A container is in the shape of an inverted right circular cone has a radius of 2 in at the top and a height of 6 in. At the instant when the water in the conatiner is 5 in deep, the surface level is falling at the rate of -.4 in/s. Find the rate at which the water is being drained.

To find the rate at which the water is being drained, we need to determine the rate at which the volume of water is changing with respect to time.

Let's start by finding the volume of water in the container at a given height. The volume of a cone is given by the formula V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.

In this case, the radius at the top of the container is 2 inches, and the height of the container is 6 inches.

At the instant when the water is 5 inches deep, the remaining height of the water in the container can be calculated as follows:

Remaining height = Total height - Depth of water
Remaining height = 6 inches - 5 inches
Remaining height = 1 inch

Now, let's find the volume of the water in the container when the depth is 5 inches:

V = (1/3)πr^2h
V = (1/3)π(2^2)(5)
V = (1/3)π(4)(5)
V = (1/3)(20π)
V = (20/3)π

Next, we need to determine how the volume is changing with respect to time. The rate of change of volume is given by dV/dt, where t represents time and dV/dt represents the derivative of the volume with respect to time.

Since the rate at which the surface level is falling is given as -.4 inches per second, the rate at which the remaining height is changing will be .4 inches per second.

Now, let's differentiate the volume equation with respect to time:

dV/dt = dV/dh * dh/dt

We already know that dh/dt = .4 inches per second and dV/dh is the derivative of the volume formula with respect to the height. Let's calculate it:

dV/dh = d/dh [(1/3)πr^2h]
dV/dh = (1/3)πr^2

Substituting the known values and the calculated derivative:

dV/dt = (1/3)π(2^2) * .4

Simplifying:

dV/dt = (1/3)π(4) * .4
dV/dt = (4/3)π * .4
dV/dt = (16/15)π

Therefore, the rate at which the water is being drained is (16/15)π cubic inches per second.

To find the rate at which the water is being drained, we need to determine the rate at which the volume of water is changing with respect to time.

Let's start by defining some variables:
- Let r represent the current radius of the water surface
- Let h represent the current height of the water column
- Let V represent the volume of water in the container

The volume of a cone is given by the formula V = (1/3) * π * r^2 * h.

Given that the container is an inverted cone with a radius of 2 in at the top and a height of 6 in, we can set up a ratio to find the relationship between r and h at any given moment:
r/h = (2/6) or r = (1/3) * h

Now, we can differentiate both sides of the equation with respect to time (t) to find the derivatives of r and h with respect to t:
dr/dt = (1/3) * dh/dt

Given that the rate of change of the surface level is -0.4 in/s (dh/dt = -0.4 in/s), we can substitute the values into the equation:
dr/dt = (1/3) * (-0.4)

Simplifying the equation, we find:
dr/dt = -0.4/3 = -0.1333 in/s

So, the rate at which the water is being drained is approximately -0.1333 in/s.

dv/dt = 3pi/27 x^2 dx/dt

-4 = pi/9 (25) dx/dt

dx/dt = -36/25pi = -.46 in/s

when the water is at height x, by similar triangles, the radius r of the surface is given by

r/x = 2/6, so r = x/3

v = 1/3 pi (x/3)^2 * x
= pi/27 x^3

dv/dt = 2pi/27 x^2 dx/dt
-4 = 2pi/27 (25) dx/dt

dx/dt = -54/25pi = -.69 in/s