Tides are cyclical phenomena caused by the gravitational pull of the sun and the moon. On a particular retaining wall, the ocean generally reaches the 3m mark at high tide. At low tide, the water reaches the 1m mark. Assume that high tide occurs at 12:00 pm and at 12:00 am, and that low tide occurs at 6:00pm and 6:00am. What is the height of the water at 10:30 am?

Question

Reiny · Accepted Answer

First problem:
The height of tides follows a sinusoidal path, that is , their height is represented a either a sine or cosine curve, there is no constant change.
Second problem: what is the difference between 12:00 pm and 12:00 am ? Both are midnight.

http://askascientist.co.uk/physics/day-midnight-belong/

(that is why countries that use the metric system also use the 24:00 hour method of measuring time)

I will assume:
0:00 ---> 3 m
6:00 ---> 1 m
12:00 ---> 3 m
18:00 ---> 1 m
24:00 ---> 3 m

So according to your data, a tide period is 12 hours, (not really but ....)
and the amplitude of the cosine curve is 1 m
2π/k = 12
k = π/6

my function is:
height = cos (π/6)t + 2 , where t is the hours after midnight.

testing:
midnight --> t =0 , height = cos 0 + 2 = 1+2 = 3
6:00 -----> t = 6, height = cosπ + 2 = -1+2 = 1
12:00 (noon) --> t = 12, height = cos2π + 2 = 3
looks good

so you want the tide at 10:30 am or t = 10.5
height = cos π(10.5)/6 + 2
= appr 2.71 m

check:
http://www.wolframalpha.com/input/?i=plot+y+%3D+cos+(%CF%80t%2F6)+%2B+2

Kolb · Answer

So this would mean that there is 6 hours in each cycle. 10:30am is 4.5 hours into the rising cycle out of 6 total hours, or 4.5/6 . This ratio is directly proportional to the tide marking assuming that this is a constant change. The change in height from 3 to 1 is 2m. Thus we should multiply our ratio by 2m. 2*(4.5/6)=4.5/3=1.5. So the tide is at 1m+1.5m=2.5m by 10:30 am.