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**Unformatted text preview: **9.4 GRAPHING IN POLAR COORDINATES 539 25. Find a formula for the distance between [r1 , θ1 ] and [r2 , θ2 ]. 26. Show that for r1 > 0, r2 > 0, |θ1 − θ2 | < π the distance formula you found in Exercise 25 reduces to the law of cosines. Find the point [r , θ ] symmetric to the given point about: (a) the x-axis; (b) the y-axis; (c) the origin. Express your answer with r > 0 and θ ∈ [0, 2π ). 27. [ 1 , 1 π ]. 26 29. [−2, 1 π ]. 3 28. [3, − 5 π ]. 4 30. [−3, − 7 π ]. 4 57. tan θ = 2. 58. r = 2 sin θ . Write the equation in rectangular coordinates and identify the curve. 6 4 . 60. r = . 59. r = 2 − cos θ 1 + 2 sin θ 2 4 . 62. r = . 61. r = 1 − cos θ 3 + 2 sin θ 63. Show that if a and b are not both zero, then the curve r = a sin θ + b cos θ is a circle. Find the center and the radius. 64. Find a polar equation for the set of points P [r , θ ] such that the distance from P to the pole equals the distance from P to the line x = −d . See the ﬁgure.
y P [r, θ ] x = –d Test the curve for symmetry about the coordinate axes and for symmetry about the origin. 31. r = 2 + cos θ . 33. r ( sin θ + cos θ ) = 1. 35. r 2 sin 2θ = 1. x = 2. 2xy = 1. x2 + ( y − 2)2 = 4. y = x. 32. r = cos 2θ . 34. r sin θ = 1. 36. r 2 cos 2θ = 1. y = 3. x2 + y2 = 9. (x − a)2 + y2 = a2 . x2 − y2 = 4. Write the equation in polar coordinates. 37. 39. 41. 43. 38. 40. 42. 44. 45. x2 + y2 + x = x2 + y2 . 46. y = mx. 48. (x2 + y2 )2 = x2 − y2 . 47. (x2 + y2 )2 = 2xy. Polar axis x Identify the curve and write the equation in rectangular coordinates. 49. r sin θ = 4. 51. θ = 1 π . 3 53. r = 2(1 − cos θ )−1 . 55. r = 3 cos θ . 50. r cos θ = 4. 52. θ 2 = 1 π 2 . 9 54. r = 4 sin (θ + π ). 56. θ = − 1 π . 2 65. Find a polar equation for the set of points P [r , θ ] such that the distance from P to the pole is half the distance from P to the line x = −d . 66. Find a polar equation for the set of points P [r , θ ] such that the distance from P to the pole is twice the distance from P to the line x = −d . 9.4 GRAPHING IN POLAR COORDINATES
We begin with the curve r = θ, θ ≥ 0. The graph is a nonending spiral, part of the famous spiral of Archimedes. The curve is shown in detail from θ = 0 to θ = 2π in Figure 9.4.1. At θ = 0, r = 0; at θ = 1 π , r = 1 π ; at θ = 1 π , r = 1 π ; and so on. 4 4 2 2 The next examples involve trigonometric functions.
Example 1
SOLUTION Sketch the curve r = 1 − 2 cos θ . Since the cosine function is periodic with period 2π , the curve r = 1 − 2 cos θ is a closed curve. We will draw it from θ = 0 to θ = 2π . The curve just repeats itself for values of θ outside the interval [0, 2π ]. 540 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS
1 π, 1 π 2 2 3 π, 3 π 4 4 1 π, 1π 4 4 Polar axis [2 π , 2π ] x [π , π ] O 5 π, 5π 4 4 3 π, 3π 2 2 7 π, 7π 4 4 r = θ, θ ≥ 0 spiral of Archimedes Figure 9.4.1 We begin by compiling a table of values: θ r 0 π/4 π/3 π/2 2π/3 3π/4 π 5π/4 4π/3 3π/2 5π/3 7π/4 2π 0 1 2 2. 41 3 2. 41 2 1 0 −0. 41 −1 −1 −0. 41 The values of θ for which r = 0 or |r | is a local maximum are as follows: r = 0 at θ = 1 π , 5 π 3 3 for then cos θ = 1 ; 2 |r | is a local maximum at θ = 0, π , 2π . These ﬁve values of θ generate four intervals: [0, 1 π ], 3 [ 1 π , π ], 3 [π , 5 π ], 3 [ 5 π , 2π ]. 3 We sketch the curve in four stages. These stages are shown in Figure 9.4.2. [–1, 0] polar axis 0 ≤ θ ≤ 3π
1 [–1, 0] [3, π ] 0 ≤θ ≤π polar axis [–1, 0] polar axis 0 ≤ θ ≤ 3π
5 polar axis 0 ≤ θ ≤ 2π Figure 9.4.2 As θ increases from 0 to 1 π , cos θ decreases from 1 to 1 and r = 1 − 2 cos θ increases 3 2 from −1 to 0. As θ increases from 1 π to π , cos θ decreases from 1 to −1 and r increases from 0 to 3. 3 2 As θ increases from π to 5 π , cos θ increases from −1 to 1 and r decreases from 3 to 0. 3 2 Finally, as θ increases from 5 π to 2π , cos θ increases from 1 to 1 and r decreases 3 2 from 0 to −1. As we could have read from the equation, the curve is symmetric about the x-axis [r (−θ ) = 1 − 2 cos (−θ ) = 1 − 2 cos θ = r (θ )].
Example 2 Sketch the curve r = cos 2θ , 0 ≤ θ ≤ 2π . 9.4 GRAPHING IN POLAR COORDINATES
SOLUTION 541 As an alternative to compiling a table of values as we did in Example 1, we refer to the graph of cos 2θ in rectangular coordinates for the values of r . See Figure 9.4.3. The values of θ for which r is zero or has an extreme value are as follows:
π π π r = 0 at θ = π , 34 , 54 , 74 ; 4 π local maxima/minima at θ = 0, π , π , 32 , 2π . 2 cos 2θ 1 π
4 5π 4 3π 2 7π 4 π
2 3π 4 π 2π θ –1 Figure 9.4.3 In Figure 9.4.4 we sketch the curve in eight stages.
θ = 4π
1 0 ≤ θ ≤ 4π 1 0 ≤ θ ≤ 2π 1 0 ≤ θ ≤ 4π 3 0 ≤θ ≤π 0 ≤ θ ≤ 4π 5 0 ≤ θ ≤ 2π 3 0 ≤ θ ≤ 4π 7 0 ≤ θ ≤ 2π Figure 9.4.4 Example 3 Figure 9.4.5 shows four cardioids, heart-shaped curves. Rotation of r = 1 + cos θ by 1 π radians, measured in the counterclockwise direction, gives 2 r = 1 + cos (θ − 1 π ) = 1 + sin θ . 2 r = 1 + cos θ r = 1 + sin θ r = 1 – cos θ r = 1 – sin θ Figure 9.4.5 Rotation by another 1 π radians gives 2 r = 1 + cos (θ − π ) = 1 − cos θ . 542 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS Rotation by yet another 1 π radians gives 2 r = 1 + cos (θ − 3 π ) = 1 − sin θ . 2 Notice how easy it is to rotate axes in polar coordinates: each change
cos θ → sin θ → − cos θ → − sin θ represents a counterclockwise rotation by 1 π radians. 2 At this point we will try to give you a brief survey of some of the basic polar curves. (The numbers a and b that appear below are to be interpreted as nonzero constants.) Lines : θ = a,
y y=x x r = a sec θ ,
y r = a csc θ .
x = –2 y y=x x x (Figure 9.4.6)
y y=2 x θ= π 1 4 θ= π 5 4 r = –2 sec θ r = 2 csc θ Figure 9.4.6 Circles : r = a, r = a sin θ , r = a cos θ . (Figure 9.4.7) r=2 r = –2 r = 4 sin θ r = –4 cos θ Figure 9.4.7 Limaçons : †r = a + b sin θ , r = a + b cos θ . (Figure 9.4.8) r = 3 + cos θ convex limaçon r = 3 + cos θ 2 limaçon with a dimple r = 1 + cos θ cardioid r = 1 + cos θ 2 limaçon with an inner loop Figure 9.4.8 The general shape of the curve depends on the relative magnitudes of |a| and |b|.
