Problem 41 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. Pressure of a Liquid The pressure exerted by a certain liquid at a given point is directly proportional to the depth of the point beneath the surface of the liquid. If the pressure exerted at 30 feet is 13 pounds per square inch, what is the pressure exerted at 70 feet?

Step-by-Step Solution

Answer

The pressure at 70 feet is approximately 30.33 PSI.

Step 1: Understanding the relationship

The problem states that the pressure exerted by the liquid is directly proportional to the depth. This means if we denote pressure by \( P \) and depth by \( d \), the relationship can be expressed as \( P = k \cdot d \), where \( k \) is a constant of proportionality.

Step 2: Establish the equation using given information

We are given that at a depth of 30 feet, the pressure is 13 pounds per square inch. So, substitute \( d = 30 \) and \( P = 13 \) into the equation \( P = k \cdot d \). This gives us: \( 13 = k \cdot 30 \).

Step 3: Solve for the constant of proportionality \( k \)

To find \( k \), rearrange the equation from Step 2 to \( k = \frac{13}{30} \). Calculate \( k \) to obtain the value needed for the next computation.

Step 4: Calculate the pressure at 70 feet depth

Now, we need to determine the pressure at 70 feet. Using the equation \( P = k \cdot d \) and the value found for \( k \), substitute \( d = 70 \). This gives us \( P = \left(\frac{13}{30}\right) \cdot 70 \).

Step 5: Compute the result

Calculate the final expression from Step 4: \( P = \left(\frac{13}{30}\right) \cdot 70 = \frac{910}{30} \), which simplifies to \( P \approx 30.3333 \). So, the pressure at 70 feet is approximately 30.33 pounds per square inch.

Key Concepts

Pressure Calculations

Pressure calculations in liquids often involve understanding the concept of direct proportionality. This concept means that as one variable increases, the other does too, at a consistent rate. In the context of the problem, the pressure ( P) exerted by the liquid is directly proportional to the depth ( d) beneath the surface. This relationship can be expressed as \( P = k \cdot d \), where \( k \) is the constant rate of increase, known as the constant of proportionality.

• In our specific exercise, we know that at 30 feet, the pressure is 13 psi (pounds per square inch).
• Using this data, we plug the values into our equation: \( 13 = k \cdot 30 \).
• Solving for \( k \), we divide both sides by 30, giving us \( k = \frac{13}{30} \).

From this point, we can use the value of \( k \) to find the pressure at any other depth, by simply multiplying \( k \) by the desired depth.

Depth-Pressure Relationship

The depth-pressure relationship is a key factor in understanding how liquids behave under pressure. This relationship, as stated, is direct: the deeper you go, the greater the pressure. This occurs because of the weight of the liquid above a given point, which increases the force exerted downward.

• The deeper the point within the liquid, the greater the pressure due to the increased weight of the liquid exerting force on that point.
• This is why pressure at 70 feet is higher than at 30 feet; simply because there is more liquid above and thus more weight creates greater pressure.

Using the problem statement, we calculate the pressure at 70 feet by multiplying both the depth in feet (70 feet) and the constant of proportionality (\(\frac{13}{30}\)). This shows how increased depth correlates to increased pressure.

Constant of Proportionality

The constant of proportionality \( k \) is a crucial element in solving problems involving direct relationships between two quantities. Here, it determines how much the pressure changes with every foot of depth in the liquid.

• The value for \( k \) we calculated is \( \frac{13}{30} \), indicating the pressure increase per foot.
• With this constant, you can compute pressures at any depth by plugging into the equation \( P = k \cdot d \).

Understanding \( k \) is essential because it gives the specific rate of change for the situation described. In the given problem, once known, \( k \) allows us to make pressure predictions accurately at any specified depth without recalculating from scratch, demonstrating its key role in depth-pressure calculations.