A spherical balloon inflated at a rate of 4pi cm^3/sec. initial volume of balloon 0, how fast is radius expanding after 9 sec (r=0.03m)?

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A spherical balloon inflated at a rate of 4pi cm^3/sec. initial volume of balloon 0, how fast is radius expanding after 9 sec (r=0.03m)?

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Answer 1 · 2018-03-14T11:38:07

$(\frac{d r}{d t}) ∣_{r = 3} = \frac{1}{9}$ $((dr)/dt)|_(r = 3) = 1/9$ (in units of $\frac{c m}{s}$ $(cm)/s$ )

Explanation:

Denoting volume by V, the volume of the sphere is

$V = \frac{4}{3} π r^{3}$ $V = 4/3 pi r^3$

where $r$ $r$ (and hence $V$ $V$ ) is a function of time.

Rearranging,

$r = {(\frac{3 V}{4 π})}^{\frac{1}{3}}$ $r = ((3V)/(4 pi))^(1/3)$

Noting

$\frac{d r}{d t} = \frac{d r}{d V} \frac{d V}{d t}$ $(dr)/dt = (dr)/(dV) (dV)/dt$ (chain rule)

and further noting

$\frac{d r}{d V} = (\frac{1}{3}) {(\frac{3 V}{4 π})}^{- \frac{2}{3}} (\frac{3}{4 π})$ $(dr)/(dV) = (1/3)((3V)/(4 pi))^(-2/3)((3)/(4 pi))$

it is given that

$\frac{d V}{d t} = 4 π$ $(dV)/(dt) = 4 pi$ (noting units for later!)

So

$\frac{d r}{d t} = \frac{d r}{d V} \frac{d V}{d t} = (\frac{1}{3}) {(\frac{3 V}{4 π})}^{- \frac{2}{3}} (\frac{3}{4 π}) (4 π)$ $(dr)/dt = (dr)/(dV) (dV)/dt = (1/3)((3V)/(4 pi))^(-2/3)((3)/(4 pi)) (4 pi)$

$= {(\frac{3 V}{4 π})}^{- \frac{2}{3}}$ $= ((3V)/(4 pi))^(-2/3)$

$= {(r^{3})}^{- \frac{2}{3}}$ $= (r^3)^(-2/3)$

$= \frac{1}{r^{2}}$ $= 1/r^2$

So (with $r$ $r$ in $c m$ $cm$ )

$(\frac{d r}{d t}) ∣_{r = 3} = \frac{1}{3^{2}} = \frac{1}{9}$ $((dr)/dt)|_(r = 3) = 1/3^2 = 1/9$ (in units of $\frac{c m}{s}$ $(cm)/s$ )

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A spherical balloon inflated at a rate of 4pi cm^3/sec. initial volume of balloon 0, how fast is radius expanding after 9 sec (r=0.03m)?

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