Question

Solve the following:

A water tank in the farm of an inverted cone is being emptied at the rate of 2 cubic feet per second. The height of the cone is 8 feet and the radius is 4 feet. Find the rate of change of the water level when the depth is 6 feet.

Answer 1

Let r be the radius of base, h be the height and V be the volume of the water level at any time t.

Since, the height of the cone is 8 feet and the radius is 4 feet , `r/h = 4/8 = 1/2`

∴ r = `h/2`                     ...(1)

Given: `(dV)/dt = (2"cu feet")/sec`

Now, V = `1/3 r^2h`

= `1/3 π(h/2)^2h`       ...[By (1)]

∴ V = `π/12 h^3`

Differentiating w.r.t. t, we get

`(dV)/dt = π/12 xx 3h^2 (dh)/dt`

= `(πh^2)/4*(dh)/dt`

∴ `(dh)/dt = 4/(πh^2)*(dV)/dt`

When h = 6, then

`(dh)/dt = 4/(π(6)^2) xx 2`

= `(2/(9pi))"feet"/sec`

Hence, the rate of change of water level is `((2)/(9pi))"feet"/sec`.