Problem 35
Question
In Exercises \(35-37,\) use the following information. When a person walks, the pressure on each boot sole varies inversely with the area of the sole. Denise is walking through deep snow, wearing boots that have a sole area of 29 square inches each. The pressure on the sole is 4 pounds per square inch when she stands on one foot. Use unit analysis to explain why the constant of variation is Denise's weight. How much does she weigh?
Step-by-Step Solution
Verified Answer
Denise's weight is \(4 \times 29\) pounds, which equals to 116 pounds.
1Step 1: Understanding the Problem
The problem describes an inverse relationship between pressure and area. According to this relationship, when Denise stands on one foot, the pressure on the boot sole (P) and the area of the boot sole (A) are related by the formula \(P = k/A\), where k is a constant. Here, Denise's weight becomes the constant of the variation, which means her weight is the same as the value of k. We are given that the area, A, is 29 square inches and the pressure, P, is 4 psi.
2Step 2: Calculating the Constant of Variation
We can use the given values for P and A along with the formula \(P = k/A\) to calculate the constant of variation. Inserting the given values into the formula gives: \(4 = k/29\). To find k, we multiply both sides by 29 which gives \(k = 4 \times 29\).
3Step 3: Finding Denise's Weight
The problem tells us that Denise's weight is equal to the constant of variation, k. Therefore, once we determine the value of k, we also find Denise's weight. Hence to get Denise's weight, we calculate \(k = 4 \times 29\).
Key Concepts
PressureAreaConstant of VariationUnit Analysis
Pressure
Pressure is a way to understand how force is distributed over an area. When we think about pressure, we're considering how much force is applied to a certain surface. In this scenario where Denise is standing on one foot in snow, the pressure is described as 4 pounds per square inch (psi).
This means that for every square inch of the boot sole in contact with the snow, there is a force equivalent to 4 pounds being applied.
Pressure is calculated using the formula:
where \( P \) is the pressure, \( F \) is the force exerted (which in this case is Denise's weight), and \( A \) is the area covered by the force.
This formula shows us why if the area decreases, the pressure increases given the same force, emphasizing the relationship between pressure and area.
This means that for every square inch of the boot sole in contact with the snow, there is a force equivalent to 4 pounds being applied.
Pressure is calculated using the formula:
- \( P = \frac{F}{A} \)
where \( P \) is the pressure, \( F \) is the force exerted (which in this case is Denise's weight), and \( A \) is the area covered by the force.
This formula shows us why if the area decreases, the pressure increases given the same force, emphasizing the relationship between pressure and area.
Area
The area here refers to the surface of the boot sole Denise is standing on. Area is all about how much surface space is available for the force or weight to be spread across.
A larger area means the weight is spread out, which in turn reduces the pressure on any given point of the surface. For this exercise, each of Denise's boot soles has an area of 29 square inches.
Understanding how area influences pressure is crucial, especially in scenarios involving inverse variations, such as:
This illustrates why, in the given context, having a large boot sole is beneficial in deep snow to avoid sinking into it due to high pressure.
A larger area means the weight is spread out, which in turn reduces the pressure on any given point of the surface. For this exercise, each of Denise's boot soles has an area of 29 square inches.
Understanding how area influences pressure is crucial, especially in scenarios involving inverse variations, such as:
- An increased area reduces the pressure because the force is spread over a larger space.
- A decreased area increases the pressure because the same force impacts a smaller space.
This illustrates why, in the given context, having a large boot sole is beneficial in deep snow to avoid sinking into it due to high pressure.
Constant of Variation
The constant of variation, denoted by \( k \), is a crucial part of inverse variation problems, serving as the link between two inversely related variables.
In this exercise, the formula \( P = \frac{k}{A} \) shows the inverse relationship between pressure (\( P \)) and area (\( A \)). Here, \( k \), the constant of variation, is Denise's weight.
To find this constant, we rearrange the formula to solve for \( k \): \[ k = P \times A \] By inputting the known values (pressure \( P = 4 \) psi and area \( A = 29 \) square inches), we calculate: \[ k = 4 \times 29 = 116 \]
This means Denise's actual weight, the constant \( k \), is 116 pounds. The constant reminds us how tightly pressure and area are linked through inverse variation.
In this exercise, the formula \( P = \frac{k}{A} \) shows the inverse relationship between pressure (\( P \)) and area (\( A \)). Here, \( k \), the constant of variation, is Denise's weight.
To find this constant, we rearrange the formula to solve for \( k \): \[ k = P \times A \] By inputting the known values (pressure \( P = 4 \) psi and area \( A = 29 \) square inches), we calculate: \[ k = 4 \times 29 = 116 \]
This means Denise's actual weight, the constant \( k \), is 116 pounds. The constant reminds us how tightly pressure and area are linked through inverse variation.
Unit Analysis
Unit analysis is a method of ensuring the consistency and correctness of an equation or a calculation by examining the units involved. It's a helpful tool for verifying the results of a problem like this one.
For Denise, applying unit analysis means checking if the units balance out correctly. In the equation \( P = \frac{k}{A} \), we should analyze the units of each part:
Since the constant of variation (\( k \)) is the same as the force (or weight) exerted, it's measured in pounds. Therefore, when applying unit analysis, the equation's units line up: \[ \text{pounds per square inch} = \frac{\text{pounds}}{\text{square inches}} \] This confirms that when the calculation results in \( k = 116 \) pounds, it's consistent with the physical meaning of Denise's weight. By confirming units match up, unit analysis ensures your solution aligns with real-world interpretations.
For Denise, applying unit analysis means checking if the units balance out correctly. In the equation \( P = \frac{k}{A} \), we should analyze the units of each part:
- Pressure \( P \) is measured in pounds per square inch (psi).
- Area \( A \) is measured in square inches.
Since the constant of variation (\( k \)) is the same as the force (or weight) exerted, it's measured in pounds. Therefore, when applying unit analysis, the equation's units line up: \[ \text{pounds per square inch} = \frac{\text{pounds}}{\text{square inches}} \] This confirms that when the calculation results in \( k = 116 \) pounds, it's consistent with the physical meaning of Denise's weight. By confirming units match up, unit analysis ensures your solution aligns with real-world interpretations.
Other exercises in this chapter
Problem 34
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