(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This is fromExploring Black Holes, Query 1, page G-3. It involves a one dimensional creature on a two dimensional circle.

Constant dl but different dx.Collect terms in equation [5]. Show that the result can be written dl[itex]^{2}[/itex] = [itex]\frac{dx^{2}}{1 - x^{2}/r^{2}}[/itex]

Equation [5], involving dl, is for finding the length of a measuring rod of length dl along the circle, laid by the one dimensional creature.

2. Relevant equations

Equations in text for the chapter so far:

[1] r[itex]^{2}[/itex] = x[itex]^{2}[/itex] + y[itex]^{2}[/itex]

[2] y[itex]^{2}[/itex] = r[itex]^{2}[/itex] - x[itex]^{2}[/itex]

[3] 2ydy = -2xdy r = constant

[4] dy = -[itex]\frac{xdx}{y}[/itex] = -[itex]\frac{xdx}{(r^{2} - x^{2})^{1/2}}[/itex] r = constant

[5] dl[itex]^{2}[/itex] = dx[itex]^{2}[/itex] + dy[itex]^{2}[/itex] = dx[itex]^{2}[/itex] + [itex]\frac{x^{2}dx^{2}}{(r^{2} - x^{2}}[/itex] r = constant

3. The attempt at a solution

I've realized fairly easily that the bottom part of the equation I'm trying to get has an r[itex]^{2}[/itex] factored from the bottom. I haven't quite figured out where it went, though. The closest I've gotten is:

dy = -[itex]\frac{xdx}{(r^{2} - x^{2})^{1/2}}[/itex]

Solving for dx, I get dx = -[itex]\frac{dy(r^{2} - x^{2})^{1/2}}{x}[/itex]

I then input this into equation [5], getting dl[itex]^{2}[/itex] = [itex]\frac{dy^{2}(r^{2} - x^{2})^{2}}{x^{2}}[/itex] + [itex]\frac{x^{2}dx^{2}}{(r^{2} - x^{2})}[/itex] = [itex]\frac{dy^{2}(r^{2} - x^{2})^{2} + x^{4}dx^{2}}{x^{2}(r^{2} - x^{2})}[/itex]

Then I moved the denominator over, dl[itex]^{2}[/itex]x[itex]^{2}[/itex](r[itex]^{2}[/itex] - x[itex]^{2}[/itex]) = dy[itex]^{2}[/itex](r[itex]^{2}[/itex] - x[itex]^{2}[/itex])[itex]^{2}[/itex] + x[itex]^{4}[/itex]dx[itex]^{2}[/itex]

Substituting the dy gives dl[itex]^{2}[/itex]x[itex]^{2}[/itex](r[itex]^{2}[/itex] - x[itex]^{2}[/itex]) = x[itex]^{2}[/itex]dx[itex]^{2}[/itex] + x[itex]^{4}[/itex]dx[itex]^{2}[/itex]

Dividing the things by the dl gives dl[itex]^{2}[/itex] = [itex]\frac{dx^{2}(1 + x^{2})}{(r^{2} - x^{2})}[/itex] = [itex]\frac{dx^{2}(1 + x^{2})}{r^{2}(1 - x^{2}/r^{2})}[/itex]

Clearly, the only stuck thing is the [itex]\frac{(1 + x^{1})}{r^{2}}[/itex]. Does this mean anything, like it equals 1 or something, and I've just forgotten an identity? I've also tried a couple of other methods, but I would say this is pretty much what I get to. It may not make perfect sense, but I was running out of ideas, and this was the last thing I tried.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: I need some help deriving the equation

**Physics Forums | Science Articles, Homework Help, Discussion**