Problem 33 - Calculus Single Variable

Question

Suppose that grain pouring from a chute forms a conical heap in such a way that the height is always \(2 / 3\) the radius of the base. At the moment when the conical heap is \(3 \mathrm{~m}\) high, its height is rising at the rate of \(1 / 2 \mathrm{~m} / \mathrm{min}\). At what rate (in \(\mathrm{m}^{3} / \mathrm{min}\) ) is the grain pouring from the chute?

Step-by-Step Solution

Answer

The grain is pouring at a rate of \( \frac{81\pi}{8} \text{ m}^3/\text{min} \).

Step 1: Understand the Relationship Between Height and Radius

The problem states that the height \( h \) of the conical heap is always \( \frac{2}{3} \) of the radius \( r \). Therefore, we have the relationship \( h = \frac{2}{3}r \). This means that there exists a constant relationship between the height and radius of the heap.

Step 2: Express Radius in Terms of Height

Since \( h = \frac{2}{3}r \), we can rewrite this relationship to express \( r \) in terms of \( h \). By solving for \( r \), we get \( r = \frac{3}{2}h \). This expression allows us to substitute \( r \) in the formula for the volume of a cone.

Step 3: Write the Volume Formula in Terms of Height

The volume \( V \) of a cone is given by the formula \( V = \frac{1}{3}\pi r^{2} h \). Substitute \( r = \frac{3}{2}h \) into the volume formula: \[ V = \frac{1}{3}\pi \left(\frac{3}{2}h\right)^{2} h = \frac{1}{3}\pi \times \frac{9}{4}h^{2} h = \frac{3\pi}{4}h^{3} \]. Now, the volume is expressed in terms of \( h \).

Step 4: Differentiate the Volume with Respect to Time

We need to find the rate at which the volume is changing, \( \frac{dV}{dt} \). By applying the chain rule, we differentiate \( V = \frac{3\pi}{4}h^{3} \) with respect to \( t \): \[ \frac{dV}{dt} = \frac{3\pi}{4} \times 3h^{2} \cdot \frac{dh}{dt} = \frac{9\pi}{4}h^{2} \cdot \frac{dh}{dt} \].

Step 5: Substitute Known Values to Find Rate of Volume Change

At the moment when \( h = 3 \) m, the height is rising at a rate of \( \frac{1}{2} \) m/min, so \( \frac{dh}{dt} = \frac{1}{2} \). Substitute \( h = 3 \) and \( \frac{dh}{dt} = \frac{1}{2} \) into the differentiated equation: \[ \frac{dV}{dt} = \frac{9\pi}{4} \times 3^2 \times \frac{1}{2} = \frac{9\pi}{4} \times 9 \times \frac{1}{2} = \frac{81\pi}{8} \].

Key Concepts

Conical Heap

A conical heap is a three-dimensional shape that looks like a cone. It is named for its resemblance to a geometric cone and is often encountered in real-life scenarios, such as piles of grain, sand, or other granular materials. This heap forms from particles falling from a single point and piling up due to gravity.
A key characteristic of the conical heap in our exercise is the consistent relationship between its height and base radius. Here, the height of the heap is always two-thirds of the radius. This relationship simplifies how we model and analyze changes in the conical heap using mathematical formulas. Understanding this relationship is crucial when we analyze how the heap grows as more material is added.

Volume of a Cone

The volume of a cone, a vital concept for understanding a conical heap's capacity, is calculated using a specific formula. This formula helps us determine the amount of material or space within the cone.
The formula to determine the volume of a cone is given by:

• \[ V = \frac{1}{3} \pi r^2 h \]

where \( V \) is the volume, \( r \) is the radius of the base, and \( h \) is the height of the cone.
In the exercise, since the radius is expressed in terms of height (\( r = \frac{3}{2} h \)), we substitute this into the formula, yielding a volume expression entirely in terms of height:

• \[ V = \frac{3\pi}{4} h^3 \]

This adapted formula allows us to easily differentiate and analyze how the volume changes as height increases.

Differentiation

Differentiation is a fundamental mathematical tool used to understand how a function changes. It gives us the rate of change of a function concerning its variables. In this exercise, we use differentiation to find how fast the volume of the conical heap is changing as the height changes over time.
By differentiating the volume formula expressed in terms of height, we can understand how the volume changes as the height increases. The derivative here tells us the rate of volume change as a function of the heap's height and the rate at which the height itself changes. This approach is essential in solving the exercise, letting us find the exact rate at which the volume of the cone is increasing with time.

Chain Rule

The chain rule is a principle in calculus that helps us find the derivative of a composite function. It is especially useful when dealing with related rates problems, where one variable affects another through a chain of dependencies.
In the exercise, to find the rate of change of the volume of the conical heap over time, we applied the chain rule to differentiate the volume in terms of the height, and then further in terms of time. The chain rule connects how the heap's height change (\( \frac{dh}{dt} \)) affects the change in volume (\( \frac{dV}{dt} \)).
Using the formula:

• \[ \frac{dV}{dt} = \frac{9\pi}{4} h^2 \cdot \frac{dh}{dt} \]

we can substitute in the specific rate of height change (\( \frac{dh}{dt} \)) at the given moment to find out the exact rate at which the volume is increasing. The chain rule is crucial to linking these dynamic changes and solving the problem accurately.