Problem 44 - Algebra 1

Question

Use the following information. When a person walks, the pressure P on each boot sole varies inversely with the area A of the sole. Denise is walking through deep snow, wearing boots that have a sole area of 29 square inches each. The boot-sole pressure is 4 pounds per square inch when she stands on one foot. If Denise wears snowshoes, each with an area 11 times that of her boot soles, what is the snowshoe pressure when she stands on one foot?

Step-by-Step Solution

Answer

The pressure on each snowshoe when Denise stands on one foot is approximately 0.36 pounds per square inch.

Step 1: Understand inverse variation

Inverse variation can be expressed as \( P = \frac{k}{A} \), where k is the constant of variation. Simply put, P (pressure) × A (area) = k (a constant).

Step 2: Calculate constant k using boot sole information

Denise's boot soles each have an area of 29 square inches and experience a pressure of 4 pounds per square inch when she stands on one foot. Using the formula from step 1, we can calculate the constant k. \( k = P \cdot A \), therefore, \( k = 4 p.s.i \cdot 29 in^2 = 116 \)

Step 3: Calculate the snowshoe pressure

Knowing that the snowshoe area is 11 times that of her boot sole, we find that the snowshoe area is \( 29 \cdot 11 = 319 in^2 \). Now, it's just a matter of substituting the values of k and the snowshoe area into the pressure formula \( P = \frac{k}{A} \) to get the new pressure. Substituting the values, the snowshoe pressure is \( P = \frac{116}{319} \approx 0.36 p.s.i \)

Key Concepts

Pressure Calculation

Pressure is a measure of how much force is applied over an area. In problems involving inverse variation, pressure (\( P \)) varies inversely with the area (\( A \)) over which it is applied. This means if the area increases, the pressure decreases, and vice versa. In mathematical terms, this relationship is shown as: - \( P = \frac{k}{A} \) - where \( k \) is a constant. For Denise's scenario, the pressure calculation helps us understand how the snowshoes distribute her weight. When Denise stands on one foot with her boots, the pressure is 4 psi with a boot sole area of 29 square inches. Using the equation, when the area changes to the larger snowshoe area, we find the corresponding pressure lessens. This inverse relationship shows why larger shoe areas, like snowshoes, are effective for walking on snow without sinking.

Area of Soles

The area of a sole, whether it be a boot or snowshoe, plays a critical role in determining the pressure exerted on a surface. The larger the sole area, the less pressure is exerted, which is crucial in minimizing how deep a person sinks into snow. In Denise's case, her boots have a sole area of 29 square inches. However, her snowshoes expand this base area drastically. With snowshoes that each have an area 11 times her boot sole, the snowshoe area becomes 319 square inches. This enlargement of area spreads out her weight over a larger surface. It demonstrates one of the principles why snowshoes are effective in snow, reducing how much pressure is exerted per square inch and enabling smooth travel over soft snow.

Constant of Variation

The constant of variation (\( k \)) in inverse variation equations is key, representing the product of pressure and area. It's a unique value for each specific situation but remains constant as long as those conditions are unchanged. For Denise's example with her boots, we calculate \( k \) using her initial conditions: a pressure of 4 psi and a sole area of 29 square inches. The calculation is straightforward: \( k = P \cdot A = 4 \text{ psi} \times 29 \text{ in}^2 = 116 \). This number acts as a bridge, allowing us to determine new pressures when areas change, such as when switching to snowshoes. As long as \( k \) is correctly calculated, various scenarios involving different areas can be explored and resolved to find the resulting pressures.