How is the magnitude of a direction vector equal to speed? In thomas calculus $12$ edition it gives the position vector along a line as
$\vec r(t) = \vec r_0  + t\vec v$
where $\vec r_0$ is the initial position vector and $\vec v$ is the direction vector.
Then it rewrites it as 
$\vec r(t) = \vec r_0 + t, |\vec v| \, \frac{\vec v}{|\vec v|}$
and it refers to $|\vec v|$ as the speed, which I don't understand. $|\vec v|$ is basically the length of the direction vector. Wouldn't speed be the $y$ component of the direction vector divided by the $x$ component, that is the slope of the position $\vec v$ time graph formed by $\vec r(t)$.
 A: The position vector is described parametrically with parameter $t$ as 
$$\vec r(t)=\vec r_0+\vec v\,t$$
Note that if we wish to measure the rate of change of position with respect to the parameter $t$, we have
$$\frac{d\vec r(t)}{dt}=\vec v$$
The magnitude of that rate of change is 
$$\left|\frac{d\vec r(t)}{dt}\right|=|\vec v|$$
Therefore, we can write
$$\begin{align}
\vec r(t)&=\vec r_0+\left(|\vec v|\,t\right)\,\left(\frac{\vec v|}{|\vec v|}\right)\\\\
&=\vec r_0 +t\,\left|\frac{d\vec r(t)}{dt}\right|\,\left(\frac{\vec v}{|\vec v|}\right)\tag 1
\end{align}$$
Note that in $(1)$, the term $\frac{\vec v}{|\vec v|}$ is a unit vector that points in the direction of $\vec v=\frac{d\vec r(t)}{dt}$ and $\left|\frac{d\vec r(t)}{dt}\right|$ is the magnitude of $\vec v$, which is the magnitude of the rate of change of the position vector with respect to the parameter $t$.
Now, interpreting $t$ as time and $\frac{d\vec r(t)}{dt}$ as velocity, we see that the position vector is given in terms of its initial position, $\vec r_0$, plus a vector that points in the direction of the velocity with magnitude equal to the product of the speed, $|\vec v|$, and time, $t$.
A: Your staring equation is a vector equation:
$$
\vec r(t)=\vec r_0+\vec v t
$$.
Now: $|\vec v|$ is the modulus of the vector $\vec v$ and this is, by definition what we call the ''speed'' and $\dfrac{\vec v}{|\vec v|}$ is the versor (unitary vector) that represents the direction of  $\vec v$.
So, to write 
$$
\vec r(t)=\vec r_0 +|\vec v|\dfrac{\vec v}{|\vec v|}
$$
is simply a way to separate the speed $|\vec v|$ by the direction $\dfrac{\vec v}{|\vec v|}$.
A: What is "the position $\vec v$ time graph formed by $\vec r(t),$"
and how should we interpret the meaning of its slope?
Take a piece of graph paper, label $x$ and $y$ axes on it, and
make a doodle by pushing the point of your pencil continuously around on the paper without lifting it.
This doodle (the mark you made on the paper while moving the pencil) shows the $x$ and $y$ coordinates of $\vec r(t)$ where $\vec r(t)$ is the position vector of the tip of the pencil at time $t.$
In other words, the doodle is the graph formed by $\vec r(t).$
Look at this doodle carefully. Is there any information there from which you can deduce exactly how fast the tip of the pencil was moving at any instant?
Couldn't someone have drawn the doodle half as fast, or slower in some parts and a little faster in others?
It is true that if you drew your doodle smoothly (with no sharp corners or kinks) then there is a line tangent to it at every point, and the angle of that line is the same as the direction of the velocity at that point.
If we divide the $y$ component of velocity, $\frac{dy}{dt},$
by the $x$ component of velocity, $\frac{dx}{dt},$
we get the slope of that tangent line, $\frac{dy}{dx}.$
So this ratio gives some indication of the direction of the velocity
(though it cannot distinguish between a velocity directly to the right and a velocity directly to the left, and it is undefined for motion straight up or down).
To get the magnitude of the velocity, that is, the speed,
we can once again consider the components of velocity in the two
axial directions, $\frac{dx}{dt}$ and $\frac{dy}{dt}.$
Using vector addition and the Pythagorean theorem, we have
$$
|\vec v| = \sqrt{\left(\tfrac{dx}{dt}\right)^2
                 + \left(\tfrac{dy}{dt}\right)^2}.
$$
