Showing that a partial derivative equation holds

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s3a
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Homework Statement


The question is attached as Question.jpg.


Homework Equations


Partial differentiation.


The Attempt at a Solution


This seems obvious to me but I don't know how to express myself mathematically. Basically, what I'd do is:

[∂(u,v)/∂(x,y)] [∂(x,y)/∂(r,s)] = [∂(u,v)/∂(r,s)] [∂(x,y)/∂(x,y)] = ∂(u,v)/∂(r,s)

as for ∂(u,v)/∂(x,y) = 1/[∂(u,v)/∂(x,y)], that's especially obvious to me! It's like saying 1/(1/2) = 1 * 2/1 = 2! Whenever I get an “It's obvious” feeling, that usually tells me that I'm not grasping the assumptions that are made when using these mathematical tools.

Any assistance in solving this problem would be greatly appreciated!

Edit: Forgot to attach the question! :P
 
Attachments
  • Printed exercise: Show that the Jacobian of (u,v) with respect to (x,y) times the Jacobian of (x,y) with respect to (r,s) equals the Jacobian of (u,v) with respect to (r,s), and thus the Jacobian of (u,v) with respect to (x,y) equals 1 divided by the Jacobian of (x,y) with respect to (u,v). Hint: |AB| = |A||B|.
    Question.jpg
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s3a said:

Homework Statement


The question is attached as Question.jpg.


Homework Equations


Partial differentiation.


The Attempt at a Solution


This seems obvious to me but I don't know how to express myself mathematically. Basically, what I'd do is:

[∂(u,v)/∂(x,y)] [∂(x,y)/∂(r,s)] = [∂(u,v)/∂(r,s)] [∂(x,y)/∂(x,y)] = ∂(u,v)/∂(r,s)

as for ∂(u,v)/∂(x,y) = 1/[∂(u,v)/∂(x,y)], that's especially obvious to me! It's like saying 1/(1/2) = 1 * 2/1 = 2! Whenever I get an “It's obvious” feeling, that usually tells me that I'm not grasping the assumptions that are made when using these mathematical tools.

Any assistance in solving this problem would be greatly appreciated!

Edit: Forgot to attach the question! :P
Remember that a Jacobian is the determinant of a matrix. Write out the matrix for [∂(u,v)/∂(x,y)] and [∂(x,y)/∂(r,s)]. What do you get if you multiply those matrices? And see if you can use the hint that |AB|=|A||B| for the determinant of a product of square matrices.
 
I attached the Jacobian stuff you asked for.

For the hint, A and B are matrices, right? What do the || mean in this case?
 
Attachments
  • Handwritten derivation of Jacobian determinants: d(u,v)/d(x,y) = -|u_x u_y; v_x v_y|, d(x,y)/d(r,s) = -|x_r x_s; y_r y_s|, d(u,v)/d(r,s) = -|u_r u_s; v_r v_s|, followed by expansion of the product (u_x v_y - v_x u_y)(x_r y_s - y_r x_s) and a chain-rule expansion in partial derivatives.
    MyWork.jpg
    45.1 KB · Views: 498
s3a said:
I attached the Jacobian stuff you asked for.

For the hint, A and B are matrices, right? What do the || mean in this case?

|A| means the determinant of A. And in your last attachment, multiply the matrices themselves before calculating any determinants. I think you will find you don't have to actually evaluate any determinants to work the problem.
 
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