Ferris wheel problem

Question

the lowest point of a ferris wheel is 2.5 meters off the ground, and it's radius is 15 m. It's takes one complete revolution every 30 seconds. Starting at the lowest point, find a cosine function that gives the height above the ground of a riding child in terms of the time t in seconds. Hint use y=acos(t-c)+d

Answer 1

Let

h = 2.5 m be the lowest point of the wheel,

r = 15 m be the radius of the wheel,

T = 30 s be the period of the wheel.

The solution is to ignore the hint.

Instead you need the function

y=acos(2πt/T - c) + d

This is the general form required for a periodic function with period T.

For the moment let's restrict ourselves to a > 0.

Negative a can also be accommodated by adjusting the phase c.

At time 0 the child is at the lowest point.

Therefore at time 0 the cos function must be at its lowest value -1,

which happens when its argument is π + 2kπ for any integer k.

You can pick any k, but let's pick k = -1. (It will make the result look more usual.)

2π×0/T - c = -π

c = π

So the function is of the form

y=acos(2πt/T - π) + d

We are left with two unknowns a, d.

For that we use the other two given quantities for two equations.

For convenience, we pick known positions of the child at time t=0 and t=T/2

h = acos(2π×0/T - π) + d = -a + d

h + 2r = acos(2π×(T/2)/T - π) + d = acos(0) + d = a + d

Solution is

d = h+r

a = r

So the function is

y = rcos(2πt/T - π) + h + r

Now let's revisit our restriction on a > 0.

It turns out that negative a will give us a simpler function

y = -rcos(2πt/T) + h + r

You can plug in actual numbers for h, r, T.