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# help

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On a summer day in Cape Cod, the depth of the water at a dock was 4ft at low tide at 2:00 AM. At high tide 5 hours later, the heist of the water at the dock rose to 14 feet. Write a cosine function to model the height of the water at the dock x hours after the day began at midnight?

Feb 10, 2020

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We will  have an equation of this form

Height  = A cos ( Bx  + C)  + D

A  =  the amplitude  =   [ 14 - 4 ] / 2  =  5

The period =  10 hrs....so   10  =  2pi / B   ⇒ B  = 2pi/10 =  pi/5

D  is the  midline   =  [ 14 + 4 ] /2  =  18 /2   = 9

C  is the phase shift  and is a little tough to figure

At 2AM,   x  = 2,  the height is  4 ft.....so we need to solve this

4  =  5 cos ( pi/5 * 2 +  C )  +  9     subtract  9 from both sides

-5  = 5 cos ( 2pi/5 + C)     divide both sides  by  5

-1  = cos  (2pi/5 + C)         we  need to take the  arccos

arccos (-1)  =  2pi/5 +C

pi  = 2pi/5  + C

C  =  pi - 2pi/5 = [  5pi  -2pi] / 5  =  3pi/5

So....the function is

Height =  5cos [ ( pi /5) x    + 3pi/5 )  +  9

Here's the graph  : https://www.desmos.com/calculator/igberusuwx   Feb 10, 2020
edited by CPhill  Feb 11, 2020
#2
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Cos

amplitude   (14 -4 ft) /2 = 5

5 cos

period  10 hours     2pi/10 = pi/5

5 cos (pi/5  x)

Shifted 7 hours right        -7 will shift it right

5  cos  (pi/5(x  - 7))

Midline is at 9   ( the graph is shifted up 9)

5 cos (pi/5 (x-7)) +9

Feb 11, 2020