SOLUTION: The average depth of water at the end of a dock is 7 feet. This varies 3 feet in both directions with the tide. Suppose there is a high tide at 4a.m. if the tide goes from low to

Algebra ->  Trigonometry-basics -> SOLUTION: The average depth of water at the end of a dock is 7 feet. This varies 3 feet in both directions with the tide. Suppose there is a high tide at 4a.m. if the tide goes from low to       Log On


   

Question 1204543: The average depth of water at the end of a dock is 7 feet. This varies 3 feet in both directions with the tide. Suppose there is a high tide at 4a.m. if the tide goes from low to high every 6hrs.
A.) Write a cosine function d(t) describing the depth of the water as a function of time with t = 4 corresponding to 4a.m.
B.) At what two times within one cycle is the tide at a depth of 8.5 feet ?
Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.
The average depth of water at the end of a dock is 7 feet.
This varies 3 feet in both directions with the tide.
Suppose there is a high tide at 4a.m. if the tide goes from low to high every 6hrs.
(A) Write a cosine function d(t) describing the depth of the water as a function of time
with t = 4 corresponding to 4a.m.
(B) At what two times within one cycle is the tide at a depth of 8.5 feet ?
~~~~~~~~~~~~~~~~~~~~~~ 

They talk about a harmonic function of the height with the average value of 7 feet; 
the maximum value of 7+3 = 10 ft at 4 am; and the period of 2*6 = 12 hours  (= one cycle).


    +------------------------------------------------------------+
    |   They ask you to write this function as a function of t,  |
    |                placing t= 4 at t= 4 am.                    |
    +------------------------------------------------------------+


From the description, it is clear that the function is cosine with the argument (t-4)
with no shift.  So we write

    h(t) = 7+%2B+3%2Acos%282%2Api%2A%28%28t-4%29%2F12%29%29 ft.


Part (A) is solved.


To solve part (B), write this equation as the problem requires

    h(7) = 8.5  ft,  or

    7+%2B+3%2Acos%282%2Api%2A%28%28t-4%29%2F12%29%29 = 8.5.


Simplify

    3%2Acos%282%2Api%2A%28%28t-4%29%2F12%29%29 = 8.5 - 7

    3%2Acos%282%2Api%2A%28%28t-4%29%2F12%29%29 =   1.5

    cos%282%2Api%2A%28%28t-4%29%2F12%29%29   = 1.5%2F3

    cos%282%2Api%2A%28%28t-4%29%2F12%29%29   = 1%2F2
    
    2%2Api%2A%28%28t-4%29%2F12%29 = pi%2F3  or  5pi%2F3.


Reduce the factor pi  in both sides to get

    2%2A%28t-4%29%2F12 = 1%2F3  or  5%2F3.


Multiply by 12 both sides

    2(t-4) =  4  or  20

    2t - 8 =  4  or  20

    2t     = 12  or  28

     t     =  6  or  14    (military time)


ANSWER.  Time then the depth is 8.5 ft  is  6 am  and/or  2 pm

         (two time moments within each 12-hours cycle).

The solution is complete - both questions are answered.