The surface area of a sphere is decreasing at the constant rate of 3π sq. cm/sec . at what rate is the volume of the sphere decreasing at the instant its radius is 2 cm ?

Whooh

SA=4pir²
take derivitive
dSA/dt=8pir dr/dt

and do the volume as well
V=(4/3)pir³
dV/dt=4pir² dr/dt

we need to solve for dV/dt
to do taht we need dr/dt
so

dSA/dt=8pir dr/dt
given dSA/dt=3pi cm/sec
r=2
3pi=8pi2 dr/dt
3=16 dr/dt
3/16=dr/dt

now do the volume
dV/dt=4pir² dr/dt
r=2
dr/dt=3/16
dV/dt=4pi2² (3/16)
dV/dt=16pi(3/16)
dV/dt=3pi

nice
the volume of the sphere is decreasing at 3pi cm/sec as well