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I'm trying to merge the ideas of single variable calculus and the hand-waving done in newtonian mechanics when differentials are used as a measure of infinitesimal quantity change $dy$ instead of a derivative function$\frac{dy}{dx}$, this quantities are then simplified and moved around as numbers. After a while this quantities are integrated to solve some diferential equation, for example $dy$ is integrated and you get $y(x)$ as a result.

$\int dy dx= y(x)$

I'm trying to formalize this concept a bit by using the definition of integral and derivative to try to get to the same results

Given the definition of a derivative

$$ \frac{dy}{dx} = \lim_{\Delta x\rightarrow0} \left( \frac{y(x+\Delta x)-y(x)}{\Delta x} \right)$$

I Could define

$$\frac{dy}{dx}dx = \lim_{\Delta x\rightarrow0} \left( \frac{y(x+\Delta x)-y(x)}{\Delta x} \Delta x \right) $$

And then define what looks like the idea used in physics, the change in the function (as opposed to the rate of change)

$$dy= \lim_{\Delta x\rightarrow0} \left( (y(x+\Delta x)-y(x)) \right) $$

I know that the area under a curve (the reinmann sum) is calculated as $$\int_a^b y(x) dx= \lim_{n\rightarrow \infty} \sum_{i=1}^n y\left((i\frac{b-a}{n})+a\right) \left(\frac{b-a}{n} \right)$$

If you where to integrate $dy$ it would look like this

to accomodate for the definition of the integral I make $\Delta x = \lim_{n\rightarrow \infty}\frac{b-a}{n}$

$$\int_a^b dy dx= \lim_{n\rightarrow \infty} \sum_{i=1}^n \left(y(x+\frac{b-a}{n})-y(x)\right) \left(\frac{b-a}{n} \right)$$

now this sum looks like an area to me, but it's kind of a funny area. let me make an abuse of notation to try to clarify what I'm reading here

$$\sum \left(y(x+\Delta x)-y(x)\right) \Delta x $$ $$\sum \left(\Delta y\right) \Delta x $$ This area kind of represents the sum of an infinite bunch of rectangles that follow the function plot between $x=a$ and $x=b$ and have $\Delta y$ height and $\Delta x$ width.

That's a funny result, I expected the sum of the instantaneous changes (the antiderivative).

what went wrong?

did any of this make any sense?

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    $\begingroup$ If $y$ Is continuous, then your definition of $dy$ just gives zero. Look instead at linear approximations to the change in $y$. $\endgroup$ – amd Dec 19 '18 at 8:18
  • $\begingroup$ Linear aproximation does a good job to explain the concept of $dy$ in the context of physics, but when I want to integrate it I dont seem to get back $y(t)$ (and that makes sense). So I cant find how to make use of linear aproximation to explain the concept of $\frac{dy}{dx}dx = dy$ and then $\int dydx = y(t)$ $\endgroup$ – Joaquin Brandan Dec 19 '18 at 14:24

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