/sci/ - Science & Math » Thread #12795360

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Anonymous Sat 06 Mar 01:49:16 2021 No.12795360 View Reply Original Report

Quoted By: >>12795371 >>12795380 >>12795499 >>12795545 >>12796558

Can someone explain what is happening on the left side of the equation? Like how multiplying by dx/dt and then integrating will give (dx/dt)^2?

Anonymous

Anonymous Sat 06 Mar 2021 01:52:52 No.12795371 Report

Quoted By:

>>12795360
It's just integration by substitution.

Anonymous

Anonymous Sat 06 Mar 2021 01:55:11 No.12795380 Report

Quoted By: >>12795482

>>12795360
The words between the equations explain it quite clearly. Just work it through yourself.

Anonymous

Anonymous Sat 06 Mar 2021 01:56:29 No.12795384 Report

Quoted By:

it's physicology
but if you're talking the last bit, it hasn't been integrated yet by the looks of it.

Anonymous

Anonymous Sat 06 Mar 2021 02:02:14 No.12795410 Report

Quoted By: >>12795496

Let $u = \frac{dx}{dt}$ , you dumb hoe.

Anonymous

Anonymous Sat 06 Mar 2021 02:27:00 No.12795482 Report

Quoted By: >>12795503 >>12795512

>>12795380
$\displaystyle m \dfrac{d^3x}{dt^3} = -kx \dfrac{dx}{dt} - c \dfrac{d^2x}{dt^2} $
>obvious
if you say so.
thanks for nothing

Anonymous

Anonymous Sat 06 Mar 2021 02:29:35 No.12795496 Report

Quoted By: >>12795513 >>12795938

>>12795410
this. then
m du/dt=-kx-cu
now derivate the thing by d/du and you have it

Anonymous

Anonymous Sat 06 Mar 2021 02:30:17 No.12795499 Report

Quoted By:

>>12795360
literally just take the derivative of the second expression to recover the first

Anonymous

Anonymous Sat 06 Mar 2021 02:31:28 No.12795503 Report

Quoted By: >>12795513

>>12795482
What the fuck
Multiply by dx/dt and just integrate, idk what the fuck you did just there

Anonymous

Anonymous Sat 06 Mar 2021 02:35:18 No.12795512 Report

Quoted By: >>12795634

>>12795482
$d^2x/dt^2 \ne (dx/dt)^2$

Anonymous

Anonymous Sat 06 Mar 2021 02:36:14 No.12795513 Report

Quoted By: >>12795527 >>12795532

>>12795503
multiply by dx/dt is not valid or makes any sense. in maths. i am >>12795496 just listen to me faggot OP

Anonymous

Anonymous Sat 06 Mar 2021 02:41:41 No.12795527 Report

Quoted By: >>12795535

>>12795513
You don't know what you're talking about
>inb4 lol im drukn

Anonymous

Anonymous Sat 06 Mar 2021 02:44:36 No.12795532 Report

Quoted By: >>12795539

>>12795513
You're thinking of multiplying by dx or dt individually (and even then it is often justified as a mnemonic shorthand as any intro course will prove)
In this case you multiply by dx/dt to see that the left hand side is the derivative of (dx/dt)^2 as the book clearly explains.

Anonymous

Anonymous Sat 06 Mar 2021 02:45:49 No.12795535 Report

Quoted By:

>>12795527
well honeslty i didn't even try to solve it lmao

Anonymous

Anonymous Sat 06 Mar 2021 02:46:51 No.12795539 Report

Quoted By:

>>12795532
yeah i fucked up. the book is pretty clear and op is gay.

Anonymous

Anonymous Sat 06 Mar 2021 02:48:01 No.12795545 Report

Quoted By: >>12795554

>>12795360
wait what this doesn't make any sense
im assuming you integrate by time, so how do you get the first term in the LHS?

Anonymous

Anonymous Sat 06 Mar 2021 02:49:38 No.12795554 Report

Quoted By:

>>12795545
oh wait nvm I got it, im just retarded

Anonymous

Anonymous Sat 06 Mar 2021 03:10:51 No.12795634 Report

Quoted By: >>12796052 >>12796059 >>12796072

>>12795512
you sure about that?

