The surface area of a spherical balloon is increasing at the rate of 2cm2 / sec. At what rate is the volume of the balloon is increasing when the radius of the balloon is 6 cm?
Let r be the radius, A be the surface area and V be the volume of the spherical ballon at time t seconds.
`r dr/dt=2/8pi=1/4pi .............(1)`
`Now , V=4/3 pir^3`
`(dV)/dt=4pi/3xx3r^2 (dr)/dt=(4pir)(r (dr)/dt)`
Given : ` r=6, (d)V/dt=?`
Therefore, the volume of the balloon is increasing at the rate of 6cm3/sec.
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