Q. 34 - Precalculus Enhanced with Graphing Utilities

Question

The length of time between consecutive high tides is $12$ hours and $25$ minutes. According to the National Oceanic and Atmospheric Administration, on Saturday, July $12, 2012$ in Sitka Sound, Alaska, high tide occurred at $2:37am$  ($2.6167$ hours) and low tide occurred at $9:12$pm ($9.2$ hours). Water heights are measured as the amounts above or below the mean lower low water. The height of the water at high tide was
$11.09$ feet, and the height of the water at low tide was $-2.49$ feet.

(a) Approximately when will the next high tide occur?
(b) Find a sinusoidal function of the form $y=A \sin (\omega x-\varphi )+B$ that models the data.
(c) Use the function found in part (b) to predict the height of the water at $6$ pm.

Step-by-Step Solution

Answer

(a) The next tide will occur at $3:03 pm$

(b) The sinusoidal function that models the data is $y=6.79\sin (0.506x+0.2445)+4.3$

(c) The height of the water at $6$ pm on July $12, 2012$ is $2.829$ feet

Part (a) Step 1. Given

The length of time between consecutive high tides is $12$ hours and $25$ minutes.

The height of the water at high tide was $11.09$ feet, and the height of the water at low tide was $-2.49$ foot.

To find the time at which the next tide will occur.

Part (a) Step 2. To find the time of next tide

The tide occurs at  $2:37pm$

The time length between two tides is $12 hours and 25 minutes$

So the next tide will occur at $3:03 pm$

Part (b) Step 1. To find the amplitude A, vertical shift B of the function

Amplitude$=\frac{Largest value-Smallest value}{2}$

$$\begin{gathered} =\frac{11.09-(-2.49)}{2} \\ =\frac{13.58}{2} \\ =6.79 \end{gathered}$$

Vertical shift$=\frac{Largest value+Smallest value}{2}$

$$\begin{gathered} =\frac{11.09-2.49}{2} \\ =\frac{8.6}{2} \\ =4.3 \end{gathered}$$

Part (b) Step 2. Determine ω

The period is the time length between two successive tides, which is $12$ hours and $25$ minutes.

Period$=\frac{2\mathrm{\pi }}{\omega }$

$$\begin{gathered} 12.4167=\frac{2\mathrm{\pi }}{\omega } \\ \omega =0.506 \end{gathered}$$

Part (b) Step 3. Determine horizontal shift

If the first high tide is at $2.6167$ hours, then the equilibrium position occurs before at $3.1$ hours, so at $-0.4833$ hour.

$$\begin{gathered} -0.4833=\frac{\varphi }{\omega } \\ -0.4833=\frac{\varphi }{0.506} \\ \varphi =-0.2445 \end{gathered}$$ 

Part (b) Step 4. Find the function

The function that models the data is $y=6.79\sin (0.506x+0.2445)+4.3$

Part (c) Step 1. Write the function

The function that models the data is $y=6.79\sin (0.506x+0.2445)+4.3$

Since $6 pm=18$ hours,

$$\begin{gathered} y=6.79\sin (0.506(18)+0.2445)+4.3 \\ =2.829 \end{gathered}$$

The height of the tide is $2.829$ feet.