## cocoapeebles3 one year ago Xavier is riding on a Ferris wheel at the local fair. His height can be modeled by the equation H(t) = 20cosine of the quantity 1 over 15 times t + 30, where H represents the height of the person above the ground in feet at t seconds. Part 1: How far above the ground is Xavier before the ride begins? Part 2: How long does the Ferris wheel take to make one complete revolution? Part 3: Assuming Xavier begins the ride at the top, how far from the ground is the edge of the Ferris wheel, when Xavier's height above the ground reaches a minimum?

1. cocoapeebles3

-.- I just needed help figuring it out I have been stuck on it for hours and I thought openstudy would help but obviously not @IrishBoy123

2. cocoapeebles3

Okay for part 1 i think you just have to substitute in 0 for t. H(0)= 20cos(pi/15(o))+30 which is 50. But I'm not sure if that's right.

3. IrishBoy123

why not just draw the equation? i can guess sorta what you are trying to do, but that really defeats the point. loads of people here want to help and learn.

4. IrishBoy123

|dw:1439938025351:dw|

5. IrishBoy123

|dw:1439938102822:dw|

6. IrishBoy123

or post a url or do a screengrab or scribble it down and scan and link it

7. IrishBoy123

|dw:1439938283787:dw|

8. IrishBoy123

i am sorry to be such a pain but, once you get this bit sorted, you will get loads of help!

9. cocoapeebles3

|dw:1439938250052:dw|

10. IrishBoy123

an example of posting a screengrab https://gyazo.com/9bd2129de27eaf26d56f85c2ffee15eb

11. IrishBoy123
12. cocoapeebles3

sorry I'm new to this

13. IrishBoy123

we all were once please stick with it your drawing is off the screen for me https://gyazo.com/944f3bbd967745366c90fa0c079f5f56

14. IrishBoy123

|dw:1439938633443:dw|

15. cocoapeebles3

16. cocoapeebles3

which I think is equal to 50

17. IrishBoy123

and where is the $$\pi$$ being brought in? it is not in the original post...

18. IrishBoy123
19. IrishBoy123

strongly suggest you close this and start a new thread with the same question i am pretty easy going but this is too much, even for me. having learned something about how to share info online, you will have better luck with a new thread, the audience will have less to guess. good luck, and wish you well!

20. cocoapeebles3

oh sorry the original equation is this$H(t)= 20\cos(\pi/15\times t) + 30$

21. cocoapeebles3

okay thanks

22. IrishBoy123

thanks for that @cocoapeebles3 if this is right, $$\large H(t)=20cos(\frac{π t}{15})+30$$ then as cos(0) = 1 you have $$h(0) = 20+30=50$$ the *period* of the wheel can be got from $$\omega \ t = \frac{π }{15}t= \frac{2 \pi}{T} t$$ $$\frac{π }{15}t= \frac{2 \pi}{T} t \implies T = 30$$ finally, $$10 \le h(t) \le 50$$ simply because $$-1 \le cos \le 1$$

23. IrishBoy123

there might be a load of jargon in there so please do persist if anything is unclear and you have to factor in errors on my part too cheers!

24. cocoapeebles3

I think I understand it all thank you so much!

25. cocoapeebles3

So for part 3 how would I word that?

26. IrishBoy123

for last part $$H = \pm 20 + 30$$ because cosine goes from -1 to +1.