lat03 04

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Chapter 4

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4.1

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Periodic Function

 A periodic function is a function f such that

f(x) = f(x + np),

for every real number x in the domain of f, every

integer n, and some positive real number p. The

smallest possible positive value of p is the

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Example: Graph

y

= 3 sin

x

compare to

y

= sin

x

.

 Make a table of values.

0

3 0

3 0

3sin x

0

1 0

1 0

sin x



3/2

 /2

0

(8)

Amplitude

 The amplitude of a periodic function is half the

difference between the maximum and minimum values.

 The graph of y = a sin x or y = a cos x, with

a  0, will have the same shape as the graph of

y = sin xor y = cos x, respectively, except the

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Example: Graph

y

= sin 2

x

 The period is 2/2 = . The graph will complete

one period over the interval [0, ].

 The endpoints are 0 and , the three middle

points are:

 Plot points and join in a smooth curve.

1 1 3

(0 ) (0 ), and (0 )

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Period

 For b > 0, the graph of y = sin bx will resemble

that of y = sin x, but with period 2/b.

 The graph of y = cos bx will resemble that of

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Graph

y

= cos 2

x

/3 over one period

 The period is 3.

 Divide the interval into four equal parts.



 Obtain key points for one period.

3 3 9

0, , , , 3

4 2 4

   _ 1 0 1 0 1 cos 2x/3

2 3/2

 /2

0 2x/3

3 9/4

3/2 3/4

0

(13)

Graph

y

= cos 2

x

/3 over one period

continued

 The amplitude is 1.

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Guidelines for Sketching Graphs of

Sine and Cosine Functions

 To graph y = a sin bx or y = a cos bx, with b

> 0, follow these steps.

 Step 1 Find the period, 2/b. Start with 0 on

the x-axis, and lay off a distance of 2/b.

 Step 2 Divide the interval into four equal

parts.

 Step 3 Evaluate the function for each of the

five x-values resulting from Step 2. The points will be maximum points,

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Guidelines for Sketching Graphs of

Sine and Cosine Functions continued

 Step 4 Plot the points found in Step 3, and

join

them with a sinusoidal curve having

amplitude |a|.

 Step 5 Draw the graph over additional

periods, to the right and to the left,

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Graph

y

=



2 sin 4

x

 Step 1 Period = 2/4 = /2. The function will

be graphed over the interval [0, /2] .

 Step 2 Divide the interval into four equal

parts.

 Step 3 Make a table of values

3

0, , , ,

8 4 8 2

    0 2 0 2 0

2 sin 4x

0

1 0

1 0

sin 4x

2 3/2

 /2

0 4x

/2 3/8

/4

/8 0

(17)

Graph

y

=



2 sin 4

x

continued

 Plot the points and join them with a sinusoidal

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4.2

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Translations

 In trigonometric functions, a horizontal

translation is called a phase shift.

 In the equation

the graph is shifted /2 units to the right.

sin

2 y  _ x   _

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Graph

y

= sin (

x





/3)

 Find the interval for one period.

0 2

3 7

3 3

 _

 

  

 

x

5 4 11 7

, , , ,

3 6 3 6 3

(21)

Graph

y

= sin (

x





/3) continued

 0 1 0 1 0

sin (x  /3)

2 3/2

 /2

0

x  /3

7/3 11/6

4/3 5/6

/3

(22)

Graph

 Find the interval.

 Divide into four equal parts.

3cos

4 







_



_





y

x

0 2 4 7 4 4  _         x x

3 5 7

, , , ,

4 4 4 4 4



(23)

Graph continued

 3 0 3 0 3 3 cos (x + /4)

1 0

1 0

1 cos(x + /4)

2 3/2

 /2

0

x + /4

7/4 5/4

3/4

(24)

Graph 2



2 sin 3

x

 The graph is translated 2 units up from the graph

y = 2 sin 3x.

2 4

2 0

2 2  2 sin 3x

0 2

0

2 0

2 sin 3x

2 3/2

 /2

0 3x

2/3

/2

/3

/6 0

(25)

Graph 2



2 sin 3

x

continued

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Further Guidelines for Sketching Graphs of

Sine and Cosine Functions

 Method 1: Follow these steps.

 Step 1 Find an interval whose length is one

period 2/b by solving the three part inequality 0  b(x  d)  2.

 Step 2Divide the interval into four equal parts.  Step 3Evaluate the function for each of the

five x-values resulting from Step 2. The points will be maximum points,

minimum points, and points that intersect the

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Further Guidelines for Sketching Graphs of

Sine and Cosine Functions continued

 Step 4Plot the points found in Step 3, and join

them with a sinusoidal curve having amplitude |a|.

 Step 5 Draw the graph over additional periods,

to the right and to the left, as needed.

