Functions Modeling Change A Preparation for Calculus Third Edition

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2 CHAPTER 6 TRIGONOMETRIC FUNCTIONS SECTION 6.5 SINUSOIDAL FUNCTIONS

3 Transformations of the sine and cosine are called sinusoidal functions, and can be expressed in the form: y asin( B( t h)) k and y acos( B( t h)) k where A, B, h, and k are constants. Their graphs resemble the graphs of sine and cosine, but may also be shifted, flipped, or stretched. Page 269 3

4 These transformations may change the period, amplitude, and midline of the function as well as its value at t = 0. y asin( B( t h)) k and y acos( B( t h)) k Page 269 4

5 We already know: The functions of y = A sin t and y = A cos t have amplitude A. The midline of the functions y = sin t + k and y = cos t + k is the horizontal line y = k. Page 269 5

6 Graph y = sin t and y = sin 2t for 0 t 2π. Describe any similarities and differences. What are their periods? Page 269 Example #1 6

7 Using our calculator, let's graph the following: y1 = sin(x) y2 = sin(2x) Page 269 Example #1 Window Value Xmin 0 Xmax 2π Xscl 1 Ymin -1.5 Ymax 1.5 Yscl 1 7

8 y1 sin( x) 0.5 y2 sin(2 x) x Page 269 Example #1 8

9 Graph y = sin t and y = sin 2t for 0 t 2π. Describe any similarities and differences. Page 269 Example #1 9

10 We already know: The functions of y = A sin t and y = A cos t have amplitude A. The midline of the functions y = sin t + k and y = cos t + k is the horizontal line y = k. Page

11 y1 sin( x) 0.5 y2 sin(2 x) x Page 269 Example #1 11

12 Amplitude = 1, Midline = y1 sin( x) 0.5 y2 sin(2 x) x Page 269 Example #1 12

13 Graph y = sin t and y = sin 2t for 0 t 2π. What are their periods? Page 269 Example #1 13

14 y1 sin( x) 0.5 y2 sin(2 x) x Page 269 Example #1 14

15 sin(x) has period = 2π, sin(2x) has 1.5 period = π 1.0 y1 sin( x) 0.5 y2 sin(2 x) x Page 269 Example #1 15

16 sin(x) has period = 2π, sin(2x) has period = π Page 269 Example #1 16

17 If f is a function and k a positive constant, then the graph of y = f(kx) is the graph of f Horizontally compressed by a factor of 1/k if k > 1, Horizontally stretched by a factor of 1/k if k < 1. Page 221 Blue Box (Section 5.4) 17

18 sin(x) has period = 2π, sin(2x) has period = π Page 269 Example #1 18

19 This is because the factor of 2 causes a horizontal compression ( ), squeezing the graph twice as close to the y-axis. Page 269 Example #1 19

20 If B > 0 the function y = sin(bt) resembles the function y = sin t except that it is stretched or compressed horizontally. The constant B determines how many cycles the function completes on an interval of length 2π. For example, we saw that the function y = sin 2t completes two cycles on the interval 0 t 2π. Page

21 Since, for B > 0, the graph of y = sin(bt) completes B cycles on the interval 0 t 2π, each cycle has length 2π/B. The period is thus 2π/B. Page

22 In our example, since sin(2x) completes 2 cycles on the interval 0 x 2π, each cycle has length 2π/2, which equals π. Therefore, the period equals: 2π/B = 2π/2 = π. Page

23 In general, for B of any sign, we have: The functions y = sin(bt) and y = cos(bt) have period P = 2π/ B. Page 270 Blue Box 23

24 In general, for B of any sign, we have: The functions y = sin(bt) and y = cos(bt) have period P = 2π/ B. The number of cycles in one unit of time is B /2π, the frequency. Page 270 Blue Box 24

25 In our example, since sin(2x) completes 2 cycles on the interval 0 x 2π, each cycle has length 2π/2, which equals π. Therefore, the period equals: 2π/B = 2π/2 = π. The number of cycles in one unit of time is B /2π, the frequency. Here we have: 2 /2π = 2/2π = 1/π Page

26 The number of cycles in one unit of time is B /2π, the frequency. Here we have: 2 /2π = 2/2π = 1/π Ask yourself: How many cycles in 2π? 2 cycles per 2π, or 1 cycle per π. Page

