Problem 23 - Algebra 1: Concepts and Skills

Question

A scuba diver starts at sea level. The pressure on the diver at a depth of \(d\) feet is given by the formula below, where \(P\) represents the total pressure in pounds per square foot. $$P=64 d+2112.$$ If the current pressure on a diver is 4032 pounds per square foot, what is the diver's current depth?

Step-by-Step Solution

Answer

The diver's current depth is 30 feet

Step 1: Set the pressure formula equal to the known pressure

The pressure formula is \(P=64d+2112\) where \(P\) represents the total pressure in pounds per square foot and \(d\) is the depth in feet. Given the current pressure \(P=4032\) pounds per square foot, we set up the equation \(64d+2112 = 4032\)

Step 2: Solve for \(d\)

To solve for \(d\), the first step is to isolate \(d\) by subtracting 2112 from both sides of the equation, we have: \(64d = 4032 - 2112\). Then, simplify the right side of the equation: \(64d = 1920\). Finally, divide both sides of the equation by 64. We get: \(d = 1920 / 64 = 30\)

Step 3: Interpret the solution

Since the depth \(d\) is calculated as 30, this indicates that the diver's current depth is 30 feet

Key Concepts

Solving Equations

Solving equations is like finding the correct combination to a lock; each step you take gets you closer to unlocking the solution. In the exercise problem, we first encountered an equation that relates pressure to depth through this formula: \( P = 64d + 2112 \), where \(P\) is the total pressure in pounds per square foot, and \(d\) represents the depth in feet.
When we want to find out how deep the diver is under the sea given a specific pressure, we need to solve the equation for \(d\). Solving involves:

• Isolating the Variable: Our goal is to get \(d\) on its own on one side of the equation.
• Arithmetic Operations: Perform basic operations such as addition, subtraction, multiplication, or division to both sides of the equation to achieve our goal.

Let's see how these steps were applied: First, the equation was set up to be \(64d + 2112 = 4032\). Next, the constant 2112 was subtracted from both sides. This shifted the equation to \(64d = 1920\). Finally, by dividing each side by 64, the depth \(d\) was isolated, providing the solution \(d = 30\). This suggests that each step uncovers more pieces of the puzzle, helping us understand and find the unknown value.

Pressure Calculations

Understanding pressure calculations, especially under water, is crucial for divers. When diving, a person experiences pressure changes due to water overhead. The equation \(P = 64d + 2112\) models this scenario, demonstrating how pressure increases with depth.
Pressure in this case is dependent on two factors:

• Depth \(d\): As the diver goes deeper, the value of \(d\) increases, resulting in a higher pressure because water exerts more force the deeper you go.
• Surface Pressure Constant \(2112\): This represents the existing pressure at sea level due to the atmosphere.

Given the equation, once the specific pressure is known — like 4032 pounds per square foot — solving for \(d\) can tell us how far down the diver is. This practical application of pressure calculations informs divers and helps ensure their safety by understanding the environmental forces experienced under water.

Algebraic Formulas

Algebraic formulas are like recipes. They help us calculate unknowns in various situations by systematically involving known quantities. The formula \(P = 64d + 2112\) used in the scuba diving problem is a typical algebraic setup that connects pressure, depth, and constants.
An important aspect of algebraic formulas is:

• Coefficient \(64\): Represents how much the pressure increases with each unit of depth. It directly defines the relationship strength between the variables \(d\) and \(P\).
• Constant Term \(2112\): Serves as a starting point in the calculation, equivalent to the pressure at the surface.

The benefit of algebraic formulas is their ability to be rearranged to solve for different quantities. In this case, rearranging the formula to make \(d\) the subject provided actionable information about the diver’s real-world depth. Therefore, algebra in these contexts isn't just abstract math, but a tool for solving practical problems, illustrating its power beyond the classroom.