# Describing the radius with density

1. Jul 13, 2013

### MathJakob

How can I describe mathematically that if you take some mass $m$ with some density $d$ as the radius tends to $0$ the density tends to $\infty$

I'm not exactly sure what it is I'm trying to describe I'm just looking for something that describes when the radius of an object decreases, the density increases.

How can I write this using mathematical notation?

2. Jul 13, 2013

### Millennial

You could say $d \propto r^{-1}$, or more accurately $d \propto v^{-1}$.

3. Jul 13, 2013

### MathJakob

Ok so how would I read that?

4. Jul 13, 2013

### nickbob00

Density is inversely proportional to volume, given constant mass.

5. Jul 15, 2013

### MathJakob

OK I think I understand. so let's just use some random figures. If I had a steel ball bearing which had a radius of 8cm, a mass of 1lb, volume of 2144.66 and a density of 2.76 g/cm3. If I decreased the radius by half, what factor does the density increase by?

6. Jul 25, 2013

### CarsonAdams

This will fully explain.

Remember,
$D=\frac{m}{V}$
and, for a sphere,
$V=\frac{4}{3}$$\pi$r2
so if
$V$1=$\frac{4}{3}$$\pi$r2
with full radius r, then
$V$2=$\frac{4}{3}$$\pi$$(\frac{r}{2})$2
with half the radius or
$V$2=$\frac{1}{4}$$(\frac{4}{3}$$\pi$r2)
which yields the proportion
$V$2=$\frac{1}{4}$$V$1

So, since
$D$1=$\frac{m}{V_{1}}$
and
$D$2=$\frac{m}{V_{2}}$
then
$D$2=$\frac{4(m)}{V_{1}}$
and
$D$2=$4D$1

Notice how they're inversely proportional by a square of the radius? (I mostly wrote this up to practice Latex, but I hope this helped)

7. Jul 25, 2013

### chiro

Hey MathJakob.

Take what the above posters have said and pose your problem in terms of mathematical limits.

8. Jul 28, 2013

### MathJakob

Well mathematically all I can think of is $$\lim_{r \to 0} d \to \infty$$ As the limit of radius goes to 0, the limit of density goes to infinity

9. Jul 28, 2013

### 1MileCrash

Syntax wise, that's not quite right.

The limit of density as radius tends to 0 is infinity (a lot of people say "grows without bound.")

But, by syntax wise, I mean that you spoke of two limits, but this is one limit. Furthermore, you would write:

$$\lim_{r \to 0} d = \infty$$

(An equals sign as opposed to a second arrow)

But this is just my personal experience, it may be the way that you wrote it is acceptable, it just looks and sounds strange to me.

10. Jul 28, 2013

### Integral

Staff Emeritus
This just does not work. You have SPECIFIED a specific and fixed density. If you compute the the mass of your sphere given the above data either the density is incorrect or the mass is you cannot have it both ways. Given your density and radius the mass must be near 6kg or 13lbs.

Why are you mixing systems? Specify your quantities in either metric or American, don't mix.

I think what you want to do is hold mass constant then shrink the radius. Now as the radius tends to zero the density will grow without bound.

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