How fast is the volume of a cylinder changing with respect to the radius when the radius is mm and the height is a constant 5mm?

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Answer 1 · 2017-12-11T14:41:55

The change will be proportional to the square of the radius.

Given two cylinders with the same height and radii r1 and r2 their volume will be:

$V_1=hpir_1^2$ and $V_2=hpir_2^2$

The ratio of volumes between them will be:

$V_2/(V_1)=(hpir_2^2)/(hpir_1^2)=(r_2^2)/(r_1^2)$

This means that these two cylinders are correlated by the square of radius.

Answer 2 · 2017-12-11T16:00:16

$(dV)/(dr) = 10pir$

The general formula for the volume of a cylinder is $V = pir^2h$

We are told that the height is constant, so this cylinder has volume $V = 5pir^2$

The rate of change of volume with respect to radius is $(dV)/(dr)$

$d/(dr)(V) = d/(dr)(5pir^2) = 5pi d/(dr)(r^2) = 5pi(2r) = 10pir$

The volume is changing at a rate of

$10pir$ $mm^3"(of Volume)"$ / $mm"(of radius)"$