Problem 103 - Algebra and Trigonometry

Question

Volume of a Rocket A rocket consists of a right circular cylinder of height 20 \(\mathrm{m}\) surmounted by a cone whose height and diameter are equal and whose radius is the same as that of the cylindrical section. What should this radius be (rounded to two decimal places) if the total volume is to be 500\(\pi / 3 \mathrm{m}^{3}\) ?

Step-by-Step Solution

Answer

The radius should be approximately 4.88 meters.

Step 1: Understand the Problem

We need to find the radius of a rocket which is a combination of a cylinder and a cone. The cylinder has a height of 20 meters, and the cone has a height equal to its diameter. The total volume of this rocket is given as \( \frac{500\pi}{3} \ \text{m}^3 \).

Step 2: Recall Formula for Volume of Cylinder and Cone

The volume of a cylinder is given by \( V_{\text{cyl}} = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. The volume of a cone is \( V_{\text{cone}} = \frac{1}{3} \pi r^2 h \). Here, the height of the cone is equal to \( 2r \) since the height equals the diameter.

Step 3: Write the Total Volume Equation

The total volume of the rocket is the sum of the volumes of the cylinder and the cone: \( \frac{500\pi}{3} = \pi r^2 \cdot 20 + \frac{1}{3} \pi r^2 \cdot 2r \).

Step 4: Simplify the Equation

Simplify the equation for total volume: \[ \frac{500\pi}{3} = 20\pi r^2 + \frac{2}{3} \pi r^3 \].Divide everything by \( \pi \) to get: \[ \frac{500}{3} = 20r^2 + \frac{2}{3} r^3 \].

Step 5: Multiply the Equation to Clear Fractions

Multiply every term by 3 to clear the fractions: \[ 500 = 60r^2 + 2r^3 \].

Step 6: Rearrange the Equation and Simplify

Rearrange to form a polynomial equation: \[ 2r^3 + 60r^2 - 500 = 0 \].

Step 7: Solve the Polynomial Equation

Solve the polynomial equation \( 2r^3 + 60r^2 - 500 = 0 \) for \( r \). Using trial and error, synthetic division, or numerical methods, find \( r \approx 5 \).

Step 8: Round the Radius to Two Decimal Places

The solutions may not always be precise psychological estimations. Therefore, ensure the radius is accurate to two decimal places, giving the final answer. \( r \approx 4.88 \).

Step 9: Verify the Calculation

Substitute \( r = 4.88 \) back into the volumes of the cylinder and cone to verify that the total volume equals \( \frac{500\pi}{3} \).

Key Concepts

Cylindrical Volume

To understand the concept of cylindrical volume, picture a can sitting upright. It has a base that is a perfect circle and is stretched up equally all around to form a height.
The formula for finding the volume of a cylinder is:

• \( V_{\text{cyl}} = \pi r^2 h \)

where:

• \( r \) represents the radius of the circular base
• \( h \) is the height of the cylinder

The equation essentially calculates how much space is contained within the cylindrical shape by multiplying the area of its base by how tall the shape extends.
This formula is foundational when solving problems that involve composite shapes, like rockets, where the total volume is a sum of individual parts. By understanding cylindrical volume, students can dissect complex shapes into manageable calculations.

Conical Volume

A cone can be visualized as an ice cream cone or a party hat. It's characterized by a circular base that converges to a point at the top, forming a three-dimensional shape.
To calculate the volume of a cone, you use:

• \( V_{\text{cone}} = \frac{1}{3} \pi r^2 h \)

where:

• \( r \) is the radius of the base
• \( h \) is the height, extending from the base to the point

The factor of \( \frac{1}{3} \) arises because a cone occupies one-third the volume of a cylinder with the same base and height.
In our rocket problem, the cone's height matches its diameter, implying that \( h = 2r \). Understanding the conical volume's calculation lets us integrate it smoothly with the cylindrical part to ascertain the total volume of hybrid shapes.

Polynomial Equations

Polynomial equations appear frequently when dealing with volumes of combined solids. They are equations involving variables raised to whole number powers, often leading to solutions that are not immediately obvious or intuitive without algebraic manipulation.
In our exercise, simplifying the total volume yields the polynomial equation:

• \( 2r^3 + 60r^2 - 500 = 0 \)

This equation is a third-degree polynomial or cubic equation. Solving such equations often requires techniques like:

• Trial and error to estimate possible solutions
• Synthetic division or
• Numerical methods to find precise values of \( r \)

It's important to recognize that polynomial equations describe relationships where variables combine in more complex ways, and finding precise solutions is essential to solving volume and geometry problems accurately. This mastery allows students to apply their skills in engineering, architecture, and beyond.