Modelling with sine/cosine

A Ferris wheel has a diameter of 30m. The bottom of the wheel is 1.5m off the ground it takes 3.5min to make one complete revolution. a person gets on ferris wheel at its lowest point at time = 0.

write an equation that represents persons height above ground h at any time t

is the height 31.5? i added 30 to 1.5

how high off the ground is the person at t = 25s?

I don't get how to solve.

How long (in one rotation) is the person above 27m?

Last post for tonight, going to bed

diameter is 30 m, so a=15
So, (recall the last question with the 18 and 19)

the min value is going to be -15 , but we want it to be 1.5 above ground , (the x-axis)
so we have to add 16.5
sofar we have
y = 15 sin k(t + d) + 16.5 , t in minutes

it takes 3.5 minutes for one rotation
period = 3.5
but 2π/k = 3.5
3.5k = 2π
k = 2π/3.5 or 4π/7

getting there ....
y = 15 sin (4π/7)(t + d) + 16.5

when t= 0 , we have the lowest point ---> (0,1.5)
1.5 = 15 sin (4π/7)(0+d) + 16.5
-1 = sin ((4π/7)(d) )
I know sin 3π/2 = -1
so (4π/7)d = (3π/2)
d = 21/8

ok, how about
y = 15 sin (4π/7)(t + 21/8) + 16.5

testing:
if t=3.5, we should get 31.5

y = 15sin(4π/7)(3.5 - 21/8) + 16.5
= 15 sin (π/2) + 16.5 = 15(1) + 16.5 = 31.5 , Wow!!

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when t=25 seconds = 25/60 min or 5/12 min

y = 15sin (4π/7)(5/12 + 21/8) + 16.5
= 15 sin( 73π/42) + 16.5
= 5.5 m high

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Here comes the hard part:

we want y to be above 27
sketch a line y = 27 to cut our cosine curve, you will see two such places in each period

Let's find those two values of t

27 = 15sin(4π/7)(t+21/8) + 16.5
sin(4π/7)(t+21/8) = .7
set your calculator to radians and do inverse sin (.7)
I get .7754

but by the Cast rule, we know the sine is positive in quadrants I and II
so (4π/7)(t+21/8) = .7754 OR (4π/7)(t+21/8) = π - .7754
t+21/8 = .43193 OR t+21/8 = 2.366
t = -2.193 or t = -.2588
don't worry about the t values being negative, the show we are on the left part of the cosine curve. We could add 3.5 to both to make them positive

So the time interval = -.2588 - (-2.193) = 1.934 minutes or 116 seconds

To model the height of a person on a Ferris wheel using sine and cosine functions, we need to consider the periodic nature of the wheel's motion.

First, let's define some variables:
- h represents the height of the person above the ground at any given time t.
- r represents the radius of the Ferris wheel, which is half of its diameter. In this case, r = 15m.
- a is the vertical offset, which is the distance between the lowest point of the wheel and the ground. In this case, a = 1.5m.
- T is the period of the wheel's revolution. Here, T = 3.5min = 210s.

To find the equation for the person's height above the ground, we can use the sine function. We know that the height of a point on a circle can be represented by the formula:
h = r * sin(2πt / T)

But since the person gets on the Ferris wheel at the lowest point at time t = 0, we need to add the vertical offset a to the equation:
h = r * sin(2πt / T) + a

To answer the specific questions:

1. Is the height 31.5m? To check, substitute t = 0 into the equation:
h = 15 * sin(0) + 1.5 = 0 + 1.5 = 1.5m
So the height is not 31.5m, but rather 1.5m, which is the initial height above the ground.

2. How high off the ground is the person at t = 25s? Substitute t = 25 into the equation:
h = 15 * sin((2π * 25) / 210) + 1.5
Calculate the value using a calculator. This will give you the height above the ground at t = 25s.

3. How long (in one rotation) is the person above 27m? To find this, we need to solve the equation for h = 27. However, since the sine function is periodic, there will be multiple solutions. We can restrict our search for the first full rotation of the wheel, which is from t = 0 to t = T.
Set up the equation:
27 = 15 * sin((2πt) / 210) + 1.5
Now, solve for t within the interval t = [0, T] to find the time it takes for the person to be above 27m for the first full rotation.