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myininaya

  • 2 years ago

Show \[r dr d \theta =dx dy\] .

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  1. myininaya
    • 2 years ago
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    Also I know one way is to use the Jacobian. I wonder if there is any other way.

  2. iPwnBunnies
    • 2 years ago
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    Idk bout Jacobian...But I gotta do alot of drawing.

  3. iPwnBunnies
    • 2 years ago
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    |dw:1396312983170:dw|

  4. iPwnBunnies
    • 2 years ago
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    arclength = r*theta dA = r*dtheta*dr dy dx, dx dy = r dr dtheta

  5. Kainui
    • 2 years ago
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    You can do it geometrically. But you can definitely do it with the jacobian. The Jacobian is a lot simpler than some people make it out to be. So if you change the integral to another coordinate set then just right away we can see: |dw:1396312970588:dw| So obviously when we scale it the squares get larger as we move out and we need to account for that. The determinant is just the area of a matrix right? Well similarly, the determinant of a matrix of partial derivatives is just saying how much a little bit of area changes at each point. In fact you were doing this with u-substitution you just didn't know it because the derivative you took was only in 1 dimension.

  6. iPwnBunnies
    • 2 years ago
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    What I drew was a polar rectangle, R... \[R = [(r,\Theta) | a \le r \le b, \alpha \le \Theta \le \beta]\]

  7. myininaya
    • 2 years ago
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    @ipwnbunnies I like your geometric picture to show this. You found the area of that one piece using the fact that looked sorta rectangular-ish (use the area of a rectangle) Yep jacobian is pretty easy. \[J(r, \theta) dr d \theta =dx dy \] \[(x_r \cdot y_\theta -x_\theta \cdot y_ r) dr d \theta = dx dy \] \[(\cos(\theta) \cdot r \cos(\theta)--rsin(\theta) \cdot \sin(\theta)) d r d \theta=dx dy \] \[r(\cos^2(\theta)+\sin^2(\theta))dr d \theta = dx dy \] \[r \cdot 1 d r d \theta= dx dy \] \[r dr d \theta=dx dy \] I think it is great you guys like math enough to discuss things like this. Kainui, I like that you posting problems to get us advanced math people thinking a little more.

  8. iPwnBunnies
    • 2 years ago
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    Not familiar with Jacobian anything...But it looks interesting.

  9. iPwnBunnies
    • 2 years ago
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    Yeah, my proof is a rectangular-like, but as the polar rectangle gets smaller, it "looks" like a rectangular rectangle. :)

  10. Kainui
    • 2 years ago
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    @iPwnBunnies Have you taken linear algebra? Basically it's just the area of a parallelogram made by two vectors in a matrix. |dw:1396314943369:dw| So if you have a matrix \[\left[\begin{matrix}u_x & u_y \\ v_x & v_y\end{matrix}\right]\] the determinant of this is the area. The jacobian just has how the coordinates change in each of the directions, so it's the area of the transformation really. I hope that sort of makes it a little more understandable. Of course if you have 3 vectors or higher the determinant is the volume, or higher dimensional analogue.

  11. iPwnBunnies
    • 2 years ago
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    Oh yeah, I know the area is the cross product of the two vectors. I guess I'm not familiar with the term 'matrix'. Just vectors and determinants lol.

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