## chaotic_butterflies one year ago Please help - In Panama City in January, high tide was at midnight. The water level at high tide was 9 feet and 1 foot at low tide. Assuming the next high tide is exactly 12 hours later and that the height of the water can be modeled by a cosine curve, find an equation for water level in January for Panama City as a function of time (t). I don't know what the formula for a cosine curve is, nor am I aware of the meanings of each part. Multiple choice answers are below:

1. chaotic_butterflies

2. anonymous

@SithsAndGiggles

3. anonymous

@ybarrap

4. chaotic_butterflies

You know they're not online right? @MLG360GABEN

5. anonymous

ybarrap is

6. Afrodiddle

I wish that question had an oxford comma in it... I had to read one of the sentences like 5 times. lol

7. chaotic_butterflies

You have no idea how confusing all these questions are... virtual school goes nuts making up scenarios and barely spends any time on teaching the content >~<

8. ybarrap

You know that the peak is 9 feet and the low point is 1 foot. What is the most that cos (x) can be ? What is the smallest cos(x) can be? That is a huge hint. Answer this and we'll go forward.

9. ybarrap

?

10. ybarrap

|dw:1433619049716:dw|

11. chaotic_butterflies

Sorry I was afk

12. ybarrap

This means that maximum that 4 cos (x) can be is 4 and the smallest it can be is -4. This means that maximum that 5 cos (x) can be is 5 and the smallest it can be is -5. Next look at the offsets. Your options are either 5 or 4. This will tell you which 2 of the 4 options to eliminate. That the maximum of the cosine term and add to this offset and it should equal 9. Take the minimum of the cosine term and add to this offset and it should be 1.

13. ybarrap

|dw:1433619397690:dw|

14. imqwerty

The equation we want is h(t) = a*cos((2pi/P)*t) + b, where a is the amplitude of motion, P is the period and b is the vertical displacement. Now a = [(max. height) - (min. height)]/2 = (9 - 1)/2 = 4, and b = [(max. height) + (min. height)]/2 = (9 + 1) / 2 = 5. The period is given as P = 12 hours, so the equation is h(t) = 4*cos((2pi/12)*t) + 5 = 4*cos((pi/6)*t) + 5.

15. ybarrap

Once you figure out which of the two to eliminate, then figure out the current frequency: $$\pi/2$$ or $$\pi/6$$

16. ybarrap

|dw:1433619618409:dw|

17. ybarrap

For $$\pi/6$$, after 12 hours, you get back to your starting point. This will help eliminate one of the two remaining options.

18. imqwerty

@chaotic_butterflies did u got the answer??

19. chaotic_butterflies

I haven't taken it yet because I'm failing to understand the process - but I appreciated both @imqwerty and @ybarrap 's willingness to help.

20. ybarrap

Say you did not have he cosine term. You just had the constant. How much would you need to add to either 5 or 4 to each to get 9? This is what I'm saying. Take the last option $$4\cos \cfrac{\pi t}{6}+5$$ What if you just had $$5$$ What do you need to add to this to get 9?

21. chaotic_butterflies

well 4 of course

22. ybarrap

Ok. So if we had 4 that would be great because that would match the peak. Next, what do we need to add to $$5$$ to get 1?

23. ybarrap

That's the low point

24. chaotic_butterflies

-4

25. chaotic_butterflies

I suppose it wouldn't work...

26. chaotic_butterflies

27. chaotic_butterflies

-3

28. chaotic_butterflies

But the minimum is -4

29. ybarrap

Back to the last option $$4\cos \cfrac{\pi t}{6} + 5$$ We have the cosine term having a max of 4 and a min of -4. When we add the max to 5 we get 9 and when we add the min we get 1. That's exactly what we want. You agree?

30. ybarrap

|dw:1433620878420:dw|

31. ybarrap

|dw:1433620951827:dw|

32. chaotic_butterflies

I see it now

33. ybarrap

Great!

34. chaotic_butterflies

Sorry for not replying, someone came to my door.

35. chaotic_butterflies

Thank you so much!

36. ybarrap

you're welcome

37. chaotic_butterflies

Thank you @imqwerty too, I wish I could also give you a metal :c

38. imqwerty

:) its ok @chaotic_butterflies btw @ybarrap really deservs the medal ;)