Does slope of $y$ vs $x$ curve tell anything about magnitude of instantaneous velocity at that point?

Question

I know that slope at any point on the trajectory of a moving body gives us the direction of its instantaneous velocity.

Does it tell us anything about magnitude of the velocity at that point? I don’t think it does, but I’m not completely sure.

Slope of $y$ vs $x$ curve means $\frac{dy}{dx}$.

We can write it as

$=>$ $\frac{dy}{dx}$ $\frac{dt}{dt}$ $=$ $\frac{v_y}{v_x}$ $=$ $c$ (Some constant)

$=>$ $v_y$ $=$ $cv_x$

But this doesn’t tell us anything about magnitude of instantaneous velocity at that point unless we know the magnitude of any one of $v_y$ or $v_x$, at that point.

Is there a way we can find out the value of magnitude of instantaneous velocity at a point on the trajectory with the help of just the slope at the point?

Answer 1

The velocity cannot be determined by $\frac{\text dy}{\text dx}$, and a simple example will show why.

Imagine we both start at the origin and move along the positive x-axis until reaching the coordinate $(1,0)$. Let's I travel at $1\,\mathrm{m/s}$ and you travel at $2\, \mathrm{m/s}$. What are both of our trajectories $y(x)$? Well we have the same trajectory $$y(x)=0 \text{ for }0\leq x\leq 1$$ and therefore the same trajectory slope $$\frac{\text dy}{\text dx}=0 \text{ for }0\leq x\leq 1$$

However, we both had different magnitudes of our velocities.

Of course, this is a simple example. But this is true for any $y(x)$ (or even if the trajectory is not a well-defined function of $x$). This is because if we have one parametrization $(x(t),y(t))$ we can easily pick a different parametrization that follows the same curve but does it at a different velocity, for example $(x(2t),y(2t))$.

Answer 2

No. On the information of the trajectory alone it cannot be done, with the exception that you can always say that the components orthogonal to the tangent are zero.

If you have some more information then yes it can. E.g. if you know the acceleration as a constant vector (say in the -ve y direction) then from the top of its trajectory to a certain $\Delta y$ lower you can determine both $t$ and $v_y$ and thence the constant $v_x$ from the slope.