27 Again, for sin(2x), how many cycles in 2π? 2 cycles per 2π, or 1 cycle per π (freq). Page

28 Another example: sin(3x) Period? Frequency? Page N/A 28

29 Another example: sin(3x) Since sin(3x) completes 3 cycles on the interval 0 x 2π, each cycle has length 2π/3, which equals (2/3) π. Therefore, the period equals: 2π/3 = (2/3) π. Page

30 The number of cycles in one unit of time is B /2π, the frequency. Here we have: 3 /2π = 3/2π Ask yourself: How many cycles in 2π? 3 cycles per 2π, or 1.5 cycles per π. Page

31 Find possible formulas for the functions f and g shown in Figures 6.56 and Page 270 Example #2 31

32 Page 270 Example #2 32

33 The graph of f resembles the graph of y = sin t except that its period is P = 4π. Page 270 Example #2 33

34 The graph of f resembles the graph of y = sin t except that its period is P = 4π. Using P = 2π/B gives Page 270 Example #2 34

35 The graph of f resembles the graph of y = sin t except that its period is P = 4π. Using P = 2π/B gives 4 2 B 4 B 2 B Page 270 Example #2 35

36 Therefore: 1 f ( t) sin 2 t Page 270 Example #2 36

37 Let's use our calculator and graph: 1 y1 sin x 2 Page 270 Example #2 Window Value Xmin 0 Xmax 14 Xscl 2 Ymin -2 Ymax 2 Yscl 1 37

38 Find possible formulas for the functions f and g shown in Figures 6.56 and Page 270 Example #2 38

39 Page 270 Example #2 39

40 The graph of g resembles the graph of y = cos t except that its period is P = 20. Using P = 2π/B gives Page 270 Example #2 40

41 The graph of g resembles the graph of y = cos t except that its period is P = 20. Using P = 2π/B gives 20 2 B 20B 2 B Page 270 Example #2 41

42 Therefore: g( t) cos 10 t Page 270 Example #2 42

43 Let's use our calculator and graph: y1 cos 10 x Page 270 Example #2 Window Value Xmin 0 Xmax 60 Xscl 20 Ymin -2 Ymax 2 Yscl 1 43

44 Horizontal Shift Figure 6.59 shows the graphs of two trigonometric functions, f and g, with period P = 12. Page

45 Horizontal Shift The graph of f resembles a sine function, so a possible formula for f is f(t) = sin Bt. Since the period of f is 12, what is B? Page

46 Horizontal Shift The graph of f resembles a sine function, so a possible formula for f is f(t) = sin Bt B 12B 2 Page 271 B

47 Horizontal Shift t f( t) sin 6 Page

48 If y = g(x) is a function and k is a constant, then the graph of y = g(x) + k is the graph of y = g(x) shifted vertically k units. If k > 0, the shift is up; if k < 0, the shift is down. y = g(x + k) is the graph of y = g(x) shifted horizontally k units. If k > 0, the shift is to the left; if k < 0, the shift is to the right. Page 196 Blue Box (Section 5.1) 48

49 Page

50 Horizontal Shift The graph of g looks like the graph of f shifted to the right by 2 units. Thus a possible formula for g is g( t) f ( t 2) Page

51 Horizontal Shift The graph of g looks like the graph of f shifted to the right by 2 units. Thus a possible formula for g is g( t) sin ( t 2) 6 (Wherever you see t, you substitute t-2.) Page

52 Horizontal Shift We can rewrite the functions as: g( t) sin t 6 3 Page

53 Horizontal Shift BUT... g( t) sin t 6 3 Page

54 Horizontal Shift BUT... g( t) sin t 6 3 the horizontal shift is not 3 Page

55 Horizontal Shift To pick out the horizontal shift from the formula, we must write the formula in factored form, that is, as sin(b(t h)). g( t) sin ( t 2) 6 Page

56 Horizontal Shift The graphs of: y = sin(b(t h)) & y = cos(b(t h)) are the graphs of y = sin Bt and y = cos Bt shifted horizontally by h units. Page 271 Blue Box 56

57 Describe in words the graph of the function: g( t) cos 3t 4 Page 272 Example #5 57

58 Write the formula for g in the form cos(b(t h)) by factoring 3 out from the expression 3t π/4 to get g(t) = cos(3(t π/12)). What is the period? Page 272 Example #5 58

59 Since cos(3x) completes 3 cycles on the interval 0 x 2π, each cycle has length 2π/3, which equals (2/3) π. Therefore, the period equals: 2π/3 = (2/3) π. Page

