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Salt is being poured onto a conical pile with a constant radius of 20 meters. The twice differentiable function $V(t)$ gives the volume of the pile in $m^3$ with $t$ measured in seconds. The table gives selected values of $V^{\prime}(t)$ at various times during the interval $0 \leq t \leq 15$. The formula for the volume of a cone is $V=\frac{1}{3} \pi r^2 h$. At what rate is the height of the pile changing when $h=12$ m? At time $t=5$ seconds, the height of the pile is 12 m.



$t$ (sec)		051015

$V^{\prime}(t)\left(m^3 / sec \right)$		4678

Verified Answer :

  • Express the volume of the cone as \(V = \frac{400\pi}{3}h\) using the constant radius of \(20\) meters.
  • Differentiate with respect to time \(t\) to get \(\frac{dV}{dt} = \frac{400\pi}{3} \frac{dh}{dt}\).
  • Use the given \(\frac{dV}{dt} = 6\) \(\text{m}^3/\text{sec}\) at \(t=5\) to solve for \(\frac{dh}{dt}\).
  • Obtain the final rate of change as \(\boxed{\frac{9}{200\pi}}\) m/sec.

Explanation

  1. Volume Expression for the Conical Pile

We are given that salt is poured into a conical pile with a constant radius \(r = 20\) meters. The volume \(V\) of a cone is given by the formula \(V = \frac{1}{3} \pi r^2 h\), where \(h\) is the height of the pile. Since the radius \(r = 20\) is constant, we can express the volume as a function of the height \(h\) by substituting \(r = 20\):
\[V = \frac{1}{3} \pi (20)^2 h = \frac{400 \pi}{3} h\]

  1. Relating the Rates of Change

To find the rate of change of the height \(h\) with respect to time \(t\), we differentiate the volume equation with respect to \(t\) using the chain rule:
\[\frac{dV}{dt} = \frac{400 \pi}{3} \frac{dh}{dt}\]
This equation relates the rate of change of the volume, \(\frac{dV}{dt}\), to the rate of change of the height, \(\frac{dh}{dt}\).

  1. Calculating the Rate of Change of Height

At the specific time \(t = 5\) seconds, we are given that the height is \(h = 12\) meters and the rate of change of the volume is \(V'(5) = 6\) \(\text{m}^3/\text{sec}\) from the provided table. Substituting these values into our derivative equation:
\[6 = \frac{400 \pi}{3} \frac{dh}{dt}\]
Now, we solve for \(\frac{dh}{dt}\):
\[\frac{dh}{dt} = 6 \times \frac{3}{400 \pi} = \frac{18}{400 \pi} = \frac{9}{200 \pi}\]
Thus, the rate of change of the height at \(t=5\) seconds is \(\frac{9}{200\pi}\) meters per second.

  1. Final Answer

The rate of change of the height of the pile when \(h = 12\) m (at \(t = 5\) sec) is \(\boxed{\frac{9}{200\pi}}\) m/sec.

Examples

This type of problem is essential in fields like industrial engineering or civil engineering, where managing the storage of bulk materials like salt, sand, or grain is common. By understanding how the height of a pile changes in relation to the volume being added, engineers can determine the capacity of silos and conveyors, or monitor how quickly a stockpile is growing or shrinking to ensure it remains stable and within designated boundaries. Mastering related rates allows professionals to make precise, real-time adjustments to their systems.

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