The following data was collected recording the height of tides in Austraila (NSW Dept of Education).

As expected, the heights seem to be periodic. This suggests that a good model might be a sine or cosine function. Since there seems to be a maximum height at time \(0\), a cosine model is a good choice.

In the following DESMOS app, select choices for \(A\), \(B\), \(C\), and \(D\) to fit the model to the data.

The best model is approximately

However, it doesn’t do such a good job. It looks like every other high tide height is missed.

If we graph more of the data, we can see this better.

There appear to be two sinusoidal waves of different heights.

A better model might involve two cosine functions.

Our current model seems to be doing half the job. It appears we need to shorten every other wave. We can do this with another cosine function with twice the period.

When the model was a single cosine function, it appear that the period was approximately \(12.2\) hours, which gives a frequency of \(\frac {2 \pi }{12.2} = \frac {\pi }{6.1} = 0.515\), which was the coefficient inside the cosine function.

Our new model begins to look like:

We have three unknowns.

We need three data points.

We’ll select the first three peaks/troughs: \((0, 1.63)\), \((6.4, 0.64)\), and \((12.2, 1.36)\).

\(\blacktriangleright \) \(first(t) = \cos \left ( \frac {\pi \, t}{6.1} \right )\) is the faster piece.

\(\blacktriangleright \) \(second(t) = \cos \left ( \frac {\pi \, t}{12.2} \right )\) is the slower piece.

The second cosine function is slower function and has double the period as the first. They both start at maximums at \(t=0\). Both functions have a value of \(1\).

Then, the first cosine function arrive at its minimum of \(-1\) at \(t=6.4\), however, the second function is slower and is at a zero when \(t=6.4\).

At \(t=12.2\) hours the first cosine is at it maximum of \(1\) and the second function has arrived at its minimum of \(-1\).

• At \((0, 1.63)\) we have \(1.63 = A + B + C\)
• At\((6.4, 0.64)\) we have \(0.64 = -A + C\)
• At \((12.2, 1.36)\) we have \(1.36 = A - B + C\)

From these three equations, we can see that \(1.63 + 1.36 = 2 \, A + 2 \, C\) or \(1.495 = A + C\).

This and the second equation tell us that \(0.64 + 1.495 = 2 \, C\) or \(C = 1.068\).

We can put this in for \(C\), in the second equation: \(0.64 = -A + 1.068\) or \(A = 0.4275\).

And, from all of this we get \(B = 0.1345\).

Much better.