Problem 26 - Calculus - AP Edition

Question

Grain pouring from a chute at the rate of \(8 \mathrm{ft}^{3} / \mathrm{min}\) forms a conical pile whose height is always twice its radius. How fast is the height of the pile increasing at the instant when the pile is \(6 \mathrm{ft}\) high?

Step-by-Step Solution

Answer

The height of the pile is increasing at a rate of \(\frac{8}{9\pi}\) ft/min when the height is 6 ft.

Step 1: Given Variables and Equations

The problem states that grain is being poured at a rate of \(\frac{dV}{dt} = 8 \, \text{ft}^3/\text{min}\). The height \(h\) is always twice the radius \(r\), so \(h = 2r\). Volume \(V\) of a cone is given by \(V = \frac{1}{3}\pi r^2 h\). We need \(\frac{dh}{dt}\) when \(h = 6\).

Step 2: Substitute Relationship into Volume Formula

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

Step 3: Differentiate the Volume Formula with Respect to Time

Differentiate the volume equation with respect to time \(t\):\[\frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt}.\]

Step 4: Solve for the Radius when Height is 6 ft

Given \(h = 2r\) and \(h = 6\), solve for \(r\):\[ r = \frac{h}{2} = \frac{6}{2} = 3. \]

Step 5: Substitute Known Values into the Derivative Equation

Substitute \(\frac{dV}{dt} = 8\), \(r = 3\) into the derivative equation:\[ 8 = 2\pi (3^2) \frac{dr}{dt}. \]

Step 6: Solve for \(\frac{dr}{dt}\)

Compute \(\frac{dr}{dt}\) using the values:\[ 8 = 2\pi (9) \frac{dr}{dt} \rightarrow \frac{dr}{dt} = \frac{8}{18\pi} = \frac{4}{9\pi}. \]

Step 7: Relate \(\frac{dh}{dt}\) and \(\frac{dr}{dt}\)

Since \(h = 2r\), we have \(\frac{dh}{dt} = 2\frac{dr}{dt}\). Substitute \(\frac{dr}{dt} = \frac{4}{9\pi}\):\[ \frac{dh}{dt} = 2 \cdot \frac{4}{9\pi} = \frac{8}{9\pi}. \]

Key Concepts

Volume of a Cone

The volume of a cone is a fundamental concept in geometry and calculus. It helps us understand how much space a conical shape can occupy. To find the volume of a cone with height \( h \) and radius \( r \), you use the formula: \[ V = \frac{1}{3} \pi r^2 h. \] This formula is derived from the volume of a cylinder, which is \( \pi r^2 h \). By considering that a cone is essentially a third of a cylinder with the same base and height, the one-third factor comes into play. Breaking down the formula:

• \( \pi r^2 \) represents the base area of the cone.
• The \( h \) indicates how tall the cone is.
• Multiplying these and dividing by three gives the cone's volume.

Understanding this helps when dealing with dynamic problems, like the conical pile problem, where the dimensions change over time.

Differentiation with Respect to Time

In related rates problems, differentiation with respect to time is a crucial technique. This is because the problem involves variables that change over time. For instance, in the conical pile problem, both the radius and height of the cone change as grain is added. To find how fast these changes occur, we differentiate the relevant equations with respect to time. Consider that for a function \( V \), which depends on \( t \), the derivative \( \frac{dV}{dt} \) informs us how \( V \) changes as time progresses. When you differentiate the volume of a cone \( V = \frac{1}{3} \pi r^2 h \) with respect to time, you apply the chain rule. This is because both \( r \) and \( h \) are functions of time as well. Here’s how it looks for the given problem, where \( V = \frac{2}{3} \pi r^3 \): \[ \frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt}. \] This derivative reflects how changes in the radius impact the volume over time.

Conical Pile Problem

The conical pile problem is a classic "related rates" problem in calculus, often used to illustrate the concept of differentiation with respect to time. The problem involves grain being poured to form a conical pile. The key aspect is that the pile's height is always twice the radius, making it stay proportional as it grows. This problem involves adjusting known volumes and growth rates to find how quickly certain dimensions are changing. Here’s the process used in this specific problem:

• We start by understanding the relationship \( h = 2r \), which simplifies calculations.
• We derive appropriate differential equations, such as \( \frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt} \), to measure changing quantities.
• Substituting known values (\( h = 6 \) thus \( r = 3 \)) into our differentiated equation allows us to solve for \( \frac{dr}{dt} \) and subsequently \( \frac{dh}{dt} \).
• Finally, using \( \frac{dh}{dt} = 2\frac{dr}{dt} \) helps to find how fast the height increases over time.

This problem is a great introduction to applying calculus to real-world scenarios where changes occur continuously.