† From the French term for “snail”. The word is pronounced with a soft c. 9.4 GRAPHING IN POLAR COORDINATES 543 Lemniscates : †r 2 = a sin 2θ ,
2 r 2 = a cos 2θ .
3 4 (Figure 9.4.9)
π
1 4 π 2 2 2 r 2 = 4 sin 2θ r 2 = 4 cos 2θ Figure 9.4.9 Petal Curves : r = a sin nθ , r = a cos nθ ,
1 3 integer n.
3 8 (Figure 9.4.10)
π π
1 6 π
1 8 π 1 1 r = sin 3θ r = cos 4θ Figure 9.4.10 If n is odd, there are n petals. If n is even, there are 2n petals. The Intersection of Polar Curves
The fact that a single point has many pairs of polar coordinates can cause complications. In particular, it means that a point [r1 , θ1 ] can lie on a curve given by a polar equation although the coordinates r1 and θ1 do not satisfy the equation. For example, the coordinates of [2, π ] do not satisfy the equation r 2 = 4 cos θ : r 2 = 22 = 4 but 4 cos θ = 4 cos π = −4. Nevertheless the point [2, π ] does lie on the curve r 2 = 4 cos θ . It lies on the curve because [2, π ] = [−2, 0] and the coordinates of [−2, 0] do satisfy the equation: r 2 = (−2)2 = 4, 4 cos θ = 4 cos 0 = 4. In general, a point P [r1 , θ1 ] lies on a curve given by a polar equation if it has at least one polar coordinate representation [r , θ ] which satisﬁes the equation. The difﬁculties are compounded when we deal with two or more curves. Here is an example.
Example 4 Find the points where the cardioids r = a(1 − cos θ ) and r = a(1 + cos θ )
(a > 0) intersect.
† From the Latin lemniscatus, meaning “adorned with pendant ribbons.” 544 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS We begin by solving the two equations simultaneously. Adding these equations, we get 2r = 2a and thus r = a. This tells us that cos θ = 0 and therefore θ = 1 π + nπ . The points [a, 1 π + nπ ] all lie on both curves. Not all of these points 2 2 are distinct:
SOLUTION for n even, [a, 1 π + nπ ] = [a, 1 π ]; 2 2 for n odd, [a, 1 π + nπ ] = [a, 3 π ]. 2 2 In short, by solving the two equations simultaneously we have arrived at two common points: [a, 1 π ] = (0, a) 2 and [a, 3 π ] = (0, −a). 2 However, by sketching the two curves (see Figure 9.4.11), we see that there is a third point at which the curves intersect; the two curves intersect at the origin, which clearly lies on both curves: for for r = a(1 − cos θ ) r = a(1 + cos θ )
θ= π
[a,
1 π] 2 1 2 take θ = 0, 2π , . . . , take θ = π , 3π , . . . . = (0, a) θ =π polar axis r = a (1 – cos θ ) r = a (1 + cos θ ) θ= π 3 2 [a, 3 π] 2 = (0, – a) Figure 9.4.11 The reason that the origin does not appear when we solve the two equations simultaneously is that the curves do not pass through the origin “simultaneously”; that is, they do not pass through the origin for the same values of θ . Think of each of the equations r = a(1 − cos θ ) and r = a(1 + cos θ ) as giving the position of an object at time θ . At the points we found by solving the two equations simultaneously, the objects collide. (They both arrive there at the same time.) At the origin the situation is different. Both objects pass through the origin, but no collision takes place because the objects pass through the origin at different times. Sketch the polar curves r = 2 sin θ and r = 2 sin 2θ , and ﬁnd their points of intersection, if any.
