Rate of change of radius of spherical balloon

Question

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:

$\frac{dV}{dt} = \frac {4}{3} \pi \cdot \frac{d}{dt}[r^2]$

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

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

• $r^2= 27$
• $r= \pm 3 \sqrt{3}$

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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?

Answer 1

$\frac{dV}{dt} = \frac {4}{3} \pi \cdot \frac{d}{dt}[r^3]$

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

---

Volume formula: 36 π = 4/3 π r^3

• $r^3= 27$
• $r= 3$

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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!

Answer 2

Just as another way to approach the calculation, since the volume of the balloon is given, rather than its radius:

You had the result from implicit differentiation, $$ \frac{dV}{dt} \ \ = \ \ \frac43 \pi \cdot \frac{d}{dt} \ [r^3] \ \ = \ \ \frac43 \pi \cdot 3r^2 · \frac{dr}{dt} \ \ = \ \ 4 \pi r^2 · \frac{dr}{dt} \ \ . $$

We might wish to have this "related-rates" equation expressed in terms of the current volume of the balloon. To do this, it will be helpful to find a constant $ \ c \ $ for which
$$ \ c·(4 \pi r^2)^{3/2} \ = \ V \ = \ \frac43 \pi r^3 \ \ \Rightarrow \ \ c · 4^{3/2} · \pi^{3/2} · r^3 \ \ = \ \ \frac43 \pi r^3 $$ $$ \Rightarrow \ \ c \ \ = \ \ \frac43 \pi · \frac{1}{4^{3/2} · \pi^{3/2}} \ \ = \ \ \frac{4 · \pi}{3 · 4^{3/2} · \pi^{3/2}} \ \ = \ \ \frac{1}{3 · 4^{1/2} · \pi^{1/2}} \ \ = \ \ \frac{1}{6 · \sqrt{\pi}} $$ $$ \Rightarrow \ \ \left(\frac{1}{6 · \sqrt{\pi}} \right)·(4 \pi r^2)^{3/2} \ = \ V \ \ \Rightarrow \ \ 4 \pi r^2 \ \ = \ \ (6 · \pi^{1/2})^{2/3}· V^{2/3} \ \ = \ \ 6^{2/3} · \pi^{1/3} · V^{2/3} \ \ . $$

The related-rates equation becomes $$ \frac{dV}{dt} \ \ = \ \ 6^{2/3} · \pi^{1/3} · V^{2/3} · \frac{dr}{dt} \ \ , $$

and inserting the given quantities leads us to

$$ 4 \ \ = \ \ 6^{2/3} · \pi^{1/3} · (36 \pi)^{2/3} · \frac{dr}{dt} \ \ \Rightarrow \ \ 4 \ \ = \ \ 6^{2/3} · \pi^{1/3} · (6^2)^{2/3} · \pi^{2/3} · \frac{dr}{dt} \ \ $$ $$ \Rightarrow \ \ 4 \ \ = \ \ 6^{2/3} · (6)^{4/3} · \pi^{1/3} · \pi^{2/3} · \frac{dr}{dt} \ \ \Rightarrow \ \ 4 \ \ = \ \ 6^2 · \pi · \frac{dr}{dt} \ \ \Rightarrow \ \ \frac{dr}{dt} \ \ = \ \ \frac{4}{36·\pi} \ \ = \ \ \frac{1}{9·\pi} \ \ . $$

[In calculus classes, instructors are generally not concerned about units, but they matter very much to scientists and engineers. Going back to the earlier equations we used to find the "proportionality constant", we have $$ \frac{dV}{dt} \ \ = \ \ 4 \pi r^2 · \frac{dr}{dt} \ \ \Rightarrow \ \ \frac{dV/dt}{dr/dt} \ \ = \ \ 4 \pi r^2 \ \frac{\text{cm}^3\text{/sec}}{\text{cm/sec}} \ \ = \ \ 4 \pi r^2 \ \text{cm}^2 $$ $$ \ c· (4 \pi r^2\ \text{cm}^2)^{3/2} \ \ = \ \ V \ \text{cm}^3 \ \ \Rightarrow \ \ c \ \ = \ \ \frac{1}{6 \sqrt{\pi}} \ \frac{\text{cm}^3}{\text{cm}^3} \ \text{ or no units} \ \ $$ $$ 4 \ \frac{\text{cm}^3}{\text{sec}} \ \ = \ \ (6 \sqrt{\pi} \ \text{no units})^{2/3} · \left(36 \pi \ \text{cm}^3 \right)^{2/3} · \frac{dr}{dt} \ \ \Rightarrow \ \ \frac{dr}{dt} \ \ = \ \ \frac{4 \ \frac{\text{cm}^3}{\text{sec}}}{36 \pi \ \text{cm}^2} \ \ = \ \ \frac{1}{9 \pi} \ \frac{\text{cm}}{\text{sec}} \ \ . $$ Since $ \ \frac{dr}{dt} \ $ could be thought of as "differential length divided by differential time", we would expect the units to be in cm./sec. ]