1
$\begingroup$

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=$\sqrt{x^2 +y^2 }$

$\endgroup$
7
  • 1
    $\begingroup$ Hi and welcome to MSE! If you want to make your question more attractive so that it gets answered as soon as possible, you might want to use $\LaTeX$... There's a fantastic tutorial here: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Dr. Mathva
    Mar 10, 2019 at 14:08
  • $\begingroup$ Fixed. Apologies. edit: Also, thanks a lot. I was looking for how to make it that good! $\endgroup$
    – ebehr
    Mar 10, 2019 at 14:17
  • $\begingroup$ That looks much better! $\endgroup$
    – Dr. Mathva
    Mar 10, 2019 at 14:18
  • 1
    $\begingroup$ You wrote, "displacement in x is $\int(\frac{dx}{dt})^2$ and displacement in y is $\int(\frac{dy}{dt})^2$". However, I believe you meant "displacement in x is $\int\frac{dx}{dt}$ and displacement in y is $\int\frac{dy}{dt}$", i.e., the integration of the change of $x$ and $y$ over time. This results in distance values. Otherwise, with integrals of the squares, the resulting values don't make much sense, plus the units would be distance squared over time which, as far as I know, doesn't have any simple physical interpretation. $\endgroup$
    – John Omielan
    Mar 10, 2019 at 21:59
  • $\begingroup$ Yea, that was a mistake. Added the square for purpose of pythagorean theorem to find diagonal, the displacement value doesn't have it. Thank you. $\endgroup$
    – ebehr
    Mar 11, 2019 at 3:27

1 Answer 1

Reset to default
2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .