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myininaya
 one year ago
Show \[r dr d \theta =dx dy\] .
myininaya
 one year ago
Show \[r dr d \theta =dx dy\] .

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myininaya
 one year ago
Best ResponseYou've already chosen the best response.1Also I know one way is to use the Jacobian. I wonder if there is any other way.

iPwnBunnies
 one year ago
Best ResponseYou've already chosen the best response.1Idk bout Jacobian...But I gotta do alot of drawing.

iPwnBunnies
 one year ago
Best ResponseYou've already chosen the best response.1dw:1396312983170:dw

iPwnBunnies
 one year ago
Best ResponseYou've already chosen the best response.1arclength = r*theta dA = r*dtheta*dr dy dx, dx dy = r dr dtheta

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1You 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 usubstitution you just didn't know it because the derivative you took was only in 1 dimension.

iPwnBunnies
 one year ago
Best ResponseYou've already chosen the best response.1What I drew was a polar rectangle, R... \[R = [(r,\Theta)  a \le r \le b, \alpha \le \Theta \le \beta]\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1@ipwnbunnies I like your geometric picture to show this. You found the area of that one piece using the fact that looked sorta rectangularish (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.

iPwnBunnies
 one year ago
Best ResponseYou've already chosen the best response.1Not familiar with Jacobian anything...But it looks interesting.

iPwnBunnies
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, my proof is a rectangularlike, but as the polar rectangle gets smaller, it "looks" like a rectangular rectangle. :)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1@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.

iPwnBunnies
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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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