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(a) r = −3 (b) θ = 4π 3 (c) r = 1 − cos(θ) 11.4 Polar Coordinates 927 Solution. 1. One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with r cos(θ) and every occurrence of y with r sin(θ) and use identities to simplify. This is the technique we employ below. (... |
origin. As before, r = 0 describes the origin, but nothing else. Consider the equation θ = − π 4. In this equation, the variable r is free,7 meaning it can assume any and all values including r = 0. If we imagine plotting points (r, − π 4 ) for all conceivable values of r (positive, negative and zero), we are essentia... |
) by cos2(θ). Doing so, we get r = sin(θ) cos2(θ), or r = sec(θ) tan(θ). As before, the r = 0 case is recovered in the solution r = sec(θ) tan(θ) (let θ = 0), so we state the latter as our final answer. 2. As a general rule, converting equations from polar to rectangular coordinates isn’t as straight forward as the reve... |
4π 3. 3. Of course, we pause a moment to wonder 3 generate the same set of points.11 x, we get y 3 or c) Once again, we need to manipulate r = 1 − cos(θ) a bit before using the conversion formulas given in Theorem 11.7. We could square both sides of this equation like we did in part 2a above to obtain an r2 on the lef... |
!), we also went from y x = 3, in which x cannot be 0, to y = x √ 3, 3 in which we assume and θ = 4π x can be 0. 11.4 Polar Coordinates 929 algebraic maneuvers which may have altered the set of points described by the original equation. First, we multiplied both sides by r. This means that now r = 0 is a viable solutio... |
cos(θ) which means that any point (r, θ) which satisfies r2 = r2 + r cos(θ)2 has a representation which satisfies r = r2 + r cos(θ), and we are done. In practice, much of the pedantic verification of the equivalence of equations in Example 11.4.3 is left unsaid. Indeed, in most textbooks, squaring equations like r = −3 t... |
− 23. 9, 7π 2 20. (−20, 3π) 24. −5, − 9π 4 26. (−117, 117π) 27. (6, arctan(2)) 28. (10, arctan(3)) 17. 5, 7π 4 21. 3 5, π 2 25. 42, 13π 6 29. −3, arctan 31. 33. 2, π − arctan −1, π + arctan 4 3 1 2 3 4 34. 2 3 35. (π, arctan(π)) 36. 13, arctan 30. 5, arctan − 4 3 32. − 1 2, π − arctan (5), π + arctan 2 √ 2 12 5 In Exe... |
. y = 4x − 19 67. x = 3y + 1 68. y = −3x2 69. 4x = y2 70. x2 + y2 − 2y = 0 71. x2 − 4x + y2 = 0 72. x2 + y2 = x 73. y2 = 7y − x2 75. x2 + (y − 3)2 = 9 74. (x + 2)2 + y2 = 4 76. 4x2 + 4 y − 2 1 2 = 1 In Exercises 77 - 96, convert the equation from polar coordinates into rectangular coordinates. 77. r = 7 81. θ = 2π 3 78... |
−1 1 2 3 x −2 −3 11.4 Polar Coordinates 933 5. 12, − 12, −, −12,, 12, 7π 6 19π 6 11π 6 17π 6 6. 3, − 3, −, −3,, 3, 5π 4 13π 4 7π 4 11π 4 √ √ 7. 2 2 √ 2, −π, −2 √ 2, −3π, 2 2, 0 2, 3π 8. 7 2 7 2, −, − 13π 23π 6 5π 6 y 6 3 −12 −9 −6 −3 x y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 3 2 1 y −3 −2 −1 1 2 3 x −1 −2 −3 4 3 2 1 y −4 −3... |
π, −2π), (π, 2π3 −2 −1 −1 −2 −3 2 1 −2 −1 −1 −2 2 1 −2 −1 −1 −2 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 936 Applications of Trigonometry √ 5 2 2, − √ 2 5 2 17. 18. 1, √ 3 19. − √ 11 2 3, 11 2 22. 2 √ 3, −2 23. (0, −9) 20. (20, 0) 24 28. √ 10, 3 √ 10 √ 26 52 √ 5 26 52, − 32. 36. (5, 12) √ 2 3, 40. 44. 1 2, 11π 6 7π 6 4 3 48. (5... |
an(2) 64. r = 5 66. r = 19 4 cos(θ)−sin(θ) 67. x = 1 cos(θ)−3 sin(θ) 68. r = − sec(θ) tan(θ) 3 69. r = 4 csc(θ) cot(θ) 70. r = 2 sin(θ) 71. r = 4 cos(θ) 72. r = cos(θ) 11.4 Polar Coordinates 937 73. r = 7 sin(θ) 74. r = −4 cos(θ) 75. r = 6 sin(θ) 76. r = sin(θ) 77. x2 + y2 = 49 78. x2 + y2 = 9 79. x2 + y2 = 2 80. y = x... |
below for completeness. 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 P (r, θ) is on the graph of an equation if and only if there is a representation of P, say (r, θ), such that r and θ satisfy the... |
