SOLUTION: At this one coastal city, the tides vary greatly during the day. During May, the predicted height of the tides are 2.5 m at 4 AM, 17 m at 10 AM, 2.5 m at 4 PM and 17 m at 10 PM. As

Algebra ->  Graphs -> SOLUTION: At this one coastal city, the tides vary greatly during the day. During May, the predicted height of the tides are 2.5 m at 4 AM, 17 m at 10 AM, 2.5 m at 4 PM and 17 m at 10 PM. As      Log On


   



Question 1182380: At this one coastal city, the tides vary greatly during the day. During May, the predicted height of the tides are 2.5 m at 4 AM, 17 m at 10 AM, 2.5 m at 4 PM and 17 m at 10 PM. Assume we start the graph at midnight. What is the height of the tide at 11 AM? Round to the nearest tenth.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
well, i had to play with it a while but i think i have it.

your equation is y = 7.25 * sin(30 * (x-7)) + 9.75

the graph of that is shown below.



the analysis was as follows:

the general form of the sine equation is y = a * sin(b * (x-c)) + d

a is the amplitude
b is the frequency
c is the horizontal displacement
d is the vertical displacement.

the high point of the wave was 17 and the low point of the wave was 2.5.

the difference was 17 minus 2.5 = 14.5

the amplitude is half of that = 7.25

that made a = 7.25

without any vertical displacement, the graph would go from -7.25 to 7.25

since you wanted the high point to be 17 and the low point to be 2.5, you needed to add 17 - 7.25 = 9.75 to the equation as vertical displacement.

the eqution became y = 7.25 * sin(b * (x - c)) + 9.75

there are 24 hours in a day.

since you needed to have two high points and two low point, then you need to have two full cycles of the sine wave in a 24 hour period.

that means one full cycle in a 12 hour period.

that';s your period.

frequency = 360 / period becomes frequency = 360 / 12 = 30.

that's your frequency.

the equation becomes y = 7.25 * sin(30 * (x - c)) + 9.75

without any horizontal displacement, the valur of y when x = 0 would be 9.75. and y = 17 at x = 3.

you 2want it to be at y = 17 when x = 10.

you need to shift the grsph 7 units to the right.

tht makes c = 7

the equation becomes y = 7.25 * sin(30 * (x - 7)) + 9.75

the horizontal shift was successful and now you have y = 17 at x = 10 and x = 22 (22 - 12 = 10 pm).

you also have y = 2.5 at x = 4 and x = 16 (16 - 12 = 10 pm)

i was using military time.
the am hours are as they are.
the pm hours are equal to what they are plus 12.
4 am = 4
10 am = 10
4 pm = 4 + 12 = 16
10 pm = 10 + 12 = 22

to find the height of the wave at 11:00 am, replace x with 11 to get:

y = 7.25 * sin(30 * (x = 7)) + 9.75 becomes:
y = 7.25 * sin(30 * (11 - 7)) + 9.75.

this makes y = 16.02868418 which rounds to y = 16.0.