Are Position and Velocity (or Velocity and Acceleration) Vectors Always Parallel? While reading Chapter 1 of an astrodynamics textbook, I came across the statement:
$$\mathbf{v}\cdot \mathbf{{\dot{v}}}=v{\dot{v}}$$
In other words, the dot product of velocity and the time-rate-of-change of velocity is simply equal to the product of the magnitudes of the velocity and acceleration vectors.  I think I see how this works mathematically:
$$\frac{1}{2}\frac{d}{dt}\left ( \mathbf{v}\cdot \mathbf{v} \right )=\frac{1}{2}\left [\mathbf{v}\cdot \frac{d\mathbf{v}}{dt}  +\mathbf{v}\cdot \frac{d\mathbf{v}}{dt}\right]=\mathbf{v}\cdot \frac{d\mathbf{v}}{dt}=\mathbf{v}\cdot \mathbf{\dot{v}}$$
Also, since a vector is always parallel to itself,
$$\mathbf{v}\cdot \mathbf{v}=vv=v^{2}$$
So,
$$\frac{1}{2}\frac{d}{dt}\left ( \mathbf{v}\cdot \mathbf{v} \right )=\frac{1}{2}\frac{d}{dt}\left (v^{2}  \right )=\frac{1}{2}\left (2v  \right )\frac{dv}{dt}=v\frac{dv}{dt}=v\dot{v}$$
My hangup is this: are a vector and the first time derivative of that same vector always parallel (and in the same direction)?  It would seem so, since the (not shown) cosine of theta term equals 1.  Seems strange to me, because an object can be traveling in a certain direction while experiencing acceleration in another direction, such as the parabolic trajectory of a projectile near the Earth's surface (ignoring drag).  Velocity will be tangent to the arc, while the only acceleration is straight down (due to gravity) and never actually parallel to velocity, so long as there is a horizontal component to the trajectory.  Am I missing somehting very obvious here?  Thank you for your time and consideration.
Citation: See page 15 (sheet 30 of 470) section 1.4.1.2 of this document. 

Some very helpful people left great answers, thank you. More webmining also yielded this thread, which provided yet another good answer to the question. Seems like that one statement is responsible for a fair bit of consternation. 
 A: Speed (or any vector for that matter) can change both in modulus and in direction. Your proof fails if $\mathbf{v}$ changes only in direction. In that case the time variation of $\mathbf{v}\cdot\mathbf{v}$ vanishes, even if the time variation of $\mathbf{v}$ does not. An example is circular motion with constant speed, in which velocity and acceleration are perpendicular.
EDIT: to clarify what I did not express very well above: the proof of
\begin{equation}d/dt (\mathbf{v}\cdot\mathbf{v})=2 \dot{\mathbf{v}} \cdot\mathbf{v}\end{equation} is correct. However if v is constant \begin{equation}
d/dt \ v^2 =0,
\end{equation}
which is still equal to $2 \dot{\mathbf{v}} \cdot\mathbf{v} $ as $\mathbf{v}$, $\dot{\mathbf{v}}$ are perpendicular.
You can think of it this way:
\begin{equation}
d/dt \ \mathbf{v}=
d/dt \ (v \hat {\mathbf{v}})=\dot v\ \mathbf{v} +v \ d/dt\ \hat{\mathbf{v}}
\end{equation}
with $\hat{\mathbf{v}}$ of unit norm. The time derivative of $\hat{\mathbf{v}}$ is perpendicular to $\hat{\mathbf{v}}$ itself as follows differentiating $\hat{\mathbf{v}}\cdot\hat{\mathbf{v}} = 1$.
A: I think that I got your mistake! Here it is
$$\left\|\frac{d\bf{v}}{dt} \right\| \ne \frac{d\bf{\left\|v\right\|}}{dt}\tag{1}$$
In fact, this is simply saying that

In general, norm and differentiation are not interchangble.

