The surface area of a spherical bubble is increasing at the rate of 2 cm2/sec. Find the rate at which the volume of the bubble is increasing at the instant if its radius is 6 cm.

£=PIEdA/dt= 4£r^2/dt=2 cm^2/s ... 1st eqn dV/dt=?In 1st multiply and divide by dr/dr.(4£r^2/dr)(dr/dt)=2 8£r dr/dt= 2 dr/dt= 1/4£r....2nd eqndV/dt= 4/3 £r^3/dt=( 4/3 £r^3/dr )dr/dt= (4£r^2)1/4£r=rHence dV/dr. = 6 cm^3/s