SOLUTION: The tide on a certain shore on a planet has a period of 36.5 hours, and the high tide level is 8 m above the low tide level. At t = 0 the water level is 2 m above the low tide leve

Question 1150036: The tide on a certain shore on a planet has a period of 36.5 hours, and the high tide level is 8 m above the low tide level. At t = 0 the water level is 2 m above the low tide level and rising. Using trigonometric functions, find a function to describe the height H(t) of the water above the low tide level
Found 3 solutions by josmiceli, Alan3354, jim_thompson5910:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!

(a) The amplitude is
(b) When , the tide is back to starting level of
(c) When , and rising
Here is the plot:

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Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
I asked the doctor for a 2nd opinion, and he said, "OK. You're ugly."
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

t = number of hours that has elapsed since some starting point
H(t) = height, in meters, of water above the low tide level
H(t) = 8 means we're 8 meters above the low tide level
H(t) = 2 means we're 2 meters above the low tide level
The smallest H(t) can get is 0. The largest is 8.
The range of this trig function is which is the same as saying if y = H(t)

General sine function is
y = A*sin(B(x-C))+D
A = amplitude
B = determined by the period
C = phase shift (aka horizontal shift)
D = vertical shift = midline

Lets find the values for A,B,C,D
A = amplitude
A = half the vertical distance between the lower and upper bounds of the range
A = (upper - lower)/2
A = (8-0)/2
A = 4

P = period = 36.5 hours
B = 2pi/P
B = 2pi/36.5

C = phase shift
We'll come back to this later

D = midline
D = midpoint of upper and lower boundaries of the range
D = (upper+lower)/2
D = (8+0)/2
D = 4

We have
A = 4
B = 2pi/36.5
C = unknown for now
D = 4

So,

turns into

Let's plug in (x,y) = (0,2) to indicate we want the starting water level to be 2 meters.
In other words, we want H(t) = 2 when t = 0.
Now solve for the variable C

We can update

into

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Final Answer:

Graph:

Notes:

x = t = elapsed time in hours

y = H(t) = height, in meters, of water above the low tide level

The sine curve stays between y = 0 and y = 8

The graph repeats itself every x = 36.5 units. The water is at a height of y = 2 when x = 0. The water level goes up, comes back down to y = 0, and goes back up again to arrive back at y = 2 when x = 36.5

The horizontal distance between any peak to its adjacent neighbor valley is x = 18.25, which is half of the period 36.5

The amplitude of 4 represents vertical distance from the midline y = 4 to either the peak (y = 8) or the valley (y = 0).

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