A circle s circumference is shrinking at a rate of 3/2pi cm/min. How fast is the area of the circle changing when the circumference is 12cm?

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answers:
CogitoErgoCogitoSum say: A = pi r^2
and C = 2 pi r

Thus A = 1/(4 pi) C^2

And so, dA/dT = 1/(2 pi) C dC/dT

dA/dT = 1/(2 pi) C (3pi/2) = (3/4) C

dA/dT(at C = 12 cm) = (3/4) 12 cm = 9 cm/min

The sign is positive because the question asks for the rate of SHRINKING. A negative shrinking rate would be a growing rate.
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khalil say: p = 2πr
r = p / 2π
A = πr²
A = π(p/2π)²
dA / dt = (1 /2π) p dp / dt
dA / dt = (1/ 2π) 12 (-3π /2)
dA / dt = -9 cm² / min
the negative sign says .. it is getting smaller
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alex say: Hint:
Area = πr²
Circumference = 2πr
find dr/dt
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rotchm say: Area = πr², where r = r(t).
You want dA/dt ; dA/dt = π-2r * dr/dt.

You are given that "3/2pi " = dC/dt = d(2πr)/dt = 2π*dr/dt and you are given the circumference (hence the radius). Surly you can finish it from here.

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.
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ted s say: dC/dt = 2 π dr/dt = 1.5 π & dA / dt = 2 π r dr/dt ====> dA/dt = 2π (12) (3/4) = 18π cm²/sec
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