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Pre-AP Precalculus Module 9 Polar Equations 9.1 Polar Coordinates When converting from a Cartesian coordinate system to a Polar one, the origin is renamed the pole, and the positive π₯-axis is called the polar axis. In addition the Cartesian coordinates (π₯, π¦) become Polar coordinates (π, π), where π is the distance between the point and the pole, and π is the angle formed by the polar axis and the ray from the pole through the point. If π < 0, then the point is plotted 180° from the stated angle. Example 1 β Plot the points with the following polar coordinates: (a)(3, 5π 3 π ), (b) (2, β 4 ), (c) (3,0), and (d) π (β2, 4 ). ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ π Example 2 β Consider a point with polar coordinates (2, 4 ). Find two polar coordinates (π, π) that map the same point, one with a positive π and one with a negative π. π Example 3 β Plot the point π with polar coordinates (3, 6 ), and find other polar coordinates (π, π) of this same point for which: (a) π > 0, 2π β€ π < 4π; (b) π < 0, 0 β€ π < 2π; and (c) π > 0, β2π β€ π < 0. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± οο± οο² οο³ οο΄ οο΅ Sullivan & Sullivan β Section 9.1 ο² ο³ ο΄ ο΅ 9.1 Polar Coordinates A point with polar coordinates (π, π), π in radians, can also be represented by either of the following: (π, π + 2ππ) or (βπ, π + π + 2ππ), π any integer. The polar coordinates of the pole are (0, π), where π can be any angle. If π is a point with polar coordinates (π, π), the rectangular coordinates (π₯, π¦) of π are given by (π cos π , π sin π). Example 4 β Find the rectangular coordinates of the points with the following polar coordinates: (a) π π (6, 6 ) and (b) (β4, β 4 ). Example 5 β Find the polar coordinates of the points whose rectangular coordinates are (a) (0,3) and (b) (β3,0). Example 6 β Find the polar coordinates of a point whose rectangular coordinates are (2, β2). Sullivan & Sullivan β Section 9.1 9.1 Polar Coordinates Example 7 β Find the polar coordinates of a point whose rectangular coordinates are (β1, ββ3). π = βπ₯ 2 + π¦ 2 and π = π¦ ο· tanβ1 (π₯ ) if (π₯, π¦) is in Quadrant I or IV ο· tanβ1 (π₯ ) + π if (π₯, π¦) is in Quadrant II or III. π¦ This is due to the range of inverse tangent. Example 8 β Transform the equation π = 6 cos π from polar coordinates to rectangular coordinates. Example 9 β Transform the equation 4π₯π¦ = 9 from rectangular coordinates to polar coordinates. Sullivan & Sullivan β Section 9.1 9.1 Polar Coordinates Complete the following exercises on a separate sheet of paper. In exercises 1 β 8, plot each point given in polar coordinates, and find other polar coordinates (π, π½) of the point for which (π) π > π, βππ β€ π½ < π, (b) π < π, π β€ π½ < ππ , and (c) π > π, ππ β€ π½ < ππ . 1. (5, 2π ) 3 5. (1, 2 ) π 2. (4, 3π ) 4 6. (2, π) π 4 3. (β2, 3π) 7. (β3, β ) 4. (β3,4π) 8. (β2, β 2π ) 3 In exercises 9 β 18, the polar coordinates of a point are given. Find the rectangular coordinates of each point. π 2 9. (3, β ) 10. 3π (4, 2 ) 14. (5,300°) 15. (β2, 3π ) 4 2π ) 3 11. (β2,0) 16. (β2, β 12. (β3, π) 17. (6.3, 3.8) 13. (6,150°) 18. (8.1, 5.2) In exercises 19 β 26, the rectangular coordinates of a point are given. Find the polar coordinates for each point with π > π and π β€ π½ < ππ . 19. (0,2) 23. (β3, 1) 20. (β1,0) 24. (β2, β2β3) 21. (1, β1) 25. (1.3, β2.1) 22. (β3,3) 26. (β0.8, β2.1) In exercises 27 β 34, convert each Cartesian equation into its corresponding polar equation. 27. 2π₯ 2 + 2π¦ 2 = 3 31. 