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

$((dr)/dt)|_(r = 3) = 1/9$ (in units of $(cm)/s$ )

Denoting volume by V, the volume of the sphere is

$V = 4/3 pi r^3$

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

Rearranging,

$r = ((3V)/(4 pi))^(1/3)$

Noting

$(dr)/dt = (dr)/(dV) (dV)/dt$ (chain rule)

and further noting

$(dr)/(dV) = (1/3)((3V)/(4 pi))^(-2/3)((3)/(4 pi))$

it is given that

$(dV)/(dt) = 4 pi$ (noting units for later!)

So

$(dr)/dt = (dr)/(dV) (dV)/dt = (1/3)((3V)/(4 pi))^(-2/3)((3)/(4 pi)) (4 pi)$

$= ((3V)/(4 pi))^(-2/3)$

$= (r^3)^(-2/3)$

$= 1/r^2$

So (with $r$ in $cm$ )

$((dr)/dt)|_(r = 3) = 1/3^2 = 1/9$ (in units of $(cm)/s$ )