Question

Statement-I: The function \( u = x^2 + y^2 \), \( v = \tan^{-1}\left(\frac{y}{x}\right) \) are functionally independent.
Statement-II: The Jacobian \( \frac{\partial(u,v)}{\partial(x,y)} \) is non-zero.

• A. Statement-I is true, but statement-II is false
• B. Statement-I is false, but statement-II is true
• C. Both statement-I and statement-II are true
• D. Both statement-I and statement-II are false

Hint:

Notice that \( u = r^2 \) and \( v = \theta \) in polar coordinates! Since polar coordinates map points uniquely to independent coordinate axes, functions of \( r \) alone and \( \theta \) alone are always functionally independent.

Answer 1

The Correct Option is C

Step 1: Recognise \( u \) and \( v \) as polar-type coordinates.
Notice \( u = x^2+y^2 = r^2 \) and \( v = \tan^{-1}(y/x) = \theta \), where \( r,\theta \) are the usual polar coordinates. So really we are asking whether \( (r^2, \theta) \) are independent functions of \( x, y \), which should hold everywhere except the origin, since polar coordinates form a genuine, invertible coordinate system there.

Step 2: Use the chain rule to split the Jacobian.
Since \( u,v \) depend on \( x,y \) through \( r,\theta \), we can write \[ \frac{\partial(u,v)}{\partial(x,y)} = \frac{\partial(u,v)}{\partial(r,\theta)} \cdot \frac{\partial(r,\theta)}{\partial(x,y)} \] The first factor is easy since \( u = r^2 \) and \( v = \theta \): \[ \frac{\partial(u,v)}{\partial(r,\theta)} = \begin{vmatrix} 2r & 0 \\ 0 & 1 \end{vmatrix} = 2r \]
Step 3: Use the known polar Jacobian.
The standard result \( \frac{\partial(x,y)}{\partial(r,\theta)} = r \) gives its inverse directly: \[ \frac{\partial(r,\theta)}{\partial(x,y)} = \frac{1}{r} \] Multiplying the two factors: \[ \frac{\partial(u,v)}{\partial(x,y)} = 2r \cdot \frac{1}{r} = 2 \] This is non-zero everywhere except at the origin, so Statement-II is true, and a non-zero Jacobian means \( u, v \) are functionally independent, so Statement-I is true as well. Both statements hold, matching option (C).