Question

Sand falls from a conveyor belt at a rate of 15 StartFraction m cubed Over min EndFraction onto the top of a conical pile. The height of the pile is always​ three-eighths of the base diameter. How fast are the height and the radius changing when the pile is 5 m​ high?

Answer 1

Answer:

dh/dt = 0.107m/min

dr/dt = 0.1435m/min

Explanation:

The volume of a cone is given by:

$V = \pi*r^2*h/3$

The relationship between height and radius is:

h = 3/8 * (2*r) = 3/4 * r

Replacing this value into the volume formula:

$V = \pi/4*r^3$

If we derive this equation:

$dV/dt =\pi/4*3*r^2*dr/dt$   Solving for dr/dt and knowing that dV/dt=15m^3/min

$dr/dt = \frac{4*dV/dt}{3*\pi*r^2}=0.1435m/min$

With the relation between height and radius:

dh/dt=3/4*dr/dt = 0.107m/min