About tides max and min height, trig problem

Question

Low tide in the ocean occurs at 8 am. The depth of the water at this time is 4 meters. High tide occurs at 4 pm and the depth of the water at this time is 38 meters. Assume that the 12:00 am (prior to the first low tide) is t=0. Sketch a sine/cosine wave that represents the height of the water as a function of time. Make a table of values for two full cycles. You do not need to record the height at every hour, but you should be recording the times for maximums and minimums and the axis of the wave. Write an equation for your graph.

Answer 1

The general form of a cosine equation is y =  Acos(ωt + Φ) + B where A is the amplitude, ω is the angular frequency and B is the vertical shift or baseline (the horizontal line that divides the wave up evenly).  We will find an equation for the wave beginning at 8am then shift that equation so that t = 0 corresponds to 12am.

 

1) Start by finding the time for one cycle or period.  For this problem we have  P = 2*(time for one half cycle) = 2*8 = 16 hours.

 

2) Find the angular frequency.  P = 1/f → f = 1/P, ω = 2πf = 2π/P = 2π/16 = π/8

 

3) Find the baseline.  B = (38 + 4)/2 = 21

 

4) Find the amplitude.  A = (38 - 4)/2 = 17

 

5) Write the equation and shift it if necessary.  Our intermediate equation is y = 17cos(πt/8) + 21.  Note that the time of a half cycle is 8 hours.  Eight hours earlier from the low tide time is 12:00 am so our equation needs no phase shift.  t is the number of hours after 12:00 am.  Make a table of values.

 

P = 16, I = 16/4 =4 (I is increment), displacement of wave is 0 (no phase shift)

 

t = 0 (12:00am) is start of wave (max y = 38), t1 = 0 + 4 = 4 (4:00am), (zero y = 0), t2 = 4 + 4 = 8 (8:00am), (min y = 4), t3 = 8 + 4 = 12 (12:00pm) (zero y = 0), t4 = 12 + 4 = 16 (4:00pm) (max y = 0).  Continue this pattern for another cycle.

 

Hannah I did the best I could.  I hope this helps.

 

In these types of problems we are trying to model the behavior of a real world phenomenon as a mathematical equation. We are trying to fit a curve to our given data. The ebb and flow of tides follow a sinusoidal pattern. We can use either a cosine or sine function as a model. I chose cosine because the tides ends up high as does the cosine wave (see figure below). The basic cosine function is y = cos(x). We are going to manipulate this equation to fit our data by vertical stretching, vertical shifting, horizontal shifting and horizontal expanding or compressing. The amplitude is the absolute value of the coefficient of cos(x) which is 1. Amplitude also called peak value is the highest positive value that the cosine reaches from the baseline. The baseline splits the wave up evenly horizontally. Graphically we have

S R
|
|* *
| * *
|-------------------*----------------------------------*----------------------- baseline
| * *
| *
Q

There are five key points to a cosine or sine wave. They are the positive and negative peaks and the zero crossings. For this wave they are (0,1), (pi/2, 0), (pi, -1) (3pi/2, 0) and (2pi, 1). Notice that these are evenly space along the x-axis which explains the increment, I = period/4. The displacement is how far the wave is shifted horizontally left or right of the origin. For the above wave it is 0. Notice that the base line is midpoint from positive peak to the negative peak which explains the formula baseline = (max + min)/2. For the above wave the baseline is at x = 0. Also notice that the amplitude is one half the distance between the positive peak and negative peak which give us the formula amplitude = (max-min)/2. For the above wave we have (1-(-1))/2 = 1. To relate the cosine function to time we use 2??t/P instead of x where t is time and P is the time for one cycle. So we have y =cos(2??t/P). The graph of this equation still looks the same as above.

In our problem low tide occurs 8:00 am which correspond to Q and our high value occurs at 4:00 pm corresponding to R. This is only one half of the wave. The time between 8:00 am to 4:00 pm is 8 hours so the period is 2*8 = 16 hours. So y = cos(??t/8). To find the time at which the first positive peak, S, occurs we go back 8 hours from 8:00 am which end up at 12:00 am. Remember these values are even spaced along the x-axis. So we see that we do not need to shift the wave left or right horizontally because 12:00 am ends up at the origin. So t = 0 corresponds to 12:00 am. Remember that t is the number of hours after 12:00 am. Since I = 4 we can add 4 to our starting t value to get the first zero crossing giving t = 4. Another way to look at this is to notice that the time the first zero occurs is half way between S and Q so t = (0+8)/2 = 4.

Now we must change the amplitude of y = cos(??t/8) to match our amplitude = (38-4)/2=17. Multiply by 17, this is vertical stretching. y = 17cos(??t/8). The graph of this equation still has the same shape as above except the amplitude is now 17. But we have the problem that our wave’s maximum is 17 instead of 38 and our minimum is -17 instead of 4 so we must vertical shift the wave up so our minimum and maximum peak matches our problem. We can do this by adding (38+4)/2 = 21. This raises the baseline from 0 to 21 so y = 17cos(??t/8) + 21. Again the graph of this equation looks like the above graph except that it is moved up by 21 units.