Example 5
SOLUTION The curve r = 2 sin θ is a circle of radius 1 and center on the ray θ = π/2. The entire circle is traced out as θ varies from 0 to π . The curve r = 2 sin 2θ is a petal curve with four petals. This curve is traced out as θ varies from 0 to 2π . The two curves are shown in Figure 9.4.12. 9.4 GRAPHING IN POLAR COORDINATES
r = 2 sin θ B A 545 r = 2 sin 2θ C Figure 9.4.12 From the ﬁgure, we see that there are three points of intersection: the origin, the point labeled A, and the point labeled B. Solving the two equations simultaneously, we have 2 sin 2θ = 2 sin θ , 2 sin θ cos θ = sin θ , sin θ (2 cos θ − 1) = 0. Setting sin θ = 0, we get θ = nπ , n an integer. We can take [0, 0] as the coordinates of this point of intersection. Setting 2 cos θ − 1 = 0, we get √= (π/3) + 2nπ and θ = (5π/3) + 2nπ . θ The point A clearly has coordinates [ 3, π/3], but the point B involves a complication. √ do The coordinates [ 3, 2π/3] satisfy the equation for the circle, but they √ not satisfy the equation for the petal curve: at θ = 2π/3, r = 2 sin 2(2π/3) = − 3 (the point labeled C in the ﬁgure). √ You can verify that both of the sets of coordinates [ 3, (2π/3) + 2nπ ] and √ √ [− 3, (5π/3) + 2nπ ] satisfy r = 2 sin θ , while only [− 3, (5π/3) + 2nπ ] satisﬁes r = 2 sin 2θ . As in Example 4, if the equations describe the positions of two objects at time θ , one moving around the circle and the other around the petal curve, then at θ = 2π/3 the object on the circle is at B and the object on the petal curve is C . However, at time θ = 5π/3 both objects are at the point B and they collide. The point √ [− 3, 5π/3] satisﬁes both equations simultaneously. Remark Problems of incidence (does such and such a point lie on the curve with the following polar equation?) and problems of intersection (where do such and such polar curves intersect?) can usually be analyzed by sketching the curves. However, there are situations where such problems can be handled more readily by ﬁrst changing to rectangular coordinates. For example, the rectangular equation of the curve r 2 = 4 cos θ discussed above is (x2 + y2 )3 = 16x2 . The point P [2, π ] has rectangular coordinates (−2, 0), and it is easy to verify that the pair x = −2, y = 0 satisﬁes this equation. In a similar manner, the symmetry properties of a curve can often be analyzed more easily in rectangular coordinates. To illustrate this, the curve C given by r 2 = sin θ 546 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS is symmetric about the x-axis: (1)
y 1 if [r , θ ] ∈ C , then [r , −θ ] ∈ C . But this is not easy to see from the polar equation because, in general, if the coordinates of the ﬁrst point satisfy the equation, the coordinates of the second point do not. One way to see that (1) is valid is to note that [r , −θ ] = [−r , π − θ ] and then verify that, if the coordinates of [r , θ ] satisfy the equation, then so do the coordinates of [−r , π − θ ]. But all this is very cumbersome. The easiest way to see that the curve r 2 = sin θ is symmetric about the x-axis is to write it as ( x 2 + y 2 )3 = y 2 . 0.5 0.5 – 0.5 x –1 Figure 9.4.13 The other symmetries of the curve, symmetric about the y-axis and symmetric about the origin, can also be seen from this equation (see Figure 9.4.13). EXERCISES 9.4
Sketch the polar curve. 1. θ = − 1 π . 4 3. r = 4. 5. r = −2 sin θ . 7. r csc θ = 3. 9. r = θ , − 1 π ≤ θ ≤ π . 2 11. r = sin 3θ . 13. r 2 = sin 2θ . 15. r = 4,
2 3 2. r = −3. 4. r = 3 cos θ . 6. θ = 2 π . 3 8. r = 1 − cos θ . 10. r sec θ = −2. 12. r = cos 2θ .
2 37. Show that the point [2, π ] lies both on r 2 = 4 cos θ and on r = 3 + cos θ . 38. Show that the point [2, 1 π ] lies both on r 2 sin θ = 4 and on 2 r = 2 cos 2θ . Sketch the curves and ﬁnd the points at which they intersect. Express your answers in rectangular coordinates. 39. r = sin θ , 40. r = sin θ ,
2 r = − cos θ . r = 2 − sin θ . r = −1. r = 2 cos θ . r = cos θ . r = sin θ . r = 1 + sin θ . 14. r = cos 2θ .
