Related rates for a filling spherical balloon

Expert intro: The sphere volume formula $V=\frac{4}{3}\pi r^3$ is the engine behind every part of this related-rates balloon problem.

Detailed walkthrough

Start with the volume formula

For a sphere, the volume is

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

Because the balloon is being filled, both $V$ and $r$ change with time. Differentiate both sides with respect to $t$:

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$.

We are given $\frac{dV}{dt}=12\text{ cm}^3/\text{s}$, so solve for $\frac{dr}{dt}$:

$\frac{dr}{dt}=\frac{12}{4\pi r^2}=\frac{3}{\pi r^2}$.

Part (a): when $r=22$ cm

Substitute $r=22$:

$\frac{dr}{dt}=\frac{3}{\pi(22)^2}=\frac{3}{484\pi}$ cm/s.

So the radius is increasing at

$\boxed{\frac{3}{484\pi}\text{ cm/s}}$.

Part (b): when $V=1500$ cm$^3$

First find the radius from the volume formula:

$1500=\frac{4}{3}\pi r^3$.

So

$r^3=\frac{4500}{4\pi}=\frac{1125}{\pi}$,

and therefore

$r=\sqrt[3]{\frac{1125}{\pi}}$.

Now substitute into the rate formula:

$\frac{dr}{dt}=\frac{3}{\pi\left(\sqrt[3]{\frac{1125}{\pi}}\right)^2}$.

That is a valid exact answer. If you want a decimal approximation, compute $r$ first, then plug it in.

Part (c): when it has been filling for 25 seconds

If the balloon starts from empty, then the volume after 25 seconds is

$V=12\cdot 25=300\text{ cm}^3$.

Now solve for the radius:

$300=\frac{4}{3}\pi r^3$

so

$r^3=\frac{900}{4\pi}=\frac{225}{\pi}$,

and

$r=\sqrt[3]{\frac{225}{\pi}}$.

Then

$\frac{dr}{dt}=\frac{3}{\pi\left(\sqrt[3]{\frac{225}{\pi}}\right)^2}$.

Why the setup matters

The derivative $\frac{dr}{dt}=\frac{3}{\pi r^2}$ shows that the radius increases more slowly as the balloon gets larger. That makes physical sense: when the sphere is bigger, the same added volume spreads over more surface expansion. Related-rates problems almost always become straightforward once the geometry formula is differentiated correctly and the changing quantity is isolated.

💡 Pitfall guide

The hardest part of $V=\frac{4}{3}\pi r^3$ related rates is not the differentiation; it is forgetting that each part of the question gives a different way to find $r$. In part (a), the radius is already known, so you substitute directly. In part (b), you must solve for $r$ from the volume first; plugging $V=1500$ straight into the derivative formula does not work because the derivative formula depends on radius, not volume alone. In part (c), another common mistake is assuming the balloon has been filling from a nonzero starting volume. If the problem does not state an initial volume, the usual assumption is that it starts empty, so volume equals rate times time. Finally, keep units consistent: $\text{cm}^3/\text{s}$ for volume rate and $\text{cm}/\text{s}$ for radius rate.

🔄 Real-world variant

If the balloon were filled at $18\text{ cm}^3/\text{s}$ instead of 12, the derivative step would become $\frac{dr}{dt}=\frac{18}{4\pi r^2}=\frac{9}{2\pi r^2}$. The same geometry formula still applies, but every numeric rate changes. Another useful variant is changing the shape from a sphere to a cylinder with fixed height. In that case, the volume formula would be different and the related-rates equation would involve the radius and height relationship for a cylinder rather than $\frac{4}{3}\pi r^3$. For a sphere, though, the strategy always stays the same: differentiate volume with respect to time, isolate $\frac{dr}{dt}$, and substitute the given condition last.

🔍 Related terms

sphere volume formula, differentiating with respect to time, rate of radius change