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Finding the Jacobian

by roam
Tags: jacobian
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roam
#1
May1-10, 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?
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jtyler05si
#2
May1-10, 02:30 AM
P: 27
Part A looks correct to me.
jtyler05si
#3
May1-10, 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.

Je m'appelle
#4
May1-10, 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 plug-in 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]
jtyler05si
#5
May1-10, 02:45 AM
P: 27
Quote Quote by Je m'appelle View Post
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 plug-in that in:

[tex]\int \int_R f(x,y) dxdy = \int \int_S f(rcos\theta,rsin\theta)\begin{bmatrix} {\partial (x,y) \\ {\partial (r,\theta)} \end{bmatrix} drd\theta [/tex]
Exactly.

Again sorry for my lack of correct notation and such.
roam
#6
May1-10, 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 xy-plane. But in this case how do I need to prove it?
The Chaz
#7
May1-10, 07:48 PM
P: 186
Quote Quote by roam View Post
... I know that a mapping is one to one if distinct points in the [tex]r, \theta[/tex] plane have distinct images in the xy-plane. But in this case how do I need to prove it?
A common strategy to show that a function (let's say f(x), for instance) is 1-1 ("injective", "invertible")is to:
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 1-1.

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 1-1
roam
#8
May1-10, 09:44 PM
P: 907
Quote Quote by The Chaz View Post
A common strategy to show that a function (let's say f(x), for instance) is 1-1 ("injective", "invertible")is to:
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 1-1.

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 1-1
I know this definition. What I don't understand is, how to apply it to this particular problem. Do we just say since the Jacobian is "J=r", then any value we substitute into r is itself, thus every element in the domain has a distinct and unique image under this transformation? I doubt this is correct though...
The Chaz
#9
May1-10, 10:03 PM
P: 186
Quote Quote by Je m'appelle View Post
[tex]x = rcos\theta [/tex]

[tex]y = rsin\theta [/tex]
This is the mapping, not J. Basically, you're trying to show that every point on the unit circle (except (1,0) ) is the mapping of exactly one angle in the interval 0 to 2π
roam
#10
May2-10, 10:46 PM
P: 907
Quote Quote by The Chaz View Post
This is the mapping, not J. Basically, you're trying to show that every point on the unit circle (except (1,0) ) is the mapping of exactly one angle in the interval 0 to 2π
I see. But how do we need to ensure this for every single point on the unit circle?
lanedance
#11
May3-10, 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 non-zero jacobian determinant the function is invertible near p.

As your jacobian is non-zero 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.
lanedance
#12
May3-10, 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
roam
#13
May4-10, 02:02 AM
P: 907
Quote Quote by lanedance View Post
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
The question asks "prove that the mapping used is 1 to 1 on region R". I see what you mean, but I'm confused because I don't know what to write down in order to "prove" that the mapping is 1 to 1. What should I write that will be sufficient as a proof?
lanedance
#14
May4-10, 03:26 AM
HW Helper
P: 3,307
so the non-zero 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
roam
#15
May5-10, 05:58 AM
P: 907
Quote Quote by lanedance View Post
so the non-zero 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
Does this show that for any given [tex] (r,\theta) [/tex] there is a uniqe (x,y):

[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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