In Daytona Beach, Florida, the first high tide was 3.99 feet at 12:03 A.M. The first low tide of 0.55 foot occurred at 6:24 A.M. The second high tide occurred at 12:19 P.M. (Lesson 6-6)
a. Find the amplitude of a sinusoidal function that models the tides.
b. Find the vertical shift of the sinusoidal function that models the tides.
c. What is the period of the sinusoidal function that models the tides?
d. Write a sinusoidal function to model the tides, using t to represent the number of hours in decimals since midnight.
e. According to your model, determine the height of the water at noon.
The Correct Answer and Explanation is:
Let’s solve each part of the problem step by step using the tide data from Daytona Beach.
Given:
- High tide #1: 3.99 ft at 12:03 A.M.
- Low tide #1: 0.55 ft at 6:24 A.M.
- High tide #2: occurs at 12:19 P.M.
We’ll use a sinusoidal function of the form:
$$
h(t) = A \cdot \cos(B(t – C)) + D
$$
Where:
- $A$ is the amplitude
- $B = \frac{2\pi}{\text{period}}$
- $C$ is the horizontal shift
- $D$ is the vertical shift
Let $t$ represent hours since midnight.
(a) Find the amplitude:
The amplitude is half the distance between the high and low tides.
$$
A = \frac{\text{high} – \text{low}}{2} = \frac{3.99 – 0.55}{2} = \frac{3.44}{2} = \boxed{1.72 \text{ ft}}
$$
(b) Find the vertical shift:
The vertical shift is the average of the high and low tide levels.
$$
D = \frac{\text{high} + \text{low}}{2} = \frac{3.99 + 0.55}{2} = \frac{4.54}{2} = \boxed{2.27 \text{ ft}}
$$
(c) Find the period:
The period is the time between two consecutive high tides.
- From 12:03 A.M. to 12:19 P.M. is 12.27 hours (12 h + 16 min = 12 + 16/60 = 12.27)
$$
\text{Period} = \boxed{12.27 \text{ hours}}
$$
(d) Write a sinusoidal function:
Let’s use a cosine function since the first high tide occurs at $t = 0.05$ hr (12:03 A.M.)
- Horizontal shift $C = 0.05$
- Amplitude $A = 1.72$
- Vertical shift $D = 2.27$
- $B = \frac{2\pi}{12.27} \approx 0.5117$
$$
h(t) = 1.72 \cdot \cos(0.5117(t – 0.05)) + 2.27
$$
(e) Find the height at noon:
Noon = 12:00 P.M. = 12 hours
Plug $t = 12$ into the function:
$$
h(12) = 1.72 \cdot \cos(0.5117(12 – 0.05)) + 2.27 = 1.72 \cdot \cos(0.5117 \cdot 11.95) + 2.27
$$
Calculate the angle:
$$
\theta = 0.5117 \cdot 11.95 \approx 6.113
$$
Now compute cosine:
$$
\cos(6.113) \approx 0.990
$$
$$
h(12) = 1.72 \cdot 0.990 + 2.27 \approx 1.703 + 2.27 = \boxed{3.97 \text{ ft (approximately)}}
$$
Conclusion (300-word explanation):
To model ocean tides mathematically, a sinusoidal function is ideal because it reflects the natural periodic rise and fall of water levels. In this problem, we’re given the times and heights of high and low tides in Daytona Beach and asked to derive such a model.
First, we found the amplitude as half the difference between the highest and lowest tide. This measures how far the tide varies from its average level. The amplitude is essential because it represents the maximum deviation from the mean sea level.
Second, we calculated the vertical shift, which is the average of the high and low tide heights. This shift translates the sinusoidal graph up or down to align with the average water level. Without it, the graph would oscillate around zero, which isn’t realistic for tides.
The period was determined from the time difference between consecutive high tides, which tells us how long one full cycle (high to low back to high) lasts. This allows us to compute the frequency part of the function, represented by $B$ in the cosine formula.
Next, we formed a sinusoidal function with known amplitude, vertical shift, period, and a phase shift (based on the time of the first high tide). We used the cosine function because high tide aligns with the cosine’s peak.
Finally, we substituted $t = 12$ to find the water height at noon. The model predicted it to be around 3.97 feet, consistent with the second high tide occurring shortly after noon.
This process demonstrates how math can effectively model and predict natural events like tides.