The high and low tides of a coastal city on May 3 2000 are s
Solution
Ans(14):
midpoint: (98 + 42) / 2 = 140 / 2 = 70
amplitude: |70 - 42| = 28
Hence basic formula in x for the height is:
28 cos(x) + 70
However, t starts at midnight, so given the value at high and low, we need to find the y intercept for t = 0, which will be the starting angle, and the cycle of the cosine wave 2 expressed as a function of t.
That will give us the exact wave
The time for a full 2 period in a cosine wave is high to low, to high, which is twice (low - high).
2 period: 2(8:27 - 2:00) = 2(6:27) = 12:54 or 12+54/60 hours = 12.9 hours
The angular movement in one hour is 2/12.9 = /6.45 = 0.487068628464
At -2 hours, it is -2/6.45 = -0.974137256927
So, at midnight, the angle is 0.974
and for each t, increases by 0.487
So the angle is 0.487t-0.974
The final equation: h(t)=27cos(0.487t-0.974)+70
now we have to find height at 9:05AM
means plug t=9+5/60=9.083
So plug t=9.083 in h(t)
h(t)=27cos(0.487*9.083-0.974)+70= 44.269
Hence answer is 44.269 cm
