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

[Veronica_Oles]

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.

 

[Mark44]

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.

 

[haruspex]

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))?

 

[Veronica_Oles]

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?

 

[haruspex]

I'm not quite sure what to do?

What value of x between 0 and 360 degrees maximises sin(x)?