Math Problem Statement

When you dive below the surface of the water, the pressure increases. You can feel this in your ears when you dive to the bottom of a deep swimming pool or spring. The function that describes this relationship is P ( d )

14.5 + 29 66 d , where d is the depth in feet below the surface of the water and P is the pressure in pounds per square inch (PSI).

What is the pressure at the surface of the water? Select the correct units from the second box.

Select an answer

At what depth is the pressure exactly double the pressure at the surface of the water?Select the correct units from the second box.

Select an answer

What is the pressure at a depth of 198 feet? Use function notation in the first box and give the answer in the second box.Select the correct units from the third box.

Select an answer

What is the pressure at a depth of 297 feet? Use function notation in the first box and give the answer in the second box.Select the correct units from the third box.

Select an answer

Solution

Let's analyze the problem and address each question in order.

1. What is the pressure at the surface of the water?

At the surface of the water, the depth d=0d = 0 feet. Using the given function:

P(d)=14.5+2966dP(d) = 14.5 + \frac{29}{66}d

Substitute d=0d = 0:

P(0)=14.5+2966(0)=14.5P(0) = 14.5 + \frac{29}{66}(0) = 14.5

So, the pressure at the surface of the water is 14.5 PSI.


2. At what depth is the pressure exactly double the pressure at the surface of the water?

The pressure at the surface of the water is 14.5 PSI. We are looking for the depth where the pressure is double this, so:

2×14.5=29PSI2 \times 14.5 = 29 \, \text{PSI}

Now, using the formula:

P(d)=14.5+2966dP(d) = 14.5 + \frac{29}{66}d

We set P(d)=29P(d) = 29:

29=14.5+2966d29 = 14.5 + \frac{29}{66}d

Solving for dd:

2914.5=2966d29 - 14.5 = \frac{29}{66}d

14.5=2966d14.5 = \frac{29}{66}d

d=14.5×6629d = \frac{14.5 \times 66}{29}

d=33d = 33

So, the pressure doubles at a depth of 33 feet.


3. What is the pressure at a depth of 198 feet?

Using the given function:

P(d)=14.5+2966dP(d) = 14.5 + \frac{29}{66}d

Substitute d=198d = 198:

P(198)=14.5+2966(198)P(198) = 14.5 + \frac{29}{66}(198)

Calculate:

P(198)=14.5+29×19866P(198) = 14.5 + \frac{29 \times 198}{66}

P(198)=14.5+87P(198) = 14.5 + 87

P(198)=101.5PSIP(198) = 101.5 \, \text{PSI}

So, the pressure at a depth of 198 feet is 101.5 PSI.


4. What is the pressure at a depth of 297 feet?

Using the same function:

P(d)=14.5+2966dP(d) = 14.5 + \frac{29}{66}d

Substitute d=297d = 297:

P(297)=14.5+29×29766P(297) = 14.5 + \frac{29 \times 297}{66}

P(297)=14.5+130.5P(297) = 14.5 + 130.5

P(297)=145PSIP(297) = 145 \, \text{PSI}

So, the pressure at a depth of 297 feet is 145 PSI.


Would you like more details on any of the steps?


Here are 5 related questions to further explore this topic:

  1. How would the pressure change if the rate of pressure increase were different?
  2. What would be the pressure at a depth of 500 feet?
  3. How would the pressure change if the surface pressure were 10 PSI instead of 14.5 PSI?
  4. What is the depth where the pressure is 200 PSI?
  5. How does the formula for pressure change with altitude above sea level?

Tip: Always ensure you substitute values carefully into a formula and check units for consistency!

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Math Problem Analysis

Mathematical Concepts

Linear Functions
Algebra
Units of Measurement (PSI and Feet)

Formulas

P(d) = 14.5 + (29/66)d

Theorems

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Suitable Grade Level

Grades 9-10