Question

sand falls from an overhead bin and accumulates in a conical pile with a radius that is always four times its height. suppose the height of the pile increases at a rate of 1 cm/s when the pile is 11 cm high. at what rate is the sand leaving the bin at that instant?
the sand is leaving the bin at a rate of
(type an exact answer, using \\( \pi \\) as needed.)

Explanation:

Step1: Write the volume formula for a cone

The volume \(V\) of a cone is given by \(V=\frac{1}{3}\pi r^{2}h\). Since \(r = 4h\), substitute \(r\) into the formula: \(V=\frac{1}{3}\pi(4h)^{2}h=\frac{16}{3}\pi h^{3}\).

Step2: Differentiate the volume formula with respect to time \(t\)

Using the chain rule \(\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}\). Differentiate \(V=\frac{16}{3}\pi h^{3}\) with respect to \(h\): \(\frac{dV}{dh}=16\pi h^{2}\).
We know that \(\frac{dh}{dt} = 1\space cm/s\).

Step3: Substitute the value of \(h\)

When \(h = 11\space cm\), \(\frac{dV}{dt}=16\pi(11)^{2}\cdot1\).
Calculate \(16\pi\times121=1936\pi\space cm^{3}/s\).

Answer:

\(1936\pi\)