axis. in the θr-plane in the xy-plane 6. In the last step, we find that as θ runs through to, r increases from 2 to 6, and we end up back where we started, 6 units from the origin on the positive x-axis. 24cosr24cosr43,324324cosr24cosr322rθ θ  43xyθ θ 43�
�rθ θ  43xyθ θ 43 354 in the θr-plane in the xy-plane Again, we invite the reader to show that plotting the curve for values of θ outside results in retracing a portion of the curve already traced. Our final graph is below. in the θr-plane in the xy-plane Example 8.3.4. Graph the polar equation. Solution. As usual, we start by graphing a fundamental cycle of in the θr-plane, which in this case occurs as θ ranges from 0 to π. We partition our interval into subintervals to help us with the graphing, namely,, and. 24cosr24cosr0,224cosr24cosr5sin2r5sin2r0,4,423,243,4rθ θ xyθ θ �
�rθ θ xyθ θ 43 23 355 1. As θ ranges from 0 to, r increases from 0 to 5. This means that the graph of in the xy-plane starts at the origin and gradually sweeps our so it is 5 units away from the origin on the line. in the θr-plane in the xy-plane 2. Next, we see that r decreases from 5 to 0 as θ runs through and, furthermore, r is heading negative as θ crosses. Hence, we draw the curve hugging the line (the y-axis) as the curve heads to the origin. in the θr-plane in the xy-plane 45sin2r45sin2r5sin2r,42225sin2r5sin2rxyθ θ xyθ θ θ θ  4 2 34  r θ θ  4 2 34 r   356 3. As θ runs from to, r becomes negative and ranges from 0 to. Since, the curve pulls away from the
negative y-axis into Quadrant IV. in the θr-plane in the xy-plane 4. For, r increases from to 0, so the curve pulls back to the origin. in the θr-plane in the xy-plane Even though we have finished with one complete cycle of, if we continue plotting beyond, we find that the curve continues into the third quadrant! Below we present a graph of a second cycle of which continues on from the first. 23450r5sin2r5sin2r3455sin2r5sin2r5sin2r5sin2rxyθ θ θ θ  4 2 34  r θ θ  4 2 34  r xyθ θ 357 in the θr-plane in the xy-plane We have the final graph below. in the θr-plane in the xy-plane Example 8.3.5. Graph. Solution. Graphing is complicated by the r2, so we solve to get How do we sketch such a curve? First off, we sketch a fundamental period of, which is in the figure below. When, is undefined, so we don’t have any values on the 5sin2r5sin2
r5sin2r5sin2r216cos2r2216cosr16cos24cos2rrcos2rcos20cos2xyθ θ θ θ   54 32 74 2 5 5xyθ θ 358 interval. On the intervals which remain, ranges from 0 to 1, inclusive. Hence, ranges from 0 to 1 as well.3 From this, we know ranges continuously from 0 to ±4, respectively. Below we graph both and on the θr-plane and use them to sketch the corresponding pieces of the curve in the xy-plane. and in the θr-plane and in the xy-plane As we have seen in earlier examples, the lines and, which are the zeros of the functions, serve as guides for us to draw the curve as it passes through the origin. As we plot points corresponding to values of θ outside of the interval parts of the curve,4 so our final answer is below., we find ourselves retracing 3 Owing to the relationship between and over, we also know wherever the former is defined. 4 In this case, we could have generated the entire graph by using just the plot, but graphed over the interval in the θr-plane. We leave the details to the reader. 3,44cos2cos24cos(2)r4cos(2
)r4cos(2)r216cos2r4cos2r=-4cos2rθ4cos2r=-4cos2rθ4344cos2r0,yxyx0,1cos(2)cos(2)4cos2r0,2xy1324θ θ 34 41234θ θ  r 4 2 34   cos2r 359 in the θr-plane in the xy-plane A few remarks are in order. 1. There is no relation, in general, between the period of the function and the length of the interval required to sketch the complete graph of in the xy-plane. As we saw at the beginning of this section, despite the fact that the period of is, we sketched the complete graph of in the xy-plane just using the values of θ as θ ranged from 0 to π. In Example 8.3.4, the period of is π, but in order to obtain the complete graph of we needed to run θ from 0 to 2π. While many of the ‘common’ polar graphs can be grouped into families,5 taking the time to work through each graph in the manner presented here is the best way to not only understand the polar coordinate system but also prepare you for what is needed in Calculus. 2. The symmetry seen in the examples is a common occurrence when graphing polar equations
. In addition to the usual kinds of symmetry discussed up to this point in the text (symmetry about each axis and the origin), it is possible to talk about rotational symmetry. Keep rotational symmetry in mind as you work through the Exercises. 5 Example 8.3.2 and Example 8.3.3 are examples of limacons. Example 8.3.4 is an example of a polar rose, and Example 8.3.5 is the famous Lemniscate of Bernoulli. 4cos2r216cos2rfrf6cosf26cosr5sin2f5sin2rθ θ  r 4 2 34  xyθ θ 34 4 360 8.3 Exercises In Exercises 1 – 20, plot the graph of the polar equation by hand, without the aid of a calculator. Carefully label your graphs. 1. Circle: 3. Rose: 5. Rose: 7. Rose: 9. Cardioid: 11. Cardioid: 13. Limacon: 15. Limacon: 17. Limacon: 2. Circle: 4. Rose: 6. Rose: 8. Rose: 10. Cardioid: 12. Cardioid: 14. Limacon: 16. Limacon: 18. Limacon: 19. Lemniscate: 20. Lemniscate: Exercises 21 – 30 give you some curves to graph using a graphing calculator or other form of technology. Notice that some of the curves have explicit bounds on θ and others do not. 21., 23., 25. 27., 29.,, 22
