Problem 91 - Introductory and Intermediate Algebra for College Students 4th

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. The record depth for breath-held diving, by Francisco Ferreras (Cuba) off Grand Bahama Island, on November \(14,1993,\) involved pressure of 201 pounds per square foot. To what depth did Ferreras descend on this illadvised venture? (He was underwater for 2 minutes and 9 seconds!

Step-by-Step Solution

Answer

The depth Francisco Ferreras descended to is calculated to be approximately 407.2 feet.

Step 1: Substitute the given pressure into the formula

We can replace \( p \) in the formula with 201: \( 201 = 15 + \frac{5d}{11} \)

Step 2: Isolate d

To isolate 'd', first we can subtract 15 from both sides of the equation: \( 201 - 15 = 15 - 15 + \frac{5d}{11} \). Simplifying, we get \( 186 = \frac{5d}{11} \). Then, we can multiply both sides by 11 to get rid of the denominator: \( 186 \cdot 11 = 5d \).

Step 3: Solve for d

Finally, to solve for 'd', divide both sides by 5: \( \frac{186 \cdot 11}{5} = d \)

Key Concepts

Pressure formula

Understanding pressure in relation to depth is essential in various fields, such as underwater diving and marine biology. In our exercise, we use the pressure formula that relates sea water pressure, \( p \), to the depth, \( d \), under the ocean surface. The specific formula given is:\[ p = 15 + \frac{5d}{11} \]Here, pressure \( p \) is measured in pounds per square foot. The formula accounts for the atmospheric surface pressure (15 pounds per square foot) and the additional pressure caused by the water column above, which is calculated by the term \( \frac{5d}{11} \). This kind of expression is crucial for safely calculating conditions in deep-sea environments. Understanding this formula can help divers and engineers predict pressure conditions at various depths.

Isolating variables

Isolating a variable involves manipulating an equation so that the variable of interest stands alone on one side of the equation, with all other terms on the opposite side. This step is key when solving for unknowns as it allows us to neatly isolate what we are solving for.In our exercise, we need to isolate \( d \) from the initial equation \( 201 = 15 + \frac{5d}{11} \). To achieve this, we start by removing constants from the side of the equation containing \( d \). We do this by subtracting 15 from both sides, resulting in:\[ 186 = \frac{5d}{11} \]Next, by knowing these steps, it becomes easier to manage equations, simplify steps, and focus on the desired variable. The skill to isolate variables is immensely useful when tackling any algebraic equation.

Solving equations

The process of solving equations is fundamental in algebra, allowing us to find the value of unknown variables. After isolating the desired variable, the goal is to simplify the equation further to solve for this variable completely.In our example, we left off with the simplified equation:\[ 186 = \frac{5d}{11} \]To solve for \( d \), we must eliminate the fraction by performing the inverse operation, which, in this case, is multiplication. Specifically, multiplying both sides by 11, we get:\[ 186 \times 11 = 5d \]Finally, to isolate \( d \), divide both sides by 5, resulting in:\[ d = \frac{186 \times 11}{5} \]With this process, we've solved the equation, finding the depth \( d \) corresponding to the pressure of 201 pounds per square foot, illustrating the practical applications of algebra in real-world scenarios.

Mathematical substitution

Mathematical substitution is a critical strategy where known values replace variables in a formula, allowing for straightforward calculation and problem-solving. This technique simplifies an equation by using given numerical values in place of the unknowns.In our exercise, substitution gave us a new equation by replacing \( p \) with 201:\[ 201 = 15 + \frac{5d}{11} \]This simplification transforms a word problem or complex formula into a more manageable arithmetic form. By systematically substituting known quantities, we bridge the gap from theoretical formulas to practical solutions, offering a powerful tool in algebra that assists with clarity and comprehension in solving mathematical problems.