Problem 37 - Precalculus Mathematics for Calculus

Question

The power \(P\) of a jet of water is jointly proportional to the cross-sectional area \(A\) of the jet and to the cube of the velocity \(v\). If the velocity is doubled and the cross-sectional area is halved, by what factor will the power increase?

Step-by-Step Solution

Answer

The power increases by a factor of 4.

Step 1: Understand Proportionality

The problem states that the power \(P\) is jointly proportional to the area \(A\) and the cube of the velocity \(v\). This can be represented as \(P = k \cdot A \cdot v^3\) where \(k\) is a proportionality constant.

Step 2: Original Power Formula

The initial power can be represented by \(P_1 = k \cdot A \cdot v^3\). This is the power before any changes to the area or velocity.

Step 3: Calculate New Variables

According to the problem, the velocity \(v\) is doubled, becoming \(2v\), and the area \(A\) is halved, becoming \(\frac{A}{2}\).

Step 4: New Power Formula

Using the new area and velocity, the new power \(P_2\) can be calculated as \(P_2 = k \cdot \left( \frac{A}{2} \right) \cdot (2v)^3\).

Step 5: Simplify the New Power

Substitute and simplify the expression for \(P_2\): \[P_2 = k \cdot \frac{A}{2} \cdot 8v^3 = 4k \cdot A \cdot v^3.\] This shows \(P_2 = 4 \cdot P_1\).

Step 6: Calculate Factor of Increase

The factor by which the power increases can be found by dividing \(P_2\) by \(P_1\): \[\text{Factor} = \frac{P_2}{P_1} = \frac{4k \cdot A \cdot v^3}{k \cdot A \cdot v^3} = 4.\]

Key Concepts

Proportionality Constant

In problems involving joint variation, the proportionality constant plays a crucial role. It connects various variables, showing how they relate to each other in a proportional manner. In our exercise, the power of the water jet is said to be "jointly proportional" to the area and the cube of velocity. This means the equation takes the form \( P = k \cdot A \cdot v^3 \) where \( k \) is the proportionality constant.

Why is \( k \) important? Because it acts as a bridge that allows the mathematical relationship between \( P \), \( A \), and \( v^3 \) to exist. Without \( k \), you couldn't calculate how changes in \( A \) and \( v \) affect \( P \). Even though we don't calculate \( k \) in this specific problem, it's essential in establishing the foundational relationship that we build upon.

Power Formula

The concept of the power formula in this exercise is key to understanding how joint variation works in practice. Initially, the formula given is \( P_1 = k \cdot A \cdot v^3 \). This represents the power before any changes occur to the water jet's velocity or area.

When solving such problems, it's helpful to identify each variable and their relationship to create the initial formula. Here, the power, \( P_1 \), depends on both the area, \( A \), and the velocity cubed, \( v^3 \), through the proportionality constant \( k \). Identifying this equation allows us to see how changes affect the entire system, providing a clear path towards solving the problem.

Velocity and Area Relationship

In the context of this exercise, understanding the relationship between velocity, area, and power is crucial. The problem involves varying two factors: velocity \( v \) and area \( A \).

Upon doubling the velocity of the water jet, it is represented as \( 2v \). Since the power is proportional to the velocity cubed (\( v^3 \)), this increase in velocity significantly impacts power because \((2v)^3 \) calculates to \(8v^3\). This step is critical as it highlights the non-linear impact of increasing velocity.

Conversely, halving the area \( A \) means replacing \( A \) with \( \frac{A}{2} \). By assessing both changes, it becomes apparent how power is influenced when these variables are adjusted in tandem.

Factor of Increase

The factor of increase gives insight into how much the power changes due to adjustments in velocity and area. Our newly calculated power formula is \( P_2 = 4 \cdot P_1 \). This was derived by substituting the changed values into the power equation.

Where was this factor derived? After altering \( v \) and \( A \), we know:

• The velocity doubling results in \( (2v)^3 = 8v^3 \), meaning the velocity contributes a factor of 8.
• Halving the area contributes a factor of \( \frac{1}{2} \).

Multiplying these factors, we get \( 8 \times \frac{1}{2} = 4 \), meaning the power increases by a factor of 4. Understanding this rise highlights how powerful joint variation equations handle multiple variable changes seamlessly.