<s>[INST] Rate of change of radius of spherical balloon 
A spherical balloon is filled with gas at a rate of $4 \text{ cm}^3/\text{s}$. What rate is the radius $r$ changing with respect to the time when the vol $V=36π \text{ cm}^3$?

[ans:$\frac 19, \frac{\pi}{9} $or 9$\pi$]
I tried this one with implicit differentiation:
$\frac{dV}{dt} = \frac {4}{3} \pi \cdot \frac{d}{dt}[r^2]$

*

*$4 = \frac {4}{3} \pi \cdot 2r \frac{dr}{dt}$

Volume formula: $36 π = \frac43 π r^2$

*

*$r^2= 27$

*$r= \pm 3 \sqrt{3}$

Substituting r into the differentiation to find $\frac{dr}{dt}$:
$4 = \frac {4}{3} \pi \cdot 2r \frac{dr}{dt}$
$4 = \frac {4}{3} \pi \cdot 2(3 \sqrt{3}) \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{1}{2\sqrt{3}\pi}$ , which seems wrong. Please can someone tell me where my knowledge gap is?
 [/INST] $\frac{dV}{dt} = \frac {4}{3} \pi \cdot \frac{d}{dt}[r^3]$

*

*$4 = \frac {4}{3} \pi \cdot 3r^2 \frac{dr}{dt}$

Volume formula: 36 π = 4/3 π r^3

*

*$r^3= 27$

*$r= 3$

Substituting r into the implicit differentiation to find $\frac{dr}{dt}$:
$4 = \frac {4}{3} \pi \cdot 3r^2 \frac{dr}{dt}$
$4 = \frac {4}{3} \pi \cdot 2(9) \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{1}{9\pi}$
Many thanks @alessandro!
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