finite length of a spiral consider a "spiral" $\alpha(t)=r(t)\left(\cos(t),\sin(t)\right)$, where $r$ is $\mathcal{C}^1$ and  $0\le r(t) \le 1$ for all $0 \le t$
Show that if $\alpha$ has finite length on $ [0,\infty)$ and $r$ is decreasing, then $r(t) \rightarrow 0$ as $t\rightarrow\infty$
im not exactly sure where to start with this.  later on in the question it asks you to prove the result without the fact that the function is decreasing and im not really sure what to do for either. i keep getting to a point where i can logically follow the result in my head but i cant put it down in a mathematical context.  any help is appreciated. thanks!
 A: Hints:  What is $\frac {ds}{dt}$?  Write the arclength formula.  As $t$ increases by $2\pi$ the spiral makes a full circle around the origin.  Given the minimum $r$ over the circle, can you bound the arc length of that term from below?
Added:  if $r \not \to 0$ there must be some $\epsilon$ such that $r \gt \epsilon$ infinitely often but $r$ must be very close to zero (less than $\epsilon/2$)most of the time or the arclength will be infinite by the first solution.  That means it must bump up from $0$ to $\epsilon$ and go back down.  Each trip has arclength greater than $\epsilon$.  Can you flesh this out?
A: For the non-decreasing version, observe that
\begin{align}
\alpha'&=r'(\cos t,\sin t)+r(-\sin t,\cos t),\\
\Vert\alpha'\Vert &=\sqrt{r'^2+r^2}.
\end{align}
So
\begin{align}
\int_0^\infty\vert r'\vert \,dt
\le
\int_0^\infty\sqrt{r'^2+r^2}\,dt
=
\int_0^\infty\Vert\alpha'\Vert\,dt<\infty,
\end{align}
where $\displaystyle\int_0^\infty\Vert\alpha'\Vert\,dt$ denotes the arclength of
$\alpha$ on $[0,\infty)$.
It follows that $\displaystyle\int_0^\infty r'\,dt$ converges. Now, consider
$r(b)=r(0)+\displaystyle\int_0^b r'(t)\,dt$, where $b\in[0,\infty)$, then
clearly $r(b)\to r_0$ as $b\to\infty$, for some number $r_0\in[0,1]$.
Suppose $r_0>0$, then we may easily conclude that the spiral $\alpha$ has no finite length. This gives a contradiction.
