Thread: Curvature of a rectangular hyperbola

I'm trying to figure out what the curvature of the hyperbola given by y = 1/x is, and I already know the answer. Unfortunately, I'm getting a strange factor of x^3 in my answer.

Essentially:

y = 1/x

Curvature formula:

|f''(x)|/((1+(f'(x))^2)^(3/2))

So

f'(x) = -1/(x^2)
f''(x) = 2/(x^3)

And therefore

k = |2/(x^3)|/((1/(x^4) + 1)^(3/2))
= 2/(|t^3| * (1 + 1/(t^4))^(3/2)) (as x is the same as the parameter t, (x, y) = (t, 1/t))

and p = 1/t = 1/2 * (1 + 1/(t^4))^(3/2) * |t^3|

But, the curvature which leads to the correct evolute for the curve (which is what I'm really looking for) is simply 1/2(1 + 1/(t^4))^(3/2), without the |t^3|. I've been working on this for an hour, and I still can't figure out why it disappears. Thanks for any help.

Originally Posted by StarWrecker

I'm trying to figure out what the curvature of the hyperbola given by y = 1/x is, and I already know the answer. Unfortunately, I'm getting a strange factor of x^3 in my answer.

... [SNIP] ...

And therefore

k = |2/(x^3)|/((1/(x^4) + 1)^(3/2))
= 2/(|t^3| * (1 + 1/(t^4))^(3/2)) (as x is the same as the parameter t, (x, y) = (t, 1/t))

and p = 1/t = 1/2 * (1 + 1/(t^4))^(3/2) * |t^3|

But, the curvature which leads to the correct evolute for the curve (which is what I'm really looking for) is simply 1/2(1 + 1/(t^4))^(3/2), without the |t^3|. I've been working on this for an hour, and I still can't figure out why it disappears.

Your formula can be written (by writing ). That is the correct answer. It is symmetric in t and t^{-1}, as it has to be on geometric grounds (just think of the graph of the hyperbola: it is symmetric about the line y=x, so it must have the same curvature at x=t^{-1} as it does at x=t).

So your calculation is absolutely correct, but unfortunately for you "the curvature which leads to the correct evolute" is not the correct curvature.

Oh, I see what I did there. I must have made a mistake elsewhere, because 2/(t^2 + t^-2)^(3/2) works for the evolute now. Thanks a lot. I didn't think of dealing with the |t^3| that way.