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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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.

Explanation:

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

$V_{1} = h π r_{1}^{2}$ $V_1=hpir_1^2$ and $V_{2} = h π r_{2}^{2}$ $V_2=hpir_2^2$

The ratio of volumes between them will be:

$\frac{V_{2}}{V_{1}} = \frac{h π r_{2}^{2}}{h π r_{1}^{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}$ $V_2/(V_1)=(hpir_2^2)/(hpir_1^2)=(r_2^2)/(r_1^2)$

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

Answer 2 · 2017-12-11T16:00:16

$\frac{d V}{d r} = 10 π r$ $(dV)/(dr) = 10pir$

Explanation:

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

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

The rate of change of volume with respect to radius is $\frac{d V}{d r}$ $(dV)/(dr)$

$\frac{d}{d r} (V) = \frac{d}{d r} (5 π r^{2}) = 5 π \frac{d}{d r} (r^{2}) = 5 π (2 r) = 10 π r$ $d/(dr)(V) = d/(dr)(5pir^2) = 5pi d/(dr)(r^2) = 5pi(2r) = 10pir$

The volume is changing at a rate of

$10 π r$ $10pir$ $m m^{3} (of Volume)$ $mm^3"(of Volume)"$ / $m m (of radius)$ $mm"(of radius)"$

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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?

2 Answers

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