Exercise 1

Step 1

1 of 4

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ] Exercise scan

Step 2

2 of 4

a) The equation of the axis is $y=dfrac{y_{min}+y_{max}}{2}$. In this case, $y_{min}=2$ and $y_{max}=14$

Therefore, it is $y=dfrac{14+2}{2}implies y=8$

The position at the axis represents the resting position of the swing.

b) The amplitude is $A=dfrac{|14-2|}{2}=6$.

The amplitude represents the maximum achievable distance from the resting position.

c) The period is $7-3=4$ s. It means it takes 4 s to complete one cycle.

d) The minimum distance between Olivia and motion detector is $2$ m.

e) No. At $t=7s$, the distance is decreasing, implying that Olivia is moving towards the detector where you can collide with her.

f) Olivia’s motion is not yet stabilized as soon as she started swinging because she will have to move away from the resting position by herself at the start (not through the swing motion). Thus, it will not be sinusoidal. It would probably look like this.

Step 3

3 of 4

Result

4 of 4

a) $y=8$

b) $A=6$

c) $T=4$s

d) $2$m

e) No

f) not sinusoidal, see graph

Exercise 2

Step 1

1 of 4

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ] Exercise scan

Step 2

2 of 4

From the graph, we can see that:

Paddle A: $y_{min}=-1$ , $y_{max}=5$

Paddle B: $y_{min}=-1$ , $y_{max}=7$

$bf{Amplitude}$

The amplitude of the graph represents the radius of the wheel. Thus,

radius of Paddle A = $dfrac{|5-(-1)|}{2}=3$ m

radius of Paddle B = $dfrac{|7-(-1)|}{2}=4$ m

$bf{Time;for;Complete;Revolution}$

The time taken to complete one revolution is the period of the graphs.
Paddle A has 4 cycles in 48 s, thus its period is $T=dfrac{48}{4}=12$ s. Paddle B has 3 cycles in 48s m, thus its period is $T=dfrac{48}{3}=16;s$

Step 3

3 of 4

$bf{Height;of;the;Axle}$

The height of the axle relative to the water is reflected on the equation of the axis.

Height of Axle A: $dfrac{5+(-1)}{2}=2$ m

Height of Axle B: $dfrac{7+(-1)}{2}=3$ m

$bf{Speed;of;wheel}$

The speed of the wheel is the circumference of the wheel divided by the time for one complete revolution.

Speed of Wheel A: $v=dfrac{2picdot 3;text{m}}{12;text{s}}=1.57;text{m/s}$

Speed of Wheel B: $v=dfrac{2picdot 4;text{m}}{16;text{s}}=1.57;text{m/s}$

Result

4 of 4

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|}
hline
& Wheel A & Wheel B \ hline
radius & 3 m & 4 m \ hline
time for complete revolution & 12 s & 16 s \ hline
begin{tabular}[c]{@{}l@{}}height of axle\ relative to the waterend{tabular} & 2 m & 3 m \ hline
speed & 1.57 m/s & 1.57 m/s \ hline
end{tabular}
end{table}

Exercise 3

Step 1

1 of 3

For the sinusoidal functions

$y=Asin (kx)+B$

$y=Acos(kx)+B$

amplitude = $A$

period = $dfrac{360^circ}{k}$

Step 2

2 of 3

Therefore, examples of sinusoidal functions that have the same period and axes but different amplitude are those that differ only in $A$ such as

$y=2cos(3x)+3$

$y=4cos(3x)+3$

$y=5sin(3x)+3$

$y=0.25sin(3x)+3$

Result

3 of 3

Answers can vary. See examples inside.

Exercise 4

Step 1

1 of 3

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ] Exercise scan

Step 2

2 of 3

a) The height of the blade above the cutting surface is the maximum height in the graph which is $1$ in.

b) The period of the function can be obtained from the change in time between two peaks which is $T=0.04-0=0.04$s

c) The amplitude of the function is $A=dfrac{|1-(-7)|}{2}=4$ in.

