Help me ! I have confusion in understanding this!

Question

Sand is pouring from a pipe at a rate of 12 cm^3/s .The falling sand forms a cone on the ground in such a way that the height of cone is always 1/6th of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Answer 1

The volume is increase at 12 cubic centimeters per second which is dV/dt = 12

R = 6H <--- height is 1/6 times the radius of the base

We seek dH/dt when h=4

V = 1/3 * pi * R^2 * h

= 1/3 * pi * (6H)^2 * h

= 1/3 * pi * 36 * H^3

= 12 * pi * H ^3

so dV/dt = 36 * pi * H^2 * dH/dt

12 = 36 * pi * 4^2 * dH/dt

12/(36*16*pi) = dH/dt

dH/dt = 0.0066287