SOLUTION: The water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 am. High tide occurs at 3:30 pm. What is a cosine function that models the vari

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Question 1194135: The water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 am. High tide occurs at 3:30 pm. What is a cosine function that models the variation in inches above and below the water level as a function of time and hours since 9:15 am?
Please tell me what should I put as answer for the a cosine function. From chegg I got the answer as f(x) = 20 (x -0.99 ) + 32. Is that correct?
Found 4 solutions by Alan3354, Theo, greenestamps, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!

The function will be he water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 am. High tide occurs at 3:30 pm. What is a cosine function that models the variation in inches above and below the water level as a function of time and hours since 9:15 am?
Please tell me what should I put as answer for the a cosine function. From chegg I got the answer as f(x) = 20 (x -0.99 ) + 32. Is that correct?
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It's obviously not correct as it's not a cosine function.
----
The peak-to-peak is 52-12 = 40 inches --> amplitude = 20 inches
The average level is (12+52)/2 = 32
The time at 32" is (1530 + 0915)/2 = 1222.5 (12:22.5) PM
The period = 2*(1530-0915) = 12.5 hours
=====================
The function will be

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
the answer from chegg is not correct.
the formula that i came up with is y = 20 * cos(28.8 * (x-3)) + 32.
the graph of that formula is shown below.

you have low tide of 12 inches at x = 9.25 and high tide of 52 inches at x = 15.5
x = 9.25 is equal to 9 hours and 15 minutes from midnight which makes it equal to 9:30 am on a 12 hour clock and 0930 on a 24 hour clock.
x = 15.5 is equal to 15 hours and 30 minutes from midnight which makes it equal to 1530 on a 24 hour clock which makes it equal to 3:30 pm on a 12 hour clock.
let me know if you have any questions.
theo

addition information to answer your questions are below.

i wanted the period to be 6.25 hours.
since this was half the cycle of the cosine function, i used the formula of frequency = 180 / 6.25 rather than frequency = 360/6.25
180/6.25 gave me a frequency of 28.8 which is what i used.

the shift of 3 units was because the cosine function, without a shift, gave me a value of y = 12 when x was 6.25.
i wanted a value of 12 when x was 9.25.
the difference was 3 units of x.
that's what i used (x-3) rather than x.

here is the graph of both functions so you can see why the shift was 3 units.



i did use the graphing software to guide me.
without that, it may not have been that straight forward.
my answer is not the standard answer, but i thought it made sense and so i gave it to you.

let me know if you have any further questions.
theo

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The starting point for the function (t=0) is at low tide. Since the maximum of the basic cosine function is at t=0, we can use a negative cosine function with no horizontal shift.

High tide is 52 inches and low tide is 12 inches. The difference is 40 inches, so the amplitude is 20; the midline is 32. So the function is of the form



The only missing part is the coefficient b, which is (2pi) divided by the period.

Half the period is between low tide and high tide, which is 6.25 hours; so the period is 12.5 hours. The desired function is then



A graph, showing low tide of 12 inches at t=0 (9:15am) and high tide of 52 inches at t=6.25 (3:30pm) -- and the next low tide 12.5 hours after the first one (9:45pm):



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Commenting on tutor @ikleyn's comments about the other responses, note that the response from alan3354 is in fact correct, because he used a horizontal shift of pi radians, making the positive coefficient on the cosine function correct.

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To the student, who asked a completely different question in a "thank you" note regarding this problem....

We generally can't answer questions that you ask in that way. You need to post your questions in the normal way.

However, there is a quick answer to the question you asked this time.

The angle t goes from 0 to 2pi; but the angle in the problem is 2t, so 2t goes from 0 to 4pi. The two answers you didn't find are each simply 2pi more than the answers you DID find, because the range is 0 to 4pi instead of 0 to 2pi.
Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

The info below is for your education


                Tides and Water Levels

            Frequency of Tides - The Lunar Day



Most coastal areas, with some exceptions, experience two high tides and two low tides every day.
Almost everyone is familiar with the concept of a 24-hour solar day, which is the time that it takes 
for a specific site on the Earth to rotate from an exact point under the sun to the same point under the sun.


Similarly, a lunar day (also known as a "tidal day") is the time it takes for a specific site on the Earth 
to rotate from an exact point under the moon to the same point under the moon. Unlike a solar day, 
however, a lunar day is 24 hours and 50 minutes. The lunar day is 50 minutes longer than a solar day 
because the moon revolves around the Earth in the same direction that the Earth rotates around its axis. 


So, it takes the Earth an extra 50 minutes to “catch up” to the moon.
Because the Earth rotates through two tidal “bulges” every lunar day, coastal areas experience 
two high and two low tides every 24 hours and 50 minutes. High tides occur 12 hours and 25 minutes apart. 
It takes six hours and 12.5 minutes for the water at the shore to go from high to low, or from low to high.


See this site with nice illustration

https://oceanservice.noaa.gov/education/tutorial_tides/tides05_lunarday.html


///////////////


Regarding the three solutions you obtained so far, the solution by @greenestamps is correct. You may use it.

The solution by @Theo is incorrect due to different reasons; in particular,
he counts the time starting from midnight, while the problem asks to count time starting from 9:15 am.
The period of the cosine function is also incorrect in the Theo' solution.

The solution by Alan has wrong sign at cosine function. 

  //  O-o-o-p-s ! Sorry. The solution by @Alan is correct.

      It was my mistake to say that it is wrong . . . Sorry for that.

      I simply ran too fast by my eyes.

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If you want to see many other similar solved problems,  look into the lesson
    - Word problems on Trigonometric functions
in this site,  and learn the subject from there.



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