Adjusting the Model: d = 12 sin (30(t-5)) + 14

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Veronica_Oles
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Homework Statement


Modify the model d = 12 sin (30(t-5)) + 14 to match the new data which is as follows; maximum water depth is 22 m minimum is 6 m, and the first high tide occurs at 5:00am.

Homework Equations

The Attempt at a Solution



The answer is y= 8 sin (30(t-2)) + 14

Ik it's 8 b/c (22-6) / 2 = 8 but the (t-2) not sure where it comes from.
 

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Veronica_Oles said:

Homework Statement


Modify the model d = 12 sin (30(t-5)) + 14 to match the new data which is as follows; maximum water depth is 22 m minimum is 6 m, and the first high tide occurs at 5:00am.

Homework Equations

The Attempt at a Solution



The answer is y= 8 sin (30(t-2)) + 14

Ik it's 8 b/c (22-6) / 2 = 8 but the (t-2) not sure where it comes from.
All that's going on here is to align the first model (d = 12 sin(30(t - 5)) + 14) so that a high point on the graph comes at 5am.
 
Veronica_Oles said:

Homework Statement


Modify the model d = 12 sin (30(t-5)) + 14 to match the new data which is as follows; maximum water depth is 22 m minimum is 6 m, and the first high tide occurs at 5:00am.

Homework Equations

The Attempt at a Solution



The answer is y= 8 sin (30(t-2)) + 14

Ik it's 8 b/c (22-6) / 2 = 8 but the (t-2) not sure where it comes from.
I assume t is measured in hours, and the 30(t-5) is in degrees.
What value of t, between 0 and 12, maximises sin(30(t-5))?
 
haruspex said:
I assume t is measured in hours, and the 30(t-5) is in degrees.
What value of t, between 0 and 12, maximises sin(30(t-5))?
I'm not quite sure what to do?