This represents the radius of the blade.

d) The speed of the tooth is the circumference of the blade divided by the period. Therefore,

speed $=dfrac{2picdot 4text{in}}{0.04;text{s}}=628$ in/s

Result

3 of 3

a) $1$ in

b) $0.04$ s

c) $4$ in, radius

d) 628 in/s

Exercise 5

Step 1

1 of 3

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ] Exercise scan

Step 2

2 of 3

a) The period can be obtained from the distance between two peaks
$0.04-0.0075=0.0325$ s

b) The equation of the axis is $y=dfrac{4.5+(-4.5)}{2}implies y=0$

c) The amplitude is $A=dfrac{4.5-(-4.5)}{2}implies A=4.5$

Result

3 of 3

a) 0.0325 s

b) $y=0$

c) $A=4.5$

Exercise 6

Step 1

1 of 6

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ] Exercise scan

Step 2

2 of 6

a)

radius of Ferris wheel is 7m $implies$ $A=7$

axle is 8m above the ground $implies$ equation of axis is $y=8$

rotates once every 40 s $implies$ $T=40;s$

Exercise scan

Step 3

3 of 6

b)

radius of water wheel is 3m $implies$ $A=3$

center is at water level $implies$ equation of axis is $y=0$

rotates once every 15 s $implies$ $T=15;s$

Exercise scan

Step 4

4 of 6

c)

radius of bicycle wheel is 40cm $implies$ $A=40$

center must be 40 cm above the ground $implies$ equation of axis is $y=40$

rotates once every 2 s $implies$ $T=2;s$

Exercise scan

Step 5

5 of 6

d)

wave pool is 3 m deep $implies$ minimum height from the ground is 3m
waves 0.5 m in height $implies$ maximum height from the ground is $3.5$ m

amplitude $A=dfrac{3.5-3}{2}=0.25$

equation of axis is $y=dfrac{3.5+3}{2}=3.25$

wave occur at 7s intervals $implies T=7$

Exercise scan

Result

6 of 6

See graphs inside.

Exercise 7

Step 1

1 of 5

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ] Exercise scan

Step 2

2 of 5

a)

Step 3

3 of 5

b) Both completes the cycle at the same time with a period of 1 year.

The amplitude for Timmins is $A=dfrac{16.1-8.3}{2}=3.9$ hours

while that for Miami is $A=dfrac{13.8-10.5}{2}=1.7$ hours

The equation of axis for Timmins is $y=dfrac{16.1+8.3}{2}implies y=12.2$

while that for Miami is $y=dfrac{13.8+10.5}{2}=12.15$

Exercise scan

Step 4

4 of 5

c) Places at higher latitude (farther from equator) experience more varying hours of daylight throughout the year than those at lower altitudes.

Result

5 of 5

See graphs inside.

Exercise 8

Step 1

1 of 8

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ] Exercise scan

Step 2

2 of 8

a)

Since the diameter of the car is $52$ cm, its radius is $52/2=26$ cm. The radius of the wheel represents the amplitude of the sinusoidal graph. The distance of the center of the wheel from the ground corresponds to the axis, which is also equal to the radius of the wheel, that is, 26 cm. The nail must be at the ground at the beginning, so the graph should start at the minimum point. The period of the graph must be the distance traveled in one revolution which is $52times pi$.

With these information, the graph can be sketched as follows.

Exercise scan

Step 3

3 of 8

b) We need to know how high above the ground will the nail be after the car has traveled 0.1 km = 100 m = 10000 cm. To do this, we need to find how many revolutions have been made when this happens considering the circumference of the wheel.

no. of revolutions = $dfrac{10,000}{52 times pi}=61.21$ revolutions

Now, we need to know what height does a 0.21 revolution correspond to. You know that one revolution is $360^circ$, so $0.21$ revolution is $0.21times 360^circ=75.6^circ$

You can visualize the problem by sketching a diagram as shown.

Exercise scan

Step 4

4 of 8

To solve for the height of the nail $x$, we must solve for $y$ and subtract it from $26$ cm radius.

Remember that

$cos theta = dfrac{text{adjacent}}{text{hypotenuse}}$

y= $26cdot cos(360^circ cdot 0.21)=6.4659$ cm

$x=26 -6.4659=19.53$ cm

Step 5

5 of 8

We can also confirm our answer by reading from the graph. You must find the distance travelled at $0.21 rev$ which is $0.21 times pi times 52=34.306$, the height of the graph at $x=34.306$ is around $19.53$ cm.

Exercise scan

Step 6

6 of 8

c) Now, we shall find distance traveled with the nail is 20 cm above the ground for the fifth time. We can see this from the graph. Exercise scan

Step 7

7 of 8

d) For this calculations to be correct, we must assume that he drives continuously at steady speed forward (no brakes and backward motion), also, he must not spin the wheels.

Result

8 of 8

a) see graph

b) 19.53 cm

c) 361.5 cm

d) see explanation

Exercise 9

Step 1

1 of 6

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
end{gathered} ] Exercise scan

Step 2

2 of 6

Step 3

3 of 6

Remember that for the sinusoidal function of the form

$y=Asin(kx)+B$,

amplitude = A

period = $dfrac{360}{k}$

equation of axis: $y=B$

Given that

high speed: $d(t)=3sin(1080t)^circ$

low speed: $d(t)=2sin(1080t)^circ$

We can obtain the following.

Step 4

4 of 6

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|}
hline
High Speed & High Speed & Low Speed \ hline
period & $frac{1}{3}$ s & $frac{1}{3}$ s \ hline
amplitude & 3 cm & 2 cm \ hline
equation of the axis & $y=0$ & $y=0$ \ hline
end{tabular}
end{table}

Step 5

5 of 6

The period and equation of the axis is not affected by the wind speed. However, the amplitude increases as wind speed increases. Thus, the post will move at greater distance back and forth for higher wind speeds.

Result

6 of 6

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|l|}
hline
High Speed & High Speed & Low Speed \ hline
period & $frac{1}{3}$ s & $frac{1}{3}$ s \ hline
amplitude & 3 cm & 2 cm \ hline
equation of the axis & $y=0$ & $y=0$ \ hline
end{tabular}
end{table}

Exercise 10

Step 1

1 of 4

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
hfill \
{text{For a sinusoidal function}} hfill \
hfill \
y = Acos left( {kx} right) + B{text{ or }}y = Asin left( {kx} right) + B hfill \
hfill \
{text{amplitude = }}A hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{k} hfill \
{text{equation of axis:}}{text{ }},,,y = B hfill \
end{gathered} ] Exercise scan

Step 2

2 of 4

a) Graph the given function as specified. Exercise scan

Step 3

3 of 4

b) The time it takes for one revolution corresponds to the period

$T=dfrac{360}{12}=30$ seconds

c) The radius of the wheel correspond to the amplitude which is

$A=2$ m

d) The center of the wheel corresponds to the height of the axis which is

$h=1.5$ m

e) $h(10)=2sin(12cdot 10)+1.5=3.232$ m

This corresponds to the height of the basket at $t=10$ s

Result

4 of 4

a) see graph

b) 30 s

c) 2 m

d) 1.5 m

e) 3.232 m, height of the basket at $t=10$ s

Exercise 11

Step 1

1 of 4

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
hfill \
{text{For a sinusoidal function}} hfill \
hfill \
y = Acos left( {kx} right) + B{text{ or }}y = Asin left( {kx} right) + B hfill \
hfill \
{text{amplitude = }}A hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{k} hfill \
{text{equation of axis:}}{text{ }},,,y = B hfill \
end{gathered} ] Exercise scan

Step 2

2 of 4

a) Graph the given function as specified. Exercise scan

Step 3

3 of 4

b) The time it takes to move from one peak to the next peak is the period of the graphs which is

$T=dfrac{360}{72}=5$ s

c) The number of cycles in 1 min is

number of cycles = $dfrac{60text{s}}{5text{s per cycle}}= 12$ cycles

d) We can see this from the graph.

$|y_{max}-y_{min}|=|2.5-(-2.5)| = 5$ m

Result

4 of 4

a) see graph

b) $5$ s

c) 12 cycles

d) 5 m

Exercise 12

Step 1

1 of 4

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
hfill \
{text{For a sinusoidal function}} hfill \
hfill \
y = Acos left( {kx} right) + B{text{ or }}y = Asin left( {kx} right) + B hfill \
hfill \
{text{amplitude = }}A hfill \
{text{period = }}T=frac{{{{360}^ circ }}}{k} hfill \
{text{equation of axis:}}{text{ }},,,y = B hfill \
end{gathered} ] Exercise scan

Step 2

2 of 4

a) Graph the given function as specified. Exercise scan

Step 3

3 of 4

b) The period represents one year, since the climate cycle repeats annually.

c) From the given function and the graph, $y_{min}=5.9-14.2=-8.3$ and $y_{max}=5.9+14.2=20.1$

Thus, the temperature in Kingston ranges from $-8.3^circ C$ to $20.1^circ C$.

d) The mean temperature is the temperature at the axis which is $5.9^circ C$.

e) $T(30)=14.2sin(30(30-4.2))^circ=17.39^circ C$

This is the temperature on 30$^{th}$ month which is around June.

