a conical tank with vertex down is 10 feet across the top and 12 feet deep. If water is flowing out of the tank at a rate of 12 cubic feet per minute, find the rate at which the water level is dropping when radius of water level is 4ft.

The water level is dropping at a rate of 0.24 ft/s

Step-by-step explanation:

Here, we simply want to calculate the change in depth (height of the cone) , given the volume change

Mathematically, we have the volume of a cone as;

V = 1/3 * π * r^2 * h

we are given dv/dt as 12 ft^3/m

dv/dh = 1/3 * π * r^2

Substituting the value for the radius, we have

dv/dh = 1/3 * 22/7 * 4^2 = 50.29

dh/dt = dh/dv * dv/dt

dh/dv = 1/(dv/dh) = 1/50.29

Thus,

dh/dt = 1/50.29 * 12

dh/dt = 0.24 ft/s

nope lol

step-by-step explanation:

agree

step-by-step explanation: