Problem 63 - Algebra and Trigonometry with Analytic Geometry

Question

As sand leaks out of a certain container, it forms a pile that has the shape of a right circular cone whose altitude is always one-half the diameter \(d\) of the base. What is \(d\) at the instant at which \(144 \mathrm{~cm}^{3}\) of sand has leaked out? Exercise 63

Step-by-Step Solution

Answer

The diameter \(d \approx 17.4 \) cm when 144 cm³ of sand has leaked out.

Step 1: Identify the Given Information

We know that the sand forms a right circular cone and is leaking in such a way that the altitude (height) of the cone is always half the diameter of the base. The volume of sand leaked out is given as 144 cm³.

Step 2: Express Height in Terms of Diameter

Since the altitude (height) is always one-half the diameter, let the diameter be \( d \). Then, the height \( h = \frac{d}{2} \).

Step 3: Write the Volume Formula for the Cone

The volume \( V \) of a cone is given by \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius of the base and \( h \) is the height. Here, \( r = \frac{d}{2} \) and \( h = \frac{d}{2} \).

Step 4: Substitute and Simplify the Volume Equation

Substitute \( r = \frac{d}{2} \) and \( h = \frac{d}{2} \) into the volume formula: \[ V = \frac{1}{3} \pi \left(\frac{d}{2}\right)^2 \left(\frac{d}{2}\right) = \frac{1}{3} \pi \frac{d^3}{8} = \frac{\pi d^3}{24} \].

Step 5: Solve for Diameter \( d \)

Set \( \frac{\pi d^3}{24} = 144 \) and solve for \( d \):1. Multiply both sides by 24: \( \pi d^3 = 3456 \).2. Divide both sides by \( \pi \): \( d^3 = \frac{3456}{\pi} \).3. Find the cube root of both sides to solve for \( d \): \( d = \sqrt[3]{\frac{3456}{\pi}} \).

Key Concepts

Geometric Problem Solving

Solving geometric problems often involves interpreting the characteristics of shapes and their dimensions. In the problem of the leaking sand forming a right circular cone, we need to extract relationships between known and unknown variables.
The given scenario provides that the volume of the cone formed is 144 cm³. The relationship between the height and diameter of the base needs to be understood to solve the problem. Here, the height of the cone is always half the diameter of the base.
For geometric problem-solving, it's crucial to:

• Translate the text into mathematical relationships.
• Understand the properties of shapes involved (in this case, the cone).
• Identify any proportions or constants, such as the constant ratio between height and diameter here.

This approach allows us to define the problem in clear mathematical terms. Knowing how to manipulate these relationships is important for finding solutions.

Right Circular Cone

Understanding the properties of a right circular cone is crucial in this task. A right circular cone is a 3-dimensional shape with a circular base and a pointed vertex directly above the center of the base. Here are some key properties:
The height of the cone (denoted as 'h') is the perpendicular distance from the base to the vertex. The radius 'r' is half the diameter 'd' of the base.

• Volume formula: The volume of the cone is given by \( V = \frac{1}{3} \pi r^2 h \).
• This formula requires you to understand that the height 'h' and the radius 'r' are directly related in our scenario by the diameter i.e., \( h = \frac{d}{2} \) and \( r = \frac{d}{2} \).

Thus, the volume relationship becomes encompassed in the terms of the diameter alone, making it simpler to solve for 'd'. This simplified understanding allows us to apply algebraic manipulation to find the solution.

Algebraic Manipulation

Algebraic manipulation is a key skill that helps transform a complex geometric problem into solvable equations. In our problem, once the relationship between 'h', 'r', and 'd' is established, these can be substituted back into the cone's volume formula.
The substitution leads to an equation involving the diameter 'd' in cubic terms:

• Substitute \( r = \frac{d}{2} \) and \( h = \frac{d}{2} \) into the volume formula to get \( V = \frac{\pi d^3}{24} \).
• Set this expression equal to the given volume, 144 cm³, to form the equation \( \frac{\pi d^3}{24} = 144 \).

From here, solving for 'd' involves manipulating the equation by:

• Multiplying through by 24,
• Dividing by \( \pi \),
• Finally finding the cube root of the result.

These steps lead us to \( d = \sqrt[3]{\frac{3456}{\pi}} \). This result is obtained through systematic algebraic manipulation, making it possible to solve for unknown quantities when given geometric relationships.