is free in the equation θ = − 3π 2. Plotting r, − 3π 2 for various values of r shows us that we are tracing out the y-axis. 940 Applications of Trigonometry y r > 0 r = 0 θ = − 3π 2 y 4 x −4 x 4 r < 0 −4 In θ = − 3π 2, r is free The graph of θ = − 3π 2 Hopefully, our experience in Example 11.5.1 makes the following re... |
the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y-axis (θ = π 2 ). The arrows drawn in the figure below are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the... |
, π] results in retracting some portion of the curve.4 We present the final graph below. r 6 3 −3 −6 π 2 π θ y 3 −3 3 x 6 r = 6 cos(θ) in the θr-plane r = 6 cos(θ) in the xy-plane Example 11.5.2. Graph the following polar equations. 1. r = 4 − 2 sin(θ) 2. r = 2 + 4 cos(θ) 3. r = 5 sin(2θ) 4. r2 = 16 cos(2θ) Solution. 1.... |
-plane pulls in from the negative y-axis to finish where we started. 2 to 2π, r decreases from 6 back to 4. The graph on the r 6 4 2 y x π 2 π 3π 2 2π θ θ runs from 3π 2 to 2π We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval [0, 2π] results in retracing portions ... |
which means the graph is heading into (and 3, a 3 as it approaches the origin. 2, 2π 3 5The ‘tangents at the pole’ theorem from second semester Calculus. 11.5 Graphs of Polar Equations 945 r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x 3, π, r ranges from 0 to −2. Since r ≤ 0, the curve passes through the On the in... |
positive x-axis. 2 to 2π, r increases from 2 out y r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ x θ runs from 3π 2 to 2π 11.5 Graphs of Polar Equations 947 Again, we invite the reader to show that plotting the curve for values of θ outside [0, 2π] results in retracing a portion of the curve already traced. Our final graph is b... |
≤ θ ≤ π, r increases from −5 to 0, so the curve pulls back to the origin. y π 4 π 2 3π 4 π θ x r 5 −5 11.5 Graphs of Polar Equations 949 Even though we have finished with one complete cycle of r = 5 sin(2θ), if we continue plotting beyond θ = π, we find that the curve continues into the third quadrant! Below we present ... |
the relationship between y = x and y = √ x over [0, 1], we also know cos(2θ) ≥ cos(2θ) wherever the former is defined. 950 Applications of Trigonometry examples, the lines θ = π serve as guides for us to draw the curve as is passes through the origin. 4, which are the zeros of the functions r = ±4 4 and θ = 3π cos(2θ),... |
] in the θr-plane. We leave the details to the reader. 9Numbers 1 and 2 in Example 11.5.2 are examples of ‘lima¸cons,’ number 3 is an example of a ‘polar rose,’ and number 4 is the famous ‘Lemniscate of Bernoulli.’ 11.5 Graphs of Polar Equations 951 coordinate system, but also prepare you for what is needed in Calculus... |
lima¸con called a ‘cardioid.’10 y 2 −2 2 x −4 r = 2 − 2 sin(θ) and r = 2 sin(θ) It appears as if there are three intersection points: one in the first quadrant, one in the second quadrant, and the origin. Our next task is to find polar representations of these points. In 10Presumably, the name is derived from its resemb... |
at the origin exactly when θ = πk for integers k. On the curve r = 2 − 2 sin(θ), however, we reach the origin when θ = π 2 + 2πk for integers k. There is no integer value of k for which πk = π 2 + 2πk which means while the origin is on both graphs, the point is never reached simultaneously. In any case, we have determ... |
From these solutions, we get 3. Proceeding as above, we first graph r = 3 and r = 6 cos(2θ) to get an idea of how many intersection points to expect and where they lie. The graph of r = 3 is a circle centered at the origin with a radius of 3 and the graph of r = 6 cos(2θ) is another four-leafed rose.12 y 6 3 −6 −3 3 x ... |
however, the equation r = 6 cos(2(θ + 2πk)) reduces to the same equation we had before, r = 6 cos(2θ), which means we get no additional solutions. Moving on to the case where r = −r, we have that θ = θ + (2k + 1)π for integers k. We look to see if we can find points P which have a representation (r, θ) that satisfies r ... |
