Applications of sinusoidal functions.

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Homework Help Overview

The discussion revolves around a sinusoidal function representing the depth of water at a seaport over time. The formula provided is h = 1.8 sin 2pi [(t - 4)/12.4] + 3.1, and participants are exploring questions related to the maximum depth of the water and the times it occurs.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to determine the maximum depth of water and the times it occurs based on the sinusoidal function. There are questions about the origin of specific values used in the equations, particularly the value 4.9, and the correctness of the initial setup.

Discussion Status

Some participants have provided insights into the maximum value of the sinusoidal function and its implications for the problem. There is ongoing clarification regarding the formulation of the equations and the interpretation of the results, with no explicit consensus reached yet.

Contextual Notes

Participants are discussing the implications of the sinusoidal function's parameters and the assumptions made in the problem statement. There are indications of potential missing information or misinterpretations that could affect the understanding of the problem.

anonymous12
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Homework Statement


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?

Homework Equations


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

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:
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anonymous12 said:

Homework Statement


At a seaport, the depth of the water h metres at time t hours during a certain day is given by this formula:




Homework Equations


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



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
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.
 
Oops. Here's the question:


Q: What is the maximum depth of the water? When does it occur?
 
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?
 
Mark44 said:
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?

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)
 
anonymous12 said:
Well the original equation is : h = 1.8 sin 2pi [(t - 4)/12.4] + 3.1
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
anonymous12 said:
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
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.
anonymous12 said:
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)
4.9 = 1.8sin 2pi [(t-4)/12.4] + 3.1
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
 

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