Problem 13 - The Calculus with Analytic Geometry

Question

The bottom of a swimming pool is an inclined plane. The pool is \(2 \mathrm{ft}\) deep at one end and \(8 \mathrm{ft}\) deep at the other. If the width of the pool is \(25 \mathrm{ft}\) and the length is \(40 \mathrm{ft}\), find the total force due to liquid pressure on the bottom.

Step-by-Step Solution

Answer

The total force due to liquid pressure on the bottom is approximately 10,036,800 lbs.

Step 1: Determine the Depth Function

The depth of the pool varies linearly from 2 ft to 8 ft over the length of 40 ft. Form the linear depth function, which is a line equation starting at (0,2) and ending at (40,8). The equation of the line is: \( d(x) = 2 + \frac{8-2}{40-0}x = 2 + \frac{6}{40}x = 2 + \frac{3}{20}x \), where \(d(x)\) is the depth at any point x along the length.

Step 2: Force Due to Liquid Pressure Formula

The force due to liquid pressure at a depth \(d(x)\) is given by \( F = \rho g d(x) A \), where \( \rho \) is the density of water (62.4 lbs/ft³), \( g \) is the acceleration due to gravity (32.2 ft/s²), \( d(x) \) is the depth, and \( A \) is the area.

Step 3: Calculate the Area Strip

Divide the pool into infinitesimal strips of width \(dx\) and length (width of pool) 25 ft. The area of each strip is \( dA = 25 \, dx \).

Step 4: Integrate the Pressure Over the Length

Since the depth varies, integrate the pressure force over the entire length from 0 to 40 ft: \( F = \int_{0}^{40} \rho g d(x) 25 \, dx \). Substitute \( d(x) = 2 + \frac{3}{20}x \) into the integral: \[ F = \int_{0}^{40} 62.4 \times 32.2 (2 + \frac{3}{20}x) 25 \, dx \].

Step 5: Compute the Integral

Evaluate the integral: \[ F = 62.4 \times 32.2 \times 25 \times \int_{0}^{40} (2 + \frac{3}{20}x) \, dx \]Compute the integral for \(2 + \frac{3}{20}x \):\[ \int_{0}^{40} (2 + \frac{3}{20}x) \, dx = \left[ 2x + \frac{3}{40}x^2 \right]_0^{40} \]\[ 2(40) + \frac{3}{40} (40)^2 = 80 + 120 = 200 \].Now, multiply back the constants: \[ F = 62.4 \times 32.2 \times 25 \times 200 \].

Step 6: Calculate the Result

Calculate the result of the multiplication:\[ F = 62.4 \times 32.2 \times 25 \times 200 \approx 10,036,800 \, \text{lbs} \].

Key Concepts

Depth Function

To understand how the depth of the pool changes, we start by thinking about the pool's shape. The depth changes from 2 feet at one end to 8 feet at the other end. This change occurs evenly over the pool's length, which is 40 feet. We can model this change with a linear function.

A linear function means the depth changes at a constant rate. At the starting point (0 feet along the pool), the depth is 2 feet. At the end (40 feet along the pool), the depth is 8 feet.

Using these points, we can write the depth function as:
\( d(x) = 2 + \frac{6}{40} \times x \)
where \( x \) is the distance from the shallow end. We simplify this to: \( d(x) = 2 + \frac{3}{20} \times x \).

This equation tells us the depth at any point along the length of the pool. For example, halfway down the pool (\( x = 20 \)):
\( d(20) = 2 + \frac{3}{20} \times 20 = 5 \) feet.

So, 20 feet along the pool, the water is 5 feet deep.

Force Due to Liquid Pressure

Liquid pressure is the force exerted by a fluid per unit area. In our case, we need to calculate the force due to the water pressing down on the pool's inclined bottom. The pressure at a certain depth increases with depth, and water has a known density.

The formula for the force due to liquid pressure is:
\[ F = \rho \times g \times d(x) \times A \]
Where:

• \( \rho \) is the density of water (62.4 lbs/ft³)
  
• \( g \) is the acceleration due to gravity (32.2 ft/s²)
  
• \( d(x) \) is the depth function we calculated earlier
  
• \( A \) is the area over which the pressure acts
  

Since the pressure varies with depth, we consider thin strips of the pool's bottom and sum the forces over the entire length.

For an infinitesimally small strip, the area of the strip is: \[ dA = 25 \times dx \] where 25 is the width of the pool and \( dx \) is the infinitesimal length along the pool,
soaking the pressure over these tiny strips of pool bottom.

Definite Integral Calculation

To find the total force on the pool bottom, we integrate the force over the entire length of the pool. This gives us the sum of the forces acting on each tiny strip.

The integral we need is:
\[ F = \rho \times g \times 25 \times \frac{(\text{depth function})}{dx} \times \frac{length}{\text{range}} \]

Substituting the depth function \( d(x) \) into the integral, we have:
\[ F = \rho \times g \times 25 \times \frac{(2 + \frac{3}{20}x)}{\text{range from 0 to 40}} \]

This simplifies to:
\[ F = 62.4 \times 32.2 \times 25 \times \frac{( \text{{\text{}}2 + \frac{3}{20}}x)}{dx} = \frac{40 - 0 \times 2.0( 0 - 40)}{\text{\text{0 from thin strips}}} \]
Now we evaluate the integral:

The integral of \( (2 + \frac{3}{20}x) \) from 0 to 40 is:
\[ 2 \times x + \frac{3}{40} \times x^2 /length of range)_{0}^{40} \rightarrow 2 (\text{{40}}) + \frac{\text{3 {{40}}} \times x^2{40}} = 80 + 120 = 2 (\times 200 conjectureinteger} \]

Lastly, we multiply by the constants:
\[ F = 62.4 \times 32.2 \times 25 \times 200 \]

Therefore, the total force is approximately \( 10, 036,800 \text{lbs} \).