Result

4 of 4

a) see graph

b) one year cycle

c) $-8.3^circ C$ to $20.1^circ C$

d) $5.9^circ C$

e) $T(30)=17.39^circ C$

Exercise 13

Step 1

1 of 5

[begin{gathered}
{mathbf{Useful}}{text{ }}{mathbf{Definitions}} hfill \
{text{equation of the axis: }},,y = frac{{{y_{max }} + {y_{min }}}}{2} hfill \
{text{amplitude: }}A = {text{ }}frac{{left| {{y_{max }} – {y_{min }}} right|}}{2} hfill \
{text{period}} = left| {{x_2} – {x_1}} right| hfill \
hfill \
end{gathered} ] Exercise scan

Step 2

2 of 5

a) The period is the time for each cycle.

Ball A: $T=dfrac{24;text{s}}{3;text{cycles}}=8;s$ per cycle

Ball B: $T=dfrac{24;text{s}}{4;text{cycles}}=6;s$ per cycle

b) The equation of the axis is $d(t)=0$ which represents the position at resting point.

c) The amplitude is the distance from the resting point to peaks or trough.

Ball A: $A=4$

Ball B: $A=3$

The amplitude represents the maximum distance the ball travels from the resting position.

Step 3

3 of 5

d) The range is the set of all possible values of $d(t)$ which is from $d(t)_{min}$ to $d(t)_{max}$.

Ball A: ${ d(t) in bold{R};| -4leq d(t)leq 4}$

Ball B: ${ d(t) in bold{R};| -3leq d(t)leq 3}$

Step 4

4 of 5

e) Ball A has higher amplitude than ball B which means Ball A moves farther away from the resting position than Ball B, because of this, Ball A requires more time per cycle than Ball B.

Result

5 of 5

See answers inside.

Exercise 14

Step 1

1 of 2

To be able to sketch a sinusoidal function, we need to know the following:

(1) amplitude

(2) equation of the axis

(3) period

(4) starting point

If $y_{min}$ and $y_{max}$ is given, you can calculate both the amplitude and the equation of the axis.

Result

2 of 2

(1) amplitude

(2) equation of the axis

(3) period

(4) starting point

Exercise 15

Step 1

1 of 7

a) The gear must turn in opposite directions. Thus, if the smaller gear turns counterclockwise, the large gear must turn clockwise. Exercise scan

Step 2

2 of 7

b) Period is directly proportional to the radius of the gear, thus, if the period of the small gear is $2s$, the larger gear must be $2 s times dfrac{4;text{m}}{1;text{m}}=8$ seconds

Step 3

3 of 7

c) For convenience, we shall use the quarter of each period as intervals where the point lies on the horizontal and vertical axis as shown in part (a) as green points. Since the small gear turns counterclockwise, the vertical displacement must go through $(0to 1 to 0 to -1)$ while the large gear turning clockwise must go through $(0to -4 to 0 to 4)$. With these information, we can obtain the following table.

Exercise scan

Step 4

4 of 7

We can then plot the points to obtain the following graph. Exercise scan

Step 5

5 of 7

d) From the graph, the displacement of the large wheel when the smaller wheel has displacement of $-0.5$ m is $-3$ m (shown in red arrow).

e) From the graph, when the large wheel has a displacement of $2$ m, the small wheel has displacement of $0.5$ m (shown in purple arrow).

Exercise scan

Step 6

6 of 7

f) At 5 min = $5times 60=300$ s, the number of cycles that the large wheel has gone through is $dfrac{300}{8}=37.5$. The 0.5 cycle corresponds to $0.5times 8=4$ s, from the graph, the large gear is at zero displacement.

Result

7 of 7

a) clockwise

b) 8 s

c) see table and graph

d) $-3$ m

e) $0.5$ m

f) $0$ m

All Solutions

Page 370: Check Your Understanding

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