College Trigonometry, 2011a

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1054 Applications of <strong>Trigonometry</strong> 3. To parametrize line segment which starts at (2, −3) and ends at (1, 5), we make use of the formulas x = x 0 +(x 1 −x 0 )t and y = y 0 +(y 1 −y 0 )t for 0 ≤ t ≤ 1. While these equations at first glance are quite a handful, 8 they can be summarized as ‘starting point + (displacement)t’. To find the equation for x, wehavethatthelinesegmentstarts at x = 2 and ends at x =1. This means the displacement in the x-direction is (1 − 2) = −1. Hence, the equation for x is x =2+(−1)t =2− t. For y, we note that the line segment starts at y = −3 and ends at y = 5. Hence, the displacement in the y-direction is (5 − (−3)) = 8, so we get y = −3+8t. Our final answer is {x =2− t, y = −3+8t for 0 ≤ t ≤ 1. To check, we can solve x =2− t for t to get t =2− x. Substituting this into y = −3+8t gives y = −3+8t = −3 + 8(2 − x), or y = −8x + 13. We know this is the graph of a line, so all we need to check is that it starts and stops at the correct points. When t =0,x =2− t =2,andwhent =1,x =2− t =1. Plugging in x = 2 gives y = −8(2) + 13 = −3, for an initial point of (2, −3). Plugging in x = 1 gives y = −8(1) + 13 = 5 for an ending point of (1, 5), as required. 4. In order to use the formulas above to parametrize the circle x 2 +2x+y 2 −4y =4,wefirstneed to put it into the correct form. After completing the squares, we get (x +1) 2 +(y − 2) 2 =9, or (x+1)2 9 + (y−2)2 9 = 1. Once again, the formulas x = h + a cos(t) andy = k + b sin(t) canbe a challenge to memorize, but they come from the Pythagorean Identity cos 2 (t)+sin 2 (t) =1. In the equation (x+1)2 9 + (y−2)2 9 = 1, we identify cos(t) = x+1 3 and sin(t) = y−2 3 . Rearranging these last two equations, we get x = −1 + 3 cos(t) andy =2+3sin(t). In order to complete one revolution around the circle, we let t range through the interval [0, 2π). We get as our final answer {x = −1 + 3 cos(t), y=2+3sin(t) for 0 ≤ t
11.10 Parametric Equations 1055 5. In the equation x2 4 + y2 9 = 1, we can either use the formulas above or think back to the Pythagorean Identity to get x = 2 cos(t) andy =3sin(t). The normal range on the parameter in this case is 0 ≤ t
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College Trigonometry Version ⌊π
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Table of Contents vii 10 Foundation
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Preface Thank you for your interest
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xi text and we have gone to great l
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Chapter 10 Foundations of Trigonome
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10.1 Angles and their Measure 695 k
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10.1 Angles and their Measure 697 2
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10.1 Angles and their Measure 701 T
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10.1 Angles and their Measure 703 y
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10.1 Angles and their Measure 705 y
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10.1 Angles and their Measure 707 h
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10.1 Angles and their Measure 711 I
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10.2 The Unit Circle: Cosine and Si
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30 ◦ 1 10.2 The Unit Circle: Cosi
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10.3 The Six Circular Functions and
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10.4 Trigonometric Identities 771 f
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10.4 Trigonometric Identities 773 2
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10.4 Trigonometric Identities 779 T
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10.4 Trigonometric Identities 781 R
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10.4 Trigonometric Identities 785 I
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10.5 Graphs of the Trigonometric Fu
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10.6 The Inverse Trigonometric Func
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10.7 Trigonometric Equations and In
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Chapter 11 Applications of Trigonom
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11.1 Applications of Sinusoids 883
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11.2 The Law of Sines 897 minimizes
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11.2 The Law of Sines 899 a angle-s
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11.2 The Law of Sines 901 determine
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11.2 The Law of Sines 903 Let x den
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11.2 The Law of Sines 905 24. Along
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11.2 The Law of Sines 907 33. The a
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11.2 The Law of Sines 909 25. (a)
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11.3 The Law of Cosines 911 of B ar
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11.3 The Law of Cosines 913 the pre
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11.3 The Law of Cosines 915 A 2 = =
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11.3 The Law of Cosines 917 22. A n
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11.4 Polar Coordinates 919 11.4 Pol
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11.4 Polar Coordinates 921 The poin
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11.4 Polar Coordinates 923 4. We mo
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11.4 Polar Coordinates 925 Solution
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11.4 Polar Coordinates 927 Solution
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11.4 Polar Coordinates 929 algebrai
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11.4 Polar Coordinates 931 45. ( )
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11.4 Polar Coordinates 933 5. ( 12,
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11.4 Polar Coordinates 937 73. r =7
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11.5 Graphs of Polar Equations 939
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11.6 Hooked on Conics Again 973 11.
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11.7 Polar Form of Complex Numbers
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College Trigonometry, 2011a

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