Problem 20 - Technical Mathematics with Calculus

Question

Sand poured on the ground at the rate of \(3.00 \mathrm{m}^{3} / \mathrm{min}\) forms a conical pile whose height is one-third the diameter of its base. How fast is the altitude of the pile increasing when the radius of its base is \(2.00 \mathrm{m} ?\)

Step-by-Step Solution

Answer

The altitude of the pile is increasing at a rate of approximately 0.143 \/ \mathrm{min}.

Step 1: Understanding the Geometrical Relationship

Recognize that the height (h) of the cone is one-third the diameter of the base. Because the diameter is twice the radius (r), we can express this relationship as h = (2/3)r. This relationship will be used to find the relationship between the rates of change of height and radius.

Step 2: Expressing Volume of the Cone

Use the formula for the volume of a cone, which is \( V = \frac{1}{3} \pi r^2 h \), to relate the volume to the height and radius of the cone. Since h is related to r, we can substitute h in terms of r, getting \( V = \frac{1}{3} \pi r^2 \cdot \frac{2}{3}r = \frac{2}{9} \pi r^3 \).

Step 3: Differentiating the Volume with Respect to Time

Differentiate the volume function \( V = \frac{2}{9} \pi r^3 \) with respect to time (t) to find the relationship between the rates of change of volume (\( \frac{dV}{dt} \)) and the radius (\( \frac{dr}{dt} \)). The resulting equation is \( \frac{dV}{dt} = \frac{2}{9} \pi \cdot 3r^2 \cdot \frac{dr}{dt} \) or \( \frac{dV}{dt} = \frac{2}{3} \pi r^2 \frac{dr}{dt} \).

Step 4: Substituting the given rate of volume increase

Substitute the given rate of volume increase \( \frac{dV}{dt} = 3.00 \mathrm{m}^3/\mathrm{min} \) and the radius value r = 2.00 m into the equation from Step 3. We get \( 3.00 = \frac{2}{3} \pi \cdot (2.00)^2 \cdot \frac{dr}{dt} \).

Step 5: Solving for \( \frac{dr}{dt} \)

Solve the equation from Step 4 to find the rate of change of the radius \( \frac{dr}{dt} \).

Step 6: Finding the rate of change of height

Use the rate of change of the radius \( \frac{dr}{dt} \) to find the rate of change of the height \( \frac{dh}{dt} \) using the derivative of the relationship h = (2/3)r with respect to time, which yields \( \frac{dh}{dt} = \frac{2}{3} \frac{dr}{dt} \). Then plug in the value of \( \frac{dr}{dt} \).

Key Concepts

Related Rates Calculus

Related rates problems in calculus involve finding the rate at which one quantity changes with respect to time, given the rate of change of another related quantity.

These problems require understanding the relationship between different variables in a geometric, physical, or other contexts, as changes occur over time. A classic example is the changing volume and dimensions of a growing conical pile, like in the sand pile problem.

When solving related rates problems, it is crucial to:

• Identify the variables that are changing with time.
• Express the interdependence of these variables through an equation.
• Differentiate this equation with respect to time to find the relationship between the rates of change.
• Substitute known values and solve for the desired rate.

This sequence of steps leads to understanding how one rate of change affects another within a given scenario.

Conical Pile Problem

The conical pile problem is a practical application of related rates calculus where we deal with a scenario of a pile shaped like a cone that is increasing in volume over time.

Imagine sand pouring onto a surface, forming a conical pile. Understanding the changing rates of the pile's dimensions can be important in various fields such as civil engineering and resource management. For instance, knowing how fast the pile's height increases can aid in planning the spacing of piles in a storage area.

In these problems, we typically know:

• The rate of volume increase.
• The relationship between the geometric dimensions of the cone (like radius and height).

The goal is to find out how quickly other aspects of the cone, such as the radius or height, are changing.

Volume of a Cone Differentiation

Differentiating the volume of a cone with respect to time is integral to solving problems related to changing conical dimensions over time.

The formula for the volume of a cone is given by \( V = \frac{1}{3} \pi r^2 h \), which relates the volume (V) to the radius of the base (r) and the height (h) of the cone. To find how these dimensions change over time, we must use differentiation—a mathematical technique that calculates the rate of change.

By substituting the height with its relationship to the radius, we differentiate the resulting volume formula with respect to time, assigning \( \frac{dV}{dt} \) to represent the rate of volume change and \( \frac{dr}{dt} \) for the rate of radius change.

This differentiation leads to an equation that, once we plug in the known rates and dimensions, allows us to solve for the unknown rate—in this case, the rate at which the pile's altitude is increasing.