A conical tank has height 3m and radius 2m at the top. Water flows in at a rate of 2m^3 /min. How fast is the water level rising when it is 2m.

Answer 1

V=PIr^2*h
dV/dt=2rPIh dr/dt + PIr^2 dh/dt

you know dV/dt, dr/dt=0, and you are solving for dh/dt

Answer 2

Since it is a cone, and the height is 3 to a radius of 2, then no matter what the height is h = (3/2) * r and thus r = (2/3) * h. For the computations we want to get rid of the r by substituting (2/3) * h for r.

Now the capacity of the cone filled to the brim is (1/3) * h * pi * r^2. Substitute (2h / 3) for r, and (4h^2 / 9) for r^2.

v = (1/3) * h * pi * (4h^2 / 9) = (4pi / 27) * h^3

Therefore dv/dt = (4pi / 27) * 3h^2 dh/dt = (4 * pi / 9) * h^2 * dh/dt

dv/dt = (4 * pi / 9) * h^2 * dh/dt................. We want to find out what dh/dt is and we know that

dv/dt is 2m^3/min and that h = 2m. Substitute those into the equation above.

2 = (4 * pi / 9) * 4 * dh/dt = [(16 * pi / 9) * dh/dt]...........Divide both sides of this equation by [16 * pi / 9]

[2 / (16 * pi / 9) ] = dh/dt


So dh/dt = [2 / (16 * pi / 9) ] which is simplified to [ 1 / (16 * pi / 9) ]

But we aren't done. Multiply the numerator and denominator of that fraction by 9, getting:

(9 / (16 * pi))

So dh/dt = (9 / (16 * pi))

The water level is rising at (9 / 16 * pi) m / min. <--------------- Answer

Answer 3

9/8pi