Which relation is correct for resultant instantaneous velocity in 2d?

Question

Please forgive me if the following question sounds silly and I can't exactly pin point where exactly the problem is but there is some problem with my understanding of vectors.

In Cartesian coordinates, $x$ and $y$ represent the position at any point of time. Now the distance of this point from origin can be written as $x^2 + y^2 = s^2$. If we differentiate both sides with respect to time, we get, $$x*Vx + y*Vy = s*Vs \tag{1}$$ where $V$ represents the instantaneous velocities at time $t.$

But, if we go by the geometric derivation of instantaneous velocity, we can also have $\delta x^2 +\delta y^2 = \delta s^2 $. Dividing by time, we get $$Vx^2 + Vy^2 = Vs^2 \tag{2}$$

I am familiar with relation (2) which is the resultant instantaneous velocity but what does the relation (1) mean?

Answer 1

The relations are both correct. the relation (1) comes from:

$$\frac{ds}{dt} = \frac{d}{dt}|\mathbf r| = \frac{d}{dt}\sqrt{x(t)^2 + y(t)^2} = \frac{1}{\sqrt{x(t)^2 + y(t)^2}}\left(x\frac{dx}{dt} + y\frac{dy}{dt}\right) = \frac{1}{|\mathbf r|}\left(x\frac{dx}{dt} + y\frac{dy}{dt}\right)$$ i.e. $s(t)= \sqrt{x(t)^2 + y(t)^2} = s(x(t),y(t))$ is a composed function in which the arguments are positions and they depend on time, it means that you have to use to so-called "chain-rule" as you can see above in order to derive $s$ with respect to time. I add...It is obvious that here appears a "combination" of position and velocity terms, it is just a consequence of the way in which you derive the expression: as you can see in the last equality written above, it appears $\frac{1}{|\mathbf r|}$ that is the inverse of a lenght, so, in order to have a correct dimension of the equation (a velocity on both sides) you must have something that is a position inside the brackets (and infact in the last equality written above appear $x$ and $y$ multiply respectively $\frac{dx}{dt}$ and $\frac{dy}{dt}$). It's just a relation that link positions and velocity along the 2 direction $x,y$, nothing deeper, and i repeat it is perfectly correct. Anyway, you can obtain the relation (2) by thinking that you have a position vector: $\vec{s}=(x,y)$ then you derive with respect to time the 2 components $$\vec{\frac{ds}{dt}}=(\frac{dx}{dt},\frac{dy}{dt}) = (v_{x},v_{y})\ $$ by doing the square modulus you can obtain your relation (2), that is also correct.