since h = 1/2 r then you'll have r as:
r = 2h
note that the formula in finding the volume of a cone is defined by:
[tex]V = \frac{1}{3} \pi r^2 h[/tex]
where r in that case is 2h. then you'll have it as:
[tex]V = \frac{1}{3} \pi (2h)^2 h[/tex]
[tex]V = \frac{1}{3} \pi (4h^2) h[/tex]
[tex]V = \frac{4}{3} \pi h^3[/tex]
since you are to find the rate at which the height is increasing then you will have to differentiate the derived equivalent formula.
[tex]V = \frac{4}{3} \pi h^3[/tex]
[tex] \frac{dV}{dt} = \frac{4}{3} \pi (3h^2) \frac{dh}{dt} [/tex]
[tex] \frac{dV}{dt} = 4 \pi h^2 \frac{dh}{dt} [/tex]
substituting the given you'll have:
Given: dV/dt = 1cu.m/min
h = 6m
[tex] \frac{dV}{dt} = 4 \pi h^2 \frac{dh}{dt} [/tex]
[tex]1 = 4 \pi (6^2) \frac{dh}{dt} [/tex]
[tex] \frac{dh}{dt} = \frac{1}{144 \pi } [/tex]
or
[tex] \frac{dh}{dt} = 0.00221 \frac{m}{min} [/tex]
