DIfferentials definition in single variable calculus using limits
$begingroup$
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
$$ frac{dy}{dx} = lim_{Delta xrightarrow0} left( frac{y(x+Delta x)-y(x)}{Delta x} right)$$
I Could define
$$frac{dy}{dx}dx = lim_{Delta xrightarrow0} left( frac{y(x+Delta x)-y(x)}{Delta x} Delta x right) $$
And then define what looks like the idea used in physics, the change in the function (as opposed to the rate of change)
$$dy= lim_{Delta xrightarrow0} left( (y(x+Delta x)-y(x)) right) $$
I know that the area under a curve (the reinmann sum) is calculated as
$$int_a^b y(x) dx= lim_{nrightarrow infty} sum_{i=1}^n yleft((ifrac{b-a}{n})+aright) left(frac{b-a}{n} right)$$
If you where to integrate $dy$ it would look like this
to accomodate for the definition of the integral I make $Delta x = lim_{nrightarrow infty}frac{b-a}{n}$
$$int_a^b dy dx= lim_{nrightarrow infty} sum_{i=1}^n left(y(x+frac{b-a}{n})-y(x)right) left(frac{b-a}{n} right)$$
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
$$sum left(y(x+Delta x)-y(x)right) Delta x $$
$$sum left(Delta yright) Delta x $$
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?
calculus ordinary-differential-equations notation physics
asked Dec 19 '18 at 4:59
Joaquin BrandanJoaquin Brandan
290110
$endgroup$
add a comment |
$begingroup$
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
$$ frac{dy}{dx} = lim_{Delta xrightarrow0} left( frac{y(x+Delta x)-y(x)}{Delta x} right)$$
I Could define
$$frac{dy}{dx}dx = lim_{Delta xrightarrow0} left( frac{y(x+Delta x)-y(x)}{Delta x} Delta x right) $$
And then define what looks like the idea used in physics, the change in the function (as opposed to the rate of change)
$$dy= lim_{Delta xrightarrow0} left( (y(x+Delta x)-y(x)) right) $$
I know that the area under a curve (the reinmann sum) is calculated as
$$int_a^b y(x) dx= lim_{nrightarrow infty} sum_{i=1}^n yleft((ifrac{b-a}{n})+aright) left(frac{b-a}{n} right)$$
If you where to integrate $dy$ it would look like this
to accomodate for the definition of the integral I make $Delta x = lim_{nrightarrow infty}frac{b-a}{n}$
$$int_a^b dy dx= lim_{nrightarrow infty} sum_{i=1}^n left(y(x+frac{b-a}{n})-y(x)right) left(frac{b-a}{n} right)$$
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
$$sum left(y(x+Delta x)-y(x)right) Delta x $$
$$sum left(Delta yright) Delta x $$
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?
calculus ordinary-differential-equations notation physics
asked Dec 19 '18 at 4:59
Joaquin BrandanJoaquin Brandan
290110
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
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
$$ frac{dy}{dx} = lim_{Delta xrightarrow0} left( frac{y(x+Delta x)-y(x)}{Delta x} right)$$
I Could define
$$frac{dy}{dx}dx = lim_{Delta xrightarrow0} left( frac{y(x+Delta x)-y(x)}{Delta x} Delta x right) $$
And then define what looks like the idea used in physics, the change in the function (as opposed to the rate of change)
$$dy= lim_{Delta xrightarrow0} left( (y(x+Delta x)-y(x)) right) $$
I know that the area under a curve (the reinmann sum) is calculated as
$$int_a^b y(x) dx= lim_{nrightarrow infty} sum_{i=1}^n yleft((ifrac{b-a}{n})+aright) left(frac{b-a}{n} right)$$
If you where to integrate $dy$ it would look like this
to accomodate for the definition of the integral I make $Delta x = lim_{nrightarrow infty}frac{b-a}{n}$
$$int_a^b dy dx= lim_{nrightarrow infty} sum_{i=1}^n left(y(x+frac{b-a}{n})-y(x)right) left(frac{b-a}{n} right)$$
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
$$sum left(y(x+Delta x)-y(x)right) Delta x $$
$$sum left(Delta yright) Delta x $$
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?
calculus ordinary-differential-equations notation physics
asked Dec 19 '18 at 4:59
Joaquin BrandanJoaquin Brandan
290110
$endgroup$
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
$$ frac{dy}{dx} = lim_{Delta xrightarrow0} left( frac{y(x+Delta x)-y(x)}{Delta x} right)$$
I Could define
$$frac{dy}{dx}dx = lim_{Delta xrightarrow0} left( frac{y(x+Delta x)-y(x)}{Delta x} Delta x right) $$
And then define what looks like the idea used in physics, the change in the function (as opposed to the rate of change)
$$dy= lim_{Delta xrightarrow0} left( (y(x+Delta x)-y(x)) right) $$
I know that the area under a curve (the reinmann sum) is calculated as
$$int_a^b y(x) dx= lim_{nrightarrow infty} sum_{i=1}^n yleft((ifrac{b-a}{n})+aright) left(frac{b-a}{n} right)$$
If you where to integrate $dy$ it would look like this
to accomodate for the definition of the integral I make $Delta x = lim_{nrightarrow infty}frac{b-a}{n}$
$$int_a^b dy dx= lim_{nrightarrow infty} sum_{i=1}^n left(y(x+frac{b-a}{n})-y(x)right) left(frac{b-a}{n} right)$$
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
$$sum left(y(x+Delta x)-y(x)right) Delta x $$
$$sum left(Delta yright) Delta x $$
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?
calculus ordinary-differential-equations notation physics
calculus ordinary-differential-equations notation physics
asked Dec 19 '18 at 4:59
Joaquin BrandanJoaquin Brandan
290110
asked Dec 19 '18 at 4:59
Joaquin BrandanJoaquin Brandan
290110
asked Dec 19 '18 at 4:59
Joaquin BrandanJoaquin Brandan
290110
asked Dec 19 '18 at 4:59
Joaquin BrandanJoaquin Brandan
290110
asked Dec 19 '18 at 4:59
Joaquin BrandanJoaquin Brandan
290110
290110
1
$begingroup$
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
add a comment |
1
$begingroup$
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
1
1
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
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1
$begingroup$
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