A conical tank is 8 meters high. The radius of the top is 2 meters. At what rate is the water running out if the depth is 3 meters and is decreasing at the rate of 0.4 meters per minute

DV/dt = 0,2355 m³/min

Step-by-step explanation:

Conical tank volume V = 1/3 * π*r²*h

r radius at the top 2 meters

when depth of water is 3 meters the radius of the level of water is:

let α angle of vertex of cone then

tan∠α = 2/8 tan∠α = 1/4 tan∠α = 0,25

At the same time when water is at 3 meters depth radius is

tan∠α = r/3 0,25*3 = r r = 0,75 m

Now

DV/dt = (1/3) * π*r²*Dh/dt

Dh/dt = 0,4 meters/min

By substitution

DV/dt = 0,2355 m³/min