The slant height of a cone is fixed at 7 ~cm . If the rate of increase of its height is 0.3 ~cm / sec , then the rate of increase of its volume when its height is 4 ~cm is

Step-by-step solution

We have, slant height of cone (l)=7 ~cm array ll & l²=h²+r² ...(i) & r²=7²-4² [when h=4 cm] & r²=33 & r= 33 ~cm array Now, differentiating Eq. (i) w.r.t. l , we get gathered 0=2 h d h d t +2 r d r d t d r d t =- h r d h d t ...(ii) Volume of the cone, V= 1 3 r² h d V d t = 1 3 [2 r h d r d t +r² d h d t ] d V d t = 1 3 [2 r h ( -h r d h d t )+r² d h d t ]( from Eq. (ii)) d V d t = 1 3 [-2 h² d h d t +r² d h d t ] d V d t = 1 3 d h d t [-2 h²+r² ] . d V d t ]_ h=4 = 1 3 (0.3) [-2 4²+( 33 )² ] = 10 [1]= 10 cc / sec g