sand is being dropped at a rate of 10cubic meters per minute onto a conical pile. if the height of the pile is always half the base radius, at what rate is the height increasing when the pile is 8m high?

Question

sand is being dropped at a rate of 10cubic meters per minute onto a conical pile. if the height of the pile is always half the base radius, at what rate is the height increasing when the pile is 8m high?

Answer 1

h = 1/2 r, so r = 2h

v = 1/3 πr^2 h = 1/3 π (2h)^2 h = 4/3 πh^3
dv/dt = 4πh^2 dh/dt
4π*8^2 dh/dt = 10
dh/dt = 10/(256π) m/min

Answer 2

To find the rate at which the height is increasing when the pile is 8m high, we need to use related rates. Let's denote the height of the cone as h, the base radius as r, and the rate at which the height is changing as dh/dt.

Given that the height is always half the base radius, we can express this as h = r/2.

We know that the rate at which sand is being dropped onto the cone is 10 cubic meters per minute. This means that the volume of the cone is increasing at a rate of dV/dt = 10 m^3/min.

The formula for the volume of a cone is V = (1/3)πr^2h. We can differentiate this equation with respect to time (t) to relate the rate of change of volume with the rates of change of radius (dr/dt) and height (dh/dt):

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

Since h = r/2, we can substitute r/2 for h in the equation above:

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

Simplifying the equation, we get:

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

Since we are given that dV/dt = 10 m^3/min, the equation becomes:

10 = (2/3)πr^2(dr/dt) + (1/3)πr^2(dh/dt)

We can further simplify the equation by factoring out πr^2:

10 = (πr^2/3)(2(dr/dt) + dh/dt)

Now we can solve for dh/dt by rearranging the equation:

dh/dt = (10 - (πr^2/3)(2(dr/dt))) / (πr^2/3)

Given that the height of the pile is 8m, we need to find the value of the radius (r) at that height. Using the information that h = r/2, we can set up the equation:

8 = r/2

Solving for r, we get:

r = 16

Now, we need to determine the rate at which the radius is changing (dr/dt) when r = 16. Unfortunately, the problem does not provide any information about this rate. Without this information, it is not possible to calculate the rate at which the height is increasing when the pile is 8m high.