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Calculation of a Jacobian

So x = rcosht, y = rsinht
d(x,y)/d(r,t) = r, correct?

But.. for d(r,t)/d(x,y) I can't get 1/r. I end up with r(x^2 - y^2)/x^2.y^2

Help :/
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Avatar for Zacken
Zacken
22
Probably best if you put up a picture of your exact question.
dx/dr = cosht
dx/dt = rsinht

dy/dr = sinht
dy/dt = rcosht

Jacobian = rcosh^2(t) - rsinh^2(t) = r

What else do you need?
Original post by Zacken
Probably best if you put up a picture of your exact question.


64.png

Original post by Math12345
dx/dr = cosht
dx/dt = rsinht

dy/dr = sinht
dy/dt = rcosht

Jacobian = rcosh^2(t) - rsinh^2(t) = r

What else do you need?


The inverse
Original post by StarvingAutist
So x = rcosht, y = rsinht
d(x,y)/d(r,t) = r, correct?

But.. for d(r,t)/d(x,y) I can't get 1/r. I end up with r(x^2 - y^2)/x^2.y^2

Help :/


Works when I do it.

For your inverses you can use:

r=x2−y2t=tanh⁡−1yx\displaystyle r=\sqrt{x^2-y^2}\\\\t=\tanh^{-1}\frac{y}{x}

What do you get for your partial derivatives?
Original post by ghostwalker
Works when I do it.

For your inverses you can use:

r=x2−y2t=tanh⁡−1yx\displaystyle r=\sqrt{x^2-y^2}\\\\t=\tanh^{-1}\frac{y}{x}

What do you get for your partial derivatives?


dt/dx = 1/(r sinht)
dt/dy = 1/(r cosht)
dr/dx = 1/cosht
dr/dy = 1/sinht
Original post by StarvingAutist
dt/dx = 1/(r sinht)
dt/dy = 1/(r cosht)
dr/dx = 1/cosht
dr/dy = 1/sinht


I would expect them to be function of x,y, not r,t.

Looks like you're just taking the reciprocal


∂r∂x=1∂x∂r\displaystyle\dfrac{\partial r}{\partial x }=\dfrac{1}{\frac{\partial x}{\partial r}}


Is that the case?
(edited 8 years ago)
Original post by ghostwalker
I would expect them to be function of x,y, not r,t.

Looks like you're just taking the reciprocal


∂x∂r=1∂r∂x\displaystyle\dfrac{\partial x}{\partial r }=\dfrac{1}{\frac{\partial r}{\partial x}}


Is that the case?


2016-04-19 16.55.50.jpg
Original post by StarvingAutist
2016-04-19 16.55.50.jpg


Starting at the top left, you're taking partial derivatives wrt x with r treated as a constant. But you require partial derivatives wrt x treating y as a constant.

With fuller notation:

You're working out (∂θ∂x)r\left(\dfrac{\partial \theta}{ \partial x}\right)_r

But you actually want (∂θ∂x)y\left(\dfrac{\partial \theta}{ \partial x}\right)_y

Hence the need to get r,theta in terms of x,y to start.
Original post by ghostwalker
Starting at the top left, you're taking partial derivatives wrt x with r treated as a constant. But you require partial derivatives wrt x treating y as a constant.

With fuller notation:

You're working out (∂θ∂x)r\left(\dfrac{\partial \theta}{ \partial x}\right)_r

But you actually want (∂θ∂x)y\left(\dfrac{\partial \theta}{ \partial x}\right)_y

Hence the need to get r,theta in terms of x,y to start.


Ahhh, thanks!
Original post by ghostwalker
]...


Now I'm stuck on the second part.
65.png
I am extremely confused about limits.
Attachment not found
2016-04-19 19.36.00-1.jpg22.2KB
Original post by StarvingAutist
Now I'm stuck on the second part.
65.png
I am extremely confused about limits.
Attachment not found


Can we have the whole of the original question. Your graph doesn't match the equations.
Original post by ghostwalker
Can we have the whole of the original question. Your graph doesn't match the equations.


There's nothing more to it.

66.png

I changed my graph now, still no ideas.
Original post by StarvingAutist
There's nothing more to it.

66.png

I changed my graph now, still no ideas.


OK. For clarity, here's an image:

Untitled.jpg

Now consider each of the 4 edges. What are their equations, and what do they become under the new coordinate system. The left/right edges are easiest, to start.
Original post by ghostwalker
OK. For clarity, here's an image:

Untitled.jpg

Now consider each of the 4 edges. What are their equations, and what do they become under the new coordinate system. The left/right edges are easiest, to start.


Got it in the end, not sure why I was confused. Anyway, does R = 0.25 (b^2 - a^2) * ln3 sound right?
Original post by StarvingAutist
Got it in the end, not sure why I was confused. Anyway, does R = 0.25 (b^2 - a^2) * ln3 sound right?


Not worked it through, but just graphed for b=5,a=3, and the area worked out to the value your formula gives.

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