**Example 5:**

The average time between a high tide and a low tide at a port is $6$ hours. The average depth of water at the port is $50$ meters. However at high tide the water level rises to $70$ meters. Find the trigonometric function that represents the depth of the water over time if it's peak high tide at $12:00 AM$. Also sketch the graph of the function.

**Solution:**

The general form of a sine curve function is like this:

$y(t)$ = $A sin(Bt + C) + D$

Here, $A$ is the amplitude of the curve. Amplitude refers to half the difference between the highest and lowest point on the curve. $B$ is the horizontal stretch or shrink that can be found from the period of the function. $C$ is the horizontal translation from the parent function. It is used to find the phase shift. $D$ is the vertical shift of the function.

The parent function that we need to start working form would be:

$y(t)$ = $sin(t)$

Now since the average depth is $50$ meters, so we know that the vertical shift is $D$ = $50$ meters. Thus, using this value of $D$ our function now becomes:

$y(t)$ = $sin(t) + 50$

Next, we are given that the difference between average depth and high tide depth is $70 - 50$ = $20$ meters. So that becomes our amplitude $A$. Thus using this our equation now becomes:

$y(t)$ = $20\ sin(t) + 50$

We are given that the time between high and low tide is $6$ hours. So the time between two high tides would be $12$ hours. So the period of the function would be $12$ hours. We know that:

$period$ = $\frac{2 \pi}{B}$

Thus using the period of $6$ hours we have:

$6$ = $\frac{2 \pi}{B}$

Solving that for $B$ we have:

$B$ = $\frac{2 \pi}{6}$ = $\frac{p}{3}$

So now our function becomes:

$y(t)$ = $20sin$ $(\frac{\pi}{3}$ $t + C) + 50$

Now we need to find the value for that $C$. We are given that at $12:00AM$ when $t$ = $0$, the tide is peak high at $70$ meters. Thus,

$y(t$ = $0)$ = $20 sin$ $(\frac{\pi}{3}$ $(0)\ +\ C)\ +\ 50$

Solving that for $y$ = $70$ we have:

$70$ = $20Sin(C)\ +\ 50$

$70 - 50$ = $20Sin(C)$

$20$ = $20Sin(C)$

$Sin(C)$ = $1$

$C$ = $\frac{p}{2}$

Thus our function is:

$y(t)$ = $20Sin$ $(\frac{\pi}{3}$ $t\ +$ $\frac{p}{2})$ $+\ 50$

Its graph would look like this: