On a typical day at an ocean port, the water has a maximum depth of 20 m at 8:00 a.m. The minimum depth of 8 m occurs 6.2 hours later. Assume that the relation between the depth of the water and time is a sinusoidal function.
Let's draw a graph!
a) What is the period of the function?
From the information we have been given...
* We can set the 8am as t = 0 hours.
* A maximum value is when t = 0 hours and when d = 20.
* A minimum value is when t = 6.2 and d = 8.
We can see from the graph that the period is 12.4 hours.
b) Write an equation for the depth of the water at any time, t hours.
cosine equation's parameters...
A = 6
B = (2pi) / 12.4 = pi/6.2
C = 0
D = 14
To get A, amplitude, calculate the distance from the sinusoidal axis to a maximum value or minimum value.
sinusoidal axis = (20+8)/2 = 14
amplitude = 14-8 = 6
B = pi/period = pi/6.2
C, the phase shift, is 0.
D is the sinusoidal axis, 14.
D(t) = 6cos [(pi/6.2)t] + 14
c) Determine the depth of the water at 10:00 a.m.
10:00 am = 2 hrs from when t = 0 or 8:00 am. Plug in the 2 as t into the equation to get the answer.
D(2) = 6 cos [(pi/6.4)2] + 14 = 17.1738 metres
d) Determine one time when the water is 10 m deep.
The wave is 10 metres deep, so the qestion is asking for what the time is when D = 10. Plug in 10 as D, then solve for t.
10 = 6cos[(pi/6.2)t] +14
-4 = 6cos[(pi/6.2)t]
-4/6 = cos[(pi/6.2)t]
arc cos(-4/6) = (pi/6.2)t
2.3005 = (pi/6.2)t
2.3005/(pi/6.2) = t
2.3005 x 6.2/pi = t
14.2632/pi = t
t = 4.5401
Convert the 4.5401 into "actual time" because we use hours:minutes:seconds to show time, so...
4.5401 hrs + 8am = 12.5401 hrs
Obviously, its not efficient to say .5401 hrs so we convert that to minutes.
0.5401 x 60 = 32.406 min.
12:32:24pm
We can round that to 12:30pm.
