Ferris Wheel (applications of trigonometric functions)
One of the most common applications of trigonometric
functions is, Ferris wheel, since the up and down motion of a
rider follows the shape of sine or cosine graph.
Equations used :
Y = aSin(bx-c)+d
or
Y = aCos(bx-c)+d
Formula used :
Amplitude, a =
|𝒎𝒂𝒙−𝒎𝒊𝒏|
𝟐𝝅
b = 𝒑𝒆𝒓𝒊𝒐𝒅
𝟐
vertical shift d=
𝒎𝒂𝒙+𝒎𝒊𝒏
𝟐
Example1. A Ferris wheel has a diameter of 30 m with its center
18 m above the ground. It makes one complete rotation every
60 seconds. Assuming rider starts at the lowest point, find the
trigonometric function for this situation and graph the function.
Solution:
Amplitude – radius of the wheel makes the amplitude so
amplitude(a) = 30/2 =15.
Period- Wheel complete one rotation in 60 seconds so period is
60 sec. Using period we can find b value as,
b=
2𝜋
𝑃
2𝜋
𝝅
= 60 = 𝟑𝟎
Phase shift- There is no phase shift for this cosine function so
no c value.
Vertical shift- Centre of wheel is 18m above the ground which
makes the mid line, so d= 18.
Lowest point would be 18-15=3m and highest point would be
18+15= 33m above the ground. So the rider will start from 3m
and reach to a height of 33 m in half the period (30 sec) and
come back to lowest point (3m) again in 60 secs. So its graph
would look like this.
As the graph start from lowest point and the pattern is upside
down so we put a negative sign in front of cos. Summing up all
the parameters above we get trigonometric function as,
𝜋
Y = - 15 cos (30t) +18
Example3.A water wheel on a paddle boat has a radius of 2 m.
The wheel rotates every 30 secs and bottom 0.6m of wheel is
submerged in water.
a) Considering the water surface as x axis , determine the
cosine equation of the graph starting from a point at the
top of wheel.
b) Graph the height of a point on the wheel relative to the
surface of water, starting from highest point.
c) How long is the point on wheel under water.
Solution: a)
Radius of wheel gives the amplitude so a= 2
Period =30 secs , so b=
2𝜋
𝑃
=
2𝜋
30
=
𝝅
𝟏𝟓
Since radius is 2 and bottom point is -0.6, mid line will be
at 2-0.6 =1.4m
Combining all the above parameters and considering top point
as starting point , we get the cosine equation as,
b)
𝜋
Y= 2cos(15 𝑡) +1.4
d) To find the time for wheel under water we need to find x
intercepts , intersection of cosine function with y=0(water
surface) using graphic calculator. We get the intersection
points as x=11.2s and x=18.8s So the total time for wheel
underwater is 18.8-11.2 = 7.6 seconds.
Example3. The earliest sunset occurs at 5:34 PM on Dec. 21 and
latest at 11:45 PM on June 21.
a) Write cosine equation of the graph.
b) Draw the graph approximating the sunrise time during the
year.
c) What is the sunset time on April 6.
d) The sunset time is earlier than 8PM for what percentage
of the year.
Solution:
First of all we need to convert time from hour/min to decimal
hour form. 5PM is equal to 17 hours and 34 min. are equal to
34/60=0.57 so we get 5:34PM equal to 17.57 hrs.
Same way we get 11:45PM equal to 23.75 hrs. because 11PM is
equal to 12+11=23hrs and 45 min = 45/60 =0.75hrs.
Using these maximum and minimum values we get amplitude
𝟐𝟑.𝟕𝟓−𝟏𝟕.𝟓𝟕
a=
𝟐
= 3.1
For these type of problems, period is taken as 365 days. so,
b=
𝟐𝝅
𝑷
𝟐𝝅
= 𝟑𝟔𝟓
Starting the graph on Jan1, max. value occurs on 21 June so
c = 31+28+31+30+31+21=172 days
Central line would be
d=
𝟐𝟑.𝟕𝟓+𝟏𝟕.𝟓𝟕
𝟐
= 20.66
Combining all above parameters , we get cosine function as,
𝟐𝝅
Y = 3.1cos[𝟑𝟔𝟓 (𝒙 − 𝟏𝟕𝟐)] +20.66
This equation can also be written as,
Y = -3.1cos[
as lowest point.
2𝜋
365
(𝑥 + 10)] +20.66 considering 21 Dec.
c) To find sunset time on April6, we find what day of year it is.
31+28+31+6 = 96
So we plug in x as 96 into the equation found in part a.
2𝜋
Y = 3.1cos[365 (96 − 172)] +20.66 = 21.46
Converting back 21.46 into decimal hour form we get,
21-12=9 and .46*60= 28 min
So sunset time on April 6 is 9:28PM.
d. Sunset is earlier than 8PM for days 0 to 68 and then again
days 276 to 365. So that total number of days are 157
which are 43 % of the year.
Example 4: The following table gives the average recorded
monthly temperature throughout the year.
Write the cosine equation for the graph corresponding to the
table given above.
Solution:
Amplitude, a =
22−(−17)
2
=
39
2
= 19.5
Period = 12 months, here months are used instead of days.
b value =
2𝜋
𝑃
2𝜋
𝜋
= 12 = 6
Since the maximum temp. occur in the month of July which is
the 7th month so there is a phase shift of 7.
c=7
Vertical shift d =
22+(−17)
2
=
5
2
=2.5
Combining all the parameters above, we get the final equation
as,
𝜋
y = 19.5cos[ 6 (𝑥 − 7)] +2.5
Where x represents number of months and y represents
approximate temperature.
Practice problems:
1. A Ferris wheel with radius 40 ft complete one revolution
every 60 seconds. The lowest point of wheel is 5 m above the
ground.
a) Draw the graph of the situation, starting with a person
getting on the bottom of the wheel at t=0 seconds.
b) Determine an equation representing the path of the
person on Ferris wheel.
c) Determine how high the person will be after riding for 40
seconds.
d) When the person first reach 50 ft.
2.The bottom of a windmill is 8m above the ground, and the
top is 22m above the ground. The wheel rotates once every 5
seconds.
a) Determine the cosine equation of the graph.
b) Draw the graph of two complete cycles.
3. The average temp. for Regina is hottest at 27℃ on July 28,
and coolest at -16℃ on January 10.
a) Draw the graph and write the cosine equation for the
graph.
b)The average temp. is higher than 23℃ for how many
days.
4. The latest sunrise occurs at 9:10 AM on Dec 21. The earliest
occurs at 3:43 AM on June 21. Write the cosine equation for
the graph.
Answers:
𝜋
1. Y= -40cos(30 𝑥)+45 , 65 ft , 16.2s
2𝜋
2. Y= -7cos( 5 𝑥) +15
2𝜋
3. Y = 21.5cos[365 (𝑥 − 209)]+5.5 , 72 days
2𝜋
4. Y = -2.73cos[365 (𝑥 − 172)]+ 6.45