Differential Geometry of Curves and Surfaces - Chapter 1 Section 3 Exercise 3

Problem:

Note the errata to this problem here. In particular, B should be on the line with OC and that's the half line r.

Solution:

Note that $\angle OCA$ is a right angle.

$\frac{CA}{2a} = \sin \theta$
$\frac{BA}{2a} = \tan \theta$
$CA^2 + CB^2 = BA^2$

$(2a \sin \theta)^2 + CB^2 = (2a \tan \theta)^2$
$OP^2 = CB^2 = 4a^2(\tan^2 \theta - \sin^2 \theta)$

With some trigonometry, we get

$\sec^2 \theta = 1 + \tan^2 \theta = 1 + t^2$
$\cos^2 \theta = \frac{1}{\sec^2 \theta} = \frac{1}{1 + t^2}$
$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{1 + t^2} = \frac{t^2}{1 + t^2}$
$\tan^2 \theta - \sin^2 \theta = t^2 - \frac{t^2}{1 + t^2} = \frac{t^4 + t^2 - t^2}{1 + t^2} = \frac{t^4}{1 + t^2}$

So we get $OP = \frac{2at^2}{\sqrt{1 + t^2}}$

$x = OP \cos \theta = \frac{2at^2}{1 + t^2}$
$y = OP \sin \theta = \frac{2at^3}{1 + t^2}$

This time around I solved the problem with a good diagram and geometrical insight without using MATLAB.

Part (b) is trivial, when $t \to \pm \infty$, $x \to 2a$, $y \to 0$,

$\alpha' = 2a\frac{1}{(1+t^2)^2}((1 + t^2)2t - t^2(2t), (1 + t^2)3t^2 - t^3(2t))$

So $\alpha'(t) = (0, 2a)$ as $t \to \pm \infty$

At this point I am still confused, look like $\alpha'(t)$ should be $(0, 2a)$ and not $(2a, 0)$?