# Sarah's question at Yahoo! Answers regarding a sinusoidal function

#### MarkFL

Staff member
Here is the question:

One of the largest ferris wheels ever built is in the British Airways London Eye which was completed in 2000. T?

One of the largest ferris wheels ever built is in the British Airways London Eye which was completed in 2000. The diameter is 135 m and passengers get on at the bottom 4 m above the ground. The wheel rotates once every three minutes.

a) Draw a graph which represents the height of a passenger in metres as a function of time in minutes.

b) Determine the equation that expresses your height h as a function of elapsed time t

c) How high is a passenger 5 minutes after the wheel starts rotating?

d) How many seconds after the wheel starts rotating is a passenger 85 m above the ground for the first time. Answer to the nearest tenth.

10 POINTS TO BEST SOLUTIONS AND ANSWER
Here is a link to the question:

One of the largest ferris wheel ever built is in the british airways london eye which was completed in 2000. T? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### MarkFL

Staff member
Hello Sarah,

I would do part b) first, and then plot the resulting function after that.

Since the wheel is circular and rotates at presumably a constant rate, the height of the passengers will vary sinusoidally with time. Since we are given that they are at the minimum height at time $t=0$, then we need a negative cosine function.

Since the radius of the wheel is 135 m, this will be the amplitude, and the period of the motion is said to be 3 minutes, hence we may find the angular velocity $\omega$ by:

$$\displaystyle \frac{2\pi}{\omega}=3\,\therefore\,\omega=\frac{2\pi}{3}$$

The vertical translation is the radius plus 4 metres, or 139 metres, hence the height $h$ (in metres) as a function of time $t$ (in minutes) is given by:

b) $$\displaystyle h(t)=-135\cos\left(\frac{2\pi}{3}t \right)+139$$

a) Here is a plot of the function we found for part b) over the first period: c) To find the passengers' height at time $t=5$, we need to evaluate:

$$\displaystyle h(5)=-135\cos\left(\frac{2\pi}{3}5 \right)+139=-135\left(-\frac{1}{2} \right)=\frac{413}{2}=206.5$$

d) To find when the height is 85 m for the first time, we need to equate the height to 85 m and solve for the smallest positive time:

$$\displaystyle 85=-135\cos\left(\frac{2\pi}{3}t \right)+139$$

Subtract through by 139, then multiply through by -1:

$$\displaystyle 135\cos\left(\frac{2\pi}{3}t \right)=54$$

Divide through by 135 and reduce:

$$\displaystyle \cos\left(\frac{2\pi}{3}t \right)=\frac{2}{5}$$

The inverse cosine function will give us the value we want by default.

$$\displaystyle \frac{2\pi}{3}t=\cos^{-1}\left(\frac{2}{5} \right)$$

Solve for $t$ by multiplying through by $$\displaystyle \frac{3}{2\pi}$$

$$\displaystyle t=\frac{3}{2\pi}\cos^{-1}\left(\frac{2}{5} \right)\approx0.5535151793483181\approx0.6$$

To Sarah and any other guests viewing this topic, I invite and encourage you to post other trigonometric questions here in our Trigonometry forum.

Best Regards,

Mark.

#### SarahJeen

##### New member
Thanks Mark would you mind helping me with some other question add me on skype sunshadersg thanks sarah.

#### MarkFL

Hey Sarah, glad you joined MHB! I don't have Skype, but I will be happy to help you here in our forums. Just pick the most appropriate sub-forum for your questions, give your topics titles which indicates the nature of the question (preferably no more than two related questions per topic), and post what you have tried so far or what your thoughts are on how to begin, and I or someone else will here help guide you. 