College Trigonometry, 2011a

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1050 Applications of <strong>Trigonometry</strong> system {x = f(t), y= g(t) to eliminate the parameter. If, on the other hand, the parametrization involves the trigonometric functions, the strategy changes slightly. In this case, it is often best to solve for the trigonometric functions and relate them using an identity. We demonstrate these techniques in the following example. Example 11.10.2. Sketch the curves described by the following parametric equations. 1. { x = t 3 y = 2t 2 for −1 ≤ t ≤ 1 3. { x = sin(t) y = csc(t) for 0
11.10 Parametric Equations 1051 2. For the system { x =2e −t ,y= e −2t for t ≥ 0, we proceed as in the previous example and graph x =2e −t and y = e −2t over the interval [0, ∞). We find that the range of x in this case is (0, 2] and the range of y is (0, 1]. Next, we plug in some friendly values of t to get a sense of the orientation of the curve. Since t lies in the exponent here, ‘friendly’ values of t involve ( natural logarithms. Starting ( with t = ln(1) = 0 we get 5 (2, 1), for t = ln(2) we get 1, 1 4) and for t = ln(3) we get 2 3 , 1 9) .Sincet is ranging over the unbounded interval [0, ∞), we take the time to analyze the end behavior of both x and y. As t →∞, x =2e −t → 0 + and y = e −2t → 0 + as well. This means the graph of { x =2e −t ,y= e −2t approaches the point (0, 0). Since both x =2e −t and y = e −2t are always decreasing for t ≥ 0, we know that our final graph will start at (2, 1) (where t = 0), and move consistently to the left (since x is decreasing) and down (since y is decreasing) to approach the origin. To eliminate the parameter, one way to proceed is to solve x =2e −t for t to get t = − ln ( ) x 2 . Substituting this for t in y = e −2t gives y = e −2(− ln(x/2)) = e 2ln(x/2) = e ln(x/2)2 = ( ) x 2 2 = x 2 4 .Or,wecould recognize that y = e −2t = ( e −t) 2 , and since x =2e −t means e −t = x 2 ,wegety = ( ) x 2 2 = x 2 4 this way as well. Either way, the graph of { x =2e −t ,y= e −2t for t ≥ 0isaportionofthe parabola y = x2 4 which starts at the point (2, 1) and heads towards, but never reaches,6 (0, 0). x 2 y y 1 1 1 1 2 t 1 2 t 1 2 x x =2e −t , t ≥ 0 y = e −2t , t ≥ 0 { x =2e −t ,y= e −2t , t ≥ 0 3. For the system {x =sin(t), y= csc(t) for 0
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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 935 13. (
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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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11.6 Hooked on Conics Again 989 2 9
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11.7 Polar Form of Complex Numbers
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College Trigonometry, 2011a

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