# Simple Displacement of Parametric Equations Dispute

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

• 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/… Mar 10, 2019 at 14:08
• Fixed. Apologies. edit: Also, thanks a lot. I was looking for how to make it that good! Mar 10, 2019 at 14:17
• That looks much better! Mar 10, 2019 at 14:18
• 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. Mar 10, 2019 at 21:59
• 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. Mar 11, 2019 at 3:27