Group 3: Sea Port Question

Group Members: Alvin, Jasmin, Jessica, Joseph, and Michelle

I didn't really communicate with most of you people. Michelle and I talked about the questions and agreed on these answers and Joseph looked over it to make sure it was all good.


a) State the: (i) Period (ii)Amplitude (iii) Phase shift

(i) Period: 12.4
(ii) Amplitude: 1.8
(iii) Phase shift: 4.00


b) What is the maximum depth of the water? When does it occur?

Average: 3.1
Amplitude: 1.8

3.1 + 1.8 = 4.9
The maximum depth is 4.9 meter.

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

Let 2π (t - 4.00)/12.4 be θ

4.9 = 1.8sinθ + 3.1
1.8 = 1.8sinθ
1 = sinθ

1.5708 = θ
2π (t - 4.00)/12.4 = 1.5708
2π (t - 4.00) = 19.4779
t - 4.00 = 3.1
t = 7.1
The water's maximum depth will occur in 7.1 hours.


c) Determine the depth of the water at 5:00 am and at 12:00 noon.

5:00am = 5 hours
h(5) = 1.8sin [2π (5 - 4.00)/12.4] + 3.1
h(5) = 1.8sin [2π ( 1 )/12.4] + 3.1
h(5) = 1.8sin [2π /12.4] + 3.1
h(5) = 3.9735
The water's depth will be approximately 3.9 meters at 5:00 am.

12:00pm = 12 hours
h(12) = 1.8sin [2π (12 - 4.00)/12.4] + 3.1
h(12) = 1.8sin [2π ( 8 )/12.4] + 3.1
h(12) = 1.8sin [16π /12.4] + 3.1
h(12) = 1.6766
The water's depth will be approximately 1.7 meters at 12:00 pm.


d) Determine one time when the water is 2.25 meters deep.

2.25 = 1.8sin [2π (t - 4.00)/12.4] + 3.1

Let 2π (t - 4.00)/12.4 be θ

2.25 = 1.8sinθ + 3.1
-0.85 = 1.8sinθ
-0.4722 = sinθ
-0.4918 = θ

2π (t - 4.00)/12.4 = -0.4918
2π (t - 4.00) = -6.0984
t - 4.00 = -0.9706
t = 3.0294
One time the water will be 2.25 meters is at approximately 3:00 am.

If I did anything wrong or if you have different answers, leave a comment and we shall have a discussion about it.