Homework Help: Pre Calculus 12

Posted by Shreya on Wednesday, January 23, 2013 at 11:32pm.

Another sine/cosine question i dont get

in vancouver on certain day, high tide is 20m at 2am, the next low tide is 8m at 6am, what an equation that represents height (h)of water at any time (t)since midnight)

what is the height of water at 8:45AM?
What is height of water at 8:30PM?

i tried attempting this but i think i only got the amplitude

I went 20-8/2 and got 6 as the amplitude

Im using the cos function so I have 6cos
and then for the phase shift i only got (t-2) And I dont get how to find the rest. Im using the original graphs of the sine and cosine to solve, i think that makes it easier.

Pre Calculus 12 - Reiny, Thursday, January 24, 2013 at 8:56am
you will need the period
high tide = 2:0 am
low tide = 6:00 am
time between is 4 hrs, which is half the period
(to get the full period you would go from high to low (4hrs) and back to high again(another 4 hrs)
period = 8
2π/k = 8
8k = 2π
k = 2π/8 = π/4

so you have y = 6 cos (π/4)t + h , the height

remember the min of the equation we have so far is -6, but we want that to be 8 , so we have to raise the curve 14 m

gives us:

y = 6 cos (π/4)t + 14
need : y = 6cos (π/4)(t + d) + 14

we know that at t=2 , y = 20

20 = 6 cos (π/4)(2-d) + 14
cos (π/4)(2-d) = (20-14)/6 = 1
but we know cos 0 = 1

so (π/4)(2-d) = 0
2-d=0
d=2 , but I just noticed you already had that

y = 6 cos (π/4)(t-2) + 14

check for other data:
at 6:00 am , h = 8 ---- t=6 should give us y = 8

y = 6 cos (π/4)(6-2) + 14
= 6 cos (π) + 14
= 6(-1) = 14 = 8
looks like we nailed it

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