Question

If somebody can point me towards a related webpage I could read on this, that would be much appreciated; Multivariable Calc, partial derivatives problem, 4 variables.

Answer 1

https://www.khanacademy.org/math/multivariable-calculus
perhaps?

Answer 2

Yeah, maybe a specific problem with four variables is more of what I'm looking for. I'll check out the videos, though; thanks.

Answer 3

No worries, good luck! =)

Answer 4

the subscript w means w is held constant, so dw = 0

Answer 5

here is a lecture that goes into some of the details
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-14-non-independent-variables/

Answer 6

And I'm also just not sure what's up with/how I should deal with the two equations. When they're not expressing some intermediate variable as a function of some independent variables or something like that, I've also never had to deal with a system of eqns dealing with partial derivatives. Thanks for the lecture, I'll take a look!

Answer 7

http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/

Answer 8

I still do not understand why I have two equations, but I'm starting to get the notation after figuring out the other kinks of misunderstanding I've had with the lecture.

Answer 9

Any remarks on the two equations bit? Not quite sure what I'm supposed to do, dealing with that.

Answer 10

@ganeshie8 Any clues here? Just don't know how I'm supposed to approach having the two eqns.

Answer 11

hint : they're two different problems

Answer 12

:P

Answer 13

Alright, I'll see what I can do based off of that, lol. I dunno? I, uh...yeah, I'll just...do something. Thank you, though.

Answer 14

i may be wrong though... @dan815

Answer 15

yes ofcuorse theyre 2 different problem :)

Answer 16

o________o

Answer 17

Somebody give me an idea of where to start? Or what problem O should be looking at first?

Answer 18

The only examples in the associated MIT page are keeping multiple variables constant, and dealing with one eqn.

Answer 19

sec i will help you as soon as this site stops lagging and delete my embarassing comments!!

Answer 20

okay do Dy/dz first and then derive with respect to w

Answer 21

they seem to be mixing two types of notations
this notation F_x and df/dx the leibniz notation

Answer 22

either that or they mean dy/dx eval at w, where w is some arb constatnt

Answer 23

No, that's the thing, this is a distinct kind of thing that I can only find in an MIT Courseware lecture; they're not actually doing that! They are...you just have to watch the lecture, that's exactly what I thought, but-yeah

Answer 24

It's specifying what is kept constant to avoid this problem: (one moment)

Answer 25

If you have some function given, and say it's a function of two variables, x and y-let's say you do a change of variables, u and v; while x = u, partial f w.r.t. x is *not* the same as partial f w.r.t. u

Answer 26

Does it at 15:00 in the vid

Answer 27

if u say f=f(x,y) and G=f(x(u),y(u,v)), y can be defined in both u and v in a different way so df/dx doest not have to equal df/du jut because u =x

Answer 28

however im not so sure about this argument if its valid

Answer 29

because im not quite sure unless you prove that you can indeed form another unique variable v as a function of x and y that will lead to a different function that what you started before

Answer 30

if u say f=f(x,y) and G=G(u(x),v(x,y))***(this makes more sense if i write like this), y can be defined in both u and v in a different way so df/dx doest not have to equal df/du jut because u =x

Answer 31

i.e
f=2x+siny=f(x,y)
where u=x
g=2u+sinv, f(u,v) <--- can we change the definition of a v such that df/dx =/= df/du

Answer 32

we can write v as a different set of functions of x

Answer 33

for example the infite taylor series of sin y = v

Answer 34

okay no that argument fails too i think xD

Answer 35

i dunno this has gotten complicated its supposed to be a simple queston, where u just differentiate with respect to some variable holding the other constat

Answer 36

OHh okay i know!! from the argument u gave

Answer 37

If you have some function given, and say it's a function of two variables, x and y-let's say you do a change of variables, u and v; while x = u, partial f w.r.t. x is *not* the same as partial f w.r.t. u

This is true because you could write y as a function of u and v, and then in that case df/du will not longer be the same as df/dx

Answer 38

ok done !! :) move on now

Answer 39

...I still have no clue how to do this, given these two functions.

Answer 40

I understand what the notation means. I just don't get to do with this particular problem.

Answer 41

:/

Answer 42

Anybody got an idea.

Answer 43

@SithsAndGiggles

Answer 44

Did I just horrifically misunderstand dan, or did he just repeat/justify what I said earlier? That doesn't help me at all in understanding the problem, which is why I'm getting progressively more confused by people either medalling him or thinking this is resolved; what did he point out that I didn't already understand? Could somebody please help?

Answer 45

It'd be cool if somebody could respond. Like, anybody, rather than just ignoring me. Kinda messed up. Even if "we're not going to elaborate any further" is the response, somebody saying something would be much appreciated.

Answer 46

I would take \(\left(\dfrac{\partial y}{\partial z}\right)_w\) to mean \(\dfrac{\partial^2 y}{\partial w~\partial z}\), but there's no way to know for sure. Have you come across anything else with this notation?

