# trigonometry problem

• Dec 5th 2008, 12:23 AM
Jasleen2008
trigonometry problem
A ferris wheel has a radius of 160 feet rotates at a constant 1.57 revolutions per minute. The lowest point in the wheel is 24 feet off the ground. From the point where you start the ride, it takes you 21 seconds to reach the highest point on the wheel.

a) How far off the ground will you be when the ride starts?

b) How far off the ground will you be after 5.6 minutes?
• Dec 5th 2008, 11:03 AM
nzmathman
This situation can be modeled by a trigonometric equation.

We will use cos.

$\displaystyle H = A\cos{(\omega t - D)} + C$

A is the amplitude; which is 160ft.

$\displaystyle \omega t$ is the period.

$\displaystyle \omega = \frac{\theta}{t}$

$\displaystyle \omega = \frac{1.57 \times 2\pi radians}{60} = 0.1644$

C is the vertical translation. The centre is now at $\displaystyle 24ft + 160ft = 184ft$ high.

So $\displaystyle C = 184$

Therefore

$\displaystyle H = 160\cos{(0.1644t-D)} + 184$

To solve D use the info given (at 21secs you are at max height 344ft)

$\displaystyle 344 = 160\cos{(0.1644\times21-D)} +184$

$\displaystyle 344 = 160\cos{(0.1644\times21-D)} +184$

$\displaystyle D= 3.4524$

The full equation is therefore:

$\displaystyle H = 160\cos{(0.1644t-3.4524)} + 184$

Q (a)

Substitute $\displaystyle t = 0$ into equation for height.

Q (b)

Substitute $\displaystyle t = 5.6 \times 60 = 336$ into equation for height.