Question

Sand is falling from a pipe at the rate of 12 cm\(^3\)/second. The falling sand forms such a cone on the ground that its height is always one-sixth of the radius of its base. At which rate is the height of the cone formed by sand increasing while its height is 4 cm?

Hint:

In related rates problems involving geometric shapes, first write down the relevant formula (e.g., volume, area). Then, use the given constraints to express this formula in terms of a single variable. This simplifies the differentiation process and avoids having to deal with multiple rates (like \(\frac{dr}{dt}\) and \(\frac{dh}{dt}\)) in the same equation.

Answer 1

Step 1: Understanding the Concept:
This is a related rates problem from the application of derivatives. We are given the rate of change of the volume of a cone and a relationship between its height and radius. We need to find the rate of change of its height at a specific instant.
Step 2: Key Formula or Approach:
1. Volume of a cone: \( V = \frac{1}{3}\pi r^2 h \).
2. Identify the given quantities: \(\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}\) and \( h = \frac{1}{6}r \).
3. Express the volume \(V\) solely in terms of the variable whose rate we want to find (in this case, \(h\)).
4. Differentiate the volume equation with respect to time \(t\).
5. Substitute the known values to solve for the unknown rate \(\frac{dh}{dt}\).
Step 3: Detailed Explanation:
We are given:
- Rate of change of volume, \(\frac{dV}{dt} = 12\).
- Relationship between height \(h\) and radius \(r\): \( h = \frac{1}{6}r \). This implies \( r = 6h \).
- We need to find \(\frac{dh}{dt}\) when \(h = 4\) cm.
The formula for the volume of a cone is \( V = \frac{1}{3}\pi r^2 h \).
To find \(\frac{dh}{dt}\), we should express \(V\) as a function of \(h\) only. Substitute \(r = 6h\) into the volume formula:
\[ V = \frac{1}{3}\pi (6h)^2 h = \frac{1}{3}\pi (36h^2)h = 12\pi h^3 \] Now, differentiate both sides with respect to time \(t\): \[ \frac{dV}{dt} = \frac{d}{dt}(12\pi h^3) \] Using the chain rule: \[ \frac{dV}{dt} = 12\pi \cdot (3h^2) \cdot \frac{dh}{dt} = 36\pi h^2 \frac{dh}{dt} \] Now, substitute the given values: \(\frac{dV}{dt} = 12\) and \(h = 4\). \[ 12 = 36\pi (4)^2 \frac{dh}{dt} \] \[ 12 = 36\pi (16) \frac{dh}{dt} \] \[ 12 = 576\pi \frac{dh}{dt} \] Solve for \(\frac{dh}{dt}\): \[ \frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi} \] Step 4: Final Answer:
The height of the sand cone is increasing at a rate of \( \frac{1}{48\pi} \) cm/second.