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Group 4: Seaport

GROUP 4: Linh, Haiyan, Roselle, Remyshire, and Oliver.

Seaport !!

At a Seaport, the depth of the water, (h) meters, at time (t) hours. During a certain day is given by the formula :

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

a) State the (i) period (ii) amplitude (iii) Phase shift

(i) Alright for the period we know that the "B" is related to the period and is NOT the period itself. In this formula the value for "B" is 12.4, therefore the period is " 2**π** /12.4 "

(ii) the amplitude is 1.8 indicated by the Parameter "A".

(iii) The phase shift is 4.00, it may look tricky because the 12.4 is underneath the " t-4.00 " rather than what we usually see and don't let it fool you.

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

The maximum depth of the water is 4.9 because we know that the sinusoidal axis is on 3.1 and the amplitude is 1.8. We know from this that these two values are closely related because from the sinusoidal axis the amplitude is the exact distance away from the maximum , and the minimum. The time it occurs is at 3.1 hours which is on the graph.c) Determine the depth of the water at 5:00 a.m. and at 12.00 noon.

You just need to substitute in for the variable "t" in your equation. To find out the time:

h(5.00) = 1.8sin[2π/12.4(5.00-4)]+3.1

h(5.00) = 3.973543533

The depth of the Water at 5:00 a.m. will be at around 3.9735 Meters

For the height at 12.00 p.m. :

h(12.00) = 1.8sin[2**π**/12.4(12.00-4.00)]+3.1

h(12.00) = 1.676603674

The depth of the Water at 12:00 p.m. will be at around 1.6766 Meters.

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

Alright for this question you will need to do a couple of steps.

t(h) = 1.8sin[2*π/12.4(t-4.00)]+3.1 *

t(2.25) = 1.8sin[2π/12.4(t-4.00)]+3.1

firstly we let [2

*π/12.4(t-4.00)] equal to *θ because it looks like a beast.

now your equation looks like : t(2.25) = 1.8sinθ+3.1

now we solve.

2.25 =

1.8sinθ+3.1

= -0.85 = 1.8sin

θ

= -0.472222222 = sinθ

= θ = -0.4918100943

since we only have to find ONE time at 2.25 meters, we'll use this number " -0.4918100943 "

now, here's the fun part, it's where we release that beast.

θ = -0.4918100943

= [2π/12.4(t-4.00)] =

-0.4918100943

= t-4.00 = -0.9705976939

= t = 3.029402396

The time in hours when the water is 2.25 meters deep is at 3.0294 hours.

## 2 comments:

i am told to find all the possible values, when the water is 2.25 m deep how do i do that?

@Anonymous That is explained in part d) above. It shows one time the water is 2.25 m deep. Can you find another? ;-)

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