[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

View post   
File: 23 KB, 632x290, aaaaa.png [View same] [iqdb] [saucenao] [google]
4652559 No.4652559 [Reply] [Original]

I can't into partial derivatives since my physics course has them and I haven't even learnt them in maths

Can somebody help?

>> No.4652571

bamp

>> No.4652588
File: 35 KB, 440x350, Lil-Wayne-bm09.jpg [View same] [iqdb] [saucenao] [google]
4652588

>>4652559
This will help. This really isn'y a "partial derivitive" per say, what you want is the "differential".

http://en.wikipedia.org/wiki/Differential_of_a_function

READ NOW!

>> No.4652614

>>4652588
read all of which i could understand,

so I take (3) and (4)

and get

f = n/2L (4T/pi*ro*d^2)^(1/2)

f= n/ld * (T/pi*ro)^(1/2)

then I don't see how to apply any of the rules. is there a better way to express f from (3) and (4)? thanks

>> No.4652629

can you post the equations for L and d? and possibly any related equations for the problem. possibly the previous question.

>> No.4652630
File: 86 KB, 640x949, batman-bataman-demotivational-poster-1221723096.jpg [View same] [iqdb] [saucenao] [google]
4652630

>>4652614
Look under the "Differentials in several variables" section. That is what you need.

You combine 3 and 4, until you have f as a fucntion of L and d. Your varibles are then L and d. You then find the differntial of f with respect to L and d. "triangle f" is esentially that same as "df" for your purposes (same with the other varibles).

>> No.4652634

>>4652629
It doesn't matter what the equations of L and d are. They may as well just be numbers. This problem can be done easily with just 3 and 4, and proper knowledge of basic calculus concepts.

>> No.4652638

>>4652559
5 is wrong. It is off by a "-" sign.
Either mutiply the left or the right by a minus sign, and then it will be correct. This is because when you take the partial derivite of f with respect to L, you get -f/L, and similarly for d you get -f/d. Make sense?

>> No.4652648

I think it was the attitude from this guy >>4652634
that got me across the line. I considered all the other variables as just 'c' and then went:

df/dl = L^-2(c)
df/dd=d^-2(c)

df/dd+df/dl= c[(1/d^2+1/l^2)] -->wherec=f

df=f(dd/d^2+dl/l^2)

>> No.4652649

thanks sci

>> No.4652655

>>4652648
c Doesn't equal f though. It equals negative f! Your stradegy is correct though, but something is amiss!

>> No.4652657

>>4652655
Scratch that, the first c is -f/L and the second is -f/d.

There are no "squares" in 5, nor should there be! AMISSS!

>> No.4652667

>>4652648
Nope

>> No.4652670

i typed it full retard for some reason

i meant to say:

df/dl = n/(d*l^2)(root T/2pi)
df/dd=n/(d^2*l)(root T/2pi)

df(1/dl+1/dd)=n/dl(root T/2pi)[1/d+1/l)

df=f( 1/d + 1/l)/(1/dl+1/dd)
df=f(dl/l+dd/d)

is there amiss anywhere there?
also both the derivatives should becomes -ve since the sign is -1 initially but do they cancel out or did my proff dun goof?