Manipal MET2015MathematicsApplication of Derivatives
A cone whose height is always equal to its diameter, is increasing in volume at the rate of 40 ~cm ^3 / s . At what rate is the radius increasing when its circular base area is 1 ~m ^2 ?
Options
- A1 ~mm / s
- B0.001 ~cm / s
- C2 ~mm / s
- D0.002 ~cm / s
Correct answer
D. 0.002 ~cm / s
Step-by-step solution
Let h be the height, r be the radius of the base and V be the volume of the cone at time t . Then, V= 1 3 r^2 h V= 2 3 r^3 [ h=2 , given ] On differentiating both sides w.r.t. t , we get d V d t =2 r^2 d r d t 40=2(10)^4 d r d t [ r^2=1 ~m ^2=10^4 ~cm ^2 . and d V d t =40 ~cm ^3 / s , given ] d r d t = 2 1000 ~cm / s =0.002 ~cm / s