A conical water tank is 24 ft high and has a radius of 10 ft at the top. If water flows into the tank at a rate of 20 cubic feet per minute, how fast is the depth of the water increasing when the water is 16 feet deep?

0.027 cubic ft/min

Explanation:

Volume of water is V=V (t) Depth of water is h=h (t)

The relationship between V and h is

V=π/3. r2. h

Let

h0 = the height of the cylinder = 24ft, r0 = the radius of the opening = 10ft.

Since water level (h) rise with respect to flow rate we find an expression for r and substitute into V

r/h = r0/h0 = 10/24 = 5/12

r = (5/12) h,

Substitute for r in V

so,

V=1/3.π. (h2) square = 1/3.π. 25/144. (h3)

When

h = 16ft

and dV/dt = 20ft3/min

,

dV/dt = 1/3*π*25/144*h3 * (dh/dt)

20ft3/min = 1/3*π * 25/144*h3 * (dh/dt)

20 = 1/3*π * (16) 3. (dh/dt)

Make dh/DT subject of formula

dh/dt = (20*3*144) / (25*π*4096) = 8640/321740.8

=

dh/dt = 0.027 ft3/min