. 24. 26. 28. 30. 6sinr2cosr2sin2r4cos2r5sin3rcos5rsin4r3cos4r33cosr55sinr22cosr1sinr12cosr12sinr234cosr35cosr35sinr27sinr2sin2r24cos2rr012lnr1120.1re0123r1.21.2sin53cosr32sincos23rar
ctanr11cosr12cosr123cosr 361 31. How many petals does the polar rose have? What about, and? With the help of your classmates, make a conjecture as to how many petals the polar rose has for any natural number n. Replace sine with cosine and repeat the investigation. How many petals does have for each natural number n? 32. In this exercise, we have you and your classmates explore transformations of polar graphs. For both parts (a) and (b), let and. (a) Using a graphing calculator or other form of technology, compare the graph of to each of the graphs of,, and. Repeat this process for. In general, how do you think the graph of compares with the graph of? (b) Using a graphing calculator or other form of technology, compare the graph of to each of the graphs of,, and. Repeat this process for. In general, how do you think the graph of compares with the graph of? (Does it matter if or?) 33. With the help of your classmates, research cardioid microphones. sin2rsin3rsin4rsin5rsinrncosrncosf2singrf4rf34rf4rf
34rfgrfrfrf2rf12rfrf3rfgrkfrf0k0k 362 8.4 Polar Representations for Complex Numbers In this section you will: Learning Objectives  Find the real part, the imaginary part, and the modulus of a complex number.  Graph complex numbers.  Learn the properties of the modulus and the argument of a complex number and be able to apply them. Complex Numbers and the Complex Plane A complex number is a number of the form where a and b are real numbers and i is the imaginary unit defined by. The number a is called the real part of z, denoted, while the number b is called the imaginary part of z, denoted. If for real numbers a, b, c and d, then and, verifying that and are well-defined.1 To start off this section, we associate each complex number with the point on the coordinate plane. In this case, the x-axis is relabeled as the real axis, which corresponds to the real number line, and the y-axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit i. The plane determined by these two axes is called the complex plane. The Complex Plane 1 ‘Well-defined’ means that no matter how we express z, the number Re(z) is always the same, and the number Im(z) is always the same. In other words, Re and Im are functions of complex numbers. zabi1iRezIm
zzabicdiacbdRezImzzabi,abReal AxisImaginary Axis i 2i 3i 4i i 2i 3i 4i 4,242zi 0,33zi 1,01z 363 Since the ordered pair gives the rectangular coordinates associated with the complex number, the expression is called the rectangular form of z. Of course, we could just as easily associate z with a pair of polar coordinates. Although it is not as straightforward as the definitions of and, we can still give r and θ special names in relation to z. The Modulus and Argument of Complex Numbers Definition. The Modulus and Argument of Complex Numbers: Let number with be a complex be a polar representation of the point with. Let and rectangular coordinates where.  The modulus of z, denoted, is defined by.  The angle θ is an argument of z. The set of all arguments of z is denoted.  If and, then θ is the principal argument of z, written. Some remarks are in order. We know from Section 8.1 that every point in the plane has infinitely many polar coordinate representations, which means it’s worth our time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-defined.  Concerning the modulus, if then the point associated with z is the origin. In this case, the only possible r-value is. Hence, for, is well-defined.  If, then the point associated with z is not the origin, and there are two possibilities for r: one positive and one negative. However, we stipulated in our definition so this pins down the value of to one and only one number
. Thus, the modulus is well-defined in this case too.2  Even with the requirement, there are infinitely many angles θ which can be used in a polar representation of a point. If then the point in question is not the origin, so all of these angles θ are coterminal. Since the coterminal angles are exactly 2π radians apart, we are guaranteed that only one of them lies in the interval, and this angle is what we call the principal argument of z,. 2 In case you’re wondering, the use of the absolute value notation \|z\| for modulus will be explained shortly. ,abzabizabi,rRezImzzabiReazImbz,r,ab0rzzrargz0zArgz,r0z0r0z0z0z0rz0r,r0z,Argz 364 In fact, the set of all arguments of z can be described using set-builder notation as. Note that since is a set, we will write ‘ ’ to mean ‘θ is in3 the set of arguments of z’.  If then the point in question is the origin, which we know can be represented in polar coordinates as for any angle θ. In this case, we have and since there is no one value of θ which lies in, we leave undefined. It is high time for an example. Example 8.4.1. For each of the following complex numbers find,,, and
