calculusplane-curves
I am working on this problem. Even some of the notation has me confused (the vectors $\vec i$ and $\vec j$).
Let $\vec r(t):=ae^{-bt}\cos(t)\vec i +ae^{-bt}\sin(t)\vec j$ where $a$ and $b$ are positive constants. The trace of $\vec r (t)$ is called the logarithmic spiral.
(a) Show that as $t \to +\infty $, $\vec r (t)$ approaches the origin.
(b) Show that $\vec r (t)$ has finite arc length on $[0,\infty)$.
You can check the length of the tangent of a function at a given point is:
$T = \left|\dfrac{y}{y'}\sqrt{y'^2+1}\right|$
For full detail check Piskunov's Calculus, Vol I, page 127.
Since in your equation the tractrix is running along the $y$ axis I'll use it running through the $x$ axis (since the formula I give you is for such cases).
So you have
$y = \sin t$
$x = \cos t +\log \tan\dfrac{t}{2}$
$\dfrac{dy}{dt} = \cos t$
$\dfrac{dx}{dt} = -\sin t +\dfrac{\sec ^2\dfrac{t}{2}}{2\tan\dfrac{t}{2}}$
Using the double angle formula you'll get
$$\frac{{dx}}{{dt}} = \frac{1}{{\sin t}} - \sin t$$
or
$$\frac{{dx}}{{dt}} = \frac{{\cos^2 t}}{{\sin t}} $$ and then
$$\frac{{dy}}{{dx}} = \frac{{\cos t}}{{\dfrac{{{{\cos }^2}t}}{{\sin t}}}} = \tan t$$
Plugging this in gives
$$T = \left| {\dfrac{{\sin t}}{{\dfrac{{\sin t}}{{\cos t}}}}\sqrt {{{\tan }^2}t + 1} } \right| = \left| {\dfrac{{\sin t}}{{\sin t}}\cos t\sec t} \right| = 1$$
Finding the formulas.
From geometry we know that
$$\tan \theta = \dfrac{y}{S_T}$$
But since $\tan \theta = \dfrac{dy}{dx} = y'$ we can put
$$y' = \dfrac{y}{S_T}$$
$$S_T = \left|\dfrac{y}{y'}\right|$$
We consider the absolut value since $\tan \theta$ isn't always positive.
But now, knowing $S_T$ we can use
$$T^2 = S_T^2 +y^2$$
$$T^2 = \dfrac{y^2}{y'^2} +y^2$$
$$T = \left|y' \sqrt{\dfrac{1}{y'^2}+1}\right|$$
$$T = \left|\dfrac{y}{y'} \sqrt{{y'^2}+1}\right|$$
Looking to the other triangle, we get
$$\tan \theta = \dfrac{S_N}{y}$$
So
$$|y y'|= S_N$$
And finally since
$$N^2 = y^2 + S_N^2$$
You get
$$N^2 = y^2 + y^2 y'^2$$
$$N = \left| y\sqrt{1+y'^2}\right|$$
These lengths are called tangent, subtangent, normal and subnormal. Remember $y = f(x)$ and $y' = f'(x)$, so you always have to plug in a number to get a value.
So it looks like the Wikipedia reference image uses 45 degree sections of these curves. You can use the equation for the spiral to give you the tangent line at the beginning and end of these curve sections. Evaluate the derivative at these two points to get the tangent line slope and then shift your line appropriately to hit the point used. The intersection Of these two lines should be your control point. Once you have found your control point you can put it in the function 'CGPathAddQuadCurveToPoint' for the cx, cy (I think) along with the point you want to go to (also from the spiral equation). For reference--check out the animation under 'quadratic curves' here
For extra speed, you only have to find 8 tangent lines max--just shift them out for the next cycle of the spiral and reuse them.
Best Answer
The length is given, as the OP wrote, by $$\int_0^\infty\sqrt{a^2b^2e^{-2bt}(\cos^2t+\sin^2t)+a^2e^{-2bt}(\cos^2t+\sin^2t)}\,\,dt=$$ $$=\int_0^\infty ae^{-bt}\sqrt{b^2+1}\,\,dt=\left.-\frac{a\sqrt{b^2+1}}{b}\,e^{-bt}\right|_0^\infty=\frac{a}{b}\sqrt{b^2+1}$$