Now on to a trig (circular functions) problem. This problem requires that you have covered year 11 trigonometry (called circular functions in some texts).

The water depth in a harbour on a particular day is modelled by the following equation:

\[D\left(t\right)=10+3\mathrm{sin}\left(\frac{\pi t}{6}\right),0≤t≤24\]where

*D*(*t*) = depth of water*t* = hours after midnight limited to be between 0 and 24 hours

A ship has to have at least 8.5 meters of water depth to use the harbour. At what times can the ship safely dock at the harbour?

Solving this algebraically, let’s first find the times when *D*(*t*) = 8.5.

So we have the requirement that the sine of “something” has to equal -0.5. When we take a function of something, that something is called the *argument* of the function. From a table of common sines and knowing that sin(-????) = -sin(????), we get that ????*t*/6 = -????/6 ⟹ t = -1. We need another angle on the unit circle that has a sine of -0.5:

So ????*t*/6=7????/6 ⟹ *t* = 7. The period of this sine function is

Now we add this period to our two core values of *t*, until we get *t* > 24:

7 + 12 = 19

-1 + 12 = 11

11 + 12 = 23

So the *t* values we have that are between 0 and 24 are 7, 11, 19, 23. As the sine function initially increase from 0, the value of *t* = 7, which is 7 AM, is when the depth is decreasing and is below 8.5 meters. The depth is increasing and above 8.5 at *t* = 11 which is 11 AM. The depth then decreases below 8.5 at *t* = 19 which is 7 PM and rises above 8.5 at *t* = 19 which is 11 PM. So the ship can dock between midnight and 7 AM or between 11 AM and 7 PM or between 11 PM and the following midnight.

I’ve plotted *D*(*t*) and the line *D*(*t*) = 8.5 to show these solutions.

In fact, doing a rough sketch of this function would help get a feel for the answers before we begin.

We do this by raising the standard sine curve up 10 units and draw 2 cycles from 0 to 24 since we can calculate that the period is 12 hours. Since the sine of anything goes from -1 to +1, the minimums are at 10 – 3 = 7 and the maximums are at 10 + 3 = 13. Drawing a line at *D*(*t*) = 8.5, we can see that we should get 4 points that will divide the intervals where the depth is below or above 8.5 meters. Once we have the points at *D*(*t*) = 8.5, we can easily see the periods where the depth is above 8.5 meters.