2π₯π¦ = 1 28. π₯ 2 + π¦ 2 = π₯ 32. 4π₯ 2 π¦ = 1 29. π₯ 2 = 4π¦ 33. π₯ = 4 30. π¦ 2 = 2π₯ 34. π¦ = β3 In exercises 35 β 42, convert each polar equation into its corresponding Cartesian equation. 35. π = cos π 37. π 2 = cos π 36. π = sin π + 1 38. π = sin π β cos π Sullivan & Sullivan β Section 9.1 9.1 Polar Coordinates 39. π = 2 3 42. π = 3βcos π 40. π = 4 4 41. π = 1βcos π Sullivan & Sullivan β Section 9.1 9.2 Polar Equations and Graphs A unique aspect to polar equations is that the input is π and the output is π. Thus, polar coordinates (π, π) are reversed from rectangular coordinates (π₯, π¦). Example 1 β Identify and graph the equation π = 3. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ π Example 2 β Identify and graph the equation π = 4 . ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Example 3 β Identify and graph the equation π sin π = 2. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Example 4 β Use a graphing calculator to graph the polar equation π sin π = 2. Sullivan & Sullivan β Section 9.2 9.2 Polar Equations and Graphs Example 5 β Identify and graph the equation π cos π = β3. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Let π be a nonzero real number. Then the graph of π sin π = π is a horizontal line |π| units above the pole if π > 0 and |π| below the pole if π < 0. The graph of π cos π = π is a vertical line |π|units to the right of the pole if π > 0 and |π| to the left of the pole if π < 0. Example 6 β Identify and graph the equation π = 4 sin π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Example 7 β Identify and graph the equation π = β2 cos π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Let π be a positive real number. ο· ο· ο· ο· π = 2π sin π graphs a circle with radius π and center (0, π) in rectangular coordinates. π = 2π cos π graphs a circle with radius π and center (π, 0) in rectangular coordinates. π = β2π sin π graphs a circle with radius π and center (0, βπ) in rectangular coordinates. π = β2π cos π graphs a circle with radius π and center (βπ, 0) in rectangular coordinates. Tests for Symmetry ο· In a polar equation, if replacing π with β π results in an equivalent equation, the graph is symmetric with respect to the polar axis. Sullivan & Sullivan β Section 9.2 9.2 Polar Equations and Graphs ο· In a polar equation, if replacing π with π β π results in an equivalent equation, the graph is symmetric with π respect to the line π = 2 . ο· In a polar equation, if replacing π with βπ results in an equivalent equation, the graph is symmetric with respect to the pole. Example 8 β Graph π = 1 β sin π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π Example 9 β Graph π = 3 + 2 cos π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Example 10 β Graph π = 1 + 2 cos π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Limaçons are characterized by equations of the form π = π + π cos π, π = π β π cos π, π = π + π sin π, and π = π β π sin π, where π > 0 and π > 0. ο· ο· ο· If π = π, the Limaçon is a cardioid. If π < π, the Limaçon has an inner loop. If π > π, the Limaçon has no inner loop. Sullivan & Sullivan β Section 9.2 9.2 Polar Equations and Graphs Example 11 β Graph π = 2 cos 2π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π Example 12 β Graph π = 2 sin 2π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Example 13 β Graph π = 2 cos 3π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π Example 14 β Graph π = 2 sin 3π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Sullivan & Sullivan β Section 9.2 9.2 Polar