Sand is pouring from a pipe at the rate of `18 cm^(3)//s`. The falling sand forms a cone on the ground in such a way that the height of the cone is one-sixth of the radius of the base. How fast is the height of the sand cone increasing when its height is 3 cm?

Answer

`(1)/(18pi) cm//s`

Explanation

`h = (r)/(6) rArr r = 6h`. So, `V = (1)/(3) pi r^(2) h = (1)/(3) pi r^(2) h = (1)/(3) pi (6h)^(2) h = 12 pi h^(3)`
`:. (dV)/(dt) = 36 pi h^(2).(dh)/(dt) rArr 18 = 36 pi h^(2).(dh)/(dt) rArr (dh)/(dt) = (18)/(36pi h^(2)) = (1)/(2pi h^(2))`
`rArr [(dh)/(dt)]_(h =3) = (1)/((2pi xx 9)) cm//s`