# Thread: Tough Trigonometric Function Question!!

1. ## Tough Trigonometric Function Question!!

Hi, I'm not sure how to approach this question.

Q: A Ferris wheel has a diameter of 56m, and one revolution took 2.5 min to complete. Riders could see Niagara Falls if they were higher than 50m above the ground. Sketch cycles of a graph that represents the height of a rider above the ground, as a function of time, if the rider gets on at a height of 0.5m at t=0 min. Then determine the time intervals when the rider could see Niagara Falls.

I have the graph sketched and the resulting function I got is y= 28sin(4pi/5(x+pi/2))+28.

If a rider gets on at 0.5m at t=0, the equation would be y=28sin(4pi/5(x+pi/2))+28.5?

Where do I go from here? Can someone please show me what I have to do?

The answers are: 0.98min<t<1.52min, 3.48min<t<4.02min, 5.98min<t<6.52min.

Thanks!!!

2. Pretty good attempt.

1) One error. You need y = 1/2 when x = 0 and you don't have that. A little experimentation shows that the phase shift should be +5/8.

2) Solve away. You will need a calculator. Set the whole expression equal to 50 and find the two solutions in the first trip around. Other solutions should follow by 2.5 minutes. I get 0.97335 and 1.52665. It appears I agree with the given solution.

3. Hello MATHDUDE2
Originally Posted by MATHDUDE2
Hi, I'm not sure how to approach this question.

Q: A Ferris wheel has a diameter of 56m, and one revolution took 2.5 min to complete. Riders could see Niagara Falls if they were higher than 50m above the ground. Sketch cycles of a graph that represents the height of a rider above the ground, as a function of time, if the rider gets on at a height of 0.5m at t=0 min. Then determine the time intervals when the rider could see Niagara Falls.

I have the graph sketched and the resulting function I got is y= 28sin(4pi/5(x+pi/2))+28.

If a rider gets on at 0.5m at t=0, the equation would be y=28sin(4pi/5(x+pi/2))+28.5?

Where do I go from here? Can someone please show me what I have to do?

The answers are: 0.98min<t<1.52min, 3.48min<t<4.02min, 5.98min<t<6.52min.

Thanks!!!
Your second equation is better than the first, assuming that the rider gets on at the lowest point of the cycle, for this is now $\displaystyle y = 0.5$, whereas it is $\displaystyle y = 0$ in your first equation. You have correctly used the period of 2.5 to give the factor $\displaystyle \frac{4\pi}{5}$, but the 'phase angle' that gives the initial value is not correct. So if we write:
$\displaystyle y = 28 \sin \Big(\frac{4\pi}{5}t+c\Big)+28.5$
When $\displaystyle t = 0, y = 0.5$, and hence:
$\displaystyle 0.5=28\sin c+28.5$

$\displaystyle \Rightarrow -1=\sin c$

$\displaystyle \Rightarrow c=-\frac{\pi}{2}$ (or any equivalent angle)
So the function is
$\displaystyle y = 28 \sin \Big(\frac{4\pi}{5}t-\frac{\pi}{2}\Big)+28.5$
So you now need to solve
$\displaystyle 50 = 28 \sin \Big(\frac{4\pi}{5}t-\frac{\pi}{2}\Big)+28.5$

$\displaystyle \Rightarrow 28 \sin \Big(\frac{4\pi}{5}t-\frac{\pi}{2}\Big)= 21.5$

$\displaystyle \Rightarrow \sin \Big(\frac{4\pi}{5}t-\frac{\pi}{2}\Big)=\frac{21.5}{28}$
$\displaystyle \Rightarrow \frac{4\pi}{5}t-\frac{\pi}{2}=0.8755, \pi - 0.8755, 2\pi +0.8755, ...$
Can you complete it now?

Grandad