Problem 97 - 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 (correct to two decimal places) if the total volume is to be 500\(\pi / 3 \mathrm{m}^{3}\) ?

Step-by-Step Solution

Answer

The radius \( r \) is 5 meters.

Step 1: Understanding the Problem

The rocket is made up of two parts: a cylindrical base and a conical top. The dimensions of the cylinder are given (height = 20 m), and we know the radius of the cone is the same as the cylinder. The volume of the entire rocket is given as \( \frac{500\pi}{3} \text{ m}^3 \). We need to find the radius \( r \).

Step 2: Formula for Volume of a Cylinder

The volume \( V_c \) of a cylinder is given by the formula \( V_c = \pi r^2 h \), where \( r \) is the radius of the base and \( h \) is the height of the cylinder. For this problem, \( h = 20 \text{ m} \), so the volume of the cylindrical part is \( V_c = \pi r^2 \times 20 \).

Step 3: Formula for Volume of a Cone

The volume \( V_{co} \) of a cone is calculated using the formula \( V_{co} = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius of the base, and \( h \) is the height of the cone. In this case, the cone's height equals its diameter, which means \( h = 2r \). So the cone's volume is \( V_{co} = \frac{1}{3} \pi r^2 \times 2r = \frac{2}{3} \pi r^3 \).

Step 4: Total Volume Equation

Add the volumes of the cylinder and the cone to get the total volume of the rocket: \( V_{total} = V_c + V_{co} = 20\pi r^2 + \frac{2}{3}\pi r^3 \). Given that the total volume is \( \frac{500\pi}{3} \), we have the equation: \[ 20\pi r^2 + \frac{2}{3}\pi r^3 = \frac{500\pi}{3} \].

Step 5: Simplifying the Equation

Divide every term by \( \pi \) to simplify: \[ 20 r^2 + \frac{2}{3} r^3 = \frac{500}{3} \]. Next, multiply the entire equation by 3 to clear the fraction: \[ 60 r^2 + 2 r^3 = 500 \].

Step 6: Solving the Cubic Equation

Rewrite the equation as \( 2r^3 + 60r^2 - 500 = 0 \). Factor out \( 2 \) from each term: \[ r^3 + 30r^2 - 250 = 0 \]. Use trial and error or numerical methods to find the root of the cubic equation. Starting with trial, try \( r = 5 \): substituting \( r = 5 \) into the equation, we find that it satisfies it.

Step 7: Conclusion

After solving the cubic equation, we determine that the radius \( r \) that satisfies the volume condition (to two decimal places) is \( 5 \text{ m} \).

Key Concepts

cylinder volume

In geometry, a cylinder is a three-dimensional shape with two parallel circular bases joined by a curved surface. Calculating the volume of a cylinder is a straightforward process.
The formula for calculating the volume (V_c) of a cylinder is V_c = \pi r^2 h, where \( r \) is the radius of the base and \( h \) is the height.
This formula helps calculate how much space is inside the cylinder. To use the formula, simply substitute the given values of the radius and height into the formula.

• For example, if the height of the cylinder is 20 m (as in the case of our rocket), and the radius is variable, we express the volume as \( V_c = \pi r^2 \times 20 \).
• The result will give you the space occupied by the cylindrical part of the rocket.

You should always have the units in mind. Since we are using meters, our volume will be in cubic meters. This formula is essential in finding out how much material is needed to create the cylinder or to calculate capacity.

cone volume

A cone is another three-dimensional geometric shape, characterized by a circular base that tapers smoothly up to a point called the apex. The volume of a cone can be calculated using a specific formula which takes into account its base radius and height.
The formula for the volume (V_{co}) of a cone is given by V_{co} = \frac{1}{3} \pi r^2 h.
In our problem, the height of the cone is equal to its diameter, meaning \( h = 2r \). Therefore, the volume of the cone becomes \( V_{co} = \frac{1}{3} \pi r^2 \times 2r = \frac{2}{3} \pi r^3 \).

• This derived formula allows us to easily compute the cone's volume once we have the radius.
• Always ensure that all measurements are in the same units, here in meters, to get the volume in cubic meters.

This understanding is critical when combining the volumes of both the cone and cylinder to find the total volume of composite structures like the rocket in the exercise.

cubic equation

A cubic equation is any equation that can be represented in the form \( ax^3 + bx^2 + cx + d = 0 \). Cubics have at least one real solution, and solving them typically involves trial and error, factoring, or using numerical methods.
In the rocket volume problem, our cubic equation derived from the total volume was 2r^3 + 60r^2 - 500 = 0. Simplifying by factoring out 2, it became r^3 + 30r^2 - 250 = 0.
This equation lacks a simple factorization, so solving by trial and error started with trying reasonable values of \( r \).

• We found \( r = 5 \) satisfies the equation, providing a practical solution for our geometry problem.
• Solving cubic equations can be complex, and sometimes numerical methods or specialized algorithms are applied when simple substitution fails.

Successfully solving cubic equations allows us to determine unknown dimensions and is useful in various applications beyond geometry.

geometry problem

Geometry problems often involve calculating various attributes of shapes and objects. They require an understanding of space, dimensions, and forms. In this problem, combining knowledge of both cylindrical and conical volumes was necessary to solve the exercise.
The challenge is to understand how to break down complex shapes into simpler components for volume calculations.

• By recognizing the rocket as a composite shape made up of a cylinder and a cone, we could apply known formulas accordingly.
• The critical step involved setting up an equation which combined the volumes to match a given total.

Tackling geometry problems involves logical thinking and applying mathematical concepts accurately. The process of deriving the radius for the rocket's shape required integrating the volumes into a single cohesive equation. This is a foundational skill in solving not only textbook exercises but also real-world engineering problems.