4. As usual, we begin by graphing r = 3 sin θ 2 presented in Example 11.5.2, we find that we need to plot both functions as θ ranges from 0 to 4π to obtain the complete graph. To our surprise and/or delight, it appears as if these two equations describe the same curve! and r = 3 cos θ 2 y 3 −3 3 x −3 r = 3 sin θ 2 and ... |
as functions of r, 3 2 2 these are two different animals. 11.5 Graphs of Polar Equations 955 √ 3 2 2, π 2 2 + πk = 3 cos θ 2 [θ + 2πk] = 3 cos θ get only one intersection point which can be represented by. We now investigate other representations for the intersection points. Suppose P is an intersection point with a re... |
π 2 = cos θ 2 = ± sin θ 2 − sin θ 2 sin (2k+1)π 2 = 0 and sin (2k+1)π (2k+1)π where the last equality is true since cos 2 2 2 [θ + (2k + 1)π] can be rewritten as r = ±3 sin θ Hence, −r = 3 cos 1 (2k+1)π = 1, and the equation −r = 3 cos 1 = sin π then sin 2 2 reduces to −r = −3 sin θ, or r = 3 sin θ 2 2 What this means ... |
the region in the xy-plane described by the following sets. 1. (r, θ) | 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. (r, θ) | 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 3. (r, θ) | 2 + 4 cos(θ) ≤ r ≤ 0, 2π 4. (r, θ) | 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 3 ≤ θ ≤ 4π 3 ∪ (r, θ) | 0 ≤ r ≤ 2 − 2 sin(θ), π 6 ≤ θ ≤ π 2 Solution. Our first step in these p... |
) but inside or on the rose (since r ≤ 6 cos(2θ)). We shade the region below and r = 6 cos(2θ) (r, θ) | 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 11.5 Graphs of Polar Equations 957 3. From Example 11.5.2 number 2, we know that the graph of r = 2 + 4 cos(θ) is a lima¸con whose ‘inner loop’ is traced out as θ runs through the given... |
, we are shading from the origin (r = 0) out to the cardioid r = 2 − 2 sin(θ) which pulls into the origin at θ = π 2. Putting these two regions together gives us our final answer. 6 and continues to sin(θ) and r = 2 sin(θ) (r, θ) | 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r, θ) | 0 ≤ r ≤ 2 − 2 sin(θ), π ∪ 6 ≤ θ ≤ π 2 958 Applicat... |
= 3 cos(θ) and r = 1 + cos(θ) 22. r = 1 + sin(θ) and r = 1 − cos(θ) 23. r = 1 − 2 sin(θ) and r = 2 24. r = 1 − 2 cos(θ) and r = 1 25. r = 2 cos(θ) and r = 2 27. r2 = 4 cos(2θ) and r = 2 √ 3 sin(θ) √ 26. r = 3 cos(θ) and r = sin(θ) 28. r2 = 2 sin(2θ) and r = 1 29. r = 4 cos(2θ) and r = 2 30. r = 2 sin(2θ) and r = 1 In ... |
, π 40. (r, θ) | 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ In Exercises 41 - 50, use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. 41. The region inside the circle r = 5. 42. The region inside the circle r = 5 which lies in Quadrant III. 43. The region inside the lef... |
we find three extra lines. In order for the calculator to be able to plot r = 3 cos(4θ) in the xy-plane, we need to tell it not only the dimensions which x and y will assume, but we also what values of θ to use. From our previous work, we know that we need 0 ≤ θ ≤ 2π, so we enter the data you see above. (I’ll say more ... |
rose r = sin(2θ) have? What about r = sin(3θ), r = sin(4θ) and r = sin(5θ)? With the help of your classmates, make a conjecture as to how many petals the polar rose r = sin(nθ) has for any natural number n. Replace sine with cosine and repeat the investigation. How many petals does r = cos(nθ) have for each natural nu... |
θ), show that f (π − θ) = f (θ) and the graph of r = 4 − 2 sin(θ) is symmetric about the y-axis, as required. (See Example 11.5.2 number 1.) 17Recall that this means f (−θ) = f (θ) for θ in the domain of f. 18Recall that this means f (−θ) = −f (θ) for θ in the domain of f. 962 Applications of Trigonometry (b) For f (θ)... |
about r = f (θ) and r = f (π − θ)? Test out your conjectures using a variety of polar functions found in this section with the help of a graphing utility. 67. With the help of your classmates, research cardioid microphones. 68. Back in Section 1.2, in the paragraph before Exercise 53, we gave you this link to a fascin... |
