+69 If a conical tank with a radius of 8 feet, and a height of 12 feet is being filled with water a rate of 2 cubic feet of water per minute, find how fast the water level is rising when the depth of the water in the tank is 2 feet.
+12530 If a conical tank with a radius of 8 feet, and a height of 12 feet is being filled with water a rate of 2 cubic feet of water per minute, find how fast the water level is rising when the depth of the water in the tank is 2 feet.
+129899 If a conical tank with a radius of 8 feet, and a height of 12 feet is being filled with water a rate of 2 cubic feet of water per minute, find how fast the water level is rising when the depth of the water in the tank is 2 feet.
Notice that the ratio of the height to the radius = 12/8 = 3/2
So....at any time, t, r = (2/3)h
And we can represent the volume as :
V = (1/3)pi * r^2 * h = (1/3) [ (2/3)h] ^2 *h = (1/3)pi (4/9)h^3 = (4/27)pi*h^3
Take the derivative with respect to time, t
dv/dt = 3*(4/27)pi*h^2* dh/dt
dv/dt = (12/27)pi* h^2 * dh/dt
dv/dt = (4/9)pi*h^2* dh/dt
And dv/dt = 2ft^3/min.....and we need to solve for dh/dt when h = 2ft
So we have
2ft^3/min = (4/9)pi (2ft)^2 * dh/dt
2ft^3/min = pi*(16/9)ft^2 * dh/dt
[2] / [pi*(16/9)] ft/min = dh/dt = 9 / [8 pi] ft/min ≈ 0.358 ft/min
