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Find the Jacobian $$\frac{\partial(x,y)}{\partial(u,v)}$$ for $x=u^2+v^2$, $y=u^2-v^2$.

My solution:

I tried solving it as it is by using the Jacobian matrix (determinant?) and got my answer to be $-8uv$. I think the answer is wrong since the answers in my book are all purely numerical. Is there something that I have to do to these equations to make them non-polynomials ?

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Perhaps the textbook is asking for the value of the Jacobian determinant at a given point? (If not, it's certainly possible that your textbook is in error.) –  Jesse Madnick Apr 20 '11 at 6:21
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Your calculation is correct, by the way. –  Jesse Madnick Apr 20 '11 at 6:22
    
well... actually that is all it says, that is the description of the problem, are you sure this is correct? –  Virtuoso Apr 20 '11 at 6:31

1 Answer 1

up vote 5 down vote accepted

The term Jacobian is ambiguous in that it may refer to both the Jacobian matrix

$$J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}$$

and the Jacobian determinant

$$\det{J} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}.$$

In your specific instance, the Jacobian matrix is

$$J = \begin{bmatrix} 2u & 2v \\\ 2u & -2v\end{bmatrix}$$

and the Jacobian determinant is $\det{J} = -8uv$, as you calculated.

(I'm giving the complete answer since I'm sure that you did the correct thing in your solution).

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