Problem 38 - Algebra Form and Function

Question

The energy, \(E,\) in foot-pounds, delivered by an ocean wave is proportional \(^{6}\) to the length, \(L,\) of the wave times the square of its height, \(h\). (a) Write a formula for \(E\) in terms of \(L\) and \(h\). (b) A 30 -foot high wave of length 600 feet delivers 4 million foot-pounds of energy. Find the constant of proportionality and give its units. (c) If the height of a wave is one-fourth the length, find the energy \(E\) in terms of the length \(L\). (d) If the length is 5 times the height, find the energy \(E\) in terms of the height \(h\).

Step-by-Step Solution

Answer

Answer: When h = L/4, the energy E in terms of length L is given by: \[E = \frac{7.407 \cdot L^3}{16}\] When L = 5h, the energy E in terms of height h is given by: \[E = 37.035h^3\]

Step 1: Write the proportionality equation between E, L, and h

The energy E is proportional to the length L times the square of its height h. We can represent this proportionality by the equation: \[E = k \cdot L \cdot h^2\] where k is the constant of proportionality.

Step 2: Find the constant of proportionality k and its units

We're given that a 30-foot high wave of length 600 feet delivers 4 million foot-pounds of energy. We can use this information to find the value of k. \[E = 4,000,000\] \[L = 600\] \[h = 30\] Plugging these values into the proportionality equation, \[4,000,000 = k \cdot 600 \cdot 30^2\] Solve for k: \[k = \frac{4,000,000}{600 \cdot 900}\] \[k = 7.407\] The units of k are foot-pounds since the energy is measured in foot-pounds.

Step 3: Find the energy E in terms of the length L when height h = L/4

We're given that the height of a wave is one-fourth the length. Therefore, h = L/4. Substitute h = L/4 in the proportionality equation: \[E = 7.407 \cdot L \cdot (\frac{L}{4})^2\] Simplify the equation: \[E = 7.407 \cdot L \cdot \frac{L^2}{16}\] \[E = \frac{7.407 \cdot L^3}{16}\]

Step 4: Find the energy E in terms of the height h when length L = 5h

We're given that the length of a wave is 5 times the height. Therefore, L = 5h. Substitute L=5h in the proportionality equation: \[E = 7.407 \cdot (5h) \cdot h^2\] Simplify the equation: \[E = 7.407 \cdot 5 \cdot h^3\] \[E = 37.035h^3\]

Key Concepts

Energy Formula

The energy of an ocean wave, often measured in foot-pounds, is directly affected by its physical characteristics. To express this energy, we use the energy formula, which relates energy to other properties of the wave. In this particular exercise, energy is described by how it varies with the wave's length and height. Specifically, energy \( E \) is proportional to the product of the wave's length \( L \) and the square of its height \( h^2 \).

This relationship is encapsulated in the formula:

• \( E = k \cdot L \cdot h^2 \)

Here, \( k \) stands for the constant of proportionality. This constant makes sure the equation balances correctly and matches real-world observations. So, when you're working with wave energy, remember that length and height don't just sum up; they multiply together with the square of height playing a crucial role in the energy output.

Constant of Proportionality

Finding the constant of proportionality \( k \) lets us transform a proportional relationship into a functional equation. In the exercise, we use known values of wave length, height, and energy to calculate this constant.

The energy, \( E = 4,000,000 \) foot-pounds, a wave of length \( L = 600 \) feet, and height \( h = 30 \) feet, are given. Plugging these into the equation:

• \( 4,000,000 = k \cdot 600 \cdot 30^2 \)

Solving for \( k \), we rearrange and calculate as follows:

• \( k = \frac{4,000,000}{600 \cdot 900} = 7.407 \)

This value, 7.407, carries the units of foot-pounds per unit of length and squared height. It's crucial as it bridges the theoretical formula and physical measurements, making sure calculations align with actual energy produced by waves.

Wave Equations

Waves can take on varied forms, and their dimensions influence their energy. Understanding how to adjust calculations based on specific dimensional relationships is essential. Let's delve into two different scenarios from the problem:

• **Height is One-Fourth the Length**:
  When the wave's height \( h \) is one-fourth its length \( L \), or \( h = \frac{L}{4} \), the energy equation adjusts to:
  \[ E = 7.407 \cdot L \cdot \left(\frac{L}{4}\right)^2 \]
  Simplifying gives:
  \[ E = \frac{7.407 \cdot L^3}{16} \]
• **Length is Five Times the Height**:
  Conversely, if the wave's length \( L \) is five times its height \( h \), we say \( L = 5h \). The equation adapts to:
  \[ E = 7.407 \cdot (5h) \cdot h^2 \]
  The simplification yields:
  \[ E = 37.035h^3 \]

These equations highlight how the proportions of different wave measurements affect energy. Mastering these variations is key for effectively applying wave energy concepts in diverse situations.