# Applications of sinusoidal functions.

1. Nov 1, 2011

### anonymous12

1. The problem statement, all variables and given/known data
At a seaport, the depth of the water h metres at time t hours during a certain day is given by this formula:

Q: What is the maximum depth of the water? When does it occur?

2. Relevant equations
h = 1.8 sin 2pi [(t - 4)/12.4] + 3.1

3. The attempt at a solution

4.9 = 1.8sin 2pi [(t-4)/12.4] + 3.1
1.8sin2pi = 0
4.9 - 3.1 = (t-4)/12.4
1.8 = (t-4)/12.4
1.8 x 12.4 = t - 4
22.32 + 4 = t
26.32 = t

That answer is wrong even when i convert from 24 hour clock to the 12 hour clock.
The correct answer is 7:06a.m and 7:30a.m

Last edited: Nov 1, 2011
2. Nov 1, 2011

### Staff: Mentor

If the correct answers are 7:06am and 7:30am, what is the question? There is nothing in your problem statement that asks a question.

3. Nov 1, 2011

### anonymous12

Oops. Here's the question:

Q: What is the maximum depth of the water? When does it occur?

4. Nov 1, 2011

### Staff: Mentor

Why does the "correct" answer not give the maximum depth?

And why is your first equation 4.9 = 1.8sin 2pi [(t-4)/12.4] + 3.1? Where did that 4.9 come from?

5. Nov 1, 2011

### anonymous12

Well the original equation is : h = 1.8 sin 2pi [(t - 4)/12.4] + 3.1
Since it's a sinusoidal function, the maximum height of that this sinusoidal function can achieve is 4.9. You get that by adding 3.1 + 1.8 = 4.9

And my first equation is 4.9 = 1.8sin 2pi [(t-4)/12.4] + 3.1 because I substituted the 4.9 as the y value since we want to find out what time the depth of the water is at its max (4.9)

6. Nov 2, 2011

### Staff: Mentor

I think if you'll check the book, you'll find that you are missing some parentheses. This should be 4.9 = 1.8sin (2pi (t-4)/12.4]) + 3.1
Then you should say something to establish this. The reason is that the maximum value of the sine function is 1, so the maximum value of 1.8*sin(whatever) + 3.1 is 4.9.