Functions Modeling Change A Preparation for Calculus Third Edition

CHAPTER 6 TRIGONOMETRIC FUNCTIONS SECTION 6.5 SINUSOIDAL FUNCTIONS

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

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

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

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

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

1.5 1.0 y1 sin( x) 0.5 y2 sin(2 x) 0 0.5 1 2 3 4 5 6 x 1.0 1.5 Page 269 Example #1 8

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

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 10

1.5 1.0 y1 sin( x) 0.5 y2 sin(2 x) 0 0.5 1 2 3 4 5 6 x 1.0 1.5 Page 269 Example #1 11

Amplitude = 1, Midline = 0 1.5 1.0 y1 sin( x) 0.5 y2 sin(2 x) 0 0.5 1 2 3 4 5 6 x 1.0 1.5 Page 269 Example #1 12

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

1.5 1.0 y1 sin( x) 0.5 y2 sin(2 x) 0 0.5 1 2 3 4 5 6 x 1.0 1.5 Page 269 Example #1 14

sin(x) has period = 2π, sin(2x) has 1.5 period = π 1.0 y1 sin( x) 0.5 y2 sin(2 x) 0 0.5 1 2 3 4 5 6 x 1.0 1.5 Page 269 Example #1 15

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

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

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

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

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 270 20

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 270 21

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 270 22

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

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

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 270 25

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 270 26

Again, for sin(2x), how many cycles in 2π? 2 cycles per 2π, or 1 cycle per π (freq). Page 270 27

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

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 270 29

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 270 30

Find possible formulas for the functions f and g shown in Figures 6.56 and 6.57. Page 270 Example #2 31

Page 270 Example #2 32

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

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

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 2 1 4 2 Page 270 Example #2 35

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

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

Find possible formulas for the functions f and g shown in Figures 6.56 and 6.57. Page 270 Example #2 38

Page 270 Example #2 39

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

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 2 20 10 Page 270 Example #2 41

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

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

Horizontal Shift Figure 6.59 shows the graphs of two trigonometric functions, f and g, with period P = 12. Page 271 44

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 271 45

Horizontal Shift The graph of f resembles a sine function, so a possible formula for f is f(t) = sin Bt. 12 2 B 12B 2 Page 271 B 2 12 6 46

Horizontal Shift t f( t) sin 6 Page 271 47

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

Page 271 49

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 271 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) sin ( t 2) 6 (Wherever you see t, you substitute t-2.) Page 271 51

Horizontal Shift We can rewrite the functions as: g( t) sin t 6 3 Page 271 52

Horizontal Shift BUT... g( t) sin t 6 3 Page 271 53

Horizontal Shift BUT... g( t) sin t 6 3 the horizontal shift is not 3 Page 271 54

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 271 55

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

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

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

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 270 59

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 6.60. Page 272 Example #5 60

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

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 1 4 62

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

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

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

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

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 272 67

Phase Shift Phase Shift (π/4) = Fraction of pd x 2π Now solve for "Fraction of pd": Fraction of pd = (π/4) / 2π Page 272 68

Phase Shift Fraction of pd = (π/4) / 2π = 1/8 (of a full period) Page 272 69

Phase Shift In example #5 the graph of f(t) = cos 3t is shifted 1/8 of its period to the right. Page 272 70

Phase Shift In example #5 the graph of f(t) = cos 3t is shifted 1/8 of its period to the right. Page 273 71

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 273 72

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 273 73

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 273 74

Phase Shift For the sinusoidal functions written in the form: y asin( Bt ) and y acos( Bt ) is the phase shift. Page 273 75

Using the Transformed Sine and Cosine Functions Page 273 76

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

Page 275 Example #9 78

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

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 = 30 15 Page 275 Example #9 80

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) 225 15 Page 275 Example #9 81

We can rewrite: f ( t) 225sin ( t 15 7.5) 225 f ( t) 225sin t 15 2 225 Page 275 Example #9 82

f ( t) 225sin t 225 15 2 Let's substitute equivalent values: 15 2.2094395102 1.570796327 Page 275 Example #9 83

225sin.2094395102 t 1.570796327 225 Page 275 Example #9 84

Page 275 Example #9 85

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

y=a * sin(bx+c)+d a=225 b=.2094395102 c=-1.570796327 d=225 Page N/A 87

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(.2094395102x -1.570796327)+225 Page N/A 88

End of Section 6.5 89