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.
The correct answer is}

• 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

Concept:
Two differentiable functions \(u(x,y)\) and \(v(x,y)\) are said to be functionally independent if their Jacobian determinant is non-zero: 

\[ J=\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{vmatrix} \neq 0 \] If the Jacobian is identically zero, the functions are functionally dependent. Thus, Statement II provides the condition needed to verify Statement I. 

Step 1: Find the partial derivatives of \(u\). 
Given, \[ u=x^2+y^2 \] Differentiating with respect to \(x\): \[ \frac{\partial u}{\partial x}=2x \] Differentiating with respect to \(y\): \[ \frac{\partial u}{\partial y}=2y \] 

Step 2: Find the partial derivatives of \(v\). 
Given, \[ v=\tan^{-1}\left(\frac{y}{x}\right) \] Differentiating with respect to \(x\): \[ \frac{\partial v}{\partial x} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial x}\left(\frac{y}{x}\right) \] \[ = \frac{x^2}{x^2+y^2} \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2+y^2} \] 
Differentiating with respect to \(y\): \[ \frac{\partial v}{\partial y} = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial y}\left(\frac{y}{x}\right) \] \[ = \frac{x^2}{x^2+y^2} \left(\frac{1}{x}\right) = \frac{x}{x^2+y^2} \] 

Step 3: Compute the Jacobian determinant. 
Using the partial derivatives, \[ \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 2x & 2y \\ -\dfrac{y}{x^2+y^2} & \dfrac{x}{x^2+y^2} \end{vmatrix} \] Evaluating the determinant, \[ \frac{\partial(u,v)}{\partial(x,y)} = (2x)\left(\frac{x}{x^2+y^2}\right) - (2y)\left(-\frac{y}{x^2+y^2}\right) \] \[ = \frac{2x^2}{x^2+y^2} + \frac{2y^2}{x^2+y^2} \] \[ = \frac{2(x^2+y^2)}{x^2+y^2} =2 \] Since the Jacobian is non-zero, \[ \frac{\partial(u,v)}{\partial(x,y)}=2\neq0 \] 

Step 4: Evaluate the Statements. 
Since the Jacobian determinant is non-zero: 

• Statement I is true because the functions \(u\) and \(v\) are functionally independent. 

• Statement II is also true because the Jacobian determinant is non-zero. 

Final Answer: 
Both Statement I and Statement II are true, and Statement II correctly explains Statement I.