# Modeling WIth Sinusoidial Functions

1. Nov 3, 2016

### Veronica_Oles

1. The problem statement, all variables and given/known data
The water depth in a harbor is 21m at high tide and 11m at low tide. Once cycle is completed every 12 hrs.
(a) Find equation for the depth as a function of time.

(b) Draw a graph for 48 hrs after low tide, which occurred at 14:00.

(c) State the times where the water depth is
(i) max
(ii) min
(iii) average value

(d) estimate the depth of the water at
(i) 17:00
(ii) 21:00

(e)
(i) 14m
(ii) 20m
(iii) at least 18 m

2. Relevant equations

3. The attempt at a solution

This question is really confusing me. I graphed it so that my y axis is (0,11) and went from there. Found the equation to be y= -5cos((π/6)x)=16 and I also got y = 5sin(π/6(x-3))+16, I check the solutions and they were correct. But now when I went to find the max, min, avg values I kept getting them wrong.
For max solutions said 08:00 and 20:00 I got 6hrs and 18hrs for max and I don't understand how this correlates?? Same problem occurred for the min and avg. As for part (d) I put 17 into the equation and got 20.33 and then I put 21 into the equation and got 16 could someone verify if this is correct? I'm still working on (e).

2. Nov 4, 2016

### cnh1995

Did you sketch the graph of depth as a function of time? At what time did the low tide occur? What point on the graph does the low-tide event represent?
You can't put 17:00 and 21:00 in the equation. Can you say why?

3. Nov 4, 2016

### ehild

Your solution for y is not correct. At time 14, y should be minimum, that is, the phase has to be -pi/2; 3pi/2 ; ....

4. Nov 4, 2016

### cnh1995

I believe OP wrote a general equation w.r.t. time which satisfies the amplitude and period conditions. The time in the OP's equation is not specified in terms of 24-hr clock. It is known that a minima occurs at time 14:00, so this instant is used as a reference time t=0. With a slight modification, the equation can be written in terms of 24-hr clock. (But the solution provided says OP's equations are correct, so I'm not sure if it is required to modify the equation ).

Last edited: Nov 4, 2016
5. Nov 6, 2016

### Veronica_Oles

Because that is not the actual time that we go by in question. The x axis is in number of hours.

6. Nov 6, 2016

Yes.

7. Nov 7, 2016

### Veronica_Oles

Figured out the answer, did the graph correctly, my equation was also correct at y=-5cos(π/6 (X)) + 16 but I realized when u actually start counting the time the y axis is actually 2 pm.

8. Nov 7, 2016

### cnh1995

Well, you can modify the equation and get y=2 pm at the start.