Problem 69 - Algebra and Trigonometry with Analytic Geometry

Question

A tidal wave of height 50 feet and period 30 minutes is approaching a sea wall that is \(12.5\) feet above sea level (see the figure). From a particular point on shore, the distance \(y\) from sea level to the top of the wave is given by $$ y=25 \cos \frac{\pi}{15} t $$ with \(t\) in minutes. For approximately how many minutes of each 30 -minute period is the top of the wave above the level of the top of the sea wall?

Step-by-Step Solution

Answer

The wave is above the sea wall for approximately 20 minutes each cycle.

Step 1: Understanding the Wave Function

The given wave function is \( y = 25 \cos \frac{\pi}{15} t \). This function represents the height of the wave above sea level over time \( t \) measured in minutes. The amplitude of the wave is 25 feet, indicating the maximum and minimum height deviation from sea level.

Step 2: Identifying the Maximum Wave Height

Since the function describes a cosine wave which oscillates between -25 and +25, the total height of the wave will vary between -25 + 25 = 0 feet and 25 + 25 = 50 feet above sea level. This matches the problem statement that the wave has a height of 50 feet.

Step 3: Calculating Height Above the Sea Wall

The sea wall height is 12.5 feet above sea level. We need to determine when the wave height \( y \) surpasses this threshold. The condition \( y = 25 \cos \frac{\pi}{15} t > 12.5 \) must be solved.

Step 4: Transforming the Inequality

To find the solutions, we simplify the inequality \( 25 \cos \frac{\pi}{15} t > 12.5 \) to \( \cos \frac{\pi}{15} t > 0.5 \). Dividing both sides by 25 simplifies the task.

Step 5: Solving the Inequality for Cosine

\( \cos \frac{\pi}{15} t = 0.5 \) represents angles whose cosine is 0.5, occurring at \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \) in one full cycle of \( 2\pi \). We solve the equation to find the time periods: \( \frac{\pi}{15} t = \frac{\pi}{3} \) and \( \frac{\pi}{15} t = \frac{5\pi}{3} \).

Step 6: Calculate Time Intervals

Solve \( \frac{\pi}{15} t = \frac{\pi}{3} \) which gives \( t = 5 \) minutes, and \( \frac{\pi}{15} t = \frac{5\pi}{3} \) which gives \( t = 25 \) minutes. Thus, the wave is above 12.5 ft from 5 to 25 minutes during the period.

Step 7: Determine Duration Above Sea Wall

The duration for which the wave is above the sea wall is \( 25 - 5 = 20 \) minutes in each 30-minute cycle.

Key Concepts

Trigonometric Inequality

When dealing with wave functions, one often encounters inequalities involving trigonometric functions like sine or cosine. These inequalities help us understand specific conditions that a wave might meet or exceed. In this problem with a tidal wave, we used a trigonometric inequality to find when the wave height exceeds the sea wall.The inequality started like this:

• We had the wave function: \( y = 25 \cos \frac{\pi}{15} t \)
• Set the condition for the wave crest surpassing the sea wall, which stands at 12.5 feet: \( 25 \cos \frac{\pi}{15} t > 12.5 \)

This was simplified by dividing the entire inequality by 25 to get: \( \cos \frac{\pi}{15} t > 0.5 \). This new inequality points out when the wave's cosine function produces a height beyond 12.5 feet, guiding us in our calculations below.

Cosine Function

The cosine function is crucial in wave analysis, and it describes how points on a circle project onto the x-axis as the circle rotates. It's periodic, oscillating smoothly due to its mathematical properties, making it perfect for modeling waveforms like tides.In our function, the cosine term is \( \cos \frac{\pi}{15} t \). Here's how it contributes:

• The cosine function varies between -1 and +1. When multiplied by 25, it ranges between -25 and +25 feet.
• At time zero \( (t=0) \), \( \cos(0) = 1 \), so the height \( y = 25 \times 1 = 25 \) feet.
• The role of \( \frac{\pi}{15} \) deals with scaling, impacting the wave's frequency by determining how quickly it completes one cycle.

The function models the back-and-forth motion of the wave over a fixed duration of time, pivotal here for understanding when it overtops the sea wall.

Amplitude of Wave

Amplitude in wave motion refers to the height from the equilibrium position to a wave crest or trough. It's an essential concept in determining a wave's energy and force.For our problem, the amplitude of the wave function \( y = 25 \cos \frac{\pi}{15} t \) is 25 feet.- This amplitude indicates the wave's maximum displacement from its mean position, meaning it can rise 25 feet above sea level.- Amplitude directly influences wave dynamics since it is closely tied to the wave height, impacting the forces exerted by the wave.In practical terms, knowing the amplitude helps describe how imposing or powerful a tidal wave is and how it might affect structures like a sea wall.

Wave Height

Wave height refers to the total vertical distance between a wave's highest (crest) and lowest (trough) points. Understanding this concept is crucial for assessing the maximal reach and impact of a wave.In the context of the tidal wave problem:

• The wave height was determined to be 50 feet.
• This height results from doubling the function’s amplitude, covering both the peak above and below the equilibrium level.
• Specifically, the height spans from -25 feet to +25 feet as outlined by \( y = 25 \cos \frac{\pi}{15} t \). This calculation correctly verifies the problem statement.

Understanding wave height is key to predicting when a wave might breach protective measures like the given 12.5-foot sea wall.

Period of Wave

The period of a wave is the duration it takes to complete one full cycle. It's a vital measure since it determines how frequently the wave pattern repeats itself over time.For the wave expressed by \( y = 25 \cos \frac{\pi}{15} t \):

• We deduce the period from the wave function's repetition interval, which here occurs every 30 minutes.
• This means, after 30 minutes, the wave’s crest and trough positions replicate the starting points.
• In terms of the cosine equation, one full cycle of \( \cos \left( \frac{\pi}{15} t \right) \) corresponds to a complete run through its path from peak through to peak again.

Knowing the period helps to predict wave interactions with the coastline and infrastructure, planning for periodic high events.