. Plot z in the complex plane. 1. 2. 3. 4. Solution. 1. For, we have and. To find, and, we need to find a polar representation with for the point associated with z. We first determine a value for r. We require, so we choose, and have. Next, we find a corresponding angle θ. Since and P lies in Quadrant IV, θ is a Quadrant IV angle. We have 3 Recall the symbol being used here,, is the mathematical symbol which denotes membership in a set. argzargArg2\| is an integerzzkkargzargz0z0,arg0,,Arg0RezImzzargzArgz3zi24zi3zi117z331ziiRe3zIm1zzargzArgz,r0r3,1P22222223142rxyrrr from 0r2r2z0r13tan332 for integerst an6kkyx
 from since is a Quadrant IV angle 365 Thus,. Of these values, only satisfies the requirement that, hence 2. The complex number has.,, and is associated with the point. Our next task is to find a polar representation for P where. Running through the usual calculations gives, so. To find θ, we get. Since and P lies in Quadrant II, we know θ is a Quadrant II angle. Thus, Hence, the requirement, so. Only. satisfies 3. We rewrite as to find and. The point in the plane which corresponds to z is and while we could go through the usual calculations to find the required polar form of this point, we can almost ‘see’ the answer. The point lies 3 units away from the origin on the positive y-axis. Hence, and for integers k. We get and. 4. As in the previous problem, we write, so and. The number corresponds to the point, and this is another instance where we can determine the polar form ‘by eye’. The point is 117 units away from the origin along the negative x-axis. Hence, and for integers k. We have. Only one of these values,, lies in the interval which means that. arg2 is an integer6zkk6Arg6z24ziRe2zIm4z2,4P,r0r25r25ztan20rarctan22 for integers or arctan22 for integers kkk
k since is a Quadrant II angle from odd property of arctangentargarctan22\| is an integerzkkarctan2Argarctan2z3zi03ziRe0zIm3z0,30,33rz22karg2 is an integer2zkkArg2z1171170ziRe117zIm0z117z117,0117,0117rz2karg2\| is an integerzkk,Argz We plot the four numbers from this example below. 366 Properties of the Modulus and Argument Now that we’ve had some practice computing
the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem. Theorem 8.3. Properties of the Modulus: Let z and w be complex numbers. is the distance from z to 0 in the complex plane, and if and only if     Product Rule:  Power Rule: for all natural numbers n  Quotient Rule:, provided To prove the first three properties in Theorem 8.3, suppose where a and b are real numbers. To determine, we find a polar representation with for the point. From Section 8.1, we know so that. Since we require, then it must be that, which means. z0z0z0z22ReImzzzzwzwnnzzzzww0wzabiz,r0r,ab222rab22rab0r22rab22zabReal AxisImaginary Axis i 2i 3i 4i i 2i 3i 4i 117 2 1 1 2 3 \| \| \| \| \| 3zi 24zi 3zi 117z 367  Using the distance formula, we find the distance from to is also, establishing the first property.4  For the second property, note that since is a distance,. Furthermore, if and only if the distance from z to 0 is 0, and the latter happens if and only if, which is what we are asked to show.5  For the third property, we note that since and,  To prove the product rule, suppose and for real numbers a, b, c and d. Then Therefore, 4 Since the absolute value \|x\| of a real
number x can be viewed as the distance from x to 0 on the number line, this first property justifies the notation \|z\| for modulus. We leave it to the reader to show that if z is real, then the definition of modulus coincides with absolute value so the notation \|z\| is unambiguous. 5 This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is true. We know if and only if if and only if, which is true if and only if. The latter happens if and only if. There. 0,0,ab22abz0z0z0zReazImbz2222ReImzabzzzabiwcdi2211izwabicdiacadibcibdiacadibcibdacbdadbci from 2222222222222222222222222222222222zwacbdadbcacabcdbdadabcdbcacadbcbdacdbcdabcdabcd�
�� after expanding terms rearranged zwzw product rule for radicals definition of and 0z220ab220ab0ab0zabi 368 Hence as required.  Now that the product rule has been established, we use it and the Principle of Mathematical Induction to prove the power rule. Let be the statement. Then is true since. Next, assume is true. That is, assume for some. Our job is to show that is true, namely. As is customary with induction proofs, we first try to reduce the problem in such a way as to use the induction hypothesis:. Hence, is true, which means is true for all natural numbers n.  Like the power rule, the quotient rule can also be established with the help of the product rule. We assume, so that, and get Hence, the proof really boils down to showing. This is left as an exercise. Next, we characterize the argument of a complex number in terms of its real and imaginary parts. zwzwPnnnzz1P11zzzPkkkzz1k1Pk11kkzzkkzz11kkkkkzzzzzzzz property of exponents product rule of modulus induction hypothesis property of exponents1Pknnzz0w0w11zzwwzw product rule of modulus11ww 369 Theorem 8.4. Properties of the Argument: Let z be a complex number. and and and  If  If   If If, then., then, then, then and... To prove Theorem 8.4, suppose for real numbers a and b. By definition, and, so the point associated with z is.  From Section 8.
1, we know that if is a polar representation for, then, provided.  If and, then z lies on the positive imaginary axis. Since we take, we have that θ is coterminal with, and the result follows.  If and, then z lies on the negative imaginary axis, and a similar argument shows θ is coterminal with.  The last property was already discussed following the definition at the beginning of this section. Polar Form of Complex Numbers Our next goal is to link the geometry and algebra of the complex numbers. To that end, consider the figure below. Re0zargzImtanRezzRe0zIm0zarg2 is an integer2zkkRe0zIm0zarg2 is an integer2zkkReIm0zz0zarg,zzabiReazImbz,Re,Imabzz,rRe,ImzzImtanRezzRe0z�
��Re0zIm0z0r2Re0zIm0z2 370 Polar coordinate associated with, with We know from Theorem 8.1 that and. Making these substitutions for a and b gives The expression is abbreviated so we can write. Since and, we get Definition. A Polar Form of a Complex Number: Suppose z is a complex number and The expression. is called a polar form for z. Since there are infinitely many choices for, there are infinitely many polar forms for z, so we used the indefinite article ‘a’ in the preceding definition. It is time for an example. Example 8.4.2. Find the rectangular form of the following complex numbers. Find and. 1. 2. 3. Solution. The key to this problem is to write out as 4.. ,rzabi0rcosarsinbrcossincossinzabirriricossinicisciszrrzargzargzciscossinzziargzRezImz24cis3z
32cis4z3cis0zcis2zciscossiniImaginary AxisReal Axis bi a 0 argz 22zabr ,,abzabir 371 1. By definition, After some simplifying, we get, so that and. 2. Expanding, we get From this, we find, so. 3. We get Writing number. 4. Lastly, we have, we get and, which makes sense seeing that 3 is a real Since, we get and. Since i is called the ‘imaginary unit’, these answers make sense. Example 8.4.3. Use the results from Example 8.4.1 to find a polar form of the following complex numbers. 1. 2. 3. 4. 24cis3224cossin33zi223ziRe2zIm23z32cis4332cossin44zi22ziRe2Imzz
3cis03cos0sin03zi330iRe3zIm0zcis2cossin22zii01iiRe0zIm1z3zi24zi3zi117z 372 Solution. To write a polar form of a complex number z, we need two pieces of information: the modulus and an argument (not necessarily the principal argument) of z. We shamelessly mine our solution to Example 8.4.1 to find what we need. 1. For, and, so. We can check our answer by converting it back to rectangular form to see that it simplifies to 2. For, and. Hence, a good exercise to actually show that this polar form reduces to... It is 3. For, and. In this case,. This can be checked geometrically. Head out 3 units from 0 along the positive real axis. Rotating radians counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at. 4. Last but not least, for, and. We get. As with the last problem, our answer is easily checked geometrically. z3zi2z62cis6z3zi24zi25zarctan225cisarctan2z24zi3zi3z2�
�3cis2z23zi117z117z117cisz 373 8.4 Exercises In Exercises 1 – 20, find a polar representation for the complex number z and then identify,,, and. These exercises should be worked without the aid of a calculator. 1. 5. 9. 13. 17. 2. 6. 10. 14. 18. 3. 7. 11. 15. 19. 4. 8. 12. 16. 20. In Exercises 21 – 40, find the rectangular form of the given complex number. Use whatever identities are necessary to find the exact values. These exercises should be worked without the aid of a calculator. 24. 28. 32. 21. 25. 29. 33. 35. 37. 39. 22. 26. 30. 23. 27. 31. 34. 36. 38. 40. RezImzzargzArgz99zi553zi6zi3232zi636zi2z3122zi33zi5zi2222zi6z37zi34zi2zi724zi26zi125zi52zi42zi13zi6cis0z2cis6z72cis4z
3cis2z24cis3z36cis4z9cisz43cis3z37cis4z313cis2z17cis24z12cis3z8cis12z72cis8z45cisarctan3z110cisarctan3z15cisarctan2z3cisarctan2z750cisarctan24z
15cisarctan212z 374 41. Complete the proof of Theorem 8.4, Properties of the Modulus, by showing that if then. 42. Recall that the complex conjugate of a complex number is denoted and is given by. (a) Prove that. (b) Prove that. (c) Show that and. (d) Show that if then. Interpret this result geometrically. (e) Is it always true that? 0w11wwzabizzabizzzzzRe2zzzIm2zzziargzargzArgArgzz 375 8.5 Products, Powers, Quotients and Roots of Complex Numbers In this section you will: Learning Objectives  Find the product, power, quotient and roots of complex number(s).  Learn and apply DeMoivre’s Theorem. Products, Powers and Quotients of Complex Numbers The following theorem summarizes the advantages of working with complex numbers in polar form. Theorem 8.5. Products, Powers and Quotients of Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms. Then and  Product Rule:  Power Rule (DeMoivre’s Theorem): for every natural number n  Quotient Rule:, provided The proof of Theorem 8.5 requires a healthy mix of definition, arithmetic and identities.  We start with the product rule. We now focus on the quantities in brackets on the right hand side of the equation. Putting this together with our earlier work, we get.  Moving right along, we take aim at the power rule, better known as DeMoivre’s Theorem. We proceed by induction on n.
Let be the sentence. Then is true, since ciszzciswwciszwzwcisnnzznciszzww0wcisciscossincoscisinszwzwzwii definition of 222-1cossincossin coscoscossinsincossinsin coscossinsinsincoscossin coscossinsinsincoscossin iiiiiiiiiii
 rearrange terms use ; factor out  cossin cisi sum identities cis definition of ciszwzwPncisnnzzn1P11ciscis1zzzz 376 We now assume is true, that is, we assume for some. Our goal is to show that is true, or that. We have Hence, assuming is true, we have that is true, so by the Principle of Mathematical Induction, for all natural numbers n.  The last property in Theorem 8.5 to prove is the quotient rule. Assuming, we have Next, we multiply both the numerator and denominator of the right hand side by to get Finally, we have, and we are done. Pkciskkzzk1k1Pk11cis1kkzzk11ciscisciscis1kkkkkzzzzkzzzkzk�
�� property of exponents induction hypothesis product rulePk1Pkcisnnzzn0wcisciscossincossinzzwwziwicossini2222cossincossincossincossincossincossincossincossincoscoscossinsincossinsincoscossinsincossinczziiwwiiiizwiiziiiwiiizw�
�22222oscossinsinsincoscossincossincossincossincis1iiziwzwciszzww 377 Example 8.5.1. Let and. Use Theorem 8.5 to find the following. 1. 2. 3. Write your final answers in rectangular form. Solution. In order to use Theorem 8.5,
we need to write z and w in polar form. For, we find If, we know Since z lies in Quadrant I, we have for integers k. Hence,. For, we have For an argument θ of w, we have Since w lies in Quadrant II, for integers k and. We can now proceed. 232zi13wizw5wzw232zi22232164zargzImtanRe22313 or 33zz26k4cis6z13wi22132w3tan13223k22cis3w 378 1. We get After simplifying, we get. 2. We use DeMoivre’s Theorem which yields Since is coterminal with, we get 3. Last, but not least, we have Since is a quadrantal angle, we can ‘see’ the rectangular form by moving out 2 units along the positive real axis, then rotating radians clockwise to arrive at the point 2 units below 0 on the imaginary axis. The long and short of it is that. 24cis2cis6328cis6358cis6558cossin66zwi
434zwi55522cis322cis531032cis3w1034354432cossin3316163wii4cis622cis342cis2632cis2zw222ziw Some remarks are in order. 379  First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers z and w in Example 8.5.1 into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out the old-fashioned way. However, Theorem 8.5 pays huge dividends when computing powers of complex numbers. Consider how we computed in Example 8.5.1 and compare that to accomplishing the same feat by expanding. With division being tricky in the best of times, we saved ourselves a lot of time and effort using Theorem 8.5 to find and simplify using their polar forms as opposed to starting with, rationalizing the denominator and so forth.  There is geometric reason for studying these polar forms.
Take the product rule, for instance. If and, the formula can be viewed geometrically as a two-step process. The multiplication of by can be interpreted as magnifying1 the distance, from 0 to z, by the factor. Adding the argument of w to the argument of z can be interpreted geometrically as a rotation of β radians counter-clockwise.2 Focusing on and from Example 8.5.1, we can arrive at the product by plotting z, doubling its distance from 0 (since ), and rotating radians counter- clockwise. The sequence of diagrams below attempts to describe this process geometrically. 1 Assuming \|w\| > 1. 2 Assuming β > 0. 23213zwii5w513izw23213iiciszzciswwciszwzwzwzw4cis6z22cis3wzw2w23 380 Multiplying z by Rotating counter-clockwise by radians We may visualize division similarly. Here, the formula may be interpreted as shrinking3 the distance from 0 to z by the factor, followed by a clockwise4 rotation of β radians. In the case of and from Example 8.5.1, we arrive at by first halving the distance from 0 to z, then rotating clockwise radians, as we visualize below. Dividing z by Rotating clockwise by Roots of Complex Numbers Our last goal of the section is to reverse DeMoivre’s Theorem to extract roots of complex numbers. 3 Again, assuming \|w\| > 1. 4 Again, assuming β > 0. 2w2Arg3wciszzwww4c
is6z22cis3wzw232w2Arg3wImaginary AxisReal Axis i 2i 3i 4i 5i 6i 4cis6z 8cis6zwImaginary AxisReal Axis i 2i 3i 12cis6zw 4cis6zImaginary AxisReal Axis i 2i 3i 4i 5i 6i 8cis6zw 28cis63pizw 23Imaginary AxisReal Axis i i 2i 12cis6zw 22cis63zw 381 Definition. Let z and w be complex numbers. If there is a natural number n such that, then w is an nth root of z. Here, we do not specify one particular principal nth root, hence the use of the indefinite article in defining w as ‘an’ nth root of z. Using this definition, both 4 and –4 are square roots of 16, while means the principal square root of 16,
as in. Suppose we wish to find the complex third (cube) roots of 8. Algebraically, we are trying to solve. We know that there is only one real solution to this equation, namely, but if we take the time to rewrite this equation as and factor, we get. The quadratic factor gives two more cube roots,, for a total of three cube roots of 8. Recall from College Algebra that, since the degree of the polynomial is three, there are three complex zeros, counting multiplicity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express in polar form. Since lies 8 units away from the origin on the positive real axis, we get. If we let be a polar form of w, the equation becomes The complex number on the left hand side of the equation corresponds to the point with polar coordinates while the complex number on the right hand side corresponds to the point with polar coordinates. Since by definition, so is, which means and are two polar representations corresponding to the same complex number, with both representations having positive r values. Thus, and for integers k.  Since is a real number, we solve by extracting the principal cube root to get. nwz1616438w382w380w22240www13wi38Pww8z8z8cis0zcisww38w3338cis8cis0cis38cis0www DeMoivre's Theorem3,3w8,00w3w
