Answer:
[tex]y=-67.5[cos(\frac{\pi}{15}t)-1][/tex]
Step-by-step explanation:
We can start solving this problem by doing a drawing of London Eye. (See attached picture).
From the picture, we can see that the tourists will start at the lowest point of the trajectory, which means we can make use of a -cos function. So the function will have the following shape:
[tex]y=-Acos(\omega t)+b[/tex]
where:
A=amplitude
[tex]\omega[/tex] = angular speed.
t= time (in minutes)
b= vertical shift.
In this case:
A= radius = 67.5 m
[tex]\omega=2\pi f[/tex]
where the frequency is the number of revolutions it takes every minute, in this case:
[tex]f=\frac{1}{30} rev/min[/tex]
so:
[tex]\omega=2\pi (\frac{1}{30})[/tex]
[tex]\omega=\frac{\pi}{15} rad/min [/tex]
and
b= radius, so
b=A
b=67.5m
so we can now build our equation:
[tex]y=-67.5cos(\frac{\pi}{15} t)+67.5[/tex]
which can be factored to:
[tex]y=-67.5[cos(\frac{\pi}{15}t)-1][/tex]
You can see a graph of what the function looks like in the end on the attached picture.
The required sinusoidal equation is [tex]y = -67.5 (cos\dfrac{\pi t}{15} -1)[/tex]
Given that,
It has a diameter of 135 meters and makes one revolution (counterclockwise) every 30 minutes.
It is constructed so that the lowest part of the Eye reaches ground level, enabling passengers to simply walk on to, and off of, the ride.
We have to find,
A sinusoidal which models the height h of the passenger above the ground in meters t minutes after they board the Eye at ground level.
According to the question,
The tourists will start at the lowest point of the trajectory, which means we can make use of a -cos function. So the function will have the following shape:
[tex]y = -Acos(wt) + b[/tex]
Where, A=amplitude , [tex]\omega[/tex] = angular speed, t= time (in minutes) , b= vertical shift.
Then,
[tex]Radius = \dfrac{diameter}{2}\\\\Radius = \dfrac{135}{2}\\\\Radius = 65.5m[/tex]
The frequency is the number of revolutions it takes every minute is determined by the formula is,
[tex]\omega = 2\pi f[/tex]
Where,
[tex]f = \dfrac{1}{30} revolution \ per \ minute[/tex]
Then,
[tex]\omega = 2\pi \times \dfrac{1}{30}\\\\\omega = \dfrac{\pi }{15}\ radian \ per\ minute[/tex]
Here, The sinusoidal equation can be written as;
[tex]y = -67.5cos\dfrac{\pi t}{15} + 67.5\\\\y = -67.5 (cos\dfrac{\pi t}{15} -1)[/tex]
Hence, The required sinusoidal equation is [tex]y = -67.5 (cos\dfrac{\pi t}{15} -1)[/tex]
To know more about Frequency click the link given below.
https://brainly.com/question/9661952
