Problem 77 - Algebra and Trigonometry

Question

Volume of Grain Grain is falling from a chute onto the ground, forming a conical pile whose diameter is always three times its height. How high is the pile (to the nearest hundredth of a foot) when it contains 1000 \(\mathrm{ft}^{3}\) of grain?

Step-by-Step Solution

Answer

The height of the pile is approximately 8.06 feet.

Step 1: Understand the relationship between diameter and height

We know that the diameter of the cone's base is always three times its height. Let's denote the height as \( h \). Thus, the diameter \( D \) is \( 3h \), and the radius \( r \) is half of that, so \( r = \frac{3h}{2} \).

Step 2: Write the volume formula for a cone

The volume \( V \) of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \). Substitute the expression for \( r \) from the previous step: \[ V = \frac{1}{3} \pi \left(\frac{3h}{2}\right)^2 h \].

Step 3: Simplify the volume formula with given expressions

Simplify the volume expression: \[ V = \frac{1}{3} \pi \left(\frac{9h^2}{4}\right) h = \frac{3\pi h^3}{4} \].

Step 4: Set up the equation with the given volume

We know the volume of the cone is 1000 \( \text{ft}^3 \). Substitute this value into the volume equation: \[ 1000 = \frac{3\pi h^3}{4} \].

Step 5: Solve for the height \( h \)

First, clear the fraction: \[ 1000 \times 4 = 3\pi h^3 \]. Thus, \[ 4000 = 3\pi h^3 \]. Divide both sides by \( 3\pi \): \[ h^3 = \frac{4000}{3\pi} \].

Step 6: Calculate \( h \)

Take the cube root of both sides: \[ h = \left( \frac{4000}{3\pi} \right)^{1/3} \]. Using a calculator, find \( h \approx 8.06 \) feet.

Key Concepts

Conical Pile

A conical pile is formed when material, such as grain or sand, is poured onto a flat surface, creating a 3-dimensional shape reminiscent of a cone. The shape has a circular base and tapers smoothly up to a point, known as the apex.
A crucial factor in understanding these piles is how the pile's dimensions relate to each other, often defined by specific relationships between diameter and height.
For example, in the given problem, the diameter is consistently three times the height. This consistent relationship ensures that as the grain continues to pour, the overall shape remains steadily conical.

• Diameter: Always across the widest part of the base.
• Height: The perpendicular distance from the base to the apex.
• Radius: Half of the diameter. Used frequently in calculations.

Understanding the conical pile structure is essential for applying geometric formulas to determine quantities such as volume.

Cone Diameter-Height Relationship

The relationship between the diameter and height is fundamental to solving problems involving cones. In geometry, the diameter refers to the distance across the circular base of the cone, while the height is the perpendicular distance from the base to the apex or tip of the cone.
In our exercise, the diameter is three times the height of the cone. Let’s break that down with mathematics:

• Let the height be represented by \( h \).
• The diameter \( D \) could then be represented as \( 3h \).
• Since the radius \( r \) is half the diameter, it is \( \frac{3h}{2} \).

This specific diameter-height ratio directly affects how you express other features, such as the radius, in terms of the height.
This interdependency helps in tailoring the formulae of volume and other characteristics for particular problems.

Cone Volume Formula

The cone volume formula is a pivotal tool for solving any problems related to conical shapes. The general formula for calculating the volume \( V \) of a cone is:\[V = \frac{1}{3} \pi r^2 h\]
This formula considers the circular base's area and the height.
Let's apply this to our problem:

• First, we substitute the expression for \( r \) derived from the diameter-height relationship: \( r = \frac{3h}{2} \).
• Plugging it into the volume formula, we get: \[V = \frac{1}{3} \pi \left( \frac{3h}{2} \right)^2 h\]
• Expand this to: \[V = \frac{1}{3} \pi \frac{9h^2}{4} h = \frac{3\pi h^3}{4}\]

This simplified formula allows for easier substitution of known volume to solve for unknown height, as shown in the solution steps. It shows how specific relationships simplify complex formulas into usable forms, crucial for practical problem-solving in diverse situations related to volume measurements of conical objects.