Equations and Graphs Rose curves are characterized by equations of the form π = π cos(ππ) and π = π sin(ππ), where π β 0 and π β 0. If π < 0 use the odd and even identities to eliminate the negative coefficient from the argument. ο· ο· ο· If π is even, the rose has 2π petals of length |π|. If π is odd , the rose has π petals of length |π|. For π = π cos(ππ) o The first petal lies along the polar axis if π > 0. o The first petal lies along π = π if π < 0. o The rest of the petals are equally spaced around the pole. o The rose is symmetric with respect to the polar axis. For π = π sin(ππ) π o The first petal lies along π = 2π if π > 0. ο· o o o π The first petal lies along π = β 2π if π < 0. The rest of the petals are equally spaced around the pole. π The rose is symmetric with respect to the line π = . 2 Example 15 β Graph π 2 = 4 sin(2π). ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 οο΅ π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π Lemniscates are characterized by equations of the form π 2 = π2 sin 2π and π 2 = π2 cos 2π, where π β 0. ο· π 2 = π2 cos 2π graphs a two-petal rose such that each petal is of length |π|, with one petal along π = 0 and the other along π = π. ο· π 2 = π2 sin 2π graphs a two-petal rose such that each petal is of length |π|, with one petal along π = 4 and π the other along π = 5π . 4 Example 16 β Graph π = π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ Sullivan & Sullivan β Section 9.2 π 0 π 12 π 6 π 4 π 3 π 2 π π 2π 3 3π 4 5π 6 π 7π 6 5π 4 π π 4π 3 3π 2 5π 3 7π 4 11π 6 2π π 9.2 Polar Equations and Graphs Example 17 β Graph π = 2 + 2 sin π. ο΅ y ο΄ ο³ ο² ο± x οο΅ οο΄ οο³ οο² οο± ο± ο² ο³ ο΄ ο΅ οο± οο² οο³ οο΄ οο΅ An important distinction must be made. π₯ 2 + π¦ 2 = 1 is an implicit equation that graphs the unit circle. To graph π₯ 2 + π¦ 2 = 1 in the calculator, two separate equations must be graphed: π¦ = ±β1 β π₯ 2 , which means a circle cannot be written as a function in the rectangular coordinates since every π₯-value in the domain could possibly yield two different π¦-values. However, in polar coordinates, the unit circle can be graphed with one equation π = 1, and since every angle π could only have one output, namely a radius of 1, the unit circle can be represented as a function in polar coordinates. We will find it useful in Calculus to represent nonfunctions in rectangular coordinates as functions in polar coordinates. Complete the following exercises on a separate sheet of paper. In exercises 1 β 8, transform each polar equation to an equation in rectangular coordinates. Identify and graph the equation by hand. 1. π = 2 5. π = 2 sin π π 2. π = β 4 6. π = β4 cos π 3. π cos π = 4 7. π csc π = 8 4. π sin π = β2 8. π sec π = β4 In exercises 9 β 20, identify and graph each polar equation by hand. 9. π = 1 + sin π 15. π = 2 sin 3π 10. π = 2 β 2 cos π 16. π = 3 cos 4π 11. π = 2 β cos π 17. π 2 = sin 2π 12. π = 4 + 2 sin π 18. π 2 = 9 cos 2π 13. π = 1 β 2 sin π 19. π = 3 + cos π 14. π = 2 + 4 cos π 20. π = 4 cos 3π In exercises 21 β 26, graph each pair of polar equations on the same polar grid. Find the polar coordinates of the point(s) of intersection and label the point(s) on the graph. 