On a summer day in Cape Cod, the depth of the water at a dock was 4ft at low tide at 2:00 AM. At high tide 5 hours later, the heist of the water at the dock rose to 14 feet. Write a cosine function to model the height of the water at the dock x hours after the day began at midnight

Answer 1

amplitude: (14-4)/2 = 5
center line: (14+4)/2 = 9
so, we can start with

y = 9cos(kx)

the period is 5+5 = 10 hours, so 2?/k = 10, making k=?/5

y = 9cos(?/5 x)

The high tide was at 7:00, rather than midnight, so

y = 9cos(?/5 (x-7))
or
y = 9cos(?/5 x - 7?/5)

see

http://www.wolframalpha.com/input/?i=9cos(%CF%80%2F5+(x-7))

Answer 2

Ummh, the amplitude is 5
We also need a vertical shift, as well as a phase shift.

I had
y = 5cos (?/5(x + 3)) + 9

test: let x =2 , (2:00 am)
y = 5cos ? + 9 = 5(-1)+9 = 4 ft,
let x = 7, (5 hours later, the max of 14)
y = 5cos (2?) + 9
= 5(1) + 9 = 14 ft

http://www.wolframalpha.com/input/?i=5+cos(1%2F5+%CF%80+(x+%2B+3))+%2B+9

Answer 3

Huh. Good thing we got a second opinion. Ahem. As I was saying,

amplitude: (14-4)/2 = 5
center line: (14+4)/2 = 9
so, we can start with

y = 9+5cos(kx)
y = 9+5cos(π/5 x)
y = 9+5cos(π/5 (x-7))

As for the shift, (x-7), (x+3)
tomato, tomahto. Since the period is 10, it's the same.