Problem 13 - Algebra 2

Question

A spherical water tank holds \(15,000 \mathrm{ft}^{3}\) of water. Find the diameter of the tank. (Hint: \(V=\frac{\pi}{6} d^{3}\))

Step-by-Step Solution

Answer

The diameter \(d\) of the tank is the cube root of \(\frac{15,000 \times 6}{\pi}\), which calculates to approximately 29.04 feet.

Step 1: Write down the given

The volume of the water tank is given as \(15,000 \mathrm{ft}^{3}\). The mathematical formula for the volume of a sphere is given as \(V=\frac{\pi}{6} d^{3}\).

Step 2: Substitute the given volume into the formula

Substitute \(V = 15,000 \mathrm{ft}^{3}\) into the formula. This gives us the equation \[15,000 = \frac{\pi}{6} d^{3}\].

Step 3: Resolve for \(d^{3}\)

To isolate \(d^{3}\), multiply both sides by \(\frac{6}{\pi}\). Simplifying gives \(d^{3} = \frac{15,000 \times 6}{\pi}\).

Step 4: Solve for \(d\)

To get \(d\), take the cube root from both sides of the equation: \[d= \sqrt[3]{\frac{15,000 \times 6}{\pi}}\].

Step 5: Calculate for \(d\)

Perform the calculation to get the numerical value of the diameter. Ensure your answer is in correct units (feet).

Key Concepts

Spherical Tank

A spherical tank is a three-dimensional round structure that usually stores liquids, typically water or other fluids. Spherical tanks are preferred in many applications due to their natural strength and ability to handle high pressure evenly distributed across the shape. This makes them economically efficient for large volume storage, as the surface area is minimized compared to the volume it can hold.
For example, spherical tanks are widely used in industries and municipal water storage systems. When dealing with these tanks, it’s essential to understand the parameters defining their shape, such as volume and surface area. The volume is calculated using a specific geometric formula, allowing us to determine other dimensions like the radius or diameter.

Diameter Calculation

In the context of solving geometric problems, particularly involving spheres, determining the diameter is an important step. The diameter refers to the longest straight line that passes through the center of a circle or a sphere, connecting two points on its surface.
The exercise provided gives a hint: the formula for the volume of a sphere is \(V=\frac{\pi}{6} d^{3}\). Using this formula, you can solve for the diameter if the volume is known. Substituting the known volume into the equation, you can isolate the diameter using algebraic manipulation.
The key steps to calculate the diameter from the volume are:

• Use the volume formula to create the equation.
• Rearrange to isolate \(d^{3}\).
• Take the cube root to solve for \(d\).

Knowing the diameter allows you to understand more about the sphere's size, making it an essential measure for comparing different spherical tanks.

Cubic Equations

Cubic equations involve variables raised to the third power, typically seen in equations of the form \( ax^3 + bx^2 + cx + d = 0 \). These equations are called "cubic" because the highest power of the variable present is three. This makes them one step more complex than quadratic equations.
The original exercise uses a reduced form of a cubic equation as part of solving for the diameter of a sphere, where the equation \( V=\frac{\pi}{6} d^{3} \) simplifies to an expression containing \(d^{3}\). Understanding cubic equations helps us see that the diameter is a root of the expression when rearranged.
To solve cubic equations, we often:

• Isolate the cubic term.
• Perform inverse operations like taking a cube root.
• Apply algebraic techniques or numerical methods if required.

Knowing how to manipulate cubic equations is beneficial when dealing with real-world geometric problems like those involving spherical tanks.

Geometry Problem Solving

Geometry problem-solving involves understanding and applying geometric principles to find unknown parameters within figures or structures. In this exercise, we apply formulas for a sphere's volume to determine the diameter, showcasing how geometric problem-solving is often formula-driven.
Successful geometry problem-solving requires:

• Knowing relevant formulas and when to use them.
• Visualizing the problem to better understand the spatial relationships.
• Using logical reasoning and algebraic techniques to manipulate formulas.

In the case of the spherical tank, we took a known volume and worked backward using mathematical operations (like cube rooting) to solve for the diameter. This approach exemplifies logical sequences of thought critical in geometrical challenges that can be applied to numerous scenarios involving spheres or other shapes.