So we have
$${\bf{v}} \cdot \frac{d\bf{v}}{dt} = \left\|\bf{v}\right\|  \left\|\frac{d\bf{v}}{dt}\right\|  \cos \theta= \left\|\bf{v}\right\| \frac{d\bf{\left\|v\right\|}}{dt}\tag{2}$$
and hence you can at most cancel $\left\|\bf{v}\right\|$ from the both sides of the equality to get
$$\left\|\frac{d\bf{v}}{dt}\right\| \cos \theta=  \frac{d\bf{\left\|v\right\|}}{dt}\tag{3}$$
Now, according to $(1)$, you cannot cancel $\left\|\frac{d\bf{v}}{dt} \right\|$ with $\frac{d\bf{\left\|v\right\|}}{dt}$ and the $\cos\theta$ won't be equal to one.
To get more enlightened, we know that the velocity vector can be written as 
$${\bf{v}}=\left\|{\bf{v}}\right\| {\bf{t}}\tag{4}$$
Now, if you take the derivative you will get
$$\frac{d{\bf{v}}}{dt}=\frac{d\left\|{\bf{v}}\right\|}{dt} {\bf{t}}+\left\|{\bf{v}}\right\| \frac{d{\bf{t}}}{dt}\tag{5}$$
Since ${\bf{t}}$ is a unit vector we can conclude
$${\bf{t}} \cdot {\bf{t}} =1 \qquad \to \qquad {\bf{t}} \cdot \frac{d{\bf{t}}}{dt} =0 \tag{7}$$
and consequently
$$\frac{d{\bf{t}}}{dt} 
= \left\| \frac{d{\bf{t}}}{dt} \right\| {\bf{n}}
= \left\| \frac{d}{dt} \frac{{\bf{v}}}{\left\|{\bf{v}}\right\|} \right\| {\bf{n}} \tag{8}$$
So the vector $\frac{d{\bf{v}}}{dt}$ can be written as 
$$\boxed{
\dfrac{d{\bf{v}}}{dt}=\frac{d\left\|{\bf{v}}\right\|}{dt} {\bf{t}}+\left\|{\bf{v}}\right\| \left\| \frac{d}{dt} \frac{{\bf{v}}}{\left\|{\bf{v}}\right\|} \right\| {\bf{n}}
}
\tag{9}$$
Finally, since ${\bf{t}} \cdot {\bf{n}}=0$ and both are unit vectors, we can compute $\left\|\frac{d\bf{v}}{dt} \right\|$ as follows
$${\left\|\frac{d\bf{v}}{dt} \right\|} = \sqrt{\left[\frac{d\bf{\left\|v\right\|}}{dt}\right]^2 + \left\|{\bf{v}}\right\|^2 \left\| \frac{d}{dt} \frac{{\bf{v}}}{\left\|{\bf{v}}\right\|} \right\|^2} \tag{10}$$
observing Eq.$(10)$, you can see that why $(1)$ is true. There is a special case where the orientation of a vector does not change which means that 
$$\mathbf{t}=\frac{{\bf{v}}}{\left\|{\bf{v}}\right\|} = \text{Constant Vector}\tag{11}$$
In that case $(10)$ simplifies to
$$\left\|\frac{d\bf{v}}{dt} \right\| = \left|\frac{d\bf{\left\|v\right\|}}{dt}\right| \tag{12}$$
A: As an easy counterexample to the claim that $\vec v\|\dot{\vec v}$, consider
$$\vec v = \begin{pmatrix} v_0\cos \omega t\\v_0\sin\omega t\\0\end{pmatrix}$$
We can verify that
$$\vec v\cdot \dot{\vec v} = \begin{pmatrix} v_0\cos \omega t\\v_0\sin\omega t\\0\end{pmatrix} \cdot \begin{pmatrix} -v_0\omega\sin \omega t\\v_0\omega\cos\omega t\\0\end{pmatrix} = 0$$
Moreover
$$v = v_0 = const. \implies \dot v = 0 \implies v\dot v=0$$
Yet $\vec v$ and $\dot{\vec v}$ are clearly not parallel; indeed, the fact that their scalar product vanishes shows that they are orthogonal.
