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