A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 ~cm ^3 / sec . The rate at which the radius is increasing when its circular base is 1 ~m ^2 is

Step-by-step solution

Given the height h of the cone is equal to its diameter d = 2r , so h = 2r . The volume of the cone is V = 1 3 r^2 h = 1 3 r^2 (2r) = 2 3 r^3 . Differentiating with respect to time t , we get dV dt = 2 3 (3r^2) dr dt = 2 r^2 dr dt . Given dV dt = 40 cm ^3/ sec . The area of the base is A = r^2 = 1 m ^2 = 10000 cm ^2 . Substituting A = r^2 = 10000 into the derivative equation: 40 = 2 (10000) dr dt . Solving for dr dt : dr dt = 40 20000 = 4 2000 = 1 500 = 0.002 cm/sec . Answer: 0.002 ~cm / sec