1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Modeling WIth Sinusoidial Functions

  1. Nov 3, 2016 #1
    1. The problem statement, all variables and given/known data
    The water depth in a harbor is 21m at high tide and 11m at low tide. Once cycle is completed every 12 hrs.
    (a) Find equation for the depth as a function of time.

    (b) Draw a graph for 48 hrs after low tide, which occurred at 14:00.

    (c) State the times where the water depth is
    (i) max
    (ii) min
    (iii) average value

    (d) estimate the depth of the water at
    (i) 17:00
    (ii) 21:00

    (e)
    (i) 14m
    (ii) 20m
    (iii) at least 18 m

    2. Relevant equations


    3. The attempt at a solution

    This question is really confusing me. I graphed it so that my y axis is (0,11) and went from there. Found the equation to be y= -5cos((π/6)x)=16 and I also got y = 5sin(π/6(x-3))+16, I check the solutions and they were correct. But now when I went to find the max, min, avg values I kept getting them wrong.
    For max solutions said 08:00 and 20:00 I got 6hrs and 18hrs for max and I don't understand how this correlates?? Same problem occurred for the min and avg. As for part (d) I put 17 into the equation and got 20.33 and then I put 21 into the equation and got 16 could someone verify if this is correct? I'm still working on (e).
     
  2. jcsd
  3. Nov 4, 2016 #2

    cnh1995

    User Avatar
    Homework Helper

    Did you sketch the graph of depth as a function of time? At what time did the low tide occur? What point on the graph does the low-tide event represent?
    You can't put 17:00 and 21:00 in the equation. Can you say why?
     
  4. Nov 4, 2016 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Your solution for y is not correct. At time 14, y should be minimum, that is, the phase has to be -pi/2; 3pi/2 ; ....
     
  5. Nov 4, 2016 #4

    cnh1995

    User Avatar
    Homework Helper

    I believe OP wrote a general equation w.r.t. time which satisfies the amplitude and period conditions. The time in the OP's equation is not specified in terms of 24-hr clock. It is known that a minima occurs at time 14:00, so this instant is used as a reference time t=0. With a slight modification, the equation can be written in terms of 24-hr clock. (But the solution provided says OP's equations are correct, so I'm not sure if it is required to modify the equation ).
     
    Last edited: Nov 4, 2016
  6. Nov 6, 2016 #5
    Because that is not the actual time that we go by in question. The x axis is in number of hours.
     
  7. Nov 6, 2016 #6

    cnh1995

    User Avatar
    Homework Helper

    Yes.
     
  8. Nov 7, 2016 #7
    Figured out the answer, did the graph correctly, my equation was also correct at y=-5cos(π/6 (X)) + 16 but I realized when u actually start counting the time the y axis is actually 2 pm.
     
  9. Nov 7, 2016 #8

    cnh1995

    User Avatar
    Homework Helper

    Well, you can modify the equation and get y=2 pm at the start.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted