2 Oscillating functions: modelling tides
We consider how the function
$\phantom{\rule{2em}{0ex}}h=3.2sin\left(2.7t+8.5\right)$
might be used to model the rise and fall of the tide in a harbour.
Figure 9 shows a graph of this function for $\left(0\ge t\ge 5\right)$ .
Figure 9
We consider some aspects of this graph and model. It seems reasonable to suppose that the tide creates an oscillation of the water level in the harbour of $h$ m about some mean value represented on the graph by $h=0$ . There seems to be a low tide near $t=1$ and another low tide just after $t=3$ . Since we expect intervals of 12 to 14 hours between low tides around the U.K., this suggests that time in this graph is specified in 6hour intervals.
Task!
Write down the amplitude, period and frequency of $h=3.2sin\left(2.7t+8.5\right)$
The amplitude of the change in water level in the harbour is 3.2 m. The period of the function is given by $2\pi \u22152.7=2.3271$ between successive high tides or successive low tides. This corresponds to $2.3271\times 6$ hours $=13.96$ hours between high tides.The frequency of the function is $2.7\u22152\pi =0.4297$ .
The peak levels of the graph correspond to times when the sine function has the value 1. The lowest points correspond to times when the sine function is $1$ . At these times the arguments of the sine function (i.e. $2.7t+8.5$ ) are an odd number of $\pi \u22152$ starting at $3\pi \u22152$ for the first low tide.
So far all of this may be deduced from the general form $y=a$ sin $\left(bx\right)$ and from the modelling context. However there is an additional term in the function being considered here. This is a constant 8.5 within the sine function. When $t=0$ the presence of this constant means that the intercept on the height axis is 3.2sin $\left(8.5\right)$ = 2.56, implying that the water level is 2.56 m above the mean value at the start of timing. The constant 8.5 has displaced the sine curve sideways. This constant is known as the phase of the function. Phase is measured in radians as it is an angle.
As remarked earlier, at $t=0$ , this function has the value $3.2sin\left(8.5\right)$ . Since $sin\left(8.5\right)=sin\left(8.52\pi \right)\approx sin\left(2.2168\right)$ , we can replace the constant 8.5 by 2.2168 without altering the values on the graph. This means that the function
$\phantom{\rule{2em}{0ex}}h=3.2sin\left(2.74t+2.2168\right)$
does just as well as the original function in representing the tidal variation in the harbour. We now rewrite this latest form of the function, representing the variation of water level in the harbour, so that time is measured in hours rather than in sixhourly invervals. The effect of changing the units of time to hours from 6 hours is to decrease the coefficient of $t$ in the sine function by a factor of 6, so that the new function is
$\phantom{\rule{2em}{0ex}}h=3.2sin\left(0.45t+2.168\right).$ See Figure 10.
Figure 10
We can use the latest form of the function to calculate the time of the first low tide assuming that $t=0$ corresponds to midnight.
At the first low tide, $h=3.2$ and $\phantom{\rule{1em}{0ex}}sin\left(0.45t+2.2168\right)=1,$
Using the fact that $sin\left(\frac{3\pi}{2}\right)=1$ , we have
$\phantom{\rule{2em}{0ex}}0.45t+2.2168=3\pi \u22152$ , giving $t=5.5458=5.55\phantom{\rule{1em}{0ex}}\text{to2d.p.}$
so the first low tide is just before 6 a.m.
Task!
For the above tide modelling situation, assume that $t=0$ corresponds to midnight. Calculate
 the time of the first high tide after midnight
 the times either side of midnight at which the water is at its mean level.
 At the first high tide, $h=3.2$ and $sin\left(0.45t+2.2168\right)=1,$ so $0.45t+2.2168=5\pi \u22152$ giving $t=12.5271$ so the first high tide is a little before 1 p.m.

