The Clear Blue Sea Port

This is Miller, representing group 5. Why'd I have to lose to Jordan at rock paper scissors. These are the answers we got.

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

1) State the a) Period b) Amplitude c) Phase Shift

a)Period
=12.4

b)Amplitude
=1.8

c) Phase Shift
=4.00

* I believe we went over this in class.

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

1.8+3.1= 4.9m = maximum depth

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

To solve these equations we let T= 5 and 12. Then we used our handy dandy formula which was kindly given to us.

T=5
h(t) = 1.8sin[2π(t–4.00)/12.4]+3.1
h(t) = 1.8sin[2π(5–4.00)/12.4]+3.1
h(t) = 1.8sin[2π(1)/12.4]+3.1
h(t) = 3.9735 meters

T=12
h(t) = 1.8sin[2π(t–4.00)/12.4]+3.1
h(t) = 1.8sin[2π(12–4.00)/12.4]+3.1
h(t) = 1.8sin[2π(8)/12.4]+3.1
h(t) = 1.6766 meters

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

For this equation we replaced h(t) with 2.25. Since thats what were looking for. Then solved the equation as such.

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

To make life easier we let [2π(t–4.00)/12.4] equal to θ.

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

Now the we have θ we let [2π(t–4.00)/12.4] = -0.4918

2π(t–4.00)/12.4 = -0.4918

(2π/12.4)t-4= -0.4918

(12.4/2π)(2π/12.4)t-4=-0.4918(12.4/2π)

t= -0.4918(12.4/2π)4
t=-0.9705+4
t=3.0294


WOOT! DONE... Hopefully I'm right please comment on any mistakes please.. Thanks