Water is draining at a rate of 2 cubic feet per minute from the bottom of a conically shaped storage tank of overall height 6 feet and radius 2 feet . How fast is the height of water in the tank changing when 8 cubic feet of water remain the the tank? Include appropraite units in your answer. (Note: The volume of a cone is given by V=(1/3)(pi)(r^2)(h)) Your answer may be expressed in terms of pi.

Question

Water is draining at a rate of 2 cubic feet per minute from the bottom of a conically shaped storage tank of overall height 6 feet and radius 2 feet . How fast is the height of water in the tank changing when 8 cubic feet of water remain the the tank? Include appropraite units in your answer. (Note: The volume of a cone is given by V=(1/3)(pi)(r^2)(h)) Your answer may be expressed in terms of pi.

Answer 1

when the water is y ft deep, the radius of the surface is y/3.

v = 1/3 pi r^2 h = 1/2 pi (y/3)^2 * y = pi/18 y^3
so, y = (18v/pi)^(1/3)

dy/dt = 1/3 * (18v/pi)^(-2/3) * 18/pi
= 1/3 * 18/pi * (pi/18*8)^(2/3)
= 4∛(2/(3pi))

Answer 2

To solve this problem, we need to relate the height of the water in the tank to its volume and then find the rate of change of the height.

First, let's express the volume of the water in terms of the height. We can use the volume formula for a cone:
V = (1/3)(pi)(r^2)(h)

where V is the volume, pi is a mathematical constant (approximately 3.14159), r is the radius of the water surface, and h is the height of the water.

We are given that water is draining from the tank at a rate of 2 cubic feet per minute. This means that the volume is decreasing at a rate of -2 cubic feet per minute.

Now, let's differentiate the volume equation with respect to time (t) to find the rate of change of the volume with respect to time:
dV/dt = (1/3)(pi)(2r)(dh/dt)

where dV/dt is the rate of change of volume with respect to time, and dh/dt is the rate of change of the height with respect to time.

We know that dV/dt = -2 cubic feet per minute, so we can substitute this value into the equation:
-2 = (1/3)(pi)(2r)(dh/dt)

We are given that when 8 cubic feet of water remain in the tank, so we can substitute V = 8 into the volume equation:
8 = (1/3)(pi)(r^2)(h)

Now, we need to find the radius (r) and the height (h) when there are 8 cubic feet of water in the tank. We can rearrange the volume equation to solve for h:
h = (3V)/(pi(r^2))

Substituting V = 8 and r = 2 into this equation, we can find the height (h):
h = (3(8))/(pi(2^2))
h = 12/pi

Now, we have all the necessary information to find the rate of change of the height (dh/dt). Substituting the values into the equation we derived earlier:
-2 = (1/3)(pi)(2(2))(dh/dt)
-2 = (4/3)(pi)(dh/dt)
dh/dt = -3/(2pi) feet per minute

Therefore, the height of the water in the tank is changing at a rate of -3/(2pi) feet per minute when 8 cubic feet of water remain in the tank.