# Ocean Word Problem: Find Equation

• Veronica_Oles
In summary: Please try again later when you have more information.In summary, the water depth in a harbour is 21m at high tide and 11m at low tide. One cycle is completed approximately every 12 hrs. Find an equation for y = 5sin(30t) + 16 when t = high tide, low tide, and period (12 hrs). The graph of y = 5sin(30t) + 16 would be shifted 3 units to the right from y = 5sin(30(t - 3)) + 16 when t = low tide.

## Homework Statement

The water depth in a harbour is 21m at high tide and 11m at low tide. One cycle is completed approximately every 12 hrs.

Find an equation.

## The Attempt at a Solution

The answer to this problem is y = 5sin 30 (t-3) + 16

A = (M - m) / 2
= (21-11) / 2
= 5

C = (M+m) / 2
= (21+11)/2
=16

Period is 12 hrs.
K value would be 30 because 360 / 12.

But I don't know why it is (t-3) I'm not sure where that came from.

Veronica_Oles said:

## Homework Statement

The water depth in a harbour is 21m at high tide and 11m at low tide. One cycle is completed approximately every 12 hrs.

Find an equation.

## The Attempt at a Solution

The answer to this problem is y = 5sin 30 (t-3) + 16

A = (M - m) / 2
= (21-11) / 2
= 5

C = (M+m) / 2
= (21+11)/2
=16

Period is 12 hrs.
K value would be 30 because 360 / 12.

But I don't know why it is (t-3) I'm not sure where that came from.
Is there a little bit more to the problem statement? What times are high tide and low tide? You need to know those times in order to get the time part of your equation correct...

Veronica_Oles said:

## Homework Statement

The water depth in a harbour is 21m at high tide and 11m at low tide. One cycle is completed approximately every 12 hrs.

Find an equation.

## The Attempt at a Solution

The answer to this problem is y = 5sin 30 (t-3) + 16

A = (M - m) / 2
= (21-11) / 2
= 5

C = (M+m) / 2
= (21+11)/2
=16

Period is 12 hrs.
K value would be 30 because 360 / 12.

But I don't know why it is (t-3) I'm not sure where that came from.
How would the graph of ##y = 5\sin(30t) + 16## look in comparison to ##y = 5\sin(30(t - 3)) + 16##?

Mark44 said:
How would the graph of ##y = 5\sin(30t) + 16## look in comparison to ##y = 5\sin(30(t - 3)) + 16##?
Wouldn't the second graph be shifted 3 units to the right?

Veronica_Oles said:
Wouldn't the second graph be shifted 3 units to the right?
Yes. So the point (0, 16) on the first graph is shifted to the right by 3 (hours?) is now at (3, 16).

Veronica_Oles said:
Wouldn't the second graph be shifted 3 units to the right?
Why is it shifted? What in the original problem statement helps you to figure out when low tide is? You cannot answer this schoolwork problem without that information. And our help is limited by the accuracy of what you post about the Problem Statement.

EDIT -- Apologies, but this is a very straightforward problem if you post all of the question information.