# Finding unit speed reparametrisations of regular curves

• April 6th 2010, 07:58 AM
millwallcrazy
Finding unit speed reparametrisations of regular curves
How can i find unit speed reparametrisations of these regular curves?

1) y(t) = (cos^2(t),sin^2(t)) -infinity < t < infinity (cos squared t, sine squared t)

2) y(t) = (t, cosh(t) -infinity < t < infinity

thanks
• April 6th 2010, 01:10 PM
Opalg
Quote:

Originally Posted by millwallcrazy
How can i find unit speed reparametrisations of these regular curves?

1) y(t) = (cos^2(t),sin^2(t)) -infinity < t < infinity (cos squared t, sine squared t)

2) y(t) = (t, cosh(t) -infinity < t < infinity

I'll show how to do 2). Then you try 1). (Though why someone from Leeds should want to help a Millwall fan I can't imagine. (Punch) )

In 2), the coordinates are x=t, y=cosh(t). The arc length is given by $u(s) = \int_0^s\sqrt{\bigl(\tfrac{dx}{dt}\bigr)^2 + \bigl(\tfrac{dy}{dt}\bigr)^2}\,dt = \int_0^s\sqrt{1+\sinh^2t}\,dt = \sinh s$. The inverse function is given by $s = \sinh^{-1}u$ , and the unit speed parametrisation (with parameter u) is given by

$\bigl(x(s(u)),y(s(u))\bigr) = \bigl(\sinh^{-1}u,\cosh(\sinh^{-1}u\bigr) = \bigl(\sinh^{-1}u,\sqrt{1+u^2}\bigr)$.