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I have
$\begin{aligned}x_{1}&=r\sin(\theta_{1}),\\ x_{2}&=r\cos(\theta_{1})\sin(\theta_{2})\\ x_{3}&=r\cos(\theta_{1})\cos(\theta_{2}). \end{aligned} $
I know how to compute the Jacobian $\frac{\partial(x_{1},x_{2},x_3)}{\partial(\theta_{1},\theta_{2},r)}$ directly.

The thing is, there is a way to get this Jacobian that involves a ratio of an upper triangular determinant and a lower triangular determinant. I just cannot figure out how to get this. I'm guessing it's some chain rule thing. Any help will be appreciated. Thanks.

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    That's what I got too - the thing is, the actual problem is for $x_{1}, \cdots, x_{n}$ - that's why that trick I was talking about would be very useful2012-09-28

1 Answers 1

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$\frac{\partial(x_1,x_2,x_3)}{\partial(\theta_1,\theta_2,r)}=\begin{bmatrix} \frac{\partial x_1}{\partial\theta_1} & \frac{\partial x_1}{\partial\theta_2} & \frac{\partial x_1}{\partial r}\\ \frac{\partial x_2}{\partial\theta_1} & \frac{\partial x_2}{\partial\theta_2} & \frac{\partial x_2}{\partial r}\\ \frac{\partial x_3}{\partial\theta_1} & \frac{\partial x_3}{\partial\theta_2} & \frac{\partial x_3}{\partial r}\\ \end{bmatrix}$ $=\begin{bmatrix} r\cos\theta_1 & 0 & \sin\theta_1 \\ -r\sin\theta_1\sin\theta_2 & r\cos\theta_1\cos\theta_2 & \cos\theta_1\sin\theta_2 \\ -r\sin\theta_1\cos\theta_2 & -r\cos\theta_1\sin\theta_2 & \cos\theta_1\cos\theta_2 \\ \end{bmatrix}$ $=r^2(\cos^3\theta_1\cos^2\theta_2+\cos^3\theta_1\sin^2\theta_2+\sin^2\theta_1\cos\theta_1\sin^2\theta_2+\sin^2\theta_1\cos\theta_1\cos^2\theta_2)$ $=r^2(\sin^2\theta_1+\cos^2\theta_1)(\sin^2\theta_2+\cos^2\theta_2)(4\cos\theta_1)$ $=4r^2\cos\theta_1$