= 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ = 5π 6 −3 −1 1 3 x −3 −1 1 3 x −1 −3 θ = 5π 3 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) y √ 2 3 + 4 θ = 5π 6 √ 2 3 √ −2 3 − 4 θ = 7π 6 √ −2 3 √ −1 −3 16. Lima¸con: r = 3 − 5 cos(θ) y 8 3 θ = arccos 3 5 −8 −2 8 x −3 −8 θ = 2π − arccos 3 5 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con... |
1 3 x −1 −3 24. r = 1 − 2 cos(θ) and r = 1 y 3 1 −3 −1 1 3 x −1 −3 25. r = 2 cos(θ) and r = 2 √ 3 sin(θ) y 4 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 −4 968 Applications of Trigonometry √ 3 10 10, arctan(3), pole √ 2,, π 6 √, 5π 6 2, √, 7π 6 2, √ 2, 11π 6 1,, π 12 1,, 5π 12 1, 13π 12, 1, 17π 12 26. r = 3 cos(θ) and r = sin(θ) ... |
(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) | 0 ≤ r ≤ 2 sin(2θ), 3 −2 −1 1 2 3 −1 −2 −3 x −2 2 x −2 35. (r, θ) | 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 4 36. (r, θ) | 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 y 4 y 3 1 −4 4 x −3 −1 1 3 x −4 −1 −3 11.5 Graphs of Polar Equations 971 37. (r, θ) | 1 + cos(θ) ≤ r ≤ 3 cos(θ), − 3 −2 −1 1 2 3 −1 −2... |
2 or (r, θ) | 0 ≤ r ≤ 2 − 2 sin(θ), 3π 47. (r, θ) | 0 ≤ r ≤ 3 cos(4θ), 0 ≤ θ ≤ π 8 or (r, θ) | 0 ≤ r ≤ 3 cos(4θ), − ≤ 5π 2 ∪ (r, θ) | 0 ≤ r ≤ 2 − 2 sin(θ), 3π 2 ≤ θ ≤ 2π ∪ (r, θ) | 0 ≤ r ≤ 3 cos(4θ), 15π 8 ≤ θ ≤ 2π 48. {(r, θ) | 3 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 49. (r, θ) | 0 ≤ r ≤ 3 cos(θ), − π 50. (r, θ) | 0 ≤ r ≤ 6 sin(2θ), ... |
consider the point P (x, y). The coordinates (x, y) are rectangular coordinates and are based on the x- and y-axes. Suppose we wished to find rectangular coordinates based on the x- and y-axes. That is, we wish to determine P (x, y). While this seems like a formidable challenge, it is nearly trivial if we use polar coo... |
x and y. Perhaps the cleanest way2 to solve this system is to write it as a matrix equation. Using the machinery developed in Section 8.4, we write the above system as the matrix equation AX = X where A = cos(θ) − sin(θ) cos(θ) sin(θ Since det(A) = (cos(θ))(cos(θ)) − (− sin(θ))(sin(θ)) = cos2(θ) + sin2(θ) = 1, the det... |
with −φ to get x and y in terms of x and y, but that seems like cheating. The matrix A introduced here is revisited in the Exercises. 11.6 Hooked on Conics Again 975 Example 11.6.1. Suppose the x- and y- axes are both rotated counter-clockwise through the angle θ = π 3 to produce the x- and y- axes, respectively. 1. L... |
equation 21x2 +10xy we substitute x = x cos π 3 − y sin π 3 √ 3+31y2 = 144 to an equation in the variables x and y, = x 2 + y and y = x sin π 3 + y cos 976 Applications of Trigonometry and simplify. While this is by no means a trivial task, it is nothing more than a hefty dose of Beginning Algebra. We will not go thro... |
Bxy + Cy2 + Dx + Ey + F = 0 and set the coefficient of the xy term equal to 0. Terms containing xy in this expression will come from the first three terms of the equation: Ax2, Bxy and Cy2. We leave it to the reader to verify that x2 = (x)2 cos2(θ) − 2xy cos(θ) sin(θ) + (y)2 sin(θ) xy = (x)2 cos(θ) sin(θ) + xy cos2(θ) − ... |
θ) = 0 and cos(2θ) = 0, we can safely assume3 sin(2θ) = 0. We get cos(2θ) B. We have just proved the following theorem. B, or cot(2θ) = A−C sin(2θ) = A−C Theorem 11.10. The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 with B = 0 can be transformed into an equation in variables x and y without any xy terms by rotating the... |
xy-coordinates (0, 2) opening in the x direction with vertices (±2, 2) (in xy-coordinates) and asymptotes y = ± 3 4 − (y−2)2 2 x + 2. We graph it below. √ √ 3The reader is invited to think about the case sin(2θ) = 0 geometrically. What happens to the axes in this case? 978 Applications of Trigonometry 24 2. From 16x2 ... |
