what rate is the area of a smoke ring changing with respect to its radius, radius 3 inches?Incrse or decrease?

Question

Recall A =πr^2

Answer 1

dA= 2*pi*r*dr
so dA/dr= 6*pi>0 increase

Answer 2

given quantity is a functionality of radius (a)V = V(R) = 4/3piR^3 fee of substitute of quantity wrt radius = dV/dR = V'(R) = 4/3pi*(3R^2) => dV/dR = 4piR^2 (b)the quantity V'(R) provides the outdoors area of the sector thinking the contraptions V = m^3 and R =m so for dV/dR we've m^3/m = m^2 the unit m^2 corresponds to area(ie., floor area of a sphere)

Answer 3

how thick is the smoke ring? how wide is the hole in the middle?