3,3w8,038w302kw38w382w 382  As for, we get for integers k. This produces three distinct points with polar coordinates corresponding to k = 0, 1, and 2: specifically, and. The corresponding complex and rectangular forms are listed in following table. Polar Coordinate Complex Number Rectangular Form The cube roots of 8 can be visualized geometrically in the complex plane, as follows. Keeping the geometric picture in mind throughout the remainder of this section will lead to an interesting observation regarding geometric properties of complex roots. While the process for finding the cube roots of 8 seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for example, all of the fifth roots of 32. (Try using factoring techniques on that!) If we start with a generic complex number in polar form and solve in the same manner as above, we arrive at the following theorem. 23k2,022,342,32,022,342,302cis0w122cis3w242cis3w02w113wi213wiciszznwzImaginary AxisReal Axis 3i 2i i i 2i 3i 23�
� 43 02w 113wi 213wi 383 Theorem 8.6. The nth Roots of a Complex Number: Let be a complex number with polar form. For each natural number n, z has n distinct nth roots, which we denote by, and they are given by the formula The proof of Theorem 8.6 breaks into two parts: first, showing that each is an nth root, and second, showing that the set consists of n different complex numbers.  To show is an nth root of z, we use DeMoivre’s Theorem to show. Since k is a whole number, and. Hence, it follows that, so, as required.  To show that the formula in Theorem 8.6 generates n distinct numbers, we assume (or else there is nothing to prove) and note that the modulus of each of the is the same, namely. Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is, if the arguments differ by an integer multiple of. Suppose k and j are whole numbers between 0 and, inclusive, with. Then. For this to be an integer multiple of, must be a multiple of n. But because of the restrictions on k and j,. (Think this through.) Hence, is a positive number less than n, so it cannot be a multiple of n. As a result, and are different complex numbers, and we are done. From College Algebra, we know there are at most n distinct solutions to, and we have just found all of them. We illustrate Theorem 8.6 in the next example. 0zciszr011,,,nwww2cisnkwrknnkw\|0,1,,1kwknkwnkwz2cis2ciscis2nnnknnwrknnrnknnr
k DeMoivre's Theoremcos2cosksin2sinkcis2ciskcisnkwrz2nkwnr21nkj222kjkjnnnnn2kj01kjnkjkwjwnwz 384 Example 8.5.2. Find the following: 1. both square roots of 2. 3. 4. the four fourth roots of the three cube roots of the five fifth roots of Solution. 1. To find both square roots of, we start by writing. To use Theorem 8.6, we identify, and. We know that z has two square roots, and in keeping with the notation in the theorem, we’ll call them and. We get and We can check each of these roots by squaring to get. 2. To find the four fourth roots of, proceeding as above,
we write z as. With, and, we get the four fourth roots of z to be Converting these to rectangular form gives,, and. 223zi16z22zi1z223zi22234cis3zi4r232n0w1w002324cis0222cis313kwi Theorem 8.6 with rectangular form112324cis12242cis313kwi Theorem 8.6 with rectangular form223zi16z1616cisz16r4n40414243216cis02cis4442316cis12cis4442516cis22cis4442716cis32cis444wwww022wi122wi�
�222wi322wi 385 3. For finding the cube roots of, we have. With, and the usual computations yield If we were to convert these to rectangular form, we would need to use either sum and difference identities or half-angle identities to evaluate and. Since we are not explicitly told to do so, we leave this as a good, but messy, exercise. 4. To find the five fifth roots of 1, we write. We have, and. Since, the roots are The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of. At this stage, we could approximate our answers using a calculator, and we leave this as an exercise. Now that we have done some computations using Theorem 8.6, we take a step back to look at things geometrically. Essentially, Theorem 8.6 says that to find the nth roots of a complex number, we first take the nth root of the modulus and divide the argument by n. This gives the first root. Each successive root is found by adding to the argument, which amounts to rotating by radians. This results in n roots, spaced equally around the complex plane. As an example of this, we plot our answers to number 2 in Example 8.5.2. 22zi2cis4z2r43n30331322cis12932cis2cis124172cis12www0w2w11cis01r05n51101234cis012cis54cis56cis58cis5wwwww�
�250w2n0w2n 386 The four fourth roots of equally spaced around the plane We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications. For now, the following exercises will have to suffice. 16z242Imaginary AxisReal Axis 2i i i 2i 0w 1w 2w 3w 387 8.5 Exercises In Exercises 1 – 12, use and to compute the quantity. Express your answers in polar form using the principal argument. These exercises should be worked without the aid of a calculator. 1. 5. 9. 2. 6. 3. 7. 4. 8. 10. 11. 12. In Exercises 13 – 24, use DeMoivre’s Theorem to find the indicated power of the given complex number. Express your final answers in rectangular form. These exercises should be worked without the aid of a calculator. 13. 17. 21. 14. 18. 22. 15. 19. 23. 16. 20. 24. In Exercises 25 – 36, find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. These exercises should be worked without the aid of a calculator. 25. the two square roots of 26. the two square roots of 27. the two square roots of 28. the two square roots of 29. the three cube roots of 30. the three cube roots of 31. the three cube roots of 32. the three cube roots of 33322zi3232wizwzwwz4z3w52zw32zw2zw2wz32zw23wz6wz3223i33i
433i43i35522i61322i33322i43133i42222i522i53i81i4zi25zi13zi55322i64z125zzi8zi 388 33. the four fourth roots of 34. the four fourth roots of 35. the six sixth roots of 36. the six sixth roots of 37. Use the four complex fourth roots of –4 to show that the factorization of over the real numbers is. You may want to refer to the Complex Factorization Theorem from College Algebra. 38. Use the 12 complex 12th roots of 4096 to factor over the real numbers, into a product of linear and irreducible quadratic factors. 39. Given any natural number, the n complex nth roots of the number are called the nth Roots of Unity. In the following exercises, assume that n is a fixed, but arbitrary, natural number such that. (a) Show that is an nth root of unity. (b) Show that if both and are nth roots of unity then so is their product. (c) Show that if is an nth root of unity then there exists another nth root of unity such that. Hint: If let. You’ll need to verify that is indeed an nth root of unity. 40. Another way to express the polar form of a complex number is to use the exponential function. For real numbers t, Euler’s Formula defines. (a) Use
Theorem 8.5 (Products Powers and Quotients of Complex Numbers in Polar Form) to show that for all real numbers x and y. (b) Use Theorem 8.5 to show that for any real number x and any natural number n. (c) Use Theorem 8.5 to show that for all real numbers x and y. (d) If is the polar form of z, show that where radians. 16z81z64z729z4()4pxx222222pxxxxx124096pxx2n1z2n1wjwkwjkwwjw'jw'1jjwwcisjw'cis2jw'cis2jwcossinitetitixyixiyeeeninxixeeixixyiyeeeciszritzret 389 (e) Show that. (This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics.) (f) Show that and that for all real numbers t. 10iecos2ititeetsin2ititeeti 390 CHAPTER 9 VECTORS AND PARAMETRIC EQUATIONS Chapter Outline 9.1 Vector Properties and Operations 9
.2 The Unit Vector and Vector Applications 9.3 The Dot Product 9.4 Sketching Curves Described by Parametric Equations 9.5 Finding Parametric Descriptions for Oriented Curves Introduction Chapter 9 introduces vectors, their many resulting applications, and parametric equations. We begin in Section 9.1 by learning about the geometric representation of vectors along with vector arithmetic, properties and applications involving bearings. Section 9.2 continues with applications by focusing on component forms of vectors in the solution process. The unit vector is introduced along with operations on vectors in an i and j format. In this section, vectors are used to model forces. The focus of Section 9.3 is the dot product – operations, properties and applications. In Section 9.4, parametric equations are defined and graphed. Graph behavior is explored through sketching both x and y as a function of t before displaying the resulting parametric graph in the xy-plane. Section 9.5 follows with the elimination of the parameter t in an effort to transform parametric equations to Cartesian equations. Thus, the correlation between Cartesian and parametric equations is established and further efforts are made to transform Cartesian equations into their parametric form. Throughout Chapter 9, the application of Trigonometry can be seen, providing further evidence of its importance in Mathematics, Engineering, Physics and real-life applications. This chapter is a good stepping off place to conclude the study of Trigonometry before moving on to Calculus. 391 9.1 Vector Properties and Operations Learning Objectives In this section you will:  Interpret vectors and vector operations geometrically.  Perform algebraic operations on vectors, including scalar multiplication, addition and determination of inverses.  Determine the component form of a vector.  Find the magnitude and direction of a vector. As we have seen numerous times in this book, mathematics can be used to model and solve real-world problems. For many applications, real numbers suffice; that is, real numbers with the appropriate units attached can be used to answer questions like “How close is the nearest Sasquatch nest?” There are other times though, when these kinds of quantities do not suffice. Perhaps it is important to know, for instance, how close the nearest Sasquatch nest is as well as the direction in which it lies. To answer questions like these which involve both a quantitative answer, or magnitude, along with a direction, we use the mathematical objects