= 4x 5. As usual, we now substitute these quantities into 16x2 + 24xy + 9y2 + 15x − 20y = 0 and simplify. As a first step, the reader can verify 3 or tan(θ) = 3 24, we choose tan(θ) = 3 5 and y = x sin(θ) + y cos(θ) = 3x and sin(θ) = sin arctan 3 4 5 and sin(θ) = 3 1−tan2(θ) = 24 2 tan(θ) 4 x2 = 16(x)2 25 − 24xy 25 + 9... |
which conic section we were dealing with based on the presence (or absence) of quadratic terms and their coefficients, Example 11.6.2 demonstrates that all bets are off when it comes to conics with an xy term which require rotation of axes to put them into a more standard form. Nevertheless, it is possible to determine w... |
To that end, we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. We leave it to the reader to show that, after gathering like terms, the coefficient A on (x)2 and the coefficient C on (y)2 are A = A cos2(θ) + B cos(θ) sin(θ) + C sin2(θ) C = A sin2(θ) − B ... |
A − C) cos(2θ)] = −2B(A − C) cos(2θ) sin(2θ) The product of the ‘last’ quantity in each factor is (B sin(2θ))(−B sin(2θ)) = −B2 sin2(2θ). Putting all of this together yields 4AC = (A + C)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) From cot(2θ) = A−C sin(2θ) = A−C twice along with the Pythagorean Ide... |
)2 − (A − C)2 − B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 − B2 cos2(2θ) + sin2(2θ) = (A + C)2 − (A − C)2 − B2 = A2 + 2AC + C2 − A2 − 2AC + C2 − B2 = 4AC − B2 Hence, B2 − 4AC = −4AC, so the quantity B2 − 4AC has the opposite sign of AC. The result now follows by applying Exercise 34 in Section 7.5. Example 11.6.3.... |
’ definition below. Definition 11.1. Given a fixed line L, a point F not on L, and a positive number e, a conic section is the set of all points P such that the distance from P to F the distance from P to L = e The line L is called the directrix of the conic section, the point F is called a focus of the conic section, and... |
point, we convert the equation r = e(d + r cos(θ)) back into a rectangular equation in the variables x and y. If e > 0, but e = 1, the usual conversion process outlined in Section 11.4 gives6 1 − e22 e2d2 x − e2d 1 − e2 2 1 − e2 e2d2 + y2 = 1 1−e2, so the major axis has length 2ed 1−e2 and the minor axis has length 2e... |
��nition 11.1 correspond with the ‘old’ definitions given in Chapter 7. 1−cos(θ) which gives the rectangular equation y2 = 2d x + d 2, 0 opening to the right. In the language of Section 7.3, 4p = 2d so p = d e2−1, so the transverse axis has length 2ed a2 − y2 (1−e2)2 = e2d2 1−e2 = e2d2 e2d d 2 ed Before we summarize our... |
a conic section with directrix x = d. the graph of r = ed 1−e sin(θ) is the graph of a conic section with directrix y = −d. the graph of r = ed 1+e sin(θ) is the graph of a conic section with directrix y = d. In each case above, (0, 0) is a focus of the conic and the number e is the eccentricity of the conic. If 0 < e... |
rix is x = −12. This means that the ellipse has its major axis along the x-axis. We can find the vertices of the ellipse by finding the points of the ellipse which lie on the x-axis. We find r(0) = 6 and r(π) = 3 which correspond to the rectangular points (−3, 0) and (6, 0), so these are our 2, 0.8 vertices. The center of... |
(0, 6) which puts the center of the hyperbola at (0, 4). Since one focus is at (0, 0), which is 4 units away from the center, we know the other focus is at (0, 8). According to Theorem 11.12, the conjugate axis has a length of 3. Putting this together with the location of the vertices, we get that √ 3 the asymptotes o... |
e ≥ 0 and φ, the graph of the equation r = 1 − e cos(θ − φ) is a conic section with eccentricity e and one focus at (0, 0). If e = 0, the graph is a circle centered at (0, 0) with radius. If e = 0, then the conic has a focus at (0, 0) and the directrix contains the point with polar coordinates (−d, φ) where d = e. min... |
π 4 The matrix A(θ) = cos(θ) − sin(θ) cos(θ) sin(θ) is called a rotation matrix. We’ve seen this matrix most recently in the proof of used in the proof of Theorem 11.9. 17. Show the matrix from Example 8.3.3 in Section 8.3 is none other than A π 4. 18. Discuss with your classmates how to use A(θ) to rotate points in t... |
