Problem 13 - Precalculus: Functions and Graphs

Question

Liquid pressure The pressure \(P\) acting at a point in a liquid is directly proportional to the distance \(d\) from the surface of the liquid to the point. (a) Express \(P\) as a function of \(d\) by means of a formula that involves a constant of proportionality \(k\). (b) In a certain oil tank, the pressure at a depth of 2 feet is \(118 \mathrm{Ib} / \mathrm{ft}^{2} .\) Find the value of \(k\) in part (a). Find the pressure at a depth of 5 feet for the oil tank in part (b). (d) Sketch a graph of the relationship between \(P\) and \(d\) for \(d \geq 0\)

Step-by-Step Solution

Answer

(a) \(P = k \times d\), (b) \(k = 59\), (c) \(P = 295 \, \mathrm{Ib/ft^2}\) at 5 ft, (d) Line through origin with slope 59.

Step 1: Understanding Proportionality

The problem states that pressure \(P\) is directly proportional to the distance \(d\). This means we can express this relationship using a constant \(k\) where \(P = k \times d\).

Step 2: Expressing Pressure as a Function of Depth

Using the relationship from Step 1, we write the function as \(P(d) = k \times d\). This formula captures the idea that if depth doubles, the pressure also doubles.

Step 3: Determining the Constant of Proportionality \(k\)

We know from the problem that at a depth of 2 feet, the pressure is \(118 \, \mathrm{Ib/ft^2}\). Use this information: \[ 118 = k \times 2 \] Solving for \(k\), we find \(k = \frac{118}{2} = 59\). This is the constant of proportionality for this specific tank.

Step 4: Calculating Pressure at 5 Feet Depth

Using the formula \(P(d) = k \times d\) and the value of \(k\) from Step 3, substitute \(d = 5\): \[ P(5) = 59 \times 5 = 295 \, \mathrm{Ib/ft^2} \]

Step 5: Sketching the Relationship

The graph of \(P = 59d\) is a straight line through the origin (\(0,0\)), as it is a linear proportional relationship. The slope of the line is 59, showing how pressure increases with depth. For \(d \geq 0\), as \(d\) increases, \(P\) increases linearly.

Key Concepts

Direct Proportionality

Direct proportionality is a simple yet powerful concept in mathematics and physics. When two quantities are directly proportional, it means if one increases, the other increases in a consistent manner. In the context of liquid pressure, it's about how pressure changes with depth.

For example, consider pressure at a point in a liquid. If the depth of the liquid increases, the pressure will increase too, at the same rate. This relationship can be represented with the equation \( P = k \times d \), where \( P \) is the pressure, \( d \) is the depth, and \( k \) is the constant of proportionality.

This means:

• If depth doubles, pressure doubles.
• If depth is halved, pressure is halved.

It's a straightforward way to understand how two variables are linked consistently.

Constant of Proportionality

The constant of proportionality \( k \) is the multiplier that connects depth to pressure in a liquid. It defines the strength of the relationship between the two variables. In our exercise, the constant \( k \) was found by observing pressure values at a known depth.

Given that the pressure at 2 feet depth is \( 118 \, \mathrm{Ib/ft^2} \), the formula for pressure becomes \( 118 = k \times 2 \). Solving this equation gives us \( k = 59 \). This value is unique to the specific fluid and tank in the problem.

The constant:

• Allows us to predict pressure at different depths for the same situation.
• Shows how sensitive the pressure is to changes in depth.

Constant \( k \) is crucial for solving these types of problems.

Depth and Pressure Relationship

The relationship between depth and pressure in a liquid is linear, meaning it follows a straight line. Pressure increases at a steady rate as depth increases. Mathematically, this is captured by the equation \( P = k \times d \).

In our exercise, this means as you dive deeper into the liquid, for every additional unit of depth, the pressure increases by \( k \) units. This is reflected by:

• At a depth of 2 feet: pressure is \( 118 \, \mathrm{Ib/ft^2} \)
• At a depth of 5 feet: pressure is calculated as \( 295 \, \mathrm{Ib/ft^2} \) using \( P(5) = 59 \times 5 \)

The deeper the liquid, the more pressure there is at the point.

Linear Graph Sketching

Understanding how to sketch graphs for relationships like pressure versus depth helps visualize how pressure changes as you go deeper. We begin by plotting the equation \( P = 59d \), indicating a straight line on a graph.

The line starts at the origin (0,0), showing that at zero depth, the pressure is zero. As depth increases from zero onward, pressure rises linearly. For every foot deeper, pressure increases by 59 units.

Key points to note when sketching:

• Slope: The slope is determined by \( k = 59 \), reflecting how sharply the pressure rises with depth.
• Intercept: Starts from the origin, meaning there's no pressure at no depth.
• Linearity: Shows a steady pressure increase with increasing depth.

Sketching such graphs helps in quickly assessing and predicting pressure values for different depths.