Velocity of a curve given by parametric equations

Question

In standard Cartesian equations, $\frac{dy}{dx}$ is the velocity function because it's the derivative of position.

$$\frac{dx}{dt} = \sin^{-1}\left(\frac{t}{1 + t}\right) \space\space\space\space\space\space\space\space \frac{dy}{dt} = \ln(t^{2} + 1)$$

Given the derivatives of the parameters, why must one calculate the magnitude of a vector at a certain time $t$ to calculate speed? Can't you just divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to get $\frac{dy}{dx}$ which is the slope and velocity at a given time. Then take the absolute value for speed.

Answer 1

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ is valid as you said.

As for the context, it's usually not a matter of why but more of what you are given and what you must do with it. Often, it may be the case that $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are easier to calculate or find than directly going for $\frac{dy}{dx}$. This is especially so if you're just given the parametric equations.

Answer 2

The problem with reasoning is not with your formulas but your intuition. If we have a particle whose position is given by $x(t)$, its derivative $x'(t)$ is the velocity. If we have a curve in the plane given by $y=f(x)$, the slope of the curve is given by $dy/dx$. The problem is that this slope is not the 'velocity' of this curve. It gives the slope of the curve at a point, which gives the direction a particle traveling along this curve is traveling at that moment, not the speed of the particle at that point. Intuitively, you can't think about speed for a curve given by $y=f(x)$ because there is no time variable $t$. Using parametric equations however, we introduce this value of $t$, given a speed and a direction to the curve.

Simply put, $dy/dx$ is not the slope and the velocity of a curve at a given point, just the slope.