x)2 counter-clockwise through θ = π 3 4 = 1 after rotating 13x2 − 34xy √ 3 + 47y2 − 64 = 0 √ x2 − 2 3xy − y2 + 8 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 becomes (y)2 = x after rotating counter-clockwise through θ = arctan 1 2 y. y 8. 8x2 + 12xy + 17y2 − 20 = 0 becomes (x)2 + (y)2 4 = 1 after rotating counter-clockwi... |
2 −1 1 2 3 4 x −1 −2 −3 −4 990 Applications of Trigonometry 13. r = 4 1+3 cos(θ) is a hyperbola directrix x = 4 center 3 conjugate axis length 2 3, vertices (1, 0), (2, 0) √ 2 2, 0, foci (0, 0), (3, 0) 14. r = 2 1−2 sin(θ) is a hyperbola directrix y = −1, vertices 0, − 2 3 center 0, − 4, foci (0, 0), 0, − 8 3 3 √ conju... |
the coordinate plane. In this case, the x-axis is relabeled as the real axis, which corresponds to the real number line as usual, 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. Imaginary A... |
which means it’s worth our time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-defined. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the only r-value which can be used here is r = 0. Hence for z = 0, |z| = 0 is well-defined. If z = 0,... |
and Arg(z). Plot z in the complex plane. 1. z = √ 3 − i Solution. 2. z = −2 + 4i 3. z = 3i 4. z = −117 √ √ √ 1. For z = 3 − i = 3 + (−1)i, we have Re(z) = 3 and Im(z) = −1. To find |z|, arg(z) and Arg(z), we need to find a polar representation (r, θ) with r ≥ 0 for the point P ( 3, −1) associated with z. We know r2 = ( ... |
π < θ ≤ π, so Arg(z) = π − arctan(2). 5, so |z| = 2 √ √ 2In case you’re wondering, the use of the absolute value notation |z| for modulus will be explained shortly. 3Recall the symbol being used here, ‘∈,’ is the mathematical symbol which denotes membership in a set. 4If we had Calculus, we would regard Arg(0) as an ‘i... |
117 −2 −1 − Real Axis 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 11.14. Properties of the Modulus: Let z and w be complex numbers. |z| is the distance from z to 0 in the complex plane |z| ≥ 0... |
� |zwac − bd)2 + (ad + bc)2 a2c2 − 2abcd + b2d2 + a2d2 + 2abcd + b2c2 Expand a2c2 + a2d2 + b2c2 + b2d2 a2 (c2 + d2) + b2 (c2 + d2) (a2 + b2) (c2 + d2) a2 + b2 c2 + d2 Factor Factor √ Rearrange terms = = |z||w| Product Rule for Radicals Definition of |z| and |w| Hence |zw| = |z||w| as required. Now that the Product Rule ... |
value so the notation |z| is unambiguous. 6This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is a2 + b2 = 0 if and only if a2 + b2 = 0, which is true if and only if a = b = 0. √ true. We know |z| = 0 if and only if The latter happens if and only if z = a + bi ... |
) = 0 and Im(z) < 0, then z lies on the negative imaginary axis, and a similar argument shows θ is coterminal with − π 2. The last property in the theorem was already discussed in the remarks following Definition 11.2. Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. To that end,... |
of the following complex numbers. (a) z = √ 3 − i Solution. (b) z = −2 + 4i (c) z = 3i (d) z = −117 1. The key to this problem is to write out cis(θ) as cos(θ) + i sin(θ). (a) By definition, z = 4cis 2π √ 3 z = −2 + 2i + i sin 2π = 4 cos 2π √ 3 3 3, so that Re(z) = −2 and Im(z) = 2 = 2 cos − 3π 4. After some simplifyin... |
is a good exercise to actually show that this polar form reduces to z = −2 + 4i. (c) For z = 3i, |z| = 3 and θ = π 2. In this case, z = 3cis π 2 geometrically. Head out 3 units from 0 along the positive real axis. Rotating π counter-clockwise lands you exactly 3 units above 0 on the imaginary axis at z = 3i.. This can... |
(β) = (cos(α) cos(β) − sin(α) sin(β)) Since i2 = −1 + i (sin(α) cos(β) + cos(α) sin(β)) Factor out i = cos(α + β) + i sin(α + β) Sum identities = cis(α + β) Definition of ‘cis’ Putting this together with our earlier work, we get zw = |z||w|cis(α + β), as required. Moving right along, we next take aim at the Power Rule, ... |
