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?


1 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.

  • $\begingroup$ Yeah, got it. When you said both are correct, the confusion cleared. Actually I mistakenly assumed that if one is true then other can't be correct since they seemed (faulty intuition) to represent separate information...But both relations represent the same information. They actually mean same thing i.e. one can be derived from the other by using geometry (similar triangles). $\endgroup$ Commented Aug 5, 2018 at 4:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.