A company that produces landscaping materials is dumping sand into a conical pile. The sand is being poured at a rate of \(5ft^3/sec\); the physical properties of the sand, in conjunction with gravity, ensure that the cone’s height is roughly \(\frac{2}{3}\) the length of the diameter of the circular base.

How fast is the cone rising when it has a height of 30 feet?

We begin by extracting all the necessary information needed to solve the problem. The information we will be using is:

Beginning with the last equation we can substitute the diameter for the radius

  1. \(\frac{3}{2}h = d = 2r\)
  2. \(\frac{3}{4}h = r\)

Next we will substitute this into the volume equation and simplify

  1. \(V=\frac{1}{3}\pi (\frac{3}{4}h)^2h\)
  2. \(V=\frac{3}{16}\pi h^3\)

With this equation we can use implicit differentiation to find the rate of change and simplify

  1. \(\frac{dV}{dt}=\frac{3}{16}\pi (3)h^2\frac{dh}{dt}\)
  2. \(\frac{80}{9\pi h^2}=\frac{dh}{dt}\)

From here we can find the rate of change at any height by substuting in the indicated height. In this case, the problem is asking for \(h=30\)

  1. \(\frac{80}{9\pi 30^2}=\frac{dh}{dt}\)
80/(9*pi*30**2)
## [1] 0.003143801

The rate of change of the height of the conical pile at a height of 30 feet is approximately \(0.003\frac{ft}{sec}\)