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Finding the Jacobianby roam
Tags: jacobian 
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#1
May110, 02:10 AM

P: 907

1. The problem statement, all variables and given/known data
Here's my question: 3. The attempt at a solution (a) [tex]\int^{2 \pi}_0 \int^1_0 [r^2(cos^2\theta + sin^2\theta)]rdr d\theta[/tex] [tex]\int^{2 \pi}_0 \int^1_0 r^3 dr d \theta[/tex] [tex]= \int^{2 \pi}_0 \frac{r^4}{a} ^1_0 d \theta[/tex] [tex]=\frac{1}{4} \theta ^{2 \pi}_0 = \frac{\pi}{2}[/tex] Is this correct? (b) So, is the Jacobian for this problem given by the following? [tex]J(x,y)= \frac{\partial(r,\theta)}{\partial (x,y)}=\begin{bmatrix} {\partial r\over \partial x} & {\partial r\over \partial y} \\ {\partial \theta\over \partial x} & {\partial \theta\over \partial y} \end{bmatrix}[/tex] If so, how can I obtain the four entries in that matrix? 


#2
May110, 02:30 AM

P: 27

Part A looks correct to me.



#3
May110, 02:37 AM

P: 27

I believe for part B.
Use x=r*cos(theta) and y=r*sin(theta) there for set up your jacobian as [(dx/dr),(dx/dtheta)];[(dy/dr),(dy/dtheta)] solve and you will get the proper conversion. You will know when you get it. sorry for the lack of proper notation, etc. I put the matrix notation as I would in a TI version calculator. semi colon means next row. 


#4
May110, 02:42 AM

P: 109

Finding the Jacobian
Yeah, part A is correct.
On part B, shouldn't the Jacobian be: [tex]J = \frac{\partial (x,y)}{\partial (r, \theta)} = \begin{bmatrix}{\partial x\over \partial r} & {\partial y\over \partial r} \\ {\partial x\over \partial \theta} & {\partial y\over \partial \theta} \end {bmatrix} [/tex] Where [tex]x = rcos\theta [/tex] [tex]y = rsin\theta [/tex] Now just plugin that in: [tex]\int \int_R f(x,y) dxdy = \int \int_S f(rcos\theta,rsin\theta)\begin {bmatrix} {\partial (x,y) \over{\partial (r,\theta)} \end {bmatrix} drd\theta [/tex] 


#5
May110, 02:45 AM

P: 27

Again sorry for my lack of correct notation and such. 


#6
May110, 07:33 PM

P: 907

Thanks guys. I found the Jacobian to be [tex]J=r[/tex]. So to answer part (b), the only point interior to R at which the Jacobian equals zero is (0,0), right?
Furthermore, the question asks: "Prove that the mapping used is 1 to 1 on region R excluding the boundary points, and excluding the origin. Note: one way to do this is to produce the inverse mapping." I know that a mapping is one to one if distinct points in the [tex]r, \theta[/tex] plane have distinct images in the xyplane. But in this case how do I need to prove it? 


#7
May110, 07:48 PM

P: 186

1) ASSUME that f(a) = f(b) (i.e. you got the same mapping/output/image) 2) Show that a (MUST) = b e.g. g(x) = x^3. Assume g(a) = g(b). That means a^3 = b^3. Taking the cube root of both sides gives a = b. g(x) is 11. h(x) = x^2 Assume h(a) = h(b). That means that a^2 = b^2. But taking the square root (or using the "square root property" as some call it) gives a = +/ b. Two possibilities, so we can't say that a "MUST" = b. h(x) is not 11 


#8
May110, 09:44 PM

P: 907




#9
May110, 10:03 PM

P: 186




#10
May210, 10:46 PM

P: 907




#11
May310, 12:00 AM

HW Helper
P: 3,307

bit hard to follow exactly where you are, but here's some comments
The inverse function theorem shows if at a given point, a continously differntiable function has a nonzero jacobian determinant the function is invertible near p. As your jacobian is nonzero everywhere but the origin it is invertible everywhere except the origin. It is only left to select the domain such that the given mapping is one to one. Hence why [0,2pi) is usually used. 


#12
May310, 12:05 AM

HW Helper
P: 3,307

If you can't see why its ono to one around a circle of raidus r, try plotting the inverse function & it should be clear



#13
May410, 02:02 AM

P: 907




#14
May410, 03:26 AM

HW Helper
P: 3,307

so the nonzero Jacobian guarantees you are locally invertible & 1:1.
Now you want to show the functions are 1:1 on the global domain, have a look at the functions each way: [tex](r,\theta) \to (x(r,\theta), y(r,\theta))[/tex] [tex](x,y) \to (r(x,y), \theta(x,y))[/tex] then to show the function is globally 1:1, can you show that for a given [itex] (r,\theta) [/itex] there is a unique solution (x,y) in the given domain and vice versa 


#15
May510, 05:58 AM

P: 907

[tex](r,\theta) \to (x(r,\theta), y(r,\theta))=((r cos \theta), (r sin \theta))[/tex] By the way, they told us one way to do this problem is to produce the inverse mapping. How do we do that? 


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