# differential geometry?

• Dec 23rd 2008, 09:54 AM
differential geometry?
hello there.

i was reading through some work on curves of pursuit and the author stated a certain equivalence without any comment as to where it came from.

Say i have a differentiable curve $C$ in the plane. Given a point $X \in C$ let the distance of the tangent line from the origin at $X$ be $p$, the angle the tangent line makes with the x-axis be $\omega$, and the length of the curve be $s$. Then we have the relation:

$p + \frac{d^2p}{d\omega^2} = \frac{ds}{d\omega}$

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this sort of thing looks like it should be intuitive but i can't see why it is at all, or at least like it should follow from some other nice results. i've been able to show it by letting $C = (x_1(t), x_2(t))$:

$p = \frac{x_1\dot{x}_2-\dot{x}_1x_2}{\sqrt{\dot{x}_1^2+\dot{x}_2^2}}$

$\omega = \text{arctan}\left(\frac{\dot{x}_2}{\dot{x}_1}\rig ht)$

$\frac{ds}{dt} = \sqrt{\dot{x}_1^2+\dot{x}_2^2}$

and then working out all the derivatives manually but that doesn't seem to help me understand it either.

i hope that someone recognizes it and can point me in the direction for a nicer explanation of it. :)