# Sunday Function

Saw “Up” yesterday. How Pixar manages to be so consistent in their astonishing quality is entirely beyond me. In a bit of a tribute, this Sunday Function is not about any dramatically important special function, but instead it’s about filling a balloon. Air or water, as your preference.

You’ll have noticed that when you start to fill a balloon, its radius expands very rapidly at first before slowing dramatically. A water balloon will go from 1 to 2 inches much faster than it will go from 6 to 7 inches. Why? Because if you fill the balloon at a constant rate, you’re increasing the volume at a constant rate. The volume and radius of a balloon don’t change at the same rate. Pretend the balloon is a sphere for simplicity. The volume is related to the radius in the usual way:

Now we can use calculus to find how that static relationship turns into a relationship between the rates of change of the volume and radius. First, remember that the chain rule will ensure that the rate of change of volume with respect to time is equal to the rate of change of volume with respect to radius times the rate of change of radius with respect to time. A little cumbersome to say in words, but easy in the language of calculus after differentiating both sides:

But the term on the left isn’t some complicated function, it’s just our constant fill rate since the change in volume is just the rate at which we add volume via whatever we’re filling. This is true for water balloons, and a reasonably good approximation for air (which is a little compressible). But since it’s a constant, why not drop the derivative notation and just give it a constant name like the Greek letter rho? Do that and rearrange a little bit and you get our Sunday function.

What it means is that the rate of change of the radius is inversely proportional to the square of the radius at that moment. When the sphere has grown by a factor of five, the rate of growth of the radius has slowed by a factor of 25.

The equation we derived is implicitly a differential equation in the time variable, but we don’t need to go through the effort of solving it to see that we’ve already figured out mathematically why the rate of growth slows. As in making a film, simplicity is often the best choice.

1. #1 kermit
June 14, 2009

Isn’t there another reason? The resistance to stretching must increase as it nears the limit of the rubber or other material. Eventually the ability of your lungs (or other compressor) would be reached, and the balloon would break. But until then, the air pressure is increasing inside, as anyone who has blown up balloons knows. That is, less volume for the amount of air put in.

2. #2 CCPhysicist
June 14, 2009

NICE example of an application of “related rates”.

Also good question from kermit@1, but my experience is that it gets easier to blow up a balloon as it inflates. Computer interfaced pressure gauge and needle valve, anyone?

The case of a water balloon is very different from air, because the weight of the water helps stretch the balloon.

3. #3 Freiddie
June 14, 2009

Quote:
“What it means is that the rate of change of the radius is inversely proportional to the square of the volume.”
I think you meant:
“… inversely proportional to the square of the radius”
right?

4. #4 Matt Springer
June 14, 2009

#1 & 2, I’m not so worried about the ease of blowing it up, just with the speed at which its radius increases assuming you can keep the volume increasing at a constant rate. This may or may not be getting easier or harder as you pump in more air. The rate at which you pump in air may itself have to be variable to keep up a constant volume, though with incompressible water the flow rate should just be constant.

#3, Exactly so. I’ve fixed it. Thanks!

5. #5 photon
June 14, 2009

My first thought on seeing this was “Huh – Pixar’s re-made a Russ Meyer film?”

I soon realised my mistake, but strangely, your balloon calculations could apply to either film. š

6. #6 Dustron
June 15, 2009

Just a couple of comments; With regards to kermit’s comment, you make a valid point. However, as Matt said, he’s not concerned with the ease of blowing up the balloon. Assuming dV/dt is constant, his analysis is correct (also assuming a balloon is a perfect sphere, but that’s reasonable). True, the resistance to stretching will likely increase as balloon stretches. However, the stress in the balloon changes as the balloon deforms. The ratio of this stress to the pressure in the balloon varies approximately as (1/R). So as the balloon expands, this ratio decreases (even faster if the thickness of the balloon is decreasing – conservation of material volume), meaning more stress per internal pressure, and thus greater yield of the balloon; this is why it gets easier to blow up, as CCPhysicist observed.

As for the compressibility effects of air, I doubt this will be very important. Despite being a gas, air is not very compressible at the pressures we’re considering. The balloon expands precisely because it is easier to deform the balloon than it is to compress the air. Just try blowing air into a rigid container (not as easy as it might seem). Thus, unless you’re considering some extremely strong balloons, accounting for the compressibility of air will probably not change the results too much.