anonymous
  • anonymous
Help with jacobian http://postimg.org/image/yutmuzm7h/
Calculus1
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
IrishBoy123
  • IrishBoy123
i reckon that you should do a little algebra first so you have \(y = \mu \omega\) and \(x + y = \mu\) from that you can get \(x = \mu - \mu \omega\) i think you are good to go from there
anonymous
  • anonymous
I'm doing good? http://postimg.org/image/a8mwkhe1h/ Thanks for your answer :D
IrishBoy123
  • IrishBoy123
i think the J is 1 i haven't looked at it otherwise but i can try if that would be helpful

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anonymous
  • anonymous
I will be happy if you do help me with this ;_; and I think that J is 1 indeed
anonymous
  • anonymous
Help please I got stucked and I don't know If I did good with my steps of the image from top D:
IrishBoy123
  • IrishBoy123
|dw:1446331632112:dw|
anonymous
  • anonymous
omg what is that Irishboy123? D:
wmj259
  • wmj259
Those are the boundaries of the limit.
wmj259
  • wmj259
|dw:1446344445047:dw|
anonymous
  • anonymous
That is the answer then? I don't get it I just want to know the answer if it's easy why nobody tells me :(
IrishBoy123
  • IrishBoy123
naru no-one said its easy :p first of all , go back check the jacobian. i was in a rush last night, now have a bit more time to work this if you are around.... i think \(J = \mu\) but do it for yourself to be sure... that drawing is the area you are integrating over on the (x,y) system now we have a transformation, we will have a new area. as we have \(x = \mu - \mu \omega\) and \(y = \mu \omega\) we can start to draw a region in \((\mu, \omega )\) co-ords |dw:1446374509977:dw| for the line \(y = 1-x\) we can say that \(\mu \omega = 1 - (\mu - \mu \omega)\) meaning \(\mu \omega = 1 - \mu + \mu \omega\) so \(\omega = 1\) can you do this for the other 2 lines that mark out the region ie , y = 0 amd x = 0?? i'll have a look at the work you posted too... hope this is of help.
IrishBoy123
  • IrishBoy123
soz!! big typo... i repeat: "for the line y=1−x we can say that \(\mu \omega=1−(\mu−\mu \omega)\) meaning \(\mu \omega=1−\mu+\mu \omega\) so \(\color{red}{\mu=1}\)" so we have the first line of our new integration region.. |dw:1446375004380:dw|
IrishBoy123
  • IrishBoy123
OK, i have loaded up your attachment.... for the Jacobian, we have \(\dfrac{\partial x}{\partial \mu} = 1 - \omega, \; \dfrac{\partial x}{\partial \omega} = -\mu, \; \dfrac{\partial y}{\partial \mu} = \omega, \; \dfrac{\partial y}{\partial \omega} = \mu\) so you are looking at this \[\left|\begin{matrix}1-\omega & -\mu \\ \omega & \mu\end{matrix}\right|\]
anonymous
  • anonymous
Thanks for your reply IrishBoy123 :D That will be the answer to all the problem? I will really get how to do jacobian of this type if I get the answer, and I got a little confused with the graphic, drawing the graphic will make the jacobian more easy to do? After doing the operations what will be the answer a numbre or e^tosomething ? Thanks for your replies ;_;
IrishBoy123
  • IrishBoy123
in the original Cartesian co-ords you also had x = 0 and y = 0 as boundaries to the Region \[x = \mu - \mu \omega\] \[y = \mu \omega\] \( y = 0 \implies \mu \omega = 0 \implies \mu = 0, \text{ or } \; \omega = 0\) or both |dw:1446676554063:dw| \( x = 0 \implies \mu (1- \omega) = 0 \implies \mu = 0, \text{ or } \; \omega = 1\) or both |dw:1446676766605:dw|
IrishBoy123
  • IrishBoy123
|dw:1446676798580:dw|
IrishBoy123
  • IrishBoy123
so now we can work the integral \[\int\limits_{x=0}^{1} \; \int\limits_{y=0}^{1-x} \; dy \; dx \qquad \left( exp \left[ \frac{y}{y+x} \right] \right)\]
anonymous
  • anonymous
The answer is 1 X 1, or maybe too e^0?
IrishBoy123
  • IrishBoy123
well we have to factor in the jacobian and the change of limits
anonymous
  • anonymous
I think I'm getting this a little more
anonymous
  • anonymous
What I did here it's wrong them D:?
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anonymous
  • anonymous
then** I'm going to try to do the problem with tihs new information o_o
anonymous
  • anonymous
Lol I got the same answer, J=1 o_o
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IrishBoy123
  • IrishBoy123
\(dx \; dy \rightarrow |J| du \; dv\) \[\large \int\limits_{x=0}^{1} \; \int\limits_{y=0}^{1-x} \; dy \; dx \qquad \left( exp \left[ \frac{y}{y+x} \right] \right) \] \[\Huge \rightarrow\] \[\large \int\limits_{\mu=0}^{1} \; \int\limits_{\omega=0}^{1} \; d\mu \; d\omega \qquad ( \mu e^{\omega } ) \] \[\large = \int\limits_{\mu=0}^{1} \mu \; d\mu \; \int\limits_{\omega=0}^{1} e^{\omega} \; d\omega \] \[\large = \dfrac{1}{2} (e - 1) \]
IrishBoy123
  • IrishBoy123
if that is nonsense, pls advise...
anonymous
  • anonymous
|dw:1446678679186:dw|
anonymous
  • anonymous
Wait it was e^(x/y+x), that change everything? Or just a typo? D:
IrishBoy123
  • IrishBoy123
it was the substitution that changed everything.... it was inspired
IrishBoy123
  • IrishBoy123
\(\dfrac{y}{y+x} = \dfrac{ \mu \omega}{ \mu \omega + \mu - \mu \omega} = \omega\) 🍧
anonymous
  • anonymous
yay!!! Then it's e^w |dw:1446679306775:dw|
anonymous
  • anonymous
I have some ???? in my head why it's 1-(x=0) and not 1-(x=u-wu)? and I have to change dwdu for dudw? :D
IrishBoy123
  • IrishBoy123
maybe we need to go back to the Jacobian ?
anonymous
  • anonymous
Thanks :D uhmmm ... uhmmm..... I don't get it T_T I will do it without understanding that and see what I get D:
IrishBoy123
  • IrishBoy123
naru keep coming back i am in the same boat about pretty much everything.
IrishBoy123
  • IrishBoy123
for what that is worth 🍀
anonymous
  • anonymous
I did it :D Or at least I think... It's okay? o_o Thanks for your help, my boat it's moving with the air I need help ;_;
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IrishBoy123
  • IrishBoy123
i still think you need \(\mu\) in your Jacobian, but that is your decision 😋
anonymous
  • anonymous
Thanks I did it again :D now it's correct? :D
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IrishBoy123
  • IrishBoy123
coooool !! 🎀
anonymous
  • anonymous
Wow :D but I still don't understanding the 1-x= 1, withouth your help I will never were going to be capable of doing in the correct way then D: Thanks, If I give you best response this thread is still available to chat? :D
IrishBoy123
  • IrishBoy123
best response is irrelevant, don't bother doing it post another question,..., in a new thread
IrishBoy123
  • IrishBoy123
🤑
anonymous
  • anonymous
I'm having problems with that exactly D: I don't know how to do a new question because always appears this one D:, I have to go by now, really thanks for your help :D I have like 3 or more questions, but I think these are more easy than this, 1 is of partial derivatives, triple integrals, and prove a limit with radicals D: , That was what I was doing but I got stucked in the yellow part, the green and blue color are irrelevant, bye bye :D
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anonymous
  • anonymous
I came back to understand this more now :D
anonymous
  • anonymous
Thaaaaaaaaaaaaaaaaaanks, Now I really got it Thanks Thanks I will go to sleep now, thanks again :D Your help means a lot for me Thanks!!!

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