How does the formula for length of a graph relate to that of parametric description? So I know how to solve for the length of a regular graph and for a parametric description.

Regular graph length formula:
$$L = \int_{x=a}^{x=b}\sqrt[]{1 + (f'(x))^2} \,\,\,dx$$

Parametric description length formula:
$$\int_{t = a}^{t=b} \sqrt[]{(x'(t))^2+(y'(t))^2} \,\,\,\,dt$$

The form of these two are so radically different that I can't see how one relates to the other. The parametric description uses the form for path velocity but I don't understand why..
If more elaboration is needed, please ask. I'll supply.
Highly appreciated,
-Bowser
 A: For a curve given by $f(x)$, it can be parametrized as $\bar{r}(t)=(t,f(t))$
Then $\bar{r'}(t)=(1,f'(t))$, 
By definition,
$$S=\int_a^b |\bar{r'}(t)|dt=\int_a^b\sqrt{1+f'^2}dt$$
Where $x'(t)=1, y'(t)=f'(t)$
Edit: the parametric form comes from the geometric idea that, when you zoom in close enough to a segment of the curve, it is basically a straight line. Then just use pythagorean theorem that $ds^2=dx^2+dy^2$
So summing up the segments $ds$ over $t=(a,b)$ will give you the arclength.
$$\int_{t=a}^{t=b} ds=\int_a^b \sqrt {dx^2 + dy^2}$$
By the fact that $dx=x'(t)dt$,
$$=\int_a^b \sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2}dt$$
A: They are radically similar ;) you forgot the integral in your first formula (edit: I see you fixed it now)
Notice that you can parametrize the graph of $f$ by $$(t, f(t))$$ Now use the second formula, you will get the first!
A: The basic idea is that the differential arc length is
$$
ds^2 = dx^2 + dy^2
$$
Your first formula then derives from
$$
y = f(x) \\
L = 
\int ds 
= \int \frac{\sqrt{dx^2 + dy^2}}{dx} \, dx
= \int \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
$$
and the second from 
$$
\left(\frac{ds}{dt}\right)^2 = \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2
$$
A: So why does the path length of a parametrised curve have the formula $\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2} dt$? To demonstrate this, it is best to go back to the very definition of integral as a limit of finite sums.
Let's say you have a parametrised path $\gamma:[a, b] \to \Bbb R^2$ given by
$$
\gamma(t) = (x(t), y(t))
$$
and you would like to know its length. You divide $[a, b]$ into $n$ equal intervals of length $\Delta t = \frac{b-a}{n}$. I will call this interval $[a_i, a_{i+1}]$, with $a_1 = a$ and $a_{n+1} = b$, and of course, $a_{i+1} = a_i + \Delta t$. This gives the following approximation for the path length of $\gamma$:
$$
\sum_{i = 1}^n \sqrt{(x(a_{i+1}) - x(a_i))^2 + (y(a_{i+1}) - y(a_i))^2}
$$
Now, for a little trick: multiplying by $\Delta t$, and dividing by $\Delta t$ at the same time. The division by $\Delta t$ I will take into the square root, and into each of the squares. We then get the following sum, which numerically has the exact same value:
$$
\sum_{i = 1}^n \Delta t\sqrt{\left(\frac{x(a_{i+1}) - x(a_i)}{\Delta t}\right)^2 + \left(\frac{y(a_{i+1}) - y(a_i)}{\Delta t}\right)^2}
$$
Remember when I said that $a_{i+1} = a_i + \Delta t$? Let's do that substitution as well. We again get a new sum that loooks slightly different, but we haven't changed the value at all:
$$
\sum_{i = 1}^n \Delta t\sqrt{\left(\frac{x(a_{i} + \Delta t) - x(a_i)}{\Delta t}\right)^2 + \left(\frac{y(a_{i} + \Delta t) - y(a_i)}{\Delta t}\right)^2}
$$
At this point we start to make $n$ really large. As $n$ tends towards infinity, two things happen: The outer sum becomes an integral, while the brackets inside the square root becomes derivatives. This results in the integral we had at the start.
