# sketching tide functions

• Oct 28th 2006, 10:44 PM
needmathshelp
sketching tide functions
Given a set date from teh calender and a starting time
Date: 21/09/06
Time: 5am

How do you equate the tides for a 24-hour period after this time and date to an equation of the form y=a cos b(x-c)+d. I know the answer should include a sketch of the function, where teh x-axis represents time in hours (0<t<24) and I am given the starting time t=0

The low tides are:
Time Height(m)
0233 0.67
0840 3.54
1439 0.71
2045 3.96

So the high tide is 3.75 m and the low tide is 0.69m

If you can help that'd be great thanx
• Oct 28th 2006, 11:46 PM
CaptainBlack
Quote:

Originally Posted by needmathshelp
Given a set date from teh calender and a starting time
Date: 21/09/06
Time: 5am

How do you equate the tides for a 24-hour period after this time and date to an equation of the form y=a cos b(x-c)+d. I know the answer should include a sketch of the function, where teh x-axis represents time in hours (0<t<24) and I am given the starting time t=0

The low tides are:
Time Height(m)
0233 0.67
0840 3.54
1439 0.71
2045 3.96

So the high tide is 3.75 m and the low tide is 0.69m

If you can help that'd be great thanx

1. Convert the times to decimal hours from 00:00.

2. The maximum of your model of tide height y=a cos b(x-c)+d is
a+d, and the mimium is -a+d as the maximum and minimum of cos
are +1 and -1. Set these equal to the average high and low tides
respectivly so:

a+d = (3.52+3.96)/2 = 3.75
-a+d=(0.67+0.71)/2 = 0.69.

Use these to solve for a and d.

3 The interval t between the maximum and minimum of the model is:

t = 2 pi/b,

so:

b = 2 pi / t,

so now compute the mean time between the high and low and low and
high tides and use that to solve for b.

4. The first maximum of the model occurs at 08:40, and this corresponds
to x-c=0, (where x is the time in decimal hours corresponding to 08:40),
so c=x.

RonL