The radius of a sphere is increasing at a constant of 2 cm/sec at the instant when the volume of the sphere is increasing at 32pi cm^3. sec, what is the surface area of the sphere?

The formula for the volume (v) of sphere is,
                         v = 4πr³ / 3
The derivative is,
                          dv/dt = 4πr²(dr/dt)
It is given that the radius changes at a rate of 2 cm/sec which means that dr/dt is 2. dv/dt is also given to be 32π cm³/sec. Substituting these to the given,
                           32π = 4πr² x 2
The value of 4πr² which is the surface area is equal to 16π. Thus, the surface area of the sphere is 16π cm².

Ithink if im not mistaking b is -3 but im not a 100% sude