called vectors1. The Geometry of Vectors A vector is represented geometrically as a directed line segment where the magnitude of the vector is taken to be the length of the line segment and the direction is made clear with the use of an arrowhead at one endpoint of the segment. A vector has an initial point, where it begins, and a terminal point, indicated by an arrowhead, where it ends. There are various symbols that distinguish vectors from other quantities:  Lower case type, boldfaced or with an arrow on top, such as,,,,,.2  Given an initial point P and a terminal point Q, a vector can be represented as. The arrow on top is what indicates that it is not just a line, but a directed line segment. Below is a typical vector with endpoints and. The point P is the initial point, or tail, of and the point Q is the terminal point, or head, of. Since we can reconstruct completely from P 1 The word ‘vector’ comes from the Latin vehere meaning to convey or to carry. 2 In this textbook, we will adopt the boldfaced type notation for vectors, without the arrow. In the classroom, your instructor will likely use arrow notation, and arrow notation should be used whenever vectors are written by hand. vuwvuwPQv1,2P4,6Qvvv 392 and Q, we write, where the order of points P (initial point) and Q (terminal point) is important. (Think about this before moving on.) While it is true that P and Q completely determine, it is important to note that since vectors are defined in terms of their two characteristics, magnitude and direction, any directed line segment with the same length and direction as is considered to be the same vector as, regardless of its initial point. In the case of our vector above, any vector which moves three units to the right and four up3 from its initial point to arrive at its terminal point is considered the same vector as. The Component Form of a Vector The notation we use to capture this idea is, the component form of the vector, where the first number, 3, is called the x-component of and the second number, 4, is called the y-component of. If we wanted to reconstruct with initial point then we would find the terminal point
of by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point, as seen to the right. 3 If this idea of over and up seems familiar, it should. The slope of the line segment containing v is 4/3. PQvPQvvvvvv3,4vvv3,4v'2,3Pv'1,7Q3,4vxyPQ(4,6)(1,2) vxyP'Q'(1,7)(-2,3)over 3up 4QP(4,6)(1,2)up 4over 3 v v 393 The component form of a vector is what ties these very geometric objects back to Algebra and ultimately Trigonometry. We generalize our example in the following definition. Definition. Suppose is represented by a directed line segment with initial point and terminal point. The component form of is given by Example 9.1.1. Consider the vector whose initial point is and terminal point is. Write in component form. Solution. Using the definition of component form, we get Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, if and only if and. (Again, think about this before reading on.) We now set about defining operations on vectors. Vector Addition Suppose we are given two vectors and. The sum, or resultant vector, is obtained geometrically as follows. First, plot. Next, plot so that its initial point is the terminal point of. To plot the vector we begin at the initial point of and end at the terminal point of. It is helpful to think of the vector as the ‘net result’ of moving along then moving along., and v00,Pxy11,Qxyv1010,PQxxyyv�
��2,3P6,4QPQv62,434,1v1212,','vvvv11'vv22'vvvwvwvwvvwvwvwvwvwvw v w vw 394 Our next example makes good use of resultant vectors and reviews bearings and the Law of Cosines.4 Example 9.1.2. A plane leaves an airport with an airspeed of 175 miles per hour at a bearing of N40°E. A 35 mile per hour wind is blowing at a bearing of S60°E. Find the true speed of the plane, rounded to the nearest mile per hour, and the true bearing of the plane, rounded to the nearest degree. Solution. For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we’ve seen a few times before in this textbook. We let denote the plane’s velocity and denote the wind’s velocity in the diagram below. The true speed and bearing is found by analyzing the resultant vector,. Before proceeding, it will be helpful to determine the angle, labeled in the diagram to the right. We extend the vectors, with initial point at origin, and with initial point coinciding with the terminal point of, to give us the three lines, and, respectively. From vertical angles, we have. With both and containing the directed line segment, we see that they are parallel lines. We then use corresponding angles of parallel lines to conclude that. 4 If necessary, review the discussion on bearings in the 7.2 Exercises. vwvwvwwv1l2l3l402l3lw60100EN 40 60 v w vwEN 40 60 w w 1l 2l 3l vw v   395 From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of, which is 175, the magnitude of, which is
35, and the magnitude of, which we’ll call. The Law of Cosines gives us This means the true speed of the plane is approximately 184 miles per hour. To find the true bearing of the plane we need to determine the angle. Using the Law of Cosines once more5, we have We use the inverse cosine, along with the value for c from our prior calculation, to find that. Given the geometry of the situation, we add to the given 40° and find the true bearing of the plane to be approximately N51°E. We next define the addition of vectors component-wise to match the geometry.6 Definition. Suppose and. The vector is defined by Example 9.1.3. Let and suppose for and. Find and interpret this sum geometrically. Solution. Before adding the vectors using the definition, we need to write in component form. We get, and then 5 Or, since our given angle, 100°, is obtuse, we could use the Law of Sines without any ambiguity here. 6 Adding vectors component-wise should look hauntingly familiar. Compare this with matrix addition. In fact, in more advanced courses such as Linear Algebra, vectors are defined as 1 by n or n by 1 matrices, depending on the situation. vwvwc22217535217535cos1003185012250cos100184ccc222222351752175cos35175cos2175cccc1112,vvv12,wwwvw1122,vwvwvw3,4vPQw3,7P2,5Q

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