Problem 92 - Introductory Algebra for College Students

Question

The formula $$p=15+\frac{5 d}{11}$$ describes the pressure of sea water, \(p,\) in pounds per square foot, at a depth of d feet below the surface. Use the formula to solve Exercises. At what depth is the pressure 20 pounds per square foot?

Step-by-Step Solution

Answer

The pressure is 20 pounds per square foot at a depth of 11 feet beneath the sea surface

Step 1: Write down the given formula

Given the formula: \(p=15+\frac{5 d}{11}\)

Step 2: Substitute the value of p from the exercise into the formula

From the exercise, it is known that the pressure \(p = 20\) pounds per square foot. Substituting this into the formula gives: \(20 = 15+\frac{5d}{11}\)

Step 3: Solve for d

Rearrange the formula for depth (d), by first subtracting 15 from both sides of the equation. The equation becomes: \(5=\frac{5d}{11}\). Next, solve for d by multiplying both sides by 11 and dividing by 5, which gives us: \(d = \frac{5*11}{5} = 11\) feet

Key Concepts

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and at least one arithmetic operation (like addition, subtraction, multiplication, division). In the context of our exercise, the expression given is \( p=15+\frac{5d}{11} \), which represents the relationship between pressure (p) and depth (d). The terms within the expression each serve a specific purpose:

• \(15\) acts as the baseline pressure at the surface,
• \(\frac{5d}{11}\) signifies the additional pressure per foot of depth below the surface.

Understanding how to manipulate this expression is key to solving for a specific variable. In algebra, we often aim to isolate our variable of interest, in this case depth (d), to find its value when given certain conditions. Algebraic expressions are at the heart of translating real-world scenarios into a language that we can compute, which is precisely what we're looking at with the pressure at certain depths in seawater.

Problem-solving in Algebra

Effective problem-solving in algebra typically involves a set of strategic steps. Here's how it applies to our exercise:

1. Understanding the problem: You identify that you need to find the depth (d) when the pressure (p) is 20 pounds per square foot.
2. Devising a plan: You decide to use the given algebraic expression to express d in terms of p.
3. Carrying out the plan: Substitute the known value of p into the expression, rearrange and resolve for d.
4. Reviewing/Extending: After finding the depth, you might check if the answer makes sense within the context or apply the formula to different values of pressure.

Our step-by-step solution provided you with a clear pathway to isolating d and calculating its value. This routine offers a structured method that, when practiced, can make problem-solving in algebra more intuitive over time.

Pressure and Depth Relationship

The relationship between pressure and depth is directly proportional in a fluid; as depth increases, so does the pressure. In our exercise, this relationship is captured by the formula \( p=15+\frac{5d}{11} \), where p is the pressure in pounds per square foot, and d is the depth in feet. The constant 15 represents the pressure at sea level or the surface, and the fraction \( \frac{5}{11} \) reflects how pressure increases with every foot below the surface.

To better grasp this concept, imagine submerging deeper into the ocean; for every foot you descend, the water above you weighs more and thus exerts more pressure on you. This concept is crucial in fields such as oceanography and engineering, where understanding how pressure changes with depth can help in designing submarines or predicting the force exerted on underwater structures.