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Question

The surface area of a sphere is decreasing at the constant rate of 5π cm2/s. At what rate is the volume of the sphere decreasing at the instant its radius is 4 cm?

Answer 1

A(r) = 4πr²

 

dA/dt = 8πr dr/dt

 

dA/dt = constant = -5π cm²/s ⇒

 

-5π = 8πr dr/dt

 

dr/dt = -5/(8r)

 

V(r) = (4/3)πr³

 

r = 4 cm ⇒

 

dV/dt = 4πr²dr/dt = -4π(4²) 5/[8(4)] cm³/s = -10π cm³/s

 

Rate of CHANGE is -10π cm³/s.  Rate of DECREASE is 10π cm³/s.