How Does a One-Dimensional Creature Measure Length on a Circle?

[MrNerd]

Homework Statement

This is from Exploring 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^{2} = \frac{dx^{2}}{1 - x^{2}/r^{2}}
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.

Homework Equations

Equations in text for the chapter so far:
[1] r^{2} = x^{2} + y^{2}
[2] y^{2} = r^{2} - x^{2}
[3] 2ydy = -2xdy r = constant
[4] dy = -\frac{xdx}{y} = -\frac{xdx}{(r^{2} - x^{2})^{1/2}} r = constant
[5] dl^{2} = dx^{2} + dy^{2} = dx^{2} + \frac{x^{2}dx^{2}}{(r^{2} - x^{2}} r = constant

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^{2} factored from the bottom. I haven't quite figured out where it went, though. The closest I've gotten is:

dy = -\frac{xdx}{(r^{2} - x^{2})^{1/2}}
Solving for dx, I get dx = -\frac{dy(r^{2} - x^{2})^{1/2}}{x}
I then input this into equation [5], getting dl^{2} = \frac{dy^{2}(r^{2} - x^{2})^{2}}{x^{2}} + \frac{x^{2}dx^{2}}{(r^{2} - x^{2})} = \frac{dy^{2}(r^{2} - x^{2})^{2} + x^{4}dx^{2}}{x^{2}(r^{2} - x^{2})}
Then I moved the denominator over, dl^{2}x^{2}(r^{2} - x^{2}) = dy^{2}(r^{2} - x^{2})^{2} + x^{4}dx^{2}
Substituting the dy gives dl^{2}x^{2}(r^{2} - x^{2}) = x^{2}dx^{2} + x^{4}dx^{2}
Dividing the things by the dl gives dl^{2} = \frac{dx^{2}(1 + x^{2})}{(r^{2} - x^{2})} = \frac{dx^{2}(1 + x^{2})}{r^{2}(1 - x^{2}/r^{2})}
Clearly, the only stuck thing is the \frac{(1 + x^{1})}{r^{2}}. 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.

 

[daveb]

Just look at [5] above and see where you can make the whole thing a single fraction.

 

[MrNerd]

Aha! I got it now!

I did:
dl^{2} = \frac{dx^{2}(r^{2} - x^{2}) + x^{2}dx^{2}}{(r^{2} - x^{2})}

dl^{2} = \frac{dx^{2}(r^{2} - x^{2} + x^{2})}{(r^{2} - x^{2})}

Canceling out the x^{2}s and dividing r^{2} from top and bottom gets the answer.

I knew there was an obvious answer, I just couldn't see it. I guess I had a brain fart. Thanks for the help.

 

[Mark44]

MrNerd - a tip
Instead of using a slew of itex /itex pairs of tags, put one itex tag at the beginning of your equation, and a /itex tag at the end of the equation.

For the 2nd equation above, I used a tex and /tex pair of tags. This makes fractions a little larger and easier to see.