Group 2 Seaport

At a seaport the depth of the water, h meters, 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 (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?

3.1 (Vertical shift) + 1.8 (Amplitude) = 4.9

When does it occur?

Bryan's Work:

1.8sin[(2π)((t-4.00)/(12.4))]+3.1
let θ = (2π)((t-4.00)/(12.4))
4.9 = 1.8sinθ +3.1
1.8 = 1.8sinθ
1 = sinθ
θ = 1.5708

1.5708 = (2π)((t-4.00)/(12.4))
0.25 = (t-4.00)/(12.4)
3.10 = t - 4.00
7.1s = t

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

Heh...guess you just had to plug in the numbers, now i feel really stupid xD Oh yeah thanks to Jasmin for helping with this part.

h(5) = 1.8sin [2π(5-4.00)/12.4]+3.1
h(5) = 1.8sin [2π/12.4] + 3.1
*punches into calculator*
h(5) = 3.9735 m

h(12) = 1.8sin [2π(12-4.00)/12.4]+3.1
h(12) = 1.8sin [16π/12.4] + 3.1
*punches into calculator again*
h(12) = 1.6766 m

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

Bryan's Work Again:

1.8sin[(2π)((t-4.00)/(12.4))]+3.1
let θ = (2π)((t-4.00)/(12.4))
2.25 = 1.8sinθ +3.1
-0.8500 = 1.8sinθ
-0.4722 = sinθ
θ = 0.4916

-0.4916 = (2π)((t-4.00)/(12.4))
-0.0782 = = (t-4.00)/(12.4)
-0.9701 = t - 4
3.0299s = t

Done~

Group Members: Alanna, Bryan, Mary Ann, Wendy, Paulo