Calculus - Rates of Change

Question

Sand falls from a drain at the rate of 260 m³/min onto the top of a conical pile. The height of the pile is always twice the radius. How fast is the radius when the pile is 20 m high?

Answer 1

The formula for the volume of a cone is:

V= (1/3)π r² h

"The height of the pile is always twice the radius."

h = 2r

r = h/2

Plug in 2r for h:

V = (1/3)π r² (2r)

V = (2/3)π r³

Both V and r are functions of time (t), so derive both sides with respect to t:

dV/dt = 2 π r² (dr/dt)

Plugin h/2 for the value of r:

dV/dt = 2 π (h/2)² (dr/dt)

Given:

dV/dt = 260 m³/min

h = 20 m

∴ 260 = 2 π (20/2)² (dr/dt)

260 = 2 π (10)² (dr/dt)

260 = 200 π (dr/dt)

dr/dt = 13/ (10 π)

dr/dt ≈ 0.41 m/min.

Answer 2

Given rate: dV/dt = 260 ; Rate we want: dr/dt = ? ; Given instant: h = 20 (so r = 10)

If we look at the given rate and the rate we want, we can conclude we need an equation for volume as a function of radius. Volume of a cone is given by V = 1/3πr²h , and since h = 2r we have ...

V(r) = 2/3πr³ We can differentiate both sides with respect to time to get ...

dV/dt = 2πr²·dr/dt= 260

dr/dt = 1.3/π m/min