Question

Compute the Jacobian $ \frac{\partial(x,y)}{\partial(u,v)} $ of the transformation: $$ x = \frac{u+v}{1 - uv}, \quad y = \tan^{-1} u + \tan^{-1} v $$

• A. -1
• B. 0
• C. 1
• D. \( \frac{1}{x} + \frac{1}{y} \)

Hint:

Jacobian zero implies transformation variables are functionally dependent.

Answer 1

The Correct Option is B

Expressing \( x \) and \( y \) in terms of \( u \) and \( v \), the Jacobian determinant evaluates to zero due to the interdependence of \( x \) and \( y \). Specifically, since \[ x = \tan(\tan^{-1} u + \tan^{-1} v) \] and \[ y = \tan^{-1} u + \tan^{-1} v, \] \( x = \tan(y) \), indicating a functional dependency and zero Jacobian.