Problem 53 - Elementary Algebra

Question

A hemispherical water tank of radius 6 feet has water dripping into it. The equation relating the volume, \(V\), of water in the tank at any time is \(V=6 \pi h^{2}-\frac{\pi}{3} h^{3},\) where \(h\) represents the depth of the water. Using 3.14 to approximate the irrational number \(\pi,\) determine the volume of water in the tank when the depth of the water is 3 feet.

Step-by-Step Solution

Answer

Answer: Approximately 141.3 cubic feet.

Step 1: Write down the given equation

The given equation relating the volume \(V\) and depth \(h\) is: \[V = 6\pi h^2 - \frac{\pi}{3} h^3\]

Step 2: Substitute the given values

We are asked to find the volume when the depth of water is 3 feet, so substitute \(h=3\) and \(\pi \approx 3.14\) in the equation: \[V = 6(3.14)(3^2) - \frac{3.14}{3}(3^3)\]

Step 3: Simplify and solve for the volume

Perform the arithmetic operations to calculate the volume: \[V = 6(3.14)(9) - (1.0467)(27)\] \[V = 169.56 - 28.2609\] \[V \approx 141.2991\]

Step 4: Interpret the result

The volume of water in the tank when the depth of the water is 3 feet is approximately 141.3 cubic feet.

Key Concepts

Hemispherical Tank

A hemispherical tank has a shape that is half of a full sphere, with a flat base and a rounded surface. This design is commonly used in storage applications for liquids, including water. The unique geometry of the hemisphere affects how its volume is calculated.
The formula for the volume of a full sphere is \( \frac{4}{3} \pi r^3 \), where \( r \) is the radius. However, since the tank is hemispherical, only half of the sphere is used. This means we consider only half of the volume of a full sphere. Therefore, it is important to adapt the formulas to account for the hemispherical shape when trying to calculate the volume of liquid such as water based on a specific depth.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operators. In the given exercise, we are working with the expression \( V = 6\pi h^2 - \frac{\pi}{3} h^3 \). This expression describes the relationship between the volume of water, \( V \), in the tank and the depth, \( h \.\)Understanding algebraic expressions allows us to find how changing one variable (such as depth, \( h \)) affects another (like volume, \( V \)).
For instance:

• \( V = 6\pi h^2 \) represents the increase in volume with the depth squared, showing a quick increase as depth increases.
• \( - \frac{\pi}{3} h^3 \) suggests a reduction proportional to the cube of the depth, impacting volume significantly as depth keeps increasing, which helps in accommodating the spherical shape.

Depth Measurement

Depth measurement plays a key role in calculating the volume of water in a tank. The given equation uses depth, \( h \), to determine how much water is present by directly plugging it into the expression.
To understand depth measurement, consider depth as the vertical distance from the water surface to the bottom of the tank. This directly influences the calculated volume using the formula \( V = 6\pi h^2 - \frac{\pi}{3} h^3 \.\)When you know the depth, you can substitute this value into the equation to find the current volume of water. This principle applies to various shapes of tanks, not just hemispherical ones, by using suitable formulas. This method ensures accurate monitoring and management of the tank's content.