Problem 1 - Active Calculus

Question

Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 15 feet high? Recall that the volume of a right circular cone with height \(h\) and radius of the base \(r\) is given by \(V=\frac{1}{2} \pi r^{2} h\) When the pile is 15 feet high, its height is increasing at _______ feet per minute.

Step-by-Step Solution

Answer

When the pile is 15 feet high, its height is increasing at \( \frac{8}{45\pi} \) feet per minute.

Step 1: Understand the Given Information

Identify that the gravel is forming a cone, and the volume is increasing at a rate of 10 cubic feet per minute with height and diameter being the same. When height is 15 feet.

Step 2: Determine the Relationship Between Radius and Height

Since the base diameter and height are the same, the radius is half of the height. Therefore, if height is denoted as h, then the radius r = h/2.

Step 3: Substitute Radius in the Volume Formula

The volume formula for a cone is given by \[ V = \frac{1}{3} \pi r^2 h \]. Substituting r = h/2, we get \[ V = \frac{1}{3} \pi (\frac{h}{2})^2 h = \frac{1}{3} \pi \frac{h^2}{4} h = \frac{1}{12} \pi h^3 \].

Step 4: Differentiate the Volume Formula with Respect to Time

Find the derivative of both sides regarding volume using the chain rule: \[ \frac{dV}{dt} = \frac{d}{dt} (\frac{1}{12} \pi h^3) \]. This will give us \[ \frac{dV}{dt} = \frac{1}{12} \pi \cdot 3h^2 \cdot \frac{dh}{dt} \]. Simplifying, we obtain \[ \frac{dV}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt} \].

Step 5: Isolate \( \frac{dh}{dt} \)

Rearrange the differentiated equation to solve for \( \frac{dh}{dt} \). We know \( \frac{dV}{dt} = 10 \) cubic feet per minute (rate of gravel being dumped). Hence, \[ 10 = \frac{1}{4} \pi h^2 \frac{dh}{dt} \]. Solving for \( \frac{dh}{dt} \), we get \[ \frac{dh}{dt} = \frac{10 \times 4}{\pi h^2} = \frac{40}{\pi h^2} \].

Step 6: Substitute Height Value

Given that height h = 15 feet, substitute h into the equation \[ \frac{dh}{dt} = \frac{40}{\pi h^2} \]. So, \[ \frac{dh}{dt} = \frac{40}{\pi (15)^2} = \frac{40}{225\pi} = \frac{8}{45\pi} \] feet per minute.

Key Concepts

Volume of Cone

To begin with, it's essential to understand how the volume of a cone is calculated. The volume formula for a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \]where:

• \(V\) is the volume
• \(r\) is the radius of the cone's base
• \(h\) is the height of the cone

In this exercise, the height and the base diameter of the cone are always the same, making the radius half of the height. By substituting the radius \(r = \frac{h}{2}\) into the volume formula, we adjust the equation to represent our specific situation: \[ V = \frac{1}{3} \pi \left( \frac{h}{2} \right)^2 h = \frac{1}{12} \pi h^3 \] This simplified formula will make it easier to calculate changes in volume as a function of time.

Derivatives

Derivatives are a key concept in calculus used to measure how a quantity changes as another quantity changes. In this problem, we need to find out how the height of the cone changes as the volume of gravel increases over time. When we talk about taking the derivative of the volume with respect to time, we are determining how the volume changes as time progresses. To achieve this, we use the formula: \[ \frac{dV}{dt} \] Here, \( \frac{dV}{dt} \) represents the rate of change of volume with respect to time. This approach will help us figure out the relationship between the increasing volume of gravel and the increasing height of the pile.

Chain Rule

The chain rule is a fundamental method in calculus used to differentiate composite functions. In this exercise, we deal with the volume \(V\), which is a function of the height \(h\), which in turn is a function of time \(t\). When differentiating \(V\) with respect to \(t\), we utilize the chain rule to express how changes in volume are connected to changes in height over time. Simply put, we need to find \( \frac{dV}{dt} \) by differentiating \[ \frac{1}{12} \pi h^3 \] with respect to \(t\). Using the chain rule, this becomes:\[ \frac{dV}{dt} = \frac{d}{dt}(\frac{1}{12}\pi h^3) = \frac{1}{12} \pi \( 3h^2 \frac{dh}{dt}\) \] This results in: \[ \frac{dV}{dt} = \frac{1}{4} \pi \ h^2 \ \frac{dh}{dt} \] In this step, \( \frac{dV}{dt} \) represents the rate at which volume is increasing, and \( \frac{dh}{dt} \) is the rate at which the height is increasing.

Differentiation with Respect to Time

Differentiation with respect to time is a vital technique used to analyze how quantities change over time. In this problem, we need to differentiate the volume formula with respect to time to determine the rate at which the height of the gravel pile increases. From the earlier derivative result: \[ \frac{dV}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt} \] We know that the volume \(V\) is increasing at a rate of 10 cubic feet per minute: \[ \frac{dV}{dt} = 10 \] feet³/minute.By rearranging the differentiated equation to isolate \( \frac{dh}{dt}\), we get: \[10 = \frac{1}{4} \pi h^2 \frac{dh}{dt} \] Solving for \( \frac{dh}{dt} \): \[ \frac{dh}{dt} = \frac{40}{\pi h^2} \]Substituting \(h = 15\) feet gives: \[ \frac{dh}{dt} = \frac{40}{\pi (15)^2} = \frac{40}{225\pi} = \frac{8}{45\pi} \] Therefore, the height of the pile is increasing at: \( \ \frac{8}{45\tpi} \) feet per minute when the height is 15 feet.