Question

sand falls from an overhead bin and accumulates in a conical pile in such a way that the height of the cone remains a constant 27 cm. suppose the radius of the pile increases at a rate of 2 cm/s when the radius of the pile is 9 cm. at what rate is the sand leaving the bin at that instant?

Answer 1

The sand is leaving the bin at a rate of 4/3 cm/s.

Step-by-step explanation:

We can solve this problem using related rates, as we are given information about the rate at which the radius is increasing and we want to find the rate at which the sand is leaving the bin.

Let's denote the radius of the pile as r and the height of the cone as h.

Since the height of the cone remains constant at 27 cm, we have:

h = 27 cm

And we are given:

dr/dt = 2 cm/s (rate at which the radius is increasing)

We want to find:

dh/dt (rate at which the sand is leaving the bin)

We can use the relationship between the radius, height, and volume of a cone to find an equation that relates r, h, and their rates of change.

The volume of a cone is given by:

V = (1/3)πr^2h

Differentiating both sides with respect to time (t), we get:

dV/dt = (1/3)(2πrh(dr/dt) + πr^2dh/dt)

Since the volume is not changing with time (as the sand is accumulating), dV/dt is 0:

0 = (2/3)πr^2(dr/dt) + πr^2dh/dt

Simplifying, we have:

dh/dt = -(2/3)(dr/dt)

Substituting the given values:

dh/dt = -(2/3)(2 cm/s)

dh/dt = -4/3 cm/s

Therefore, the sand is leaving the bin at a rate of 4/3 cm/s.