When the water level is at the mean value,
$\phantom{\rule{2em}{0ex}}sin\left(0.45t+2.2168\right)=0.$
At the mean level before midnight, using the fact that $sin\left(0\right)=0$ we have
$\phantom{\rule{2em}{0ex}}0.45t=2.2168$ so $t=4.9262=4.93\phantom{\rule{1em}{0ex}}\text{to2d.p.}$
So this mean level occurs nearly 5 hours before midnight, i.e. about 7 p.m. the previous day. The next mean level will occur one period, or 13.963 hours, later, at approximately 9 a.m.
There are various rules connected with sine and cosine functions that can be summarised at this point.
 Placing a multiplier before $sinx$ or $cosx$ (e.g. $2sinx$ ) changes the amplitude without changing the period.
 Placing a multiplier before $x$ in $sinx$ or $cosx$ , (e.g. $sin3x$ ), changes the period or frequency without changing the amplitude.
 As with any function, the addition of a constant (e.g. $4+sinx$ ) raises or lowers the whole graph of the sine or cosine function. It alters the mean value without changing the amplitude.
 Changing the sign within a cosine function has no effect, (e.g. $cos\left(x\right)$ = $cosx$ ).
 Changing the sign within a sine function changes the sign of the function, (e.g. $sin\left(x\right)$ = $sinx$ ).
 Placing a constant or altering the constant $b$ in $sin\left(ax+b\right)$ or $cos\left(ax+b\right)$ changes the phase and shifts the sine or cosine function along the $x$ axis.
Task!
 Write down the amplitude and period of $y$ = $sin\left(3x\right)$
 Write down the amplitude and frequency of $y$ = 3 $sin\left(2x\right)$
 Write down the amplitude, period and frequency of $y$ = $asin\left(bx\right)$

Write down the amplitude, period, frequency and phase of
$\phantom{\rule{2em}{0ex}}y=4sin\left(2x+7\right).$
 Write down an equivalent expression to that in (4) but with the phase less than $2\pi $ .
 amplitude = 1 period $=2\pi \u22153$
 amplitude = 3 frequency $=\frac{2}{2\pi}=\frac{1}{\pi}$
 amplitude = $a$ period $=2\pi \u2215b$ frequency $=b\u22152\pi $
 amplitude = 4 period $=\frac{2\pi}{2}=\pi $ frequency $=\frac{1}{\pi}$ phase $=7$
 $y$ = 4 $sin\left(2x+72\pi \right)=4sin\left(2x+0.7168\right)$
Task!
Write down a function relating water level ( $L$ m) in a harbour to time ( $T$ hours), starting when the level is equal to the mean level of 5 m, that has an amplitude of 2 m and has a period of twelve hours.
In the general form $y$ = $asin\left(bx+c\right)+d$ , the phase $c$ = 0, the period $\frac{2\pi}{b}$ = 12, so $b=\frac{\pi}{6}$ the amplitude $a$ = 2, the mean value $d$ = 5.
$L=2sin\left(\frac{\pi}{6}T\right)+5\phantom{\rule{2em}{0ex}}\left(T\ge 0\right)$
Task!
The diagram shows a graph of a typical variation of the depth ( $d$ metres) of water in a particular harbour with time ( $t$ hours) as the depth changes with the tide.

Find a suitable equation for the curve in the diagram:
Equation is of the form
$\phantom{\rule{2em}{0ex}}h=a+bcos\left(\omega t\right)$ (or $h=a+bsin\left(\omega t+\frac{\pi}{2}\right)$ )
By inspection, $a=5$ and $b=3$ .
The period $\phantom{\rule{1em}{0ex}}T=12.5=\frac{2\pi}{\omega}$ so $\omega =\frac{4\pi}{25}\left(=0.502655\right)$
so the equation of the curve is $h=5+3cos\left(\frac{4\pi}{25}t\right)$

A boat enters the harbour in late morning on a day when the high tide is at 2 p.m. The boat needs a water depth of 4 m to sail safely. What advice would you give to its pilot about when to leave the harbour if the boat is not to be forced to wait in the harbour through the evening low tide?
Put $h=4$ into the equation:
$\phantom{\rule{2em}{0ex}}4=5+3cos\left(\frac{4\pi}{25}t\right)$ implying $\frac{1}{3}=cos\left(\frac{4\pi}{25}t\right)$
Now, inverting the cosine:
$\phantom{\rule{2em}{0ex}}\frac{4\pi}{25}t={cos}^{1}\left(\frac{1}{3}\right)=1.91063$ giving $t=3.80108$ hours.
So the advice to the pilot should be that he needs to be clear of the harbour by 5:45 pm at the very latest  and that he should allow a safety margin.

State two modelling assumptions you have made:
Assumptions likely are:
The tide on the day in question is typical .
No waves.
A sinusoidal function accurately models the effect of the tide on sea level.