# Change of variables

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#### GreenGoblin

##### Member
$x=u^{2} - v^{2}$
$y=2uv$

Show the lines in the u,v plane where x and y are constant? (what?)
Is this a valid change of coordinate on the whole x,y plane? (what?)

By letting u = rcost, v=rsint, show that $x=r^{2}cos(2t), y=r^{2}sin(2t)$. Hence show that the map is 2 - 1 (what?) and show that the change of coordinates maps the upper half plane to the whole plane?

Verify that (d(x,y)/d(u,v))(d(u,v)/d(r,t)) = d(x,y)/(d(r,t) (isn't this just basic crossmultiplying?) and find d(r,t)/d(x,y)

So, how do I do this and how do you do these derivatives where its just got two variables? What is d(x,y)/d(u,v) mean?

#### HallsofIvy

##### Well-known member
MHB Math Helper
$x=u^{2} - v^{2}$
$y=2uv$

Show the lines in the u,v plane where x and y are constant? (what?)
Graph the curves (they are not straight lines) $u^2- v^2= C$ and $2uv= C$ for different values of C. (They will be hyperbolas.)

Is this a valid change of coordinate on the whole x,y plane? (what?)
Does every point in the x,y plane have a unique pair of (u, v) values?

By letting u = rcost, v=rsint, show that $x=r^{2}cos(2t), y=r^{2}sin(2t)$. Hence show that the map is 2 - 1 (what?)
Replace u and v in $x= u^2- v^2$ and $y= 2uv$ with those and do the algebra!

and show that the change of coordinates maps the upper half plane to the whole plane?
For the upper half plane, r can be any number while t must be between 0 and $\pi$. What x and y values does that give you?

Verify that (d(x,y)/d(u,v))(d(u,v)/d(r,t)) = d(x,y)/(d(r,t) (isn't this just basic crossmultiplying?) and find d(r,t)/d(x,y)
No, it isn't. Do you understand that d(x,y)/d(u,v), d(u,v)/d(r,t), and d(x,y)/d(r,t) are determinants?

So, how do I do this and how do you do these derivatives where its just got two variables? What is d(x,y)/d(u,v) mean?
$\frac{d(x,y)}{d(u,v)}= \left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{array}\right|$
seem to be saying you do not know what any of these things mean! I strongly suggest you talk to your teacher about this.

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#### GreenGoblin

##### Member
Graph the curves (they are not straight lines) $u^2- v^2= C$ and $2uv= C$ for different values of C. (They will be hyperbolas.)

Does every point in the x,y plane have a unique pair of (u, v) values?

Replace u and v in $x= u^2- v^2$ and $y= 2uv$ with those and do the algebra!

For the upper half plane, r can be any number while t must be between 0 and $\pi$. What x and y values does that give you?

No, it isn't. Do you understand that d(x,y)/d(u,v), d(u,v)/d(r,t), and d(x,y)/d(r,t) are determinants?

$\frac{d(x,y)}{d(u,v)}= \left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{array}\right|$
seem to be saying you do not know what any of these things mean! I strongly suggest you talk to your teacher about this.
No, sorry, not getting it at all. How should I start? What do you mean curves = C? Graph x and y or u and v? I dont get it.

Also, did I say I have a teacher? Did I at any point say that? Am I student? Do you know my circumstances?

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#### CaptainBlack

##### Well-known member
Also, did I say I have a teacher? Did I at any point say that? Am I student? Do you know my circumstances?
If this is related to your employment, then stop working on it at once and go and get someone who does understand this stuff to do this work.

If you are studying this on your own behalf you are not ready for this work, check on the prerequisites and fill in those that you are not familiar with.

CB

#### GreenGoblin

##### Member
No no, I am perfectly content with asking. There's always a point where you don't know something and a point where you learn. That's why I'm asking. When it's explained I can solve for myself. If I could do it all by myself then I wouldn't need to ask now would I? Now please, back to business..

I said I didn't know how to do it. If someone says how to do it.. then I can do it... that's what 'help' means last I checked this forum's name is math help boards I'm pretty sure this is math so I am asking for help.

#### Jameson

Staff member
No no, I am perfectly content with asking. There's always a point where you don't know something and a point where you learn. That's why I'm asking. When it's explained I can solve for myself. If I could do it all by myself then I wouldn't need to ask now would I? Now please, back to business..

I said I didn't know how to do it. If someone says how to do it.. then I can do it... that's what 'help' means last I checked this forum's name is math help boards I'm pretty sure this is math so I am asking for help.
Let's be reasonable about this. GreenGoblin, yes this is a free math help site so we are glad to have you here but you have to comply with our policies. Sometimes people might say a comment that at first will offend you but if you think about it more you'll see that it's just bluntly honest. Mostly this is due to the fact that people are very busy as well as sometimes it's best just to get to the point. With math people online, I think it's good to develop thick skin as they usually won't sugar coat their opinions. It's nothing personal against you, trust me.

I suggest we stick to the OP's original question in this thread. As always, anyone can PM me with their comments and problems if he or she wishes to talk in private; otherwise it's best to make a thread in the Feedback forum.

#### GreenGoblin

##### Member
But the thing is the guy made assumptions without knowing that's all. All I'm doing is asking for assistance, I need a little more help than usual since I don't know this topic well! But I am capable of understanding feedback if someone explains, so, I think I am fine with posting the topic to ask for help! If it's about something I can never understand, ok, I won't bother asking since any reply is meaningless.

No I don't have a teacher, I'm not learning this from someone, online is my only chance to find it out and this is one place I ask (I ask other places too but this is usually the best response place). So I don't have another resource I can 'go back and learn the basic' since this is the basic. I'm learning many new thing so I can take exam when I do know it..

If people don't want to help they can leave it for someone else, if no one will help will again that's all their choice. But a comment just to offend is not neccessary.

I can comprehend the material but some things is new terminology, see I don't know if it means the two lines are plot together or what. I get the first bit now ok, you choose different constants and get different curves, ok. "Is this a valid change of coordinate?" see I don't know how to solve this. I don't have any resource which tells me how to solve this. I don't know which book will tell me this. So I ask how to solve it, this is fine, no? If you tell me is not fine to ask this, I won't ask it, it's your website I know. I don't know all this notation for example the way is displayed in the question, its doesn't look like Jacobian in same way I know Jacobian but now I look is clear is Jacobian.

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#### Jameson

Staff member
I'm telling you that HallsofIvy was not trying to offend you. He spends a lot of time helping people with math on a lot of websites and that's just his style of writing - blunt, honest and to the point. You need to just get over it. I am American and grew up in the South where everyone is super polite all the time, so at first comments like those offended me as well but then I stopped worrying about being offended and just took the advice or didn't. If you want to discuss this further please PM me or I will have to close this thread. It's already way off topic.

#### GreenGoblin

##### Member
I don't mind bluntless, what I don't like is an assumption without any foundation!
But as you say I wont need to talk about this part any more.. anyway, yes, please I don't want to derail because I do like to continue topic of help, thanks,

Please what I need help with is, the 'is this a valid change.. ' part because I dont know how I can proof this... to me no it does not seem valid... but am I wrong?

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