The surface area of a melting snowball decreases at a rate of

Question

3.4cm^2/min. Find the rate at which its diameter decreases when the diameter is 24cm . (Round your answer to three decimal places if required)

cm/min

Answer 1

SA = 4π r^2

dSA/dt = 8π r dr/dt

when diameter = 24, r = 12
so just plug in your values.

Answer 2

Could you explain more? Your answer is very confusing to me.

Answer 3

SA = 4π r^2 <---------- You must know that formula for the surface area

dSA/dt = 8π r dr/dt <----- I assumed you take Calculus

when diameter = 24, r = 12 <---- The question gave us that

so just plug in your values. <----- You should at least do some of the work.

Answer 4

at the end, double dr/dt, since the diameter d = 2r
or, start with A = πd^2
dA/dt = 2πd dd/dt