Q. 33 - 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 $21,2012,$ in Charleston, South Carolina, high tide occurred at $11:30$ am ($11.5$ hours) and low tide occurred at $5:31$ pm ($17.5167$ 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 $5.84$ feet, and the height of the water at low tide was $-0.37$ foot.

(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 $3$ pm on July $21,2012$.

Step-by-Step Solution

Answer

(a) The next tide will occur at $11:55pm$

(b) The sinusoidal function that models the data is $y=6.79\sin (0.506x-4.2504)+2.735$

(c) The height of the water at $3$pm on July $21, 2012$ is $2.124$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 $5.84$ feet, and the height of the water at low tide was $-0.37$ 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 $11:30am$ 

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

So the next tide will occur at $11:55pm$

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

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

$$\begin{gathered} =\frac{5.84-(-0.37)}{2} \\ =\frac{6.21}{2} \\ =3.105 \end{gathered}$$

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

$$\begin{gathered} =\frac{5.84-0.037}{2} \\ =\frac{5.47}{2} \\ =2.735 \end{gathered}$$

Part (b) Step 2. Determine ω

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

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

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

Part (b) Step 3. Determine horizontal shift

If the first high tide is at $2.6167$ hours, then the equilibrium position occurs at $3.1$ hour, so the shift is $8.4$

Vertical shift$=\frac{\varphi }{\omega }$

$$\begin{gathered} 8.4=\frac{\varphi }{0.506} \\ \varphi =8.4\times 0.506 \\ =4.2504 \end{gathered}$$

Part (b) Step 4. Find the function

The function that models the data is $y=3.105\sin (0.506x-4.2504)+2.735$

Part (c) Step 1. Write the function

The function that models the data is $y=3.105\sin (0.506x-4.2504)+2.735$

Part (c) Step 2. Find the height of the tide.

Since $3pm=15 hours,$
$$\begin{gathered} y=3.105\sin (0.506(15)-4.2504)+2.735 \\ =2.124 \end{gathered}$$

The height of the tide is $2.124$ feet.