Difference between vector and parametric differentiation

[Mr Davis 97]

This might seem like a naive question to ask, but a full explanation of why these two concepts are different would be welcome. I am confused because parametric equations are $y = 8t^2$ and $x = 5t$, but at the same time, these two equations can describe the $x$ and $y$ components of a vector. Parametric differentiation is defined as $\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}$, while vector differentiation is defined as $\frac{\mathrm{d} \vec{r}}{\mathrm{d} t} = \left \langle \frac{\mathrm{d} x}{\mathrm{d} t}, \frac{\mathrm{d} y}{\mathrm{d} t} \right \rangle$

 

[Svein]

parametric equations are y=8t2y = 8t^2 and x=5tx = 5t, but at the same time, these two equations can describe the xx and yy components of a vector.

Not exactly. As t varies, the coordinates describe a curve in the xy -plane.

Parametric differentiation is defined as dydx=dydtdxdt\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}, while vector differentiation is defined as dr⃗ dt=⟨dxdt,dydt⟩\frac{\mathrm{d} \vec{r}}{\mathrm{d} t} = \left \langle \frac{\mathrm{d} x}{\mathrm{d} t}, \frac{\mathrm{d} y}{\mathrm{d} t} \right \rangle

If we assume that t stands for time, \frac{d\vec{r}}{dt} represents the velocity at a give time (velocity has a magnitude and a direction). This quantity is fundamental in classical differential geometry.

 

[LCKurtz]

This might seem like a naive question to ask, but a full explanation of why these two concepts are different would be welcome. I am confused because parametric equations are $y = 8t^2$ and $x = 5t$, but at the same time, these two equations can describe the $x$ and $y$ components of a vector. Parametric differentiation is defined as $\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}$, while vector differentiation is defined as $\frac{\mathrm{d} \vec{r}}{\mathrm{d} t} = \left \langle \frac{\mathrm{d} x}{\mathrm{d} t}, \frac{\mathrm{d} y}{\mathrm{d} t} \right \rangle$

If you eliminate the parameter $t$ and get $y$ in terms of $x$ then $\frac{dy}{dx}$ represents the slope of the curve. That is what you are calculating when you do$$
\frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$That is not the same thing as $\frac{dr}{dt}$ which is the velocity the point moves along the curve as a function of $t$. Different things, different formulas.