Group 6: Sea Port Function

At a sea port, the depth of the water, h meter, at time, t hours, during a certain day is given by this formula.

h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1

(a)State the: (i)period=12.4 (ii)amplitude=1.8 (iii)phase shift=4.00

(b)What is the maximum depth of the water?
3.1 + 1.8 = 4.9
The maximum depth of the water is 4.9m.

When does it occur?

h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
4.9=1.8sin[2π(t-4.00)/12.4]+ 3.1

Let θ = 2π(t-4.00)/12.4

4.9=1.8sin θ + 3.1
1.8=1.8sin θ
1 = sin θ
1.5708 = θ

(12.4/2π)2π(t-4.00)/12.4 = 1.5708 (12.4/2π)
t-4.00 = 3.1
t=7.1
The maximum depth of the water occurs at 7:06am.

(c)Determine the depth of the water at 5:00am and at 12:00noon.
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1

For 5:00am
h(5)=1.8sin[2π(5-4.00)/12.4]+ 3.1
= 3.9735
The depth of the water at 5:00am is 3.9735m.

For 12:00noon.
h(12)=1.8sin[2π(12-4.00)/12.4]+ 3.1
=1.6766
The depth of the water at 12:00noon is 1.6766m.

(d)Determine one time where the water is 2.25 meters deep.
h(t)=1.8sin[2π(t-4.00)/12.4]+ 3.1
2.25=1.8sin[2π(t-4.00)/12.4]+ 3.1

Let θ = 2π(t-4.00)/12.4

2.25=1.8sin θ + 3.1
-.85=1.8sin θ
-.4722 = sin θ
-.4918 = θ

(12.4/2π)2π(t-4.00)/12.4 = -.4918 (12.4/2π)
t-4.00=-.9706
t=3.0294
The water is 2.25 meters deep at around 3:00am.


Group 6 Members: Luis, Joe, Sharmaine, Precious, Ivanna, Roslyn
Correct me guys if I'm wrong.