ator and denominator of the right hand side by (cos(β) − i sin(β)) which is the complex conjugate of (cos(β) + i sin(β)) to get z w = |z| |w| cos(α) + i sin(α) cos(β) + i sin(β) · cos(β) − i sin(β) cos(β) − i sin(β) If we let the numerator be N = [cos(α) + i sin(α)] [cos(β) − i sin(β)] and simplify we get N = [cos(α) +... |
3 + 2i and w = −1 + i √ 3. Use Theorem 11.16 to find the following. 1. zw 2. w5 Write your final answers in rectangular form. 3. z w √ √ √ 3)2 + (2)2 = Solution. In order to use Theorem 11.16, we need to write z and w in polar form. For z = 2 3+2i, √ 3 3. Since 3, we find |z| = (2 z lies in Quadrant I, we have θ = π 16 =... |
w = −2i. 3, we get w5 = 32 cos 4π z 4cis( π 6 ) 2cis( 2π 3 ) w 2 cis π = Some remarks are in order. 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 num... |
wise. The sequence of diagrams below attempts to describe this process geometrically. Imaginary Axis Imaginary Axis 6i 5i 4i 3i 2i i z|w| = 8cis π 6 z = 4cis π 6 zw = 8cis π 6 + 2π 3 6i 5i 4i 3i 2i i z|w| = 8cis Real Axis −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 Real Axis Multiplying z by |w| = 2. Rotating counter-clockwis... |
of z’. Using this definition, both 4 and −4 are 12Again, assuming |w| > 1. 13Again, assuming β > 0. 11.7 Polar Form of Complex Numbers 1001 √ √ 16 means the principal square root of 16 as in square roots of 16, while 16 = 4. Suppose we wish to find all complex third (cube) roots of 8. Algebraically, we are trying to sol... |
is a real number, √ we solve |w|3 = 8 by extracting the principal cube root to get |w| = 3 8 = 2. As for α, we get α = 2πk for integers k. This produces three distinct points with polar coordinates corresponding to 3 k = 0, 1 and 2: specifically (2, 0), 2, 2π. These correspond to the complex numbers 3 w0 = 2cis(0), w1 ... |
number, cos(θ + 2πk) = cos(θ) and sin(θ + 2πk) = sin(θ). Hence, it follows that cis(θ + 2πk) = cis(θ), so (wk)n = rcis(θ) = z, as required. To show that the formula in Theorem 11.17 generates n distinct numbers, we assume n ≥ 2 (or else there is nothing to prove) and note √ that the modulus of each of the wk is the sa... |
11.17, we identify r = 4, 3 and n = 2. We know that z has two square roots, and in keeping with the notation = 2cis π 3. In rectangular form, the two square roots of 3. We can check our answers by squaring them and 2 + 2π 2 (1) 3 and w1 = −1 − i = 2cis 4π 3 √ (2π/3) √ 2 + 2π 2 (0) and w1 = (2π/3) 4cis 4cis √ √ z are w... |
3 2cis 3π 4. 1 = 1, the roots are w0 = cis(0) = 1, w1 = cis 2π 4. To find the five fifth roots of 1, we write 1 = 1cis(0). We have r = 1, θ = 0 and n = 5. and. The situation here is even graver than in the previous example, since we have 5. At this stage, we √ Since 5 w4 = cis 8π not developed any identities to help us d... |
z). 2. z = 5 + 5i √ 3 3. z = 6i 4. z = −3 √ √ 2 2+3i 1. z = 9 + 9i 5. z = −6 √ 3 + 6i 9. z = −5i 13. z = 3 + 4i 6. z = −2 7. z = − √ √ 2 − 2i 2 11. z = 6 10. z = 2 √ 14. z = 2 + i 15. z = −7 + 24i 16. z = −2 + 6i √ 3 2 − 1 2 i 8. z = −3 − 3i √ 12. z = i 3 7 17. z = −12 − 5i 18. z = −5 − 2i 19. z = 4 − 2i 20. z = 1 − 3i... |
. z = 7cis − 3π 4 33. z = 8cis π 12 35. z = 5cis arctan 11.7 Polar Form of Complex Numbers 1005 49. w z2 50. z3 w2 51. w2 z3 52. 6 w z In Exercises 53 - 64, use DeMoivre’s Theorem to find the indicated power of the given complex number. Express your final answers in rectangular form. 53. −2 + 2i √ 33 √ 54. (− 3 − i)3 57 ... |
factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. Now that we can compute the complex fourth roots of −4 directly, we can simply apply the Complex Factorization Theorem, Theorem 3.14, to obtain the linear factorization p(x) = (x − (1 + i))(x − (1 − i))... |
wj = 1. Hint: If wj = cis(θ) let wj = cis(2π − θ). You’ll need to verify that wj = cis(2π − θ) is indeed an nth root of unity. 83. 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 eit = cos(t) + i sin(t). (a) Use Theorem 11.16 to sh... |
