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# Calculation of a Jacobian Watch

1. 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 :/
2. Probably best if you put up a picture of your exact question.
3. 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?
4. (Original post by Zacken)
Probably best if you put up a picture of your exact question.

(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
5. (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:

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

For your inverses you can use:

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
7. (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

Is that the case?
8. (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

Is that the case?
9. (Original post by StarvingAutist)
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

But you actually want

Hence the need to get r,theta in terms of x,y to start.
10. (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

But you actually want

Hence the need to get r,theta in terms of x,y to start.
Ahhh, thanks!
11. (Original post by ghostwalker)
]...
Now I'm stuck on the second part.

I am extremely confused about limits.
Attachment 524183524185
Attached Images

12. (Original post by StarvingAutist)
Now I'm stuck on the second part.

I am extremely confused about limits.
Attachment 524183524185
Can we have the whole of the original question. Your graph doesn't match the equations.
13. (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.

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

I changed my graph now, still no ideas.
OK. For clarity, here's an image:

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.
15. (Original post by ghostwalker)
OK. For clarity, here's an image:

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?
16. (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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