Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always twotwo times its height. Suppose the height of the pile increases at a rate of 33 cm divided by scm/s when the pile is 1010 cm high. At what rate is the sand leaving the bin at that instant?

-423 m³/s

Explanation:

Volume of a cone is:

V = ⅓ π r² h

Given r = 2h:

V = ⅓ π (2h) ² h

V = ⁴/₃ π h³

Taking derivative with respect to time:

dV/dt = 4π h² dh/dt

Given h = 1010 cm and dh/dt = 33 cm/s:

dV/dt = 4π (1010 cm) ² (33 cm/s)

dV/dt ≈ 4.23*10⁸ cm³/s

dV/dt ≈ 423 m³/s

The pile is growing at 423 m³/s, so the bin is draining at - 423 m³/s.