Question

I have absolutely no idea how to start on this (Multivariable Calc - Substitution of Variables).

Answer 1

(Trying to attach a file, one moment.)

Answer 2

(?)

Answer 3

... too bad wolfram alpha no longer shows steps for free

Answer 4

:/

Answer 5

Ahhh this is making my head hurt!
I can't quite figure this out.
I'll at least share my thoughts on it though.

Answer 6

So if we make the substitution, we want to think of our function of x and y as a function of u(x,y) and v(x,y).\[\Large\rm f\left[x,y\right]=f\left[u(x,y),~v(x,y)\right]\]Taking the derivative of our function, applying the chain rule we learned in multi-variable calculus,\[\Large\rm \color{orangered}{f_x=f_u u_x+f_v v_x}\]\[\Large\rm \color{royalblue}{f_y=f_u u_y+f_v v_y}\]We can find the derivative of u and v,

Answer 7

It does, the main thing I've been looking for this whole time is a change of variables problem where both the substituted variables are a function of both x and y, and they are some kind of a product of those variables, for whatever reason, that seems really horrible/difficult.

Answer 8

\[\Large\rm \color{orangered}{f_x=y f_u -\frac{y^2}{x^2}f_v}\]\[\Large\rm \color{royalblue}{f_y=x f_u+\frac{2y}{x}f_v }\]

Answer 9

\[\Large\rm x \color{orangered}{f_x}-y \color{royalblue}{f_y}=y^3\]Then we can plug those in...
and uhhh.. hmm try to do something..

Answer 10

I'm a little bit confused, though, I thought the whole premise of change of variables was to eliminate x and y entirely? The link you posted...ugh

Answer 11

Oh oh I forgot to sub into the right side!
That's the problem I was having.
I see it now!

Answer 12

Lemme take a quick shot at this.

Answer 13

\[\Large\rm v=\frac{y^2}{x}\]Multiply each side by y and x,\[\Large\rm (xy)v=y^3\]Sub u in,\[\Large\rm u v=y^3\]

Answer 14

You can get rid of x and y completely that way!
Ah nice

Answer 15

Man, that was kind of a ridiculous (but totally valid) way to substitute, lol, thanks! Lemme take a shot at this, moving from here.

Answer 16

Cool :)
You should end up with an equation involving \(\Large\rm f_v\) and only u and v.
Having only the one partial makes it much easier to deal with.

Answer 17

Also, what LaTeX commands/how did you get it both plaintext and large?

Answer 18

plaintext? :o
Like how did I type it `in-line`?

Answer 19

I mean, like, upright text, sorry, lol, not too savvy on jargon

Answer 20

(Also, that thing, whatever you just did, hah)

Answer 21

lol :)
This little trick is just the 'accents key', not the apostrophe, but the tilde key, surround your text to get the grey box.

'fun stuff'
`fun stuff`

Answer 22

(So how did you make your text big and upright in the eqn editor/just plain typing it out?)

Answer 23

For text size in latex, you can use a few different options:\[\large\text{\large} \qquad \Large\text{\Large}\qquad \LARGE\text{\LARGE}\qquad \huge\text{\huge}\qquad \Huge\text{\Huge}\]

Answer 24

Yes I type all of my stuff out :o

Answer 25

And then there are some shortcuts for different fonts,
\sf is for sans serif, looks kind of nice
I use \rm for times new roman, I really like how that looks ^^

Answer 26

Alright, cool. Might be a second on working out the problems, I write stuff out in Illustrator and just screencap it but am having technical issues atm.

Answer 27

Ah >.<

Answer 28

Lol...a nearby electromagnet was interfering with the pen and causing the cursor to freak out. Does all this look sensible so far?

Answer 29

I also don't understand something else, lemme write it out.

Answer 30

\[\Large\rm x \color{orangered}{f_x}-y \color{royalblue}{f_y}=uv\]Ok looks good so far.
Gotta get these orange and blue parts plugged in yes?

Answer 31

One sec, lol, sorry.

Answer 32

Yeah...I'm not too familiar with this yet (this is a six-week Calc III course and I also have a full-time job)-I don't know how to do that, I obviously know how to take partial derivatives and how to make substitutions and get the notation, but I don't get how we should, uh, get fx and fy in this case, without using x or y.

