Problem 60
Question
Refer to the following: The height of the water in a harbor changes with the tides. The height of the water at a particular hour during the day can be determined by the formula \(h(x)=5+4.8 \sin \left[\frac{\pi}{6}(x+4)\right]\) where \(x\) is the number of hours since midnight and \(h\) is the height of the tide in feet. What is the height of the tide at 5.00 A.M.?
Step-by-Step Solution
Verified Answer
The height of the tide at 5:00 A.M. is 0.2 feet.
1Step 1: Identify the value of x
First, we need to determine the value of \(x\) for 5:00 A.M. Since \(x\) represents the number of hours since midnight, 5:00 A.M. corresponds to \(x = 5\).
2Step 2: Substitute x into the formula
Now, substitute \(x = 5\) into the function for the height of the tide: \[ h(5) = 5 + 4.8 \sin \left[ \frac{\pi}{6}(5+4) \right] \].
3Step 3: Simplify the expression within the sine
Calculate the expression inside the sine function: \(5 + 4 = 9\), then \(\frac{\pi}{6} \times 9 = \frac{9\pi}{6} = \frac{3\pi}{2}\).
4Step 4: Evaluate the sine function
Find the sine of \(\frac{3\pi}{2}\). This angle corresponds to 270° on the unit circle, where the sine value is \(-1\).
5Step 5: Complete the formula substitution
Substitute \(-1\) into the height formula: \[ h(5) = 5 + 4.8 (-1) \].
6Step 6: Calculate the height of the tide
Perform the arithmetic to find \( h(5) = 5 - 4.8 = 0.2 \) feet.
Key Concepts
Sine FunctionTide Height CalculationUnit Circle
Sine Function
The sine function is an essential component of trigonometry, widely used to model periodic phenomena such as sound waves or tidal movements.
- The sine function is denoted as \( \sin(\theta) \), where \( \theta \) is an angle that can be measured in degrees or radians.
- It is a periodic function with a period of \( 2\pi \) radians (or 360°), meaning its values repeat every \( 2\pi \) radians.
Tide Height Calculation
Calculating the height of the tide using a trigonometric function involves understanding how the sine wave is applied in the context of time.
- The formula provided was: \[ h(x)=5+4.8 \sin \left[\frac{\pi}{6}(x+4)\right] \]
- This represents a combination of amplitude scaling and vertical shifts based on the sine wave.
Unit Circle
The unit circle is a fundamental concept for understanding trigonometric functions, especially useful for visualizing the sine function in this context.
- A unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane.
- The angle \( \theta \) is measured from the positive x-axis, counter-clockwise around the circle.
Other exercises in this chapter
Problem 60
In Exercises \(57-66,\) state the domain and range of the functions. $$y=-4 \sec (3 x)$$
View solution Problem 60
In Exercises \(49-60\), state the amplitude, period, and phase shift (including direction) of the given function. $$y=-5 \sin [-\pi(x+1)]$$
View solution Problem 61
In Exercises \(57-66,\) state the domain and range of the functions. $$y=2-\csc \left(\frac{1}{2} x-\pi\right)$$
View solution Problem 61
In Exercises \(61-66,\) sketch the graph of the function over the indicated interval. $$y=\frac{1}{2}+\frac{3}{2} \cos (2 x+\pi),\left[-\frac{3 \pi}{2}, \frac{3
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