The model d = 12 sin(30(t-5)) + 14 was modified to y = 8 sin(30(t-2)) + 14 to accurately reflect a maximum water depth of 22 m and a minimum of 6 m, with the first high tide occurring at 5:00 AM. The amplitude was calculated as 8, derived from (22-6)/2. The adjustment of (t-2) aligns the model's peak with the specified high tide time.
PREREQUISITESStudents in mathematics, particularly those studying trigonometry, as well as educators and anyone involved in modeling periodic phenomena such as tides or waves.
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.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'm not quite sure what to do?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))?
What value of x between 0 and 360 degrees maximises sin(x)?Veronica_Oles said:I'm not quite sure what to do?