sand is being dragged at the rate of 10m^3/min into a conical pile. if the height of the pile is always twice the base radius, at what rate is the height increasing when the pile is 2m high?

The volume of anything pointy with straight sides is (1/3) base area * height

so
V = (1/3) pi r^2 h
but h = 2 r
so
V = (2/3) pi r^3
find dr/dV, worry about dh/dV later (note h = 2 r)

dV = (2/3) pi (3 r^2) dr = 2 pi r^2 dr
dV/dt = 2 pi r^2 dr/dt = 10 m^3/min
but r = h/2 and dr/dt = (1/2)dh/dt
so
10 = 2 pi h^2/4 (1/2) dh/dt

40 = pi h^2 dh/dt
when h = 2
40 = 4 pi dh/dt
so
dh/dt = 10/pi meters/minute

To find the rate at which the height of the pile is increasing, we can use related rates. We need to determine an equation that relates the variables involved. In this problem, the volume of the sand and the dimensions of the conical pile are involved.

Let's denote the variables as follows:
- V: volume of the sand in the pile (in cubic meters)
- r: radius of the base of the conical pile (in meters)
- h: height of the conical pile (in meters)

We know that the volume of a cone can be given by the formula V = (1/3) * π * r^2 * h. However, we need to express the volume in terms of the changing variable (height) to solve this problem.

Given that the height of the pile is always twice the base radius, we can express the relationship as: h = 2r.

Now, rewrite the volume equation using the relationship between h and r: V = (1/3) * π * r^2 * (2r).

Now, we can differentiate both sides of the equation with respect to time (t) to find the related rates:

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

Here, dV/dt represents the rate at which the sand is being dragged into the pile, which is given as 10 m^3/min.

So, we need to find dh/dt, the rate at which the height is increasing when the pile is 2 meters high.

Given:
- dV/dt = 10 m^3/min
- h = 2 meters

Now, let's substitute these values into the equation and solve for dh/dt:

10 = (1/3) * π * (2r * dr/dt * r^2 + 2r^2 * dh/dt).

We are given that h = 2r, so substitute for h:

10 = (1/3) * π * (2r * dr/dt * r^2 + 2r^2 * dh/dt)
10 = (1/3) * π * (2r * dr/dt * r^2 + 2r^2 * (dh/dt))
10 = (1/3) * π * (2r * dr/dt * r^2 + 4r^3 * (dh/dt))
10 = (2/3) * π * (r * dr/dt * r^2 + 2r^3 * (dh/dt))
15 = r * dr/dt * r^2 + 2r^3 * (dh/dt)
15 = r * dr/dt * r^2 + 2r^3 * (dh/dt)
15 = r * dr/dt * r^2 + 2r^3 * (dh/dt)

Now, we need to find the value of dr/dt, the rate at which the radius is changing. However, this information is not provided in the problem statement, so we cannot determine the value of dh/dt without this additional information.