60 g(t) = cos(3(t π/12)). The graph is the graph of f = cos 3t shifted π/12 units to the right, as shown in Figure Page 272 Example #5 60

61 g(t) = cos(3(t π/12)) Page 272 Example #5 61

62 Let's use our calculator and graph: y1 cos(3 x) and y2 cos 3x Page 272 Example #5 Window Value Xmin -π/2 Xmax 3 Xscl 1 Ymin -1.5 Ymax 1.5 Yscl

63 Page 272 Example #5 Now add: y3 cos 3 x 12 Window Value Xmin -π/2 Xmax 3 Xscl 1 Ymin -1.5 Ymax 1.5 Yscl 1 63

64 Since: cos( B( t h)) cos 3 x 12 B 3, h 12 Page 272 Example #5 64

65 Summary of Transformations The parameters a, B, h, and k determine the graph of a transformed sine or cosine function. Page 272 Blue Box 65

66 Summary of Transformations For the sinusoidal functions y asin( B( t h)) k and y acos( B( t h)) k a is the amplitude 2π/ B is the period h is the horizontal shift y = k is the midline B /2π is the frequency; that is, the number of cycles completed in unit time. Page 272 Blue Box 66

67 Phase Shift In Example 5, we factored (3t π/4) to write the function as g(t) = cos(3(t π/12)). This allowed us to recognize the horizontal shift, π/12. However, in most physical applications, the quantity π/4, known as the phase shift, is more important than the horizontal shift. We define: Phase Shift = Fraction of pd x 2π Page

68 Phase Shift Phase Shift (π/4) = Fraction of pd x 2π Now solve for "Fraction of pd": Fraction of pd = (π/4) / 2π Page

69 Phase Shift Fraction of pd = (π/4) / 2π = 1/8 (of a full period) Page

70 Phase Shift In example #5 the graph of f(t) = cos 3t is shifted 1/8 of its period to the right. Page

71 Phase Shift In example #5 the graph of f(t) = cos 3t is shifted 1/8 of its period to the right. Page

72 Phase Shift In example #5 the graph of f(t) = cos 3t is shifted 1/8 of its period to the right. cos 3 x cos 3x 12 4 Horizontal shift Phase shift Page

73 Phase Shift Phase shift is significant because in many applications, such as optical interference, we want to know if two waves reinforce or cancel each other. Page

74 Phase Shift For two waves of the same period, a phase shift of 0 or 2π tells us that the two waves reinforce each other; a phase shift of π tells us that the two waves cancel. Thus, the phase shift tells us the relative positions of two waves of the same period. Page

75 Phase Shift For the sinusoidal functions written in the form: y asin( Bt ) and y acos( Bt ) is the phase shift. Page

76 Using the Transformed Sine and Cosine Functions Page

77 Use the sinusoidal function f(t) = a sin(b(t h)) + k to represent your height above ground at time t while riding the ferris wheel. Page 275 Example #9 77

78 Page 275 Example #9 78

79 Summary of Transformations For the sinusoidal functions y asin( B( t h)) k a is the amplitude 2π/ B is the period h is the horizontal shift y = k is the midline B /2π is the frequency; that is, the number of cycles completed in unit time. Page 272 Blue Box 79

80 The diameter of the ferris wheel is 450 feet, so the midline is k = 225 and the amplitude, A, is also 225. The period of the ferris wheel is 30 minutes, so B 2 = Page 275 Example #9 80

81 The sine graph is shifted 7.5 minutes to the right because we reach y = 225 (the 3 o'clock position) when t = 7.5. Thus, the horizontal shift is h = 7.5, so f ( t) 225sin ( t 7.5) Page 275 Example #9 81

82 We can rewrite: f ( t) 225sin ( t ) 225 f ( t) 225sin t Page 275 Example #9 82

83 f ( t) 225sin t Let's substitute equivalent values: Page 275 Example #9 83

84 225sin t Page 275 Example #9 84

85 Page 275 Example #9 85

86 You will recall from Section 6.1 our sine regression experiment: Page 275 Example #9 86

87 y=a * sin(bx+c)+d a=225 b= c= d=225 Page N/A 87

88 We utilzed the sin regression function to fit a sin curve to the set of points. Here is the result: y=a * sin(bx+c)+d y = 225sin( x )+225 Page N/A 88

89 End of Section

Section 8.4: The Equations of Sinusoidal Functions

Section 8.4: The Equations of Sinusoidal Functions In this section, we will examine transformations of the sine and cosine function and learn how to read various properties from the equation. Transformed