3 π. 4 41. r = cos2 θ , 42. r = 2 sin θ , 0≤θ ≤ 16. r = sin θ . 17. r = 9r . 18. θ = −1π, 4 1 ≤ r < 2. 20. r 2 = 4r .
1 π. 2 1 π. 2 43. r = 1 − cos θ , 44. r = 1 − cos θ , 45. r = sin 2θ , 46. r = 1 − cos θ , 19. r = −1 + sin θ . 21. r = sin 2θ . 22. r = cos 3θ , 23. r = cos 5θ , 24. r = eθ , 27. r = tan θ . 29. r = 2 + sec θ . 31. r = −1 + 2 cos θ . 33. r 2 cos θ = 1; 35. r = sin 1 θ ; 3 36. r 2 = sin 3θ ; [1, π ]. [ 1 , 1 π ]. 22 [1, − 5 π ]. 6 25. r = 2 + sin θ . 0≤θ ≤ 0≤θ ≤ r = sin θ . −π ≤ θ ≤ π (logarithmic spiral). 26. r = cot θ . 28. r = 2 − cos θ . 30. r = 3 − csc θ . 32. r = 1 + 2 sin θ . 34. r 2 = cos 2θ ; [1, 1 π ]. 4 47. Show that the polar equation r = 2a sin θ + 2b cos θ represents a circle. Find the center and radius of the circle and sketch its graph. c 48. a. Use a graphing utility to draw the curves r = 1 + cos (θ − 1 π ) 3 and r = 1 + cos (θ + 1 π ). 6 Determine whether the point lies on the curve. How do your curves compare with the graph of r = 1 + cos θ ? (See Figure 9.4.5) b. In general, what is the relationship between the graph of r = f (θ − α ) and the graph of r = f (θ )? c 49. a. Use a graphing utility to draw the curves
r = 1 + sin θ and r 2 = 4 sin 2θ in the same coordinate system. 9.4 GRAPHING IN POLAR COORDINATES 547 b. Use a CAS to ﬁnd the points of intersection of the two curves. c 50. Repeat Exercise 49 for the pair of equations r = 1 + cos θ and r = 1 + cos ( 1 θ ). 2 Use a graphing utility to draw the equipotential lines for m = 1, 2, 3. (c) Draw the curves r = 2 sin2 θ and r 2 = 2 cos θ in the same coordinate system and estimate the coordinates of their points of intersection. Use four decimal place accuracy. c 51. Repeat Exercise 49 for the pair of equations
r = 2 cos θ and r = 2 sin 2θ . c 55. Use a graphing utility to draw the curves
r = 1 + sin k θ + cos2 (2k θ ) for k = 1, 2, 3, 4, 5. If you were asked to give a name to this family of curves, what would you suggest? c 52. Repeat Exercise 49 for the pair of equations
r=2 and r = 2 sin 3θ . c 53. Repeat Exercise 49 for the pair of equations
r = 1 − 3 cos θ and r = 2 − 5 sin θ . c 54. (a) The electrostatic charge distribution consisting of a charge q (q > 0) at the point [r , 0] and a charge −q at [r , π ] is called a dipole. The lines of force for the dipole are given by the equations
r = k sin θ .
2 c 56. Use a graphing utility to draw the graph r = ecos θ − 2 cos 4θ . What name would you suggest for this curve? c 57. The graphs of r = A cos k θ and r = A sin k θ are called petal curves (see Figure 9.4.10). Use a graphing utility to draw the curves
r = 2 cos k θ and r = 2 sin k θ for k = 3 and k = 5 . Can you predict what the curve will 2 2 look like for k = m/2 for any odd integer m? Use a graphing utility to draw the lines of force for k = 1, 2, 3. (b) The equipotential lines (the set of points with equal electric potential) for the dipole are given by the equations r 2 = m cos θ . c 58. Use a graphing utility to draw the curves
r = 2 cos k θ and r = 2 sin k θ for k = 4 and k = 5 . Can you predict what the curve will 3 3 look like for k = m/3 for any even integer m that is not a multiple of 3? For any odd integer m that is not a multiple of 3? PROJECT 9.4 The Conic Sections in Polar Coordinates
In Section 9.1, we deﬁned a parabola in terms of a focus and directrix, but our deﬁnitions of the ellipse and hyperbola in Section 9.2 were given in terms of two foci; there was no mention of a directrix for either of these conics. In this project, we give a uniﬁed approach to the conic sections that involves a focus and directrix for all three cases. In the plane, let F be a ﬁxed point (the focus) and l a ﬁxed line (the directrix) which does not pass through F . Let e be a positive number (the eccentricity) and consider the set of points P that satisfy (1) distance from P to F = e. distance from P to l
y P r x=d θ
F r cos θ x In the ﬁgure, we have superimposed a polar and rectangular coordinate system and, without loss of generality, we have taken F as the origin and l as the vertical line x = d , d > 0. Problem 1. Show that the set of all points P that satisfy (1) is described by the polar equation: (2) r= ed . 1 + e cos θ Problem 2. Show that: a. If 0 < e < 1, the equation is an ellipse of eccentricity e with right focus at the origin, major axis horizontal: y2 ( x + c )2 +2 =1 a2 a − c2 with a= ed , 1 − e2 c = ea. 548 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS b. If e = 1, the equation represents a parabola with focus at the origin and directrix x = d : y2 = −4 d 2 x− d 2 . Consider the polar equation r = α/(1 + β sin θ ), α , β > 0. Since r= α α = 1 + β sin θ 1 + β cos (θ − π/2) c. If e > 1, the equation represents a hyperbola of eccentricity e with left focus at the origin, transverse axis horizontal: y2 ( x − c )2 −2 =1 a2 c − a2 with a= ed , e2 − 1 c = ea. we can conclude that r = α/(1 + β sin θ ) is simply the conic section r = α/(1 + β cos θ ) rotated π/2 radians in the counterclockwise direction. Problem 4. In terms of rotations, relate the polar equations r= α 1 − β cos θ and r= α , 1 − β sin θ α, β > 0 Problem 3. Identify each of the following conic sections and write the equation in rectangular coordinates. a. r = 8 . 4 + 3 cos θ b. r = 6 . 1 + 2 cos θ c. r = 6 . 2 + 2 cos θ to the conic section r= α . 1 + β cos θ 9.5 AREA IN POLAR COORDINATES
θ =β r = ρ (θ ) Here we develop a technique for calculating the area of a region the boundary of which is given in polar coordinates. As a start, we suppose that α and β are two real numbers with α < β ≤ α + 2π . We take ρ as a function that is continuous on [α , β ] and keeps a constant sign on that interval. We want the area of the polar region generated by the curve
θ =α Γ r = ρ (θ ), α ≤ θ ≤ β. O polar axis Such a region is portrayed in Figure 9.5.1. In the ﬁgure ρ (θ ) remains nonnegative. If ρ (θ ) were negative, the region would appear on the opposite side of the pole. In either case, the area of is given by the formula
(9.5.1) Figure 9.5.1 A= β α 1 [ρ (θ )]2 d θ . 2 β = θn θi θ i –1 α = θo
Ri ri polar axis PROOF We consider the case where ρ (θ ) ≥ 0. We take P = {θ0 , θ1 , . . . , θn } as a partition of [α , β ] and direct our attention to what happens from θi−1 to θi . We set ri = min value of ρ on [θi−1 , θi ] and Ri = max value of ρ on [θi−1 , θi ]. The part of that lies between θi−1 and θi contains a circular sector of radius ri and central angle θi = θi − θi−1 and is contained in a circular sector of radius Ri with the same central angle θi . (See Figure 9.5.2.) Its area Ai must therefore satisfy the inequality
12 r 2i θi ≤ Ai ≤ 1 R2 2i θi . † O Figure 9.5.2 By summing these inequalities from i = 1 to i = n, we can see that the total area A must satisfy the inequality (1) Lf (P ) ≤ A ≤ Uf (P ) † The area of a circular sector of radius r and central angle α is 1 r 2 α . 2 ...

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