Finding Jacobian for Homework Statement

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In summary, the homework statement is trying to find a solution for an equation involving a Jacobian, but does not state how to obtain the four entries in the Jacobian matrix.
  • #1
roam
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Homework Statement


Here's my question:

[PLAIN]http://img140.imageshack.us/img140/1500/89319562.gif [Broken]


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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  • #2
Part A looks correct to me.
 
  • #3
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
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]
 
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  • #5
Je m'appelle said:
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.
 
  • #6
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? :uhh:

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?
 
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  • #7
roam said:
... 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
 
  • #8
The Chaz said:
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...
 
  • #9
Je m'appelle said:
[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
 
  • #10
The Chaz said:
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? :confused:
 
  • #11
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.
 
  • #12
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
lanedance said:
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?
 
  • #14
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
 
  • #15
lanedance said:
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?
 

1. What is a Jacobian?

A Jacobian is a mathematical tool used in multivariate calculus to determine the relationship between two coordinate systems. It is often used in the study of change of variables in integration and optimization problems.

2. Why is finding Jacobian important?

Finding Jacobian is important because it allows us to transform a problem from one coordinate system to another, making it easier to solve. It also helps us understand the relationship between different variables and how they affect each other.

3. How do you find the Jacobian of a function?

To find the Jacobian of a function, you need to take the partial derivatives of the function with respect to each variable and then arrange them in a matrix. The determinant of this matrix is the Jacobian.

4. What is the difference between a Jacobian and a gradient?

A Jacobian is a matrix that describes the relationship between two coordinate systems, while a gradient is a vector that describes the direction and magnitude of the steepest ascent of a scalar function. They have different uses and interpretations in mathematics and physics.

5. Can the Jacobian be negative?

Yes, the Jacobian can be negative. This indicates that the transformation between the two coordinate systems is orientation-reversing, meaning that the direction of positive and negative values is reversed. This often happens when flipping the axes or using trigonometric functions in the transformation.

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