Curvature of a rectangular hyperbola

[StarWrecker]

Homework Statement

The hyperbola y = 1/x in the first quadrant can be given the parametric definition (x, y) = (t, 1/t), t>0.

Find the corresponding parametric form of its evolute, and sketch both curves in the region 0<x<10, 0<y<10

Homework Equations

Curvature formula:

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

The Attempt at a Solution

I've worked through this in full, but there is a single term in my curvature that is throwing off my end result. My calculation of this curvature is as follows:

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.

 

[Tocqueville]

Hi,StarWrecker,I am 100 percent sure "p = 1/t = 1/2 * (1 + 1/(t^4))^(3/2) * |t^3|"
is correct.
And I am also sure that p(t) leads to the correct evolute for the curve.
Check your answer again and again,you will find out your errors on computation.

 

[StarWrecker]

Yeah, I made a mistake in my later calculations for the evolute that I ironed out after trying my solution for the curvature again. Thanks.