SOLUTION: Water is pouring into an inverted cone at the rate of 8 cubic feet per minute. If the height of the cone is 12 ft and the radius of its base is 6 ft, how fast is the water level ri

Algebra ->  Finance -> SOLUTION: Water is pouring into an inverted cone at the rate of 8 cubic feet per minute. If the height of the cone is 12 ft and the radius of its base is 6 ft, how fast is the water level ri      Log On


   

Question 1197602: Water is pouring into an inverted cone at the rate of 8 cubic feet per minute. If the height of the cone is 12 ft and the radius of its base is 6 ft, how fast is the water level rising when the water is 4 ft deep?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The volume is increasing at a rate of 8 cubic feet per minute; we want to know how fast the water level (height) is changing when the height is 8 feet.

Since we are given the rate of change of the volume and we want to find the rate of change of the height, we need a formula for the volume in terms of the height.

Volume of a cone: %281%2F3%29%28base%29%28height%29=%281%2F3%29%28pi%2Ar%5E2%29%28h%29

The height of the cone is 12 feet and the radius of its base is 6 feet. Since the sides of the cone are straight, the ratio 6/12 = 1/2 of radius to height is constant. We want our formula in terms of height, so

r=%281%2F2%29h

And the volume of the cone in terms of the height is

V=%281%2F3%29%28pi%2A%28h%2F2%29%5E2%29%28h%29=%28pi%2F12%29h%5E3

Now we are ready to use our calculus for this related rates problem.

dV%2Fdt=%28pi%2F12%29%283h%5E2%29%28dh%2Fdt%29

dV%2Fdt=%28pi%2F4%29%28h%5E2%29%28dh%2Fdt%29

dh%2Fdt=%284%2F%28%28h%5E2%29pi%29%29%28dV%2Fdt%29

We want the rate of change of the height, dh%2Fdt, when the height h is 4, knowing that the rate of change in the volume, dV%2Fdt, is 8:

dh%2Fdt=%284%2F%28%28%284%29%5E2%29pi%29%29%288%29=32%2F%2816pi%29=2%2Fpi

ANSWER: The water level is rising at a rate of (2/pi) feet per second when the water is 4 feet deep.