Derive the density relation to thermal expansion

[Ryker]

Homework Statement

Show that it follows from

$$\frac{dV}{V} = \beta dT$$

that, for small changes, density has a similar linear relationship to thermal expansion, namely

$$\frac{d\rho }{\rho } = - \beta dT$$

Homework Equations

The Attempt at a Solution

Well, at first glance this one seems ridiculously simple, but I tried setting up $$dV$$ as $$V_{1} - V$$, then inserting $$V_{1} = \frac{m}{\rho _{1}}$$ and similar for $$V$$, but when I go through the whole mathematical process that leaves me with the following equation:

$$\frac{d\rho }{\rho _{1}} = - \beta dT$$

as opposed to:

$$\frac{d\rho }{\rho } = - \beta dT$$

I'm not that good with LaTeX, so I haven't sketched the whole mathematical process I went through - there's not much to it anyway, to be honest - but does someone know what one needs to do to get the proper equation and where I went wrong? Is there some trick to it that I'm missing?

The final result for $$dV$$ is namely

$$\frac{-md\rho }{\rho \rho _{1}}$$

and I was wondering if you could express $$\rho _{1} = \rho + d\rho$$, multiply and then dismiss the product $$\rho d\rho$$ as so small as to be irrelevant. Then you would namely get the right result, but I'm not that sure that product really is that small in comparison to the other one.

And sorry for any screw-ups in LaTeX, like I mentioned I'm not well versed in it (yet) and I needed almost half an hour to type even this up.

 

[hikaru1221]

When $$d\rho << \rho$$, you can write: $$\rho _1 \approx \rho$$

 

[Ryker]

God damn it, and this is something I literally lose hours over, despairing how stupid I am to not be able to get the right solution and lose motivation to do the stuff altogether! Thanks to you I can now finally get back to work, though