Problem 39 - Precalculus Mathematics for Calculus

Question

The drag force \(F\) on a boat is jointly proportional to the wetted surface area \(A\) on the hull and the square of the speed \(s\) of the boat. A boat experiences a drag force of 220 Ib when traveling at \(5 \mathrm{mi} / \mathrm{h}\) with a wetted surface area of \(40 \mathrm{ft}^{2} .\) How fast must a boat be traveling if it has \(28 \mathrm{ft}^{2}\) of wetted surface area and is experiencing a drag force of 175 Ib?

Step-by-Step Solution

Answer

The boat must travel at approximately 5.33 mi/h.

Step 1: Understand Joint Proportionality

The problem states that the drag force \( F \) is jointly proportional to the wetted surface area \( A \) and the square of the speed \( s \). In mathematical terms, this can be expressed as \( F = k \cdot A \cdot s^2 \), where \( k \) is the constant of proportionality.

Step 2: Use Given Information to Find Constant k

Given: \( F = 220 \) Ib, \( s = 5 \) mi/h, and \( A = 40 \) ft\(^2\). Substitute these values into the equation \( F = k \cdot A \cdot s^2 \) to find \( k \):\[ 220 = k \cdot 40 \cdot 5^2 \]\[ 220 = k \cdot 40 \cdot 25 \]\[ 220 = 1000k \]\[ k = 0.22 \]

Step 3: Set up Equation for New Conditions

Given new conditions: \( F = 175 \) Ib, \( A = 28 \) ft\(^2\). We need to find the speed \( s \). Substitute \( k = 0.22 \), given \( F \) and \( A \) into the formula:\[ 175 = 0.22 \cdot 28 \cdot s^2 \]

Step 4: Solve for Speed s

Rearrange the equation to solve for \( s^2 \):\[ 175 = 0.22 \cdot 28 \cdot s^2 \]\[ s^2 = \frac{175}{0.22 \cdot 28} \]Calculate \( s \):\[ s^2 = \frac{175}{6.16} \approx 28.41 \]\[ s = \sqrt{28.41} \approx 5.33 \] mi/h.

Key Concepts

Drag Force Calculation

When calculating drag force, it's important to understand how it depends on multiple factors. The drag force (F) on a moving object, such as a boat, involves both its wetted surface area and speed. This relationship is described through joint proportionality.
You can think about it like this:

• The greater the wetted surface area, the more contact the hull has with water, leading to increased resistance and drag force.
• The speed of the boat also plays a crucial role because the force increases with the square of speed. This means the faster the boat travels, the exponential growth in drag force.

In our exercise, joint proportionality means these two factors – the wetted surface area and the speed squared – both influence drag force. The equation becomes:
\[ F = k \cdot A \cdot s^2 \]
Where \( F \) is the drag force, \( A \) is the wetted surface area, \( s \) is speed, and \( k \) is the constant of proportionality. This setup allows us to calculate drag force based on known or estimated variables.

Wetted Surface Area

The concept of the wetted surface area is simple yet significant in calculating drag force. It refers to the part of a boat's hull that is underwater and directly interacts with the water. This interaction causes resistance.
The wetted surface area can vary, affecting how much resistance and drag force is produced. For instance:

• A larger wetted surface area means more water resistance, leading to higher drag force.
• Conversely, a smaller wetted surface area results in less resistance and a lower drag force.

In the exercise provided, you were given different scenarios with different wetted surface areas. These values directly impacted the calculations of the drag force experienced by the boat as the speed changed, exemplifying their practical importance.

Proportionality Constant

The proportionality constant (\( k \)) is crucial in determining the exact relationship between the drag force and other variables like speed and wetted surface area. Think of \( k \) as a binding factor that ensures all parts of the equation fit neatly together.
Knowing this constant allows you to predict how changes in speed and wetted surface area will impact the drag force. It's calculated from known values to match observed conditions. The original conditions in the problem, with a drag force of 220 Ib at a wetted area of 40 ft² and speed of 5 mi/h, are used to find \( k \):

• By substituting the known values into\[ F = k \cdot A \cdot s^2 \]
• We solve for \( k \) and find it as 0.22

This constant helped solve for the new condition, and without it, we couldn't accurately adjust for the changes in wetted surface area or force.