Modeling WIth Sinusoidial Functions

[Veronica_Oles]

Homework Statement

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

Homework Equations

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The Attempt at a Solution

[/B]
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).

 

[cnh1995]

For max solutions said 08:00 and 20:00

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?

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?

You can't put 17:00 and 21:00 in the equation. Can you say why?

 

[ehild]

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.

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 ; ...

 

[cnh1995]

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 ; ...

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 ).

 

[Veronica_Oles]

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?
I graphed it.
You can't put 17:00 and 21:00 in the equation. Can you say why?

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

 

[cnh1995]

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

Yes.

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 ).

 

[Veronica_Oles]

Yes.

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.

 

[cnh1995]

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.

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