21. π = 8 cos π ; π = 2 sec π 24. π = 3; π = 2 + 2 cos π 22. π = 8 sin π ; π = 4 csc π 25. π = 1 + sin π ; π = 1 + cos π 23. π = sin π ; π = 1 + cos π 26. π = 1 + cos π ; π = 3 cos π Sullivan & Sullivan β Section 9.2 9.3 The Complex Plane and cis Notation An imaginary number is the square root of a negative number. Using π for the unit imaginary number ββ1, we get ββπ2 = ππ. A complex number is the sum of a real number and an imaginary number, π + ππ. Real numbers and imaginary numbers fall within the set of complex numbers. You can represent complex numbers as points on a Cartesian coordinate system called the complex plane with the horizontal real axis and vertical imaginary axis. We can also represent complex numbers in polar form as well. ο· ο· ο· ο· ο· ο· π = ββ1 π 2 = β1 π§ = π + ππ is a complex number with π being the real part of π§ and π being the imaginary part of π§. π§ = π + ππ is plotted as (π, π) on the complex plane. π β ππ is the complex conjugate of π + ππ. (π + ππ)(π β ππ) = π2 + π 2 is always real. Since π + ππ can be graphed as a point with rectangular coordinates (π, π), it can also be graphed as (π, π) in polar coordinates, such that π = π cos π and π = π sin π. Thus π§ = π + ππ = π cos π + ππ sin π = π(cos π + π sin π) = π cis π. We refer to π cis π as the polar form of a complex number or cis notation. π is called the modulus or magnitude, and π is called the argument. The absolute value of a complex number is the modulus of that number: |π§| = |π + ππ| = π. ο· π§ = π + ππ = π cis π o π = π cos π o π = π sin π o π 2 = π2 + π 2 o π π π π π = tanβ1 ( ) or π = tanβ1 ( ) + π depending on which quadrant π§ falls into. Example 1 β Transform the complex number π§ = β5 + 7π into polar form. Example 2 β Transform π§ = 5 cis 144° into Cartesian form. Sullivan & Sullivan β Section 9.3 Foerster β Section 13.4 9.3 The Complex Plane and cis Notation Example 3 β If π§1 = 3 cis 83° and π§2 = 2 cis 41°, find the product π§1 π§2 . If π§1 = π1 cis π1 and π§2 = π2 cis π2 , then π§1 π§2 = π1 π2 cis(π1 + π2 ). Example 4 β Find the reciprocal of π§ = 2 cis 29°. 1 1 If π§ = π cis π, then π§ = π cis(βπ). π§ Example 5 β If π§1 = 5 cis 71° and π§2 = 2 cis 29°, find π§1. 2 π§ π If π§1 = π1 cis π1 and π§2 = π2 cis π2 , then π§1 = π1 cis(π1 β π2 ). 2 Sullivan & Sullivan β Section 9.3 Foerster β Section 13.4 2 9.3 The Complex Plane and cis Notation Example 6 β If π§ = 2 cis 29°, find π§ 5 . DeMoivreβs Theorem If π§ = π cis π, then π§ π = π π cis ππ. π π π If π§ = π cis π, then βπ§ = βπ cis (π + 360° π π), for all integers 0 β€ π < π. There will be π distinct roots. 3 Example 7 β If π§ = 8 cis 60°, find the cube roots of π§, βπ§. Complete the following exercises on a separate sheet of paper. In exercises 1 β 12, write the complex number in polar form, π πππ π½. 1. β1 + π 7. 5 + 7π 2. 1 β π 8. β11 β 2π 3. β3 β π 9. 1 4. 1 + πβ3 10. π 5. β4 β 3π 11. βπ 6. β3 + 4π 12. β8 In exercises 13 β 22, write the complex number in Cartesian form, π + ππ. 13. 8 cis 34° 15. 6 cis 120° 14. 11 cis 247° 16. 8 cis 150° Sullivan & Sullivan β Section 9.3 Foerster β Section 13.4 9.3 The Complex Plane and cis Notation 17. β2 cis 225° 20. 9 cis 90° 18. 3β2 cis 45° 21. 3 cis 270° 19. 5 cis 180° 22. 2 cis 0° π In exercises 23 β 26, find (a) ππ ππ , (b) ππ, (c) πππ , and (d) πππ . π 23. π§1 = 3 cis 47°, π§2 = 5 cis 36° 25. π§1 = 4 cis 238°, π§2 = 2 cis 51° 24. π§1 = 2 cis 154° , π§2 = 3 cis 27° 26. π§1 = 6 cis 19°, π§2 = 4 cis 96° In exercises 27 β 36, find all of the indicated roots. 27. Cube roots of 27 cis 120° 32. Square roots of βπ 28. Cube roots of 8 cis 15° 33. Cube roots of 8 29. Fourth roots of 16 cis 80° 34. Cube roots of β27 30. Fourth roots of 81 cis 64° 35. Sixth roots of β1 31. Square roots of π 36. Tenth roots of 1 Sullivan & Sullivan β Section 9.3 Foerster β Section 13.4 Module 9 β Selected Solutions
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