Answer 47

Yeah, the only instance where I see it used in a different way from what everybody is saying, which is a mixture of partial derivative notation, is in the MIT lecture, like I pointed out earlier. It means taking that partial derivative w.r.t. z, as usual, but specifying that w is to be treated as a constant, and thus dw is going to be zero. I just still am not sure what to do with this. I don't understand why the notation really exists, am still quite a bit confused by the lecture, IDK. Thanks everybody for trying to help so far.

Answer 48

if w is constant, what about x ?

Answer 49

It's also briefly, briefly, briefly mentioned in this sense in the Wikipedia page for Partial Derivatives:

Answer 50

That's it, I don't really know what it means, then, because the MIT lecture only dealt with problems of three variables. I'm not sure what about x. I still don't get this.

Answer 51

they should have specified both `w,x` in the subscripts

Answer 52

Hmm, well I'm unfamiliar with the notation and not entirely sure myself of what that's supposed to mean - it's been a while since I've taken multivariate differential calculus. If I had the time, I'd take a look at the lecture, but it's getting late here... Good luck

Answer 53

Nonetheless, thank you for responding.

Answer 54

Yeah, that's what I think, too, ganeshie, I'm just more confused by it. I guess I'm going to have to give up on this one and move elsewhere for the moment, but I feel that it may have actually been correctly written, and we're just missing something. For now, if it's unfamiliar, I'm just going to move on. Sorry, and thank you, guys.

Answer 55

\(x^2 - 2y + w^2 = z\)

\(\implies 0-2 \dfrac{\partial y}{\partial z} + 0 = 1\)

\(\implies \dfrac{\partial y}{\partial z} = -\dfrac{1}{2}\)

Answer 56

I think u get that if u dont bother about the subscripts ^

Answer 57

Yeah, but I'm pretty sure the subscripts matter. I understand what you just did. Thank you, again. Just not sure how the heck to deal with those subscripts. I'm gonna call it a day on this problem and move onto others.

This could have also just been a poorly designed problem; this particular professor seems a little bit merciless both with abstract notation and teaching things entirely out of the course curriculum, despite it being a 6 week course, so I'm thinking it's more of an instructor thing. Oh well.

Answer 58

holdup for 2 more minutes

Answer 59

actually it is a single problem, and x is not held constant from the subscript notation - only w is held constant

Answer 60

eliminate x and take the partial :

\(\large (e^z - 3y - 4w)^2 - 2y + w^2 = z\)

Answer 61

thats it i guess we had exhausted all the permutations for interpreting the notation :P

Answer 62

Alright, I'm taking a look at this. Thank you so much.

Answer 63

When you say "eliminate x", you just mean eliminating it algebraically by combining the two functions in some way, subtraction or addition or otherwise, right?

Answer 64

Oh, nevermind.

Answer 65

yeah x depends on w, y and z so we cannot treat x as constant...

Answer 66

as x was not specified in the subscript..

Answer 67

Alright, this is what I have so far. Hesitating for a moment because this whole thing is making me very nervous. Taking a shot at it again.

Answer 68

Wait, before I take the partial, should I try to isolate y to one side?

Answer 69

not needed, we can do it implicitly
just treat y and z as variables.

Answer 70

\(\large (e^z - 3y - 4w)^2 - 2y + w^2 = z\)

\(\large \implies 2(e^z - 3y - 4w)(e^z - 3 \dfrac{\partial y}{\partial z} - 0) - 2 \dfrac{\partial y}{\partial z} + 0 =1\)

Answer 71

isolate the partial

Answer 72

Alright, cool, one sec! Thank you so much for helping me figure this out! This has literally taken me searching the web all day, and I still couldn't figure it out w/the resources in front of me, just one sec.

Answer 73

Okay, either we will be right or we will be wrong in interpreting the notation -

but either ways the recent interpretation looks correct intuitively and also it matches with the basics we knw about partials....So... lets go wid it :)

Answer 74

(Sorry, taking time to write everything out, lmao)

Answer 75

haha instead of \partial, use \(\large y_z\) maybe...

Answer 76

Apologies for taking so long, I just froze in the middle because this problem has freaked me out so much, lol.

Answer 77

\(\large (e^z - 3y - 4w)^2 - 2y + w^2 = z\)

\(\large \implies 2(e^z - 3y - 4w)(e^z - 3 \dfrac{\partial y}{\partial z} - 0) - 2 \dfrac{\partial y}{\partial z} + 0 =1\)

\(\large \implies 2(e^z - 3y - 4w)(e^z) - 6 (e^z - 3y - 4w) \dfrac{\partial y}{\partial z} - 2 \dfrac{\partial y}{\partial z} =1\)

Answer 78

\(\large \implies 2(e^z - 3y - 4w)(e^z) -1 = \dfrac{\partial y}{\partial z}\left(6 (e^z - 3y - 4w) + 2\right) \)

Answer 79

\(\large \implies \dfrac{2(e^z - 3y - 4w)(e^z) -1}{\left(6 (e^z - 3y - 4w) + 2\right)} = \dfrac{\partial y}{\partial z} \)

Answer 80

ur attached work freaked me out lol, so... had to do it for myself again ;)

Answer 81

Here is one way to do this (after the first example)

Answer 82

I'm going to try this problem by using differentials/taking the total derivative to see if I get the same answer.