The discussion revolves around modifying the model d = 12 sin (30(t-5)) + 14 to fit new data regarding water depth, specifically a maximum of 22 m and a minimum of 6 m, with the first high tide occurring at 5:00 am.
Participants are actively questioning the adjustments made to the model and exploring the implications of these changes. There is a focus on understanding the reasoning behind the phase shift and the conditions for maximizing the sine function, indicating a productive exploration of the problem without reaching a consensus.
There is an assumption that t is measured in hours, and the sine function's argument is in degrees. Participants are also considering the constraints of the problem, such as the specified range for t and the need to align the model with the observed data.
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?