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Chapter 1: Problem 69

GENERAL: Water Pressure At a depth of \(d\) feet underwater, the water pressure is \(p(d)=0.45 d+15\) pounds per square inch. Find the pressure at: a. The bottom of a 6 -foot-deep swimming pool. b. The maximum ocean depth of 35,000 feet.

Short Answer

Expert verified
At 6 feet, pressure is 17.7 psi; at 35,000 feet, it's 15,765 psi.

Step by step solution

01

Understand the Pressure Formula

We are given a formula for water pressure underwater: \( p(d) = 0.45d + 15 \), where \( d \) is the depth in feet, and \( p(d) \) is the pressure in pounds per square inch.
02

Calculate Pressure at 6 Feet

To find the pressure at the bottom of a 6-foot-deep swimming pool, substitute \( d = 6 \) into the formula: \( p(6) = 0.45 \times 6 + 15 \).
03

Solve for Pressure at 6 Feet

Calculate \( 0.45 \times 6 = 2.7 \). Add this result to 15: \( 2.7 + 15 = 17.7 \). Thus, the pressure at 6 feet is 17.7 pounds per square inch.
04

Calculate Pressure at 35,000 Feet

To find the pressure at the maximum ocean depth of 35,000 feet, substitute \( d = 35,000 \) into the formula: \( p(35,000) = 0.45 \times 35,000 + 15 \).
05

Solve for Pressure at 35,000 Feet

Calculate \( 0.45 \times 35,000 = 15,750 \). Add this result to 15: \( 15,750 + 15 = 15,765 \). Thus, the pressure at 35,000 feet is 15,765 pounds per square inch.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Water Pressure Calculation
When diving into the world of water pressure, it's crucial to understand how it changes with depth. Water pressure is the force exerted by water per unit area. As you go deeper underwater, the pressure increases due to the weight of the water above you. This is why the formula given in the exercise, \( p(d) = 0.45d + 15 \), helps us calculate the pressure based on depth \( d \) in feet.

The constant \( 0.45 \) represents how much the pressure increases for each foot of depth. Meanwhile, the constant 15 represents the baseline pressure that exists even at the surface, before considering depth. This is essential for calculations at shallow depths or when accounting for atmospheric pressures close to the surface. By applying this formula, we can easily determine the pressure at any given underwater depth by substituting \( d \) into the equation.
Mathematical Formulas
Mathematical formulas are expressions that represent relationships between different variables. In our exercise, the formula \( p(d) = 0.45d + 15 \) directly helps us understand how depth relates to water pressure. This formula is a linear equation, meaning it graphs as a straight line, and it shows a direct proportion between depth \( d \) and pressure \( p(d) \).

Formulas like this use coefficients, which are the numbers multiplying the variables, to describe rates of change. Here, the coefficient 0.45 describes how rapidly pressure increases with depth. To use the formula, simply substitute the known value for the variable \( d \), and perform the arithmetic operation to find \( p(d) \).

Working with formulas requires careful attention to detail to put the right values in place, ensuring that the calculations lead to an accurate understanding of the scenario.
Function Substitution
Function substitution involves replacing a variable in an equation or formula with a specific value to calculate a result. In this exercise, we substitute different values for \( d \) in the formula \( p(d) = 0.45d + 15 \) to find the water pressure at various depths. Function substitution is straightforward: simply take your known value, like a given depth, and insert it into the function.

For example, to find the pressure at a depth of 6 feet, we set \( d = 6 \) in the formula, calculate the product \( 0.45 \times 6 = 2.7 \), and then add 15 to find \( p(6) = 17.7 \). Similarly, to find the pressure at 35,000 feet, we substitute \( d = 35,000 \), compute \( 0.45 \times 35,000 = 15,750 \), and add 15, resulting in \( p(35,000) = 15,765 \).

Function substitution is a fundamental technique in calculus and algebra, allowing for precise calculations and problem-solving.

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