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

Concept: Two differentiable functions \( u(x,y) \) and \( v(x,y) \) are said to be functionally dependent if their Jacobian determinant evaluates identically to zero across a region. Conversely, they are defined to be functionally independent if and only if their Jacobian determinant is non-zero: \[ J = \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \neq 0 \] Therefore, Statement-II functions as the direct mathematical definition/condition for establishing Statement-I. Let us calculate the partial derivatives to check the value.

Step 1: Determining the partial derivatives of \( u \).
The first given function expression is: \[ u = x^2 + y^2 \] Differentiating \( u \) partially with respect to the variable \( x \) (treating \( y \) as a constant): \[ \frac{\partial u}{\partial x} = 2x \] Differentiating \( u \) partially with respect to the variable \( y \) (treating \( x \) as a constant): \[ \frac{\partial u}{\partial y} = 2y \]

Step 2: Determining the partial derivatives of \( v \).
The second given function expression is: \[ v = \tan^{