Question

The water level at a local pier rises and falls with the tide. Yesterday, the maximum depth of the water

Answer 1

Answer:

The function  is  $D = 2sin ( \frac{\pi}{2} - \frac{\pi}{12} t) + 6$

Step-by-step explanation:

From the question we are told that

    The  maximum depth is $d = 8 \ ft$

      The  minimum depth is  $d_i = 4 \ ft$

Generally the average depth is mathematically represented as

       $d_a = \frac{8 + 4}{2}$

=>   $d_a = 6 \ ft$

Generally  the amplitude is mathematically represented as

      $A = d - d_a$

=>    $A = 8 - 6$

=>    $A = 2$

Generally the period is  24 hours given that the the interval between the maximum depth and the minimum depth is half a day

Generally the period is mathematically represented as

         $T = \frac{2 \pi }{w}$

here w is the angular  frequency

So

     $w = \frac{2 \pi}{24}$

      $w = \frac{\pi}{12}$

Generally the depth can be modeled with a sin function as follows

     $D = Acos (wt) + d_a$

Now  from co-function identity we have  that $for \ cos (z) = sin (\frac{\pi}{2} - z)$

So  

     $D = Asin ( \frac{\pi}{2} - wt) + d_a$

     $D = 2sin ( \frac{\pi}{2} - \frac{\pi}{12} t) + 6$