Confusion about successive derivatives of position in circular motion Suppose we define a unit vector $\vec r$ along radial direction for a particle in uniform circular motion at an angular frequency $\omega$. Then we can write:
$$\vec r = \cos(\omega t)\hat i + \sin(\omega t)\hat j$$
And the modulus of this vector is one. If we differentiate this to get the velocity we get:
$$\vec v = \omega(-\sin(\omega t)\hat i + \cos(\omega t)\hat j)$$
And the modulus of the velocity is $\omega$. Differentiate again to get the acceleration and we get:
$$\vec a = \omega^2(-\cos(\omega t)\hat i - \sin(\omega t)\hat j)$$
And the modulus of the velocity is $\omega^2$. In general we find:
$$ \left|\frac{d^n\vec r}{dt^n} \right| = \omega^n $$
This feels weird, because for any $\omega > 1$ the modulus of each successive derivative keeps increasing without limit as we increase $n$, and yet that's what the calculation tells us. Is there some way to understand intuitively what it means for the successive derivatives to increase all the way to infinity for $n \to \infty$?
 A: I think there is a teachable moment here, in that I think we can get some insight through a different way to frame your question.
The first thing to note is that $\omega$ has dimensions of $[{\rm time}]^{-1}$ (in SI units, ${\rm s}^{-1}$). Therefore, it does not make sense to compare $\omega$ to a dimensionless number, like $1$, so saying $\omega>1$ or $\omega<1$ is not a well-formed statement. The numerical value of $\omega$ can be larger than, less than, or equal to one, depending on the unit system chosen.
In fact, for uniform circular motion, we can always choose units where $\omega=1 /({\rm unit\ of\ time})$. This amounts to saying that we choose our unit of time so that the particle makes one revolution of the circle in one unit of time -- or in more physical terms, we use the particle itself as our clock (think of the particle like a second hand on a stopwatch).
In these units, all of the derivatives take on the same value, since $\omega=1$ in our units. So this should indicate to you that nothing is actually "blowing up" physically. What you are seeing is a kind of self-similarity, where the motion and all of its derivatives is constantly changing.
The apparent growth you were finding at larger and larger values of $n$ is an artifact of the units you use to describe the system. Let's consider the first derivative. The x component of the velocity has to change from $v_x = 2\pi R \omega $ to 0 over the span of one quarter of an orbit around the circle. The time it takes for a quarter cycle is $T_{1/4}=1/(4\omega)$, since $1/\omega$ is by definition the time for one cycle. Therefore the acceleration over this period needs to be $a_x=-2\pi R \omega / (1/4\omega) = - 8\pi R \omega^2$. There are two powers of $\omega$ in this expression: one coming from the original x component of the velocity which needed to be lost over a quarter-cycle, and another  power accounting for the unit of time over which the deceleration took place. This explains why there is a factor of $\omega^2$ in the acceleration (2nd derivative), but note it does not imply that there is any divergence. If we chose to measure the motion of the particle in different units, the numerical value of the acceleration could be larger or smaller or stay the same as the velocity.
A similar argument works for every successive derivative of motion.
A: you start with $\vec v_1$
$$\vec v_1= \left[ \begin {array}{c} \rho\,\cos \left( \varphi _{{0}} \right) 
\\ \rho\,\sin \left( \varphi _{{0}} \right) 
\end {array} \right]
$$
the first time derivative is :
$$\vec v_2=\vec {\dot{v}}_1=\omega\,S(\frac \pi2)\,\vec v_1$$
with $$S(\varphi)=\left[ \begin {array}{cc} \cos \left( \varphi  \right) &-\sin \left( 
\varphi  \right) \\ \sin \left( \varphi  \right) &
\cos \left( \varphi  \right) \end {array} \right] 
$$
$$\vec v_2=\left[ \begin {array}{c} -\omega\,\rho\,\sin \left( \varphi _{{0}}
 \right) \\ \omega\,\rho\,\cos \left( \varphi _{{0}}
 \right) \end {array} \right] 
$$
where $\vec{v}_2 \perp\vec{v}_1$
the second  time derivative is :
$$\vec v_3=\vec {\ddot{v}}_1=\omega\,S(\frac \pi2)\,\vec v_2= \left[ \begin {array}{c} -{\omega}^{2}\rho\,\cos \left( \varphi _{{0}
} \right) \\ -{\omega}^{2}\rho\,\sin \left( \varphi 
_{{0}} \right) \end {array} \right] 
$$
where $\vec{v}_3 \perp\vec{v}_2$
$$\vec v_n= {\frac{d^n}{dt^n}}\vec{v}_1=\omega\,S(\frac \pi2)\,\vec v_{n-1}$$
where
$\vec{v}_n \perp\vec{v}_{n-1}$
conclusion : a new derivative vector is obtain by rotate the pervious  derivative vector with rotation angle  $\pi/2$ and multiplying  with $\omega$


A: 
Is there some way to understand intuitively what it means for the successive derivatives to increase all the way to infinity

Similar phenomenon goes on for derivatives of position in a linear movement.
Fourth derivative of position is snap, and such movement is described by equation :
$$ {\vec {r}}={\vec {r}}_{0}+{\vec {v}}_{0}t+{\frac {1}{2}}{\vec {a}}_{0}t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}t^{3}+{\frac {1}{24}}{\vec {s}}t^{4} $$
So $\vec r = f(t^1,t^2,t^3,t^4)$ is polynomial equation of degree 4.
Fifth derivative of position is crackle, and such movement is described by :
$$ {\vec {r}}={\vec {r}}_{0}+{\vec {v}}_{0}\,t+{\frac {1}{2}}{\vec {a}}_{0}\,t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}\,t^{3}+{\frac {1}{24}}{\vec {s}}_{0}\,t^{4}+{\frac {1}{120}}{\vec {c}}\,t^{5} $$
Position vector is described by 5-th degree polynomial equation, $\vec r = f(t^1,t^2,t^3,t^4,t^5)$
Last, but not the end,- Sixth derivative of position is pounce, with movement equation :
$$ {\vec {r}}={\vec {r}}_{0}+{\vec {v}}_{0}\,t+{\frac {1}{2}}{\vec {a}}_{0}\,t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}\,t^{3}+{\frac {1}{24}}{\vec {s}}_{0}\,t^{4}+{\frac {1}{120}}{\vec {c}}_{0}\,t^{5}+{\frac {1}{720}}{\vec {p}}\,t^{6} $$
This is 6-th degree polynomial. In general it seems that position vector can be described by n-th degree polynomial with respect to time :
$$\vec r = \vec r(t^1,t^2,\cdots,t^n)$$
Or as expressed as series :
$$ \vec {r}={\vec {r}}_{0}+\sum_{k=0}^{n} \frac{1}{(k+1)!} \vec x_k\,t^{k+1}
$$
As about Physical interpretation of increasing polynomial degree in monotonic way,- hard to say. Probably that there's no technical limits of what can change.
Higher order derivatives indicates that something in a system is changing at a rate a lower-order derivative is not able to cope with.
