A container is the shape of an inverted cone. Its height is 6 ~m and radius is 4 ~m at the top. If it is filled with water at the rate of 3 ~m ^3 / min then the rate of change of height of water (in mt / min ) when the water level is 3 ~m , is

Step-by-step solution

Let V be the volume, r be the radius and h be the height of an inverted cone at any time t . Then Given, d v d t =3 ~m ^3 / min We know that, From similarity of triangle aligned & B O B A = O D A C & 6 h 4 r r= 2 3 h aligned From Eq. (i), V= 1 3 ( 2 3 h )^2 h= 4 27 h^3 On differentiating w.r.t. t , we get aligned & d v d t = 4 27 3 h^2 d h d t & 3= 4 27 3 h^2 d h d t & d h d t = 3 27 4 3 h^2 & ( d h d t )_ h=3 = 3 27 4 27 = 3 4 mt / min aligned