###
Group 3: Sea Port Question

Group Members: Alvin, Jasmin, Jessica, Joseph, and MichelleI 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] + 3.1Let 2π (t - 4.00)/12.4 be θ2.25 = 1.8sinθ + 3.1-0.85 = 1.8sinθ-0.4722 = sinθ-0.4918 = θ2π (t - 4.00)/12.4 = -0.49182π (t - 4.00) = -6.0984t - 4.00 = -0.9706t = 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.
## No comments:

Post a Comment