# Thread: Need help with Ferris Wheel exercise

1. ## Need help with Ferris Wheel exercise

The third largest Ferris Wheel in the world is the London Eye. The height of a rider after t minutes can be described as h (t)= 67sin (12(t+0.223))+70.

a) what is the diameter of the wheel? 67×2
B) where is the rider ato t=0? At the initial point.
C) how high off the ground is the rider at the top of the wheel? 70
D) at what times will the rider be at the bottom? 12 times
E) how long does it take for the Ferris wheel to go through one rotation? 12×the

I did this with a Maths teacher, but I do not know whether it is correct or not.

2. ## Re: Need help with Ferris Wheel exercise

I really don't know what you mean by "did it with a Maths teacher". The largest possible value of "sine" is 1 so the largest possible value of h is 67(1)+ 70= 137. The smallest possible value of "sine" is -1 so the least possible value of of h is 67(-1)+ 70= 3. The diameter is 137- 3= 134. That is, as you say 67x2.

When t= 0, 12(t+ 0.223)= 12(0.223)= 2.676 above the ground (you don't say what the units are; probably meters). You say "at the initial point" which is obviously true- that is what "t= 0" means- but I doubt that's what they want.

As before, the largest "sine" can be is 1 so the largest h can be is 67+ 70= 134. "70" is the height of the axis of the wheel.

"At what times will the rider be at the bottom?" You say "12 times". I don't even know what you mean by that! The are asking for specific time, in minutes after the beginning of the ride, not a number of times the rider goes around. The rider will be at the bottom when "sine"= -1. That happens with the argument of sine, here 12(t+ 0.223) equals an odd multiple of $\pi$. $12(t+ 0.223)= (2k+ 1)\pi$ where k can be any integer. Solve that for t which will of course have many solutions, one for each possible integer value of k.

"How long does it take for the Ferris wheel to go through one rotation?" You have "12 x the". I don't know what that means! "sine" has a period of $2\pi$. Starting at t= 0 the wheel will have made one rotation when $12(t+ 0.223)= 2\pi$. Again solve that for t.

I don't see any 'trigonometry' involved in this.