Unit 3

Logarithms

Practice 3.2B Solutions

Use these solutions to correct your work. When finished, give yourself a grade using the Practice Assessment rubric.

Pages 250 to 254, questions 1c, 2a, 3a part ii, 4a to 4e, 5a to 5d, 7a, 9, 24a to 24e.

Question 1c, page 250

Phase shift

= \frac{2 π}{3}

units to the left.

Vertical displacement

= 5

units up.

Additional Notes

Compare the function

y = \sin (x + \frac{2 π}{3}) + 5

y = a \sin b (x - c) + d

. The values of

c

and

d

represent the phase shift and the vertical displacement, respectively.

c = - \frac{2 π}{3}

, which indicates a horizontal phase shift

\frac{2 π}{3}

units to the left.

The vertical displacement is represented by

d

d = 5

, which indicates a vertical displacement

5

units up.

Question 2a, page 250

Phase shift

= 30 °

to the right.

Vertical displacement

= 12

units up.

Additional Notes

Compare

y = \cos (x - 30 °) + 12

y = a \cos b (x - c) + d

c = 30 °

, which indicates a horizontal phase shift

30 °

to the right.

d = 12

, which indicates a vertical displacement

12

units up.

Question 3a part ii, page 250

\{y ∣ - 5 \leq y \leq - 1, y \in R\}

Solution

The range is the possible

y

-values. In the case of a sine function, the possible

y

-values are based on the amplitude and the vertical displacement.

a = - 2

, which means there was a reflection in the

x

-axis and there was a vertical stretch about the

x

-axis by a factor of

2

.

The vertical displacement is

- 3

, which means there was a vertical shift downwards of

3

units.

The function

y = \sin x

has a maximum value of

1

, which is stretched about the

x

-axis by a factor of

2

, reflected in the

x

-axis, and then shifted down

3

units.

1 \to 2 \to - 2 \to - 5

The minimum value of

- 1

undergoes the same transformations.

- 1 \to - 2 \to 2 \to - 1

Therefore, the range is

\{y ∣ - 5 \leq y \leq - 1, y \in R\}

Questions 4a to 4e, page 250

Matches D

Matches C

Matches B

Matches A

Matches E

Additional Notes

Compare each given function to

y = a \sin b (x - c) + d

y = a \cos b (x - c) + d

|a| = 2

Amplitude refers only to the vertical stretch factor. If

a

is negative, there is also a reflection in the

x

-axis.

b = 2

The value of

|b|

can be used to determine the period of the function.

\begin{array}{rcl} period & = & \frac{2 π}{|b|} \\ = & \frac{2 π}{2} \\ = & π \end{array}

c = - 4

The phase shift is

4

units to the left.

d = - 1

The vertical displacement is

1

unit down.

This function matches the description in D.

a = 2

The amplitude is

2

b = 2

The period is

π

c = 4

The phase shift is

4

units right.

d = - 1

The vertical displacement is

1

unit down.

Matches C

a = 2

The amplitude is

2

b = 2

The period is

π

.

The equation was not given in the form

y = a \sin b (x - c) + d

. The value of

b

must be factored out so the value of

c

can be determined.

\begin{array}{l} y = 2 \sin (2 x - 4) - 1 \\ y = 2 \sin 2 (x - 2) - 1 \end{array}

c = 2

The phase shift is

2

units to the right.

d = - 1

The vertical displacement is

1

unit down.

Matches B

a = 3

The amplitude is

3

b = 3

The period is

\frac{2 π}{3}

\begin{array}{rcl} period & = & \frac{2 π}{b} \\ = & \frac{2 π}{3} \end{array}

Factor out

b

, so

c

can be determined.

\begin{array}{l} y = 3 \sin (3 x - 9) - 1 \\ y = 3 \sin 3 (x - 3) - 1 \end{array}

c = 3

The phase shift is

3

units to the right.

d = - 1

The vertical displacement is

1

unit down.

Matches A

a = 3

The amplitude is

3

b = 3

The period is

\frac{2 π}{3}

.

Factor out

b

, so

c

can be determined.

\begin{array}{l} y = \sin (3 x + π) - 1 \\ y = 3 \sin 3 (x + \frac{π}{3}) - 1 \end{array}

c = - \frac{π}{3}

The phase shift is

\frac{π}{3}

units to the left.

d = - 1

The vertical displacement is

1

unit down.

Matches E

Questions 5a to 5d, page 250

Matches D

Matches B

Matches C

Matches A

Additional Notes

In the equation of the function,

c = \frac{π}{4}

, which indicates a phase shift of

\frac{π}{4}

units to the right.

d = 0

indicates there is no vertical displacement.

The graphs of A and C show vertical displacements, so they are not possible matches. Note the midline for each is not the

x

-axis.

The graphs of B and D show no vertical displacement. The midline is still the

x

-axis in each case.

The point shown in red on graph D would sit at the origin for the graph of the function

y = \sin x

. It has shifted to the right

\frac{π}{4}

units. Therefore, the answer is D.

In the equation of the function,

c = - \frac{π}{4}

, which indicates a phase shift of

\frac{π}{4}

units to the left.

d = 0

indicates there is no vertical displacement. The point shown in red on graph B would sit at the origin for the graph of the function

y = \sin x

. It has shifted to the left

\frac{π}{4}

units. Therefore, the answer is B.

The function indicates a vertical displacement

1

unit down. The midline of graph C is at

y = - 1

.

The function matches C.

The function indicates a vertical displacement

1

unit up. The midline of graph A is at

y = 1

.

The function matches A.

Question 7a, page 251

\begin{array}{l} a = 3 \\ b = \frac{1}{2} \\ c = - 2 \\ d = 3 \end{array}

The equation of the function is

y = 3 \cos \frac{1}{2} (x + 2) + 3

Additional Notes

The vertical stretch is referred to as the amplitude and is represented by

a

in the equation of the function.

The horizontal stretch affects the period and is represented by

\frac{1}{b}

in the equation of the function.

The horizontal translation is referred to as the phase shift and is represented by

c

in the equation of the function.

The vertical translation is referred to as the vertical displacement and is represented by

d

in the equation of the function.

Question 9, page 251

The parameter that would be affected by the speed of the piston is

b

. As the piston moves faster, the entire cycle occurs more often, which corresponds to a shorter period.

Questions 24a to 24e, page 254

4

seconds

One full cycle is the period of the function.

\begin{array}{rcl} period & = & \frac{2 π}{|b|} \\ = & \frac{2 π}{|\frac{π}{2}|} \\ = & 2 π (\frac{2}{π}) \\ = & 4 \end{array}

15

cycles per minute

\frac{60 \sec / \min}{4 \sec / cycle} = 15 cycles / \min

The airflow is

0 L / s

When

t = 30,

r = 1.75 \sin \frac{π}{2} (30)

r = 0

. The lungs are either completely full, which means no inhaling can happen, or completely empty, which means no exhaling. This is an instantaneous reading because immediately after, either the lungs will begin to empty (exhale) or begin to fill (inhale).

- 1.23 L / s

t = 7.5, r = 1.75 \sin \frac{π}{2} (7.5), r = - 1.23

This means the air is flowing from the lungs because it is negative airflow.

L3.2 B3 Practice Solutions

Unit 3

Logarithms

Practice Solutions

Practice 3.2B Solutions

Question 1c, page 250

Question 2a, page 250

Question 3a part ii, page 250

Questions 4a to 4e, page 250

Questions 5a to 5d, page 250

Question 7a, page 251

Question 9, page 251

Questions 24a to 24e, page 254