Answer 33

PDEs in Calc III? oof

Answer 34

My thoughts exactly, I'm not sure why we should be learning about PDE's right, now, the prof is going well out of the curriculum in the book and teaching his own stuff, but w/e, gotta do what you gotta do, lol.

Answer 35

Look back at the blue and orange equations I wrote out.
Do you understand how I got those?
Remember your multi-variable chain rule for derivative?

Answer 36

\[\Large\rm \color{orangered}{f_x=f_u u_x+f_v v_x}\]We can plug in stuff for u_x and v_x using our substitutions.

\[\Large\rm u=xy, \qquad \qquad u_x=y\]Yes?

Answer 37

(I can write out that f-sub-x is, e.g: f-sub-x equals that partial derivative of f with respect to u times the partial derivative of u with respect to x, plus the partial derivative of f with respect to v times the partial derivative of v with respect to x. I just, I get whjy that is, but IDK, I don't get where to go from here.)

Answer 38

Oh, okay, one sec.

Answer 39

So now just do that for f-sub-y, and plug in?

Answer 40

Why y_o ? 0_o Instead of just y..?

Answer 41

yes

Answer 42

IDK, it's a habit I carried over from treating them like constants, lol. I don't need to do it but it makes me feel better, hahaha

Answer 43

It's also how my book does it when they first talk about partial derivatives

Answer 44

Ohhh I see.. that was your partial with respect to x. clever :3

Answer 45

\[\Large\rm x \color{orangered}{\left(y f_u -\frac{y^2}{x^2}f_v\right)}-y \color{royalblue}{\left(x f_u+\frac{2y}{x}f_v \right)}=uv\]Do some stuff,
plug the stuff in,
some stuff should cancel out..

Answer 46

Whoops, clearly not right, sorry, just trying to figure this out, my bad

Answer 47

( -3v f_v = uv, -3 f_v = u?)

Answer 48

Mmm yes there we go!
That looks better.

Answer 49

Thanks so much! :D I was trying to figure this out for so long.

Answer 50

(Is that it? I always feel nervous with this stuff......I'm....I'm done with the problem, right? lmao.)

Answer 51

They want us to solve it.
So let's uhhhh...


We have it written in terms of a single partial, so it's easier to work with now.
\[\Large\rm f_v=-\frac{1}{3}u\]Integrating with respect to v,\[\Large\rm f=-\frac{1}{3}uv+g(u)\]Normally you would get a +c at the end, right?
But we're integrating with respect to v, and our function is a function of u and v.

Answer 52

I think I'm going to go read for a while, because a fair amount of this stuff (namely multivariable integration or whatever you're about to do) we just haven't learned about yet; this is a homework due Thursday (three classes in between), just trying to get it done early as possible. I'm kind of not read enough at the moment to make this a good learning exchange, so I think I'm just gonna go figures this stuff out for some time. Thanks so much for your help, and sorry for anybody lingering and watching, lol.

Answer 53

Hehe yah it's a lot to take in :)

Answer 54

Interestingly, Wolfram Alpha is capable of interpreting PDEs, so if you'd like to check your answer:

http://www.wolframalpha.com/input/?i=x*D%5Bf%28x%2Cy%29%2Cx%5D-y*D%5Bf%28x%2Cy%29%2Cy%5D%3Dy%5E3

By the way, I've been looking for something to do for the summer, so I think I'll give learning how to solve PDEs a shot...

Answer 55

Alright, I'm lost...our book simply doesn't cover PDE's.....y'know, because we're not supposed to be learning how to solve them yet....so I can't really "read up" without being contextually lost. Does anybody mind if I start a new question for the solution of the PDE itself, or does anybody have some useful resources/links for learning about how to solve this kind of PDE?

Answer 56

I was hoping to brush up on the topic myself, and I came across some lessons on youtube:

General strategy: https://www.youtube.com/watch?v=VIqKKq_jZu0

Example: https://www.youtube.com/watch?v=HouEinUtASE

I found it somewhat easy to follow, considering it was my first time with the subject matter. Hope it helps.