Problem 78 - Algebra and Trigonometry

Question

Radius of a Tank A spherical tank has a capacity of 750 gallons. Using the fact that one gallon is about \(0.1337 \mathrm{ft}^{3},\) find the radius of the tank (to the nearest hundredth of a foot).

Step-by-Step Solution

Answer

The radius of the tank is approximately 2.88 feet.

Step 1: Convert Gallons to Cubic Feet

First, we need to convert the volume of the tank from gallons to cubic feet using the given conversion factor. Since 1 gallon is approximately equal to 0.1337 cubic feet, we can find the volume in cubic feet by multiplying the volume in gallons by this conversion factor. Thus, the volume of the tank in cubic feet is given by \( 750 \times 0.1337 = 100.275 \ ext{cubic feet} \).

Step 2: Use the Volume Formula for a Sphere

The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( V \) is the volume and \( r \) is the radius. We need to solve for \( r \) given that \( V = 100.275 \, \text{cubic feet} \).

Step 3: Rearrange the Formula

To find \( r \), we need to rearrange the formula as follows: \( r^3 = \frac{3V}{4\pi} \). Substituting \( V = 100.275 \), we get: \( r^3 = \frac{3 \times 100.275}{4\pi} \).

Step 4: Calculate Radius Cubed

Continuing from the previous step, calculate \( r^3 \): \( r^3 = \frac{300.825}{4\pi} \approx \frac{300.825}{12.566} \approx 23.94 \).

Step 5: Calculate the Radius

Now solve for \( r \) by taking the cube root of \( 23.94 \). This gives us \( r \approx 2.88 \).

Step 6: Round the Radius

Finally, we round the radius to the nearest hundredth, obtaining \( r \approx 2.88 \).

Key Concepts

Volume Conversion

To solve problems involving volume, it's often necessary to convert measurements from one unit to another. When working with liquid volumes such as gallons, conversion to cubic measurements is crucial in solid geometry contexts. Given that 1 gallon is approximately 0.1337 cubic feet, you multiply the volume in gallons by this factor to convert to cubic feet.
For instance, if a spherical tank holds 750 gallons, the conversion to cubic feet is accomplished by:

• Calculating the volume in cubic feet:
  • \(750 imes 0.1337 = 100.275 \) cubic feet

This conversion helps in applying geometric formulas that require volume in cubic feet. Being familiar with such conversions is essential for accurate calculations.

Volume of a Sphere

The volume of a sphere is determined by a specific formula that involves its radius.
The formula is:

• \( V = \frac{4}{3} \pi r^3 \)

where \( V \) is the volume and \( r \) is the radius.
In this problem, to find the radius of the tank, we need to know the sphere's volume in cubic feet. Once you have the volume, you substitute it into the formula and rearrange to solve for the radius. Understanding how to manipulate this formula is key in geometry, helping explore the relationship between a sphere's size and its volume.

Cubic Feet

Cubic feet is a standard unit of volume measurement used particularly in contexts involving three-dimensional spaces. It's a useful unit when dealing with volumes of solid objects, like tanks or rooms.
When the problem states a volume in gallons, converting to cubic feet is often necessary. After conversion, it provides a straightforward method to apply formulas, such as the volume formula for a sphere.

• Knowing the volume as 100.275 cubic feet in this context makes calculation practical and feasible.

Understanding how to work with cubic feet is crucial in scientific and engineering disciplines.

Radius Calculation

Finding the radius of a sphere when given its volume involves algebra and proportional reasoning. Starting with the spherical volume formula,

• \( V = \frac{4}{3} \pi r^3 \)

we first rearrange to solve for \( r \) (the radius):

• \( r^3 = \frac{3V}{4\pi} \)

Substitute the given volume (e.g., 100.275 cubic feet) into the equation to calculate \( r^3 \). The next step involves calculating the cube root of the result to find \( r \).

• Taking the cube root of \( 23.94 \), you obtain \( r \approx 2.88 \).

Finally, you round the answer as necessary.
This step-by-step logic will help tackle complex volume problems involving spherical shapes.