# Plotting Volume as a function of density, limit of this

Gold Member

## Homework Statement

The density of an object is given by its mass divided by its volume: ##p=\frac{m}{V}##
Use a calculator to plot the volume as a function of density (##V=\frac{m}{p}##), assuming a mass of 8kg (m=8).

In the follow-up question (part b): Evaluate ##\lim_{p \rightarrow 0} {V\left(p\right)}##

## The Attempt at a Solution

I just plain don't understand what this is asking me at all.
I started off by stating that volume as a function of density ⇒ V(p)

Then I made a table. The left column as density, and the right column as ##V(p)=\frac{8}{p}##

Then I got lost in thinking what values I should plug in.

As a thought process, I noted that as the density increases, the volume decreases and vice versa.

I'm not sure if I'm understanding what the question is asking me.

Mentor
2021 Award
I guess they want to see the graph ##p \longmapsto V(p)##. Do you have an idea how it will look like and what happens at ##p=0\,?##

Gold Member
Well at p=0, the function would be undefined (##\frac{8}{0}##).
As for the graph, I'm not so sure. If I think of p as any other input such as x, and move things around, I could get ##V(p)=8\frac{1}{p}## which is a simple reciprocal function with the output values multiplied by 8. In picturing this, it partly makes sense. If it is in fact a reciprocal function, as the density approaches zero, the volume goes to infinity. And as the volume goes to infinity, the density approaches zero. But the part of the graph across the origin doesn't make any sense in this situation.

Mentor
2021 Award
Well at p=0, the function would be undefined (##\frac{8}{0}##).
As for the graph, I'm not so sure. If I think of p as any other input such as x, and move things around, I could get ##V(p)=8\frac{1}{p}## which is a simple reciprocal function with the output values multiplied by 8. In picturing this, it partly makes sense. If it is in fact a reciprocal function, as the density approaches zero, the volume goes to infinity. And as the volume goes to infinity, the density approaches zero. But the part of the graph across the origin doesn't make any sense in this situation.
No, it isn't defined at ##p=0##, that's why the limit came into play ##\lim_{p \to 0} V(p)##. The question is not, that it makes no sense, the question is what happens when you approach zero. Again, forget notations like ##\frac{8}{0}##, they are nonsense. However, your description was right:
as the density approaches zero, the volume goes to infinity
The graph of ##p \longmapsto \dfrac{8\,kg}{V}## or ##f(x) \sim \dfrac{1}{x}## has a certain name. It is called a hyperbola.

Gold Member
No, it isn't defined at p=0p=0, that's why the limit came into play limp→0V(p)\lim_{p \to 0} V(p). The question is not, that it makes no sense, the question is what happens when you approach zero. Again, forget notations like 80\frac{8}{0}, they are nonsense. However, your description was right:

I feel pretty confident in saying that as x approaches zero, there is no limit that exists as on one side, it goes to minus infinity and on the other to positive infinity. However, in approaching zero from the right, the graph does indeed go to infinity.

Two questions:

1) How can we use a hyperbola to model this function if the graph of the hyperbola has values at negative infinity? We cannot have a negatively infinite density. Is this a situation where we just use judgement to say that we are only concerned with the positive values?

2) So in using a calculator the plot the volume as a function of density (given problem), what does this even mean? Plug in values, say from 0 to 1 for the input, calculate them, and note what is happening to the function?

Mentor
2021 Award
1) The formula works for negative values as well, so the function is mathematically both branches of the hyperbola. Of course it doesn't make sense physically, so only the positive branch is considered, but it's still a hyperbola.

2) Yes, e.g. values between ##0## and ##1## in order to draw ##\{\,(p,V)\,|\,V=\frac{8}{p}\,\} \subseteq \mathbb{R}^2##, and you have noted what happens. If you like you can try and formally prove that the function value ##V(p)## increases beyond all limits if you approach ##p=0## (from the right).

http://www.wolframalpha.com/input/?i=V=8/p

• opus
Gold Member
Ok thank you. I'll have a look at that website and see what I can make of it.