/sci/ - Science & Math » Thread #13941545

4KiB, 640x220, change-of-variable.png

View Same Google iqdb SauceNAO

change of variable

Anonymous Thu 09 Dec 05:43:13 2021 No.13941545 View Reply Original Report

Quoted By: >>13941906 >>13941919 >>13942051 >>13942130 >>13944408 >>13945708

I know a derivative is not a fraction (but rather the limit of a fraction)
so what is the justification for this? It's from my Physics.

Anonymous

Anonymous Thu 09 Dec 2021 05:45:04 No.13941554 Report

Quoted By:

it just werks

Anonymous

View Same Google iqdb SauceNAO integrand.png, 7KiB, 1080x270

Anonymous Thu 09 Dec 2021 07:10:46 No.13941906 Report

Quoted By:

>>13941545
This is all I could think of, for the integrand.
Is it legit, or shit? I don't like the idea of "symbol pushing",
but I can't think of anything else.
v = velocity
s = position
t = time

Anonymous

Anonymous Thu 09 Dec 2021 07:15:13 No.13941919 Report

Quoted By: >>13941935 >>13941980

>>13941545
So it has to do with the chain rule. dv/dt = (dv/ds)(ds/dt) = (dv/dt)(dt/ds)(ds/dt)

And then due to the commutative property you move (dt/ds) to the right. Then you do a change of variables of u = t(s) (making du=dt =dt/ds ds). And do that again for the last portion.

It's just abuse of the chain rule and u substitution and hides the fact that most of these variables are actually functions of eachother.

Anonymous

Anonymous Thu 09 Dec 2021 07:19:47 No.13941935 Report

Quoted By: >>13941980

>>13941919
The wild thing is, that these techniques also work for constants. A.k.a. you can set 2 = a and take derivatives with respect to it as long as you set it back = to a at the end. Physicists do these kind of things all the time.

OP Thu 09 Dec 2021 07:35:53 No.13941980 Report

Quoted By:

>>13941919
>>13941935
thx Anon

Anonymous

Anonymous Thu 09 Dec 2021 08:07:37 No.13942051 Report

Quoted By: >>13942097

>>13941545
??dvds = ??dsdv isn't controversial.
If s is differentiable with respect to t, then
??dvds = ??dv(ds/dt)dt
If v is differentiable with respect to t, then
??dsdv = ??ds(dv/dt)dt
Removing the ?dt (equivalently differentiating with respect to t) gives equality.

A more intuitive way to think about these things is taylor series.
v + dv = v + dt*v' + (dt)^2*v''/2 + (dt)^3*v'''/6 ...
s + ds = s + dt*s'+ (dt)^2*s''/2 + (dt)^3*s'''/6 ...
Integration only "compares" terms with the same order
dv*ds =
dv*(dt*s'+ (dt)^2*s''/2 + (dt)^3*s'''/6 ...) =
ds*(dt*v' + (dt)^2*v''/2 + (dt)^3*v'''/6 ...) =
(dt*v' + (dt)^2*v''/2 + (dt)^3*v'''/6 ...)*(dt*s'+ (dt)^2*s''/2 + (dt)^3*s'''/6 ...)
Equating second order terms gives dv*ds = dv*dt*s' = ds*dt*v' = dt*dt*s'*v'
Dividing by dt gives the result.

OP Thu 09 Dec 2021 08:29:30 No.13942097 Report

Quoted By:

>>13942051
>taylor series
I never considered that.
thx Anon

Anonymous

Anonymous Thu 09 Dec 2021 08:47:08 No.13942130 Report

Quoted By: >>13942330 >>13944382

>>13941545
you took two lines for a single step you fucking retard. did you know that multiplication commutes?

Anonymous

Anonymous Thu 09 Dec 2021 10:58:42 No.13942330 Report

Quoted By:

>>13942130
where to?

Anonymous

Anonymous Thu 09 Dec 2021 22:37:22 No.13944382 Report

Quoted By:

>>13942130
>two lines for a single step
ohh howw terrrible forr pooor babby too haave too reead aan addditional liine

Anonymous

View Same Google iqdb SauceNAO 140f0f71c76e953f7c00b0f7bd06cb64.jpg, 136KiB, 899x884

Anonymous Thu 09 Dec 2021 22:45:09 No.13944408 Report

Quoted By:

>>13941545
This is all just an application of the fact that the Leibnizian notation for derivative can be rearranged like a ratio in an integral formula: $ds = (ds/dt) dt$ . Just switch the terms after and then you can cancel the derivative of v instead and end up with the same integral.

Anonymous

Anonymous Fri 10 Dec 2021 06:03:45 No.13945708 Report

Quoted By:

>>13941545
It comes from the fundamental theorem of calculus
$\int_a^b dy = b - a = y(t_2) - y(t_1) = \int_{t_1}^{t_2} \frac{dy}{dt}{dt}$
given $y(t_1) = a$ and $y(t_2) = b$

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches