Question

at a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic feet per minute. the diameter of the base of the cone is approximately three times the altitude. at what rate is the height of the pile changing when the pile is 22 feet high? (hint: the formula for the volume of a cone is v

Answer 1

The rate of changing the height of the pile = 0.00292 ft/minute

The shape of the pile is a cone.

Let V be the volume of the conical pile.

Rate of change of sand falling onto the conical pile, dV/dt = 10 cubic feet/ minute

Base of a cone is in circular shape.

Let d be the diameter of the base circle and h be the altitude(height) of the conical pile.

Then diameter of the base of the cone is three times the altitude

⇒ d = 3h

Rate of change in height of the pile = dh/dt

Volume of a cone, V = πr²h/3,

where r is the radius of the cone.

Since r = d/2,

r = 3h/2

Therefore, V = π x (3h/2)² x h /3

                     = 3πh³/4

Then, dV/dt = (3π/4) x 3 x h² x (dh/dt)

When h = 22 feet,

dV/dt = (3π/4) x 3 x 22² x (dh/dt)

⇒ dV/dt = 1089π x (dh/dt)

⇒ 10 = 1089π x (dh/dt)

Then rate of change of height of the pile , dh/dt = 10/(1089π)

                                                                                = 0.00292 ft/minute

Learn more about rate of change at https://brainly.com/question/25184007

#SPJ4