Tangent of a parameterized curve and the derivative of a function

Question

A parametrized curve is a function $\gamma : (\alpha,\beta)\to \mathbb R^n$.
The parametrized curve for a straight line with slope $m$ passing through origin is $\gamma(\alpha,\beta)={\{(t,mt)}\}$ where $t\in(\alpha,\beta)$
$\dot\gamma(t)=(1,m)$
$||\dot\gamma||=\sqrt{1+m^2}$
If we see the equation $y=mx$ the image of the set of points follow this equation also represent the straight line curve.
Also, $\frac{dy}{dx}=m$

Basically, $||\dot\gamma||=\sqrt{1+m^2}$ is the magnitude of speed with which the point travels.
I have two questions-

1. The question is that what does $\frac{dy}{dx}=m$ represents? As it also represents the rate at which y changes with x. Then how it is different from $||\dot\gamma||=\sqrt{1+m^2}$.
   Basically what is the difference between $m$ and $||\dot\gamma||$? Why there values are different?
2. I want to know what is the difference between $y=mx$ and $\gamma(t)$ if $t\in(-\infty,\infty)$?

I have studied about only real analysis and basic mathematics till. So please explain accordingly.

Answer 1

1. This is a consequence of Pythagoras. Draw a triangle of base $1$ and height $m$ equal to the slope. $\sqrt{1+m^2}$ is the hypothenuses.

2. There is no difference (but $\gamma$ is parameterized).