Is the derivative of the magnitude of a position vector the speed? Is the integral of the speed the magnitude of a position vector? I'm reading a book titled "Introduction to Mechanics" by Kleppner, and I came across an example: A bead moves along the spoke of a wheel at constant speed $u$ meters per second. The wheel rotates with uniform angular velocity $\dfrac{d\theta}{dt}=\omega$ radians per second about an axis fixed in space. At $t=0$ the spoke is along the $x$-axis, and the bead is at the origin. Find the velocity at time $t$ in polar coordinates.
The solution includes inferring that $\dfrac{dr}{dt}=u$ and $r=ut$. But since $r$ is the magnitude of a position vector, it just means that its derivative is the speed. I tried to derive this mathematically but I just can't get to the desired result. Is there a mathematical explanation regarding this?
PS. I posted a similar question here. But I deleted it since my question was wrong.
 A: 
Is the derivative of the magnitude of a position vector the speed? Is the integral of the speed the magnitude of a position vector?

Not in general. Consider an object undergoing uniform circular motion about the origin. The magnitude of its position vector is constant (it is the radius of the circle) so the time derivative of the magnitude is zero, but the speed of the object is not zero.
In other words, in general
$$\displaystyle \frac {d |\vec r|}{dt} \ne \left | \frac {d \vec r}{dt} \right |$$
where $\vec r(t)$ is a position vector.
A: Since speed is a scalar quantity, its value should be independent of our choice of coordinate system. In particular, it should not depend on our choice of origin. In other words, if we replace the position vector $\vec x(t)$ with $\vec x(t)+\vec x_0$, for some arbitrary $\vec x_0$, the speed should not change. However, if we look at the derivative of the magnitude of the position, we find that this does depend on our choice of origin
$$\frac{\mathrm{d}\lVert\vec x(t)\rVert}{\mathrm{d}t} \ne \frac{\mathrm{d}\lVert\vec x(t)+\vec x_0\rVert}{\mathrm{d}t}.$$
This means that this will not work as a definition for speed. Now, this may lead you to think that this would work if we simply replaced position with displacement, effectively fixing the origin. However, with displacement, there still is implicitly a choice in the origin that comes from the choice in the initial time, since current speed shouldn't depend on what time you started keeping track of things.
Instead, speed is defined as the magnitude of the derivative of position, which is independent of origin, because $\frac{\mathrm{d}\vec x_0}{\mathrm{d}t}=0$.
$$\left\lVert\frac{\mathrm{d}\vec x(t)}{\mathrm{d}t}\right\rVert = \left\lVert\frac{\mathrm{d}(\vec x(t)+\vec x_0)}{\mathrm{d}t}\right\rVert$$
