A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 cm3/sec. At what rate is the radius increasing when its circular base area is 1 m2? (a) 1 mm/sec (b) 0.001 cm/sec (c) 2 mm/sec (d) 0.002 cm/sec

Options

A. 1 mm/sec
B. 0.001 cm/sec
C. 2 mm/sec
D. 0.002 cm/sec

Answer

D

Explanation

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)pir^(2)h`
`implies V=(2)/(3)pi r^(3)" "[because h = 2r]`
`implies (dV)/(dt)=2pir^(2)(dr)/(dt)`
`implies 40=2(10^(4))(dr)/(dt)" "[because pir^(2)=1m^(2)" (given)"=10^(4) cm^(2) and (dV)/(dt)=40cm^(3)//sec]`