Anonymous

Anonymous Sat 06 Mar 2021 04:38:03 No.12795936 Report

Quoted By:

For fuck's sake. This is undergraduate level mechanics, and 95% of /sci/ doesn't know what the fuck is going on. Let
$v = \frac{dx}{dt}$
Then
$m \frac{d^2x}{dt^2} \frac{dx}{dt} = m v\frac{dv}{dt}$
The integral is trivial.
Unironically, if you couldn't do this on your own, you're ngmi and should literally switch majors.

Anonymous

Anonymous Sat 06 Mar 2021 04:39:22 No.12795938 Report

Quoted By: >>12798172

>>12795496
>derivate
kys

Anonymous

Anonymous Sat 06 Mar 2021 05:05:15 No.12796039 Report

Quoted By:

Bro I don't even study physics and haven't studied mechanics in years and years and I got it.

Anonymous

Anonymous Sat 06 Mar 2021 05:10:43 No.12796052 Report

Quoted By:

>>12795634
not him but gb2 calculus 1

Anonymous

Anonymous Sat 06 Mar 2021 05:12:16 No.12796059 Report

Quoted By: >>12796072 >>12796085

>>12795634
Bro. x(t) = x.
x' = 1. (x')^2 = 1
x'' = 0
Does 0 = 1, motherfucker?

Anonymous

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Anonymous Sat 06 Mar 2021 05:16:01 No.12796072 Report

Quoted By: >>12796085

>>12795634
Man you just got btfo'd by this mans counterexample >>12796059

Anonymous

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Anonymous Sat 06 Mar 2021 05:20:29 No.12796085 Report

Quoted By: >>12796095

>>12796059
>>12796072
>x(t) = x

Anonymous

Anonymous Sat 06 Mar 2021 05:22:29 No.12796095 Report

Quoted By:

>>12796085
are you retarded my friend? nobody below 50 iq should post here
GTFO!!!!

Anonymous

Anonymous Sat 06 Mar 2021 08:54:51 No.12796558 Report

Quoted By: >>12797392 >>12797968 >>12799158

>>12795360
$ m \frac{d^2 x}{dt^2}\frac{dx}{dt} = -kx\frac{dx}{dt}-c(\frac{dx}{dt})^2\\ mu\frac{du}{dt}=-kx\frac{dx}{dt}-c(\frac{dx}{dt})^2\\ \text{remember } \frac{d(u^2)}{dt} = u\dot{u} + u\dot{u} = 2u\dot{u}\\ \int mu\frac{du}{dt}dt=-\int kx\frac{dx}{dt} dt-c\int (\frac{dx}{dt})^2 dt\\ \frac{m}{2} \int \frac{d(u^2)}{dt} dt = -\frac{k}{2} \int \frac{d(x^2)}{dt}dt-c\int (\frac{dx}{dt})^2dt\\ \frac{m}{2} u^2 +\frac{k}{2}x^2 = -c\int (\frac{dx}{dt})^2dt $
E_0 is missing because I couldn't bother with the boundaries. Integrate from 0 to t instead.

Anonymous

Anonymous Sat 06 Mar 2021 15:31:49 No.12797392 Report

Quoted By: >>12798183

>>12796558
Bro wtf. Just cancel the dt and int u du= u^2/2. Dumbass.

Anonymous

Anonymous Sat 06 Mar 2021 18:26:27 No.12797968 Report

Quoted By:

>>12796558
trivial

Anonymous

Anonymous Sat 06 Mar 2021 19:12:48 No.12798142 Report

Quoted By:

d/dt is a fraction

Anonymous

Anonymous Sat 06 Mar 2021 19:20:49 No.12798172 Report

Quoted By:

>>12795938
That's a perfectly fine expression, pseud.

Anonymous

Anonymous Sat 06 Mar 2021 19:24:10 No.12798183 Report

Quoted By:

>>12797392
>Just cancel the dt
lmao

Anonymous

Anonymous Sat 06 Mar 2021 22:26:51 No.12799158 Report

Quoted By: >>12799185

>>12796558
this is correct. product rule was the onl importnt part tho

Anonymous

Anonymous Sat 06 Mar 2021 22:31:05 No.12799185 Report

Quoted By:

>>12799158
Better to overexplain than to underexplain and get questions

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