Problem 43 - Algebra 2

Question

The space shuttle has an external tank for the fuel that the main engines need for the launch. This tank is shaped like a capsule, a cylinder with a hemispherical dome at either end. The cylindrical part of the tank has an approximate volume of 336\(\pi\) cubic meters and a height of 17 meters more than the radius of the tank. \(\left(\text {Hint: } V(r)=\pi r^{2} h\right)\). What are the dimensions of the cylindrical part of the tank?

Step-by-Step Solution

Answer

The radius is 6 meters and the height is 23 meters.

Step 1: Understand the Problem

We are given a cylindrical part of a fuel tank with known volume and height in relation to its radius. The volume of the cylindrical part is 336\(\pi\) cubic meters, and the height is 17 meters more than the radius. We are tasked with finding the radius and height of the cylindrical part.

Step 2: Write Down the Formula

The volume \(V\) of a cylinder is given by the formula \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. We know that \(V = 336\pi\). So we equate \(\pi r^2 h = 336\pi\).

Step 3: Relate Height to Radius

We know that the height \(h\) is 17 meters more than the radius \(r\). Therefore, we can express the height as \(h = r + 17\). Substitute this relation into the volume equation: \(\pi r^2 (r + 17) = 336\pi\).

Step 4: Simplify and Solve for Radius

Cancel \(\pi\) from both sides: \(r^2 (r + 17) = 336\). Expand the left side to get \(r^3 + 17r^2 = 336\). We need to solve for \(r\). This can be achieved by trial and error, or using algebraic methods such as factoring or numerical approaches if needed.

Step 5: Solve the Cubic Equation

By trial, or solving numerically, let's find a value of \(r\) that satisfies the equation. After testing several values, we find that \(r = 6\) satisfies \(6^3 + 17(6)^2 = 336\).

Step 6: Calculate the Height

Since \(r = 6\), we can find \(h = r + 17 = 6 + 17 = 23\). Thus, the height of the cylindrical part is 23 meters.

Key Concepts

Volume of a Cylinder

The volume of a cylinder is an essential concept in geometry, especially when dealing with three-dimensional figures. To find the volume, we use the formula \( V = \pi r^2 h \), where:

• \( V \) is the volume,
• \( r \) is the radius of the base of the cylinder, and
• \( h \) is the height of the cylinder.

The base of a cylinder is a circle, and the formula for its area is \( \pi r^2 \). The volume, therefore, is the circular area multiplied by the height.

In our exercise, the volume of the cylinder is given as 336\(\pi\) cubic meters. This tells us how much space exists inside this cylindrical part of the fuel tank. Understanding this formula allows you to easily calculate the volume provided you have the radius and height, or find either if the volume is given. This is why rearranging the formula is a crucial skill when handling different problems related to cylinders.

Radius and Height Relationship

The relationship between the radius and the height of a cylinder can significantly simplify solving problems related to its dimensions. In particular scenarios, like in our exercise, the height \( h \) is related to the radius \( r \) through a specific relationship. Here, the height is described as being 17 meters more than the radius, or mathematically expressed as \( h = r + 17 \).

• This relationship allows a single variable, \( r \), to be involved in both the height and volume equation.
• Once you establish this, you can leverage it to find the radius first, knowing the translated height will follow directly.

Understanding how to relate different dimensions of the cylinder can make solving such problems more systematic and logical. By substituting the relationship for height into the volume equation, we simplify the expression into a more manageable form that directly involves only one unknown variable.

Solving Cubic Equations

Solving cubic equations is integral to finding dimensions in geometric problems involving three-dimensional shapes like cylinders. In this exercise, substitutions led us to a cubic equation: \( r^3 + 17r^2 = 336 \). This type of equation can appear complex, but there are methods to simplify solving.

• Trial and Error: This involves testing possible values that satisfy the equation. While it may lack systematic predictability, it works efficiently if the range is small or clear.
• Factoring: If the cubic equation can be simplified into products of lower-degree polynomials, factoring becomes useful. This method, however, is more complex and isn't always applicable.
• Numerical Methods: Options like the Newton-Raphson method provide approximate solutions, suitable for technologically equipped solutions.

In the given exercise, trying values like \( r = 6 \) lead to a solution because substituting back into the equation balances both sides. Understanding how to solve these equations ensures that dimensions like radius (and consequently height) can be determined effectively.