Parametrization and computing distance traveled I have no idea how to get started on this problem, and would greatly appreciate being pointed in the right direction: 
A car travels at a constant speed of $1 miles per hour$ from point $A(1,1)$ to point $B(2,4)$. Find the parametrization of its motion and use it to compute the distance traveled. 
 A: The goal of constant-velocity parameterizations is to have, well, constant velocity:  for the chosen velocity $v$, the parameterization must meet the constraint (in 2 dimensions) $$\sqrt{x'(t)^2+y'(t)^2}=v$$
Then, since we know the velocity is constant, the distance travelled since the start ($t=0$) is simply
$$d=vt$$
In this case, we know that $v=1$ and the parameterizations are linear, thus
$$x(t)=v_xt+1$$ $$y(t)=v_yt+1$$
and $v_y=3v_x$, so
$$y(t)=3v_xt+1$$
This gives for our velocities
$$x'(t)=v_x$$
$$y'(t)=3v_x$$
$$v = \sqrt{v_x^2+9v_x^2}=\sqrt{10v_x^2}$$
$$v_x=\frac{v}{\sqrt{10}}$$
So the parameterization is
$$x(t)=\frac{1}{\sqrt{10}}t+1$$
$$y(t)=\frac{3}{\sqrt{10}}t+1$$
This conveniently makes the car's starting point the function's starting point.  Now, to find the distance.
We need to know when the car reaches the ending point.  This happens when $x(t)=2$, so
$$\begin{align}2&=\frac{1}{\sqrt{10}}t+1
\\1&=\frac{1}{\sqrt{10}}t
\\t&=\sqrt{10}\end{align}$$
Now check that solution to see if we hit the point in the $y$ direction as well.  At this point, it's possible that the $x$ value gets hit multiple times, possibly at unfeasible $t$, and that only one/some of them reach the point in the $y$ direction as well; in this case this is slightly superfluous but for more advanced problems it is necessary.
$$y\left(\sqrt{10}\right)=\frac{3}{\sqrt{10}}\sqrt{10}+1=3+1=4$$
Yep.  So $t=\sqrt{10}$.  Since $v=1$, $d=1\cdot\sqrt{10}=\sqrt{10}$.
...I guess this goes a little beyond "pointed in the right direction".  Ah well...