 Method 2 First graph the basic circular function. The

amplitude of the function is |a|, and the period is 2/b. Then use translations to graph the desired function. The vertical

translation is c units up if c > 0 and |c| units down if c < 0. The horizontal translation

(phase shift) is d units to the right if d > 0 and |

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Graph

y

=



1 + 2 sin (4

x

+



)

 Write the expression in

the form c + a sin b(x  d) by rewriting 4x +  as

 Step 1

 Step 2: Divide the

interval.

 Step 3 Table

4 . 4 x   _     

0 4 2

4 0 4 2 4 4 x x x  _        _  _         

, , 0, , 4 8 8 4

   

(29)

Graph

y

=



1 + 2 sin (4

x

+



) continued

1

3

1 1

1

1 + 2sin(4x + )

2

2 0

2 0

2 sin 4(x + /4)

0

1 0

1 0

sin 4(x + /4)

2

3/2

 /2

0 4(x + /4)

/2 3/8

/4

/8 0

x + /4

(30)

Graph

y

=



1 + 2 sin (4

x

+



) continued

 Steps 4 and 5

 Plot the points found in the table and join then

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4.3

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Guidelines for Sketching Graphs of

Cosecant and Secant Functions

 To graph y = csc bx or y = sec bx, with b > 0,

follow these steps.

 Step 1 Graph the corresponding reciprocal

function as a guide, using a dashed curve.

y = cos bx y = a sec bx

y = a sin bx y = a csc bx

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Guidelines for Sketching Graphs of

Cosecant and Secant Functions continued

 Step 2Sketch the vertical asymptotes. They

will have equations of the form x = k, where k is an x-intercept of the graph of the guide function.

 Step 3Sketch the graph of the desired

function by drawing the typical U-shapes branches between the adjacent asymptotes. The branches will be

above the graph of the guide function when the guide function values are positive and below the graph of the

(36)

Graph

y

= 2 sec

x

/2

 Step 1 Graph the corresponding reciprocal

function y = 2 cos x/2.

 The function has amplitude 2 and one period of

the graph lies along the interval that satisfies the inequality

1

0 2 .

2 x 

 

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Graph

y

= 2 sec

x

/2 continued

 Sketch the vertical asymptotes. These occur at

x-values for which the guide function equals 0,

such as x = 3, x = 3, x = , x = 3.

 Sketch the graph of y = 2 sec x/2 by drawing the

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Guidelines for Sketching Graphs of Tangent

and Cotangent Functions

 To graph y = tan bx or y = cot bx, with b > 0,

follow these steps.

 Step 1 Determine the period, /b. To locate

two adjacent vertical asymptotes solve

the following equations for x:

For tan : and

2 2

For cot : 0 and .

y a bx bx bx

 



   

(42)

Guidelines for Sketching Graphs of Tangent

and Cotangent Functions continued

 Step 2 Sketch the two vertical asymptotes found in

Step 1.

 Step 3Divide the interval formed by the vertical

asymptotes into four equal parts.

 Step 4Evaluate the function for the first-quarter

point, midpoint, and third-quarter point, using the x-values found in Step 3.

 Step 5Join the points with a smooth curve,

approaching the vertical asymptotes. Indicate

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Graph

y

= tan 2

x

 Step 1 The period of the function is /2. The

two asymptotes have equations x = /4 and x = /4.

 Step 2 Sketch two vertical asymptotes

x =  /4.

 Step 3 Divide the interval into four equal

parts. This gives the following key x-values.

 First quarter: /8

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Graph

y

= tan 2

x

continued

 Table of values

1 0

1 tan 2x

/4 0

/4 2x

/8 0

/8

(45)

Graph

y

= 2 + tan

x

 Every y value for this

function will be 2 units more than the

corresponding y in

y = tan x, causing the graph to be translated 2 units up compared to

(46)

4.4

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Simple Harmonic Motion

 The position of a point oscillating about an

equilibrium position at time t is modeled by either

 where a and are constants, with The

amplitude of the motion is |a|, the period is

and the frequency is

( )

cos

or ( )

sin

s t



a



t

s t



a



t

   0.

2



. 2

(48)

Example

 Suppose that an object is attached to a coiled

spring such the one shown (on the next slide). It is pulled down a distance of 5 in. from its

equilibrium position, and then released. The time for one complete oscillation is 4 sec.

 a) Give an equation that models the position of

the object at time t.

 b) Determine the position at t = 1.5 sec.

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Example continued

 When the object is

released at t = 0, distance the object of the object

from its equilibrium position 5 in. below equilibrium.

 We use

 a = 5 

( ) cos

s t  t

2

4, or

2

 _ 

(50)

Example continued

 The motion is modeled by

 b) After 1.5 sec, the position is

Since 3.54 > 0, the object is above the equilibrium position.

 c) The frequency is the reciprocal of the

period,

of ¼.

5cos . 2 t





(1.5) 5cos (1.5) 2.54 in. 2

s   _ _ 