z) = π 2. Im(z) = 6, 4. z = −3 √ √ 2 + 3i 2 = 6cis 3π 4, Re(z) = −3 √ 2, Im(z) = 3 √ 2, |z| = 6 arg(z) = 3π √ 5. z = −6 arg(z) = 5π 4 + 2πk | k is an integer and Arg(z) = 3π 4., Re(z) = −6 6 + 2πk | k is an integer and Arg(z) = 5π 6. 3 + 6i = 12cis 5π 6 √ 3, Im(z) = 6, |z| = 12 6. z = −2 = 2cis (π), Re(z) = −2, Im(z) =... |
cis (0), Re(z) = 6, Im(z) = 0, |z| = 6 arg(z) = {2πk | k is an integer} and Arg(z) = 0. √ Im(z) = 3, Re(z) = 0, √ 12. z = i 3 7, 7cis π 2 √ |z| = 3 7 2 + 2πk | k is an integer and Arg(z) = π 2. √ 7 = 3 arg(z) = π 13. z = 3 + 4i = 5cis arctan 4 3, Re(z) = 3, Im(z) = 4, |z| = 5 arg(z) = arctan 4 3 + 2πk | k is an integer... |
= arctan 5 12 18. z = −5 − 2i = √ 29cis π + arctan 2 5, Re(z) = −5, Im(z) = −2, arg(z) = π + arctan 2 5 5cis arctan − 1 2 19. z = 4 − 2i = 2 √ + 2πk | k is an integer and Arg(z) = arctan 2 5 √, Re(z) = 4, Im(z) = −2, |z| = 2 arg(z) = arctan − 1 2 + 2πk | k is an integer and Arg(z) = arctan − 1 2 − π. √ |z| = 29 − π. 5... |
� 5 √ 38. z = 34. z = 2cis 7π 8 √ 36 10cis arctan 1 3 √ 3cis arctan − 2 = 1 − i √ 2 39. z = 50cis π − arctan 7 24 = −48 + 14i 40. z = 1 2 cis π + arctan 5 12 = − 6 13 − 5i 26 11.7 Polar Form of Complex Numbers 1009 In Exercises 41 - 52, we have that z = − 3 we get the following. √ 3 2 + 3 2 i = 3cis 5π 6 41. zw = 18cis... |
cis 3π 2 we have w0 = 5cis 3π w1 = 2cis 5π 4 √ = − √ 2 − i 2 w1 = 5cis 7π 67. Since z = 1 + i √ 3 = 2cis π 3 we have w0 = √ 2cis w1 = √ 2cis 7π 68. Since z = 5 2 − 5 3 2 i = 5cis 5π 3 we have √ w0 = √ 5cis 5π 6 = − √ 15 2 + √ 5 2 i w1 = √ 5cis 11π 6 = √ 15 2 − √ 5 2 i 69. Since z = 64 = 64cis (0) we have w0 = 4cis (0) ... |
w1 = 2cis π w4 = 2cis − 2π 3 3 = 1 + 3i = −1 − √ w2 = 2cis 2π 3i w5 = 2cis − π = −1 + √ = 1 − 3 3i 3 √ 3i 76. Since z = −729 = 729cis(π) we have w0 = 3cis w1 = 3cis π 2 = 3i w2 = 3cis 5π w3 = 3cis 7π w4 = 3cis − 3π 2 = −3i w5 = 3cis − 11π 77. Note: In the answers for w0 and w2 the first rectangular form comes from appl... |
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... |
by adding 3 to the x-coordinate and adding 4 to the y-coordinate to obtain the terminal point Q(1, 7), as seen below. 1The word ‘vector’ comes from the Latin vehere meaning ‘to convey’ or ‘to carry.’ 2Other textbook authors use bold vectors such as v. We find that writing in bold font on the chalkboard is inconvenient ... |
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. 4If necessary, review page 905 and Section 11.3. 5That is, the speed of the plane relative to the air a... |
= w1, w2. The vector v + w is defined by v + w = v1 + w1, v2 + w2 Example 11.8.2. Let v = 3, 4 and suppose w = and interpret this sum geometrically. −−→ P Q where P (−3, 7) and Q(−2, 5). Find v + w Solution. Before can add the vectors using Definition 11.6, we need to write w in component form. Using Definition 11.5, we ... |
should be 0, it is not clear what its direction is. As we shall see, the direction of 0 is in fact undefined, but this minor hiccup in the natural flow of things is worth the benefits we reap by including 0 in our discussions. We have the following theorem. Theorem 11.18. Properties of Vector Addition Commutative Propert... |
2. Hence, v has an additive inverse, and moreover, it is unique and can be obtained by the formula −v = −v1, −v2. Geometrically, the vectors v = v1, v2 and −v = −v1, −v2 have the same length, but opposite directions. As a result, when adding the vectors geometrically, the sum v + (−v) results in starting at the initial... |
We define scalar multiplication for vectors in the same way we defined it for matrices in Section 8.3. Definition 11.7. If k is a real number and v = v1, v2, we define kv by kv = k v1, v2 = kv1, kv2 Scalar multiplication by k in vectors can be understood geometrically as scaling the vector (if k > 0) or scaling the vector... |
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