A parametric equation with $\frac{dx}{dt}$ = something, $\frac{dx}{dt}$ = something, has a resultant velocity vector by pythagorean theorem to be $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$. Calculus theorem dictates that integral of velocity over interval equals displacement over interval. Thus it must be true that d=$\int\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$.
However, my dispute (wrong for some reason) is that if we treat velocity separate, displacement in x is $\int\frac{dx}{dt}$ and displacement in y is $\int\frac{dy}{dt}$ . If you know displacement along x, and displacement along y, wouldn't your total displacement from origin be $\sqrt{x^2 +y^2 }$ as it is the diagonal length of the parallelogram formed by lengths x and y?
Now I've read before something like diagonal length can't be approximated treating x and y separate or something. It would be like breaking down a diagonal into infinitely tiny x and y steps, instead of just drawing a diagonal line. I don't really know, but can someone explain this better? thank you a lot
EDIT: This is even more confusing for me now as I remember how we are taught early on in maths that d=$\sqrt{x^2 +y^2 }$
Best Answer
I understand this actually now. The "paradox" I was referring to with infinite x steps and y steps is called the staircase paradox and can be used to falsely claim pi is 4. By doing what I did with the diagonal of the x coordinate and y coordinate you are finding the distance (length) of a path that follows straight from the origin to the diagonal.
The actual arc length of a path doesn't just take into account final x and y coordinate and their formed diagonal's distance from origin. Multiple paths (e.g. a squiggly line going back and forth vs a straight line) can both reach the same coordinates. The arc length formula takes into account the true limit of a curve and thus finds its distance using integrals of velocity functions, rather than just "approximating" the shape of the distance with staircases.
EDIT: There are more proper mathematical solutions to my problem, but my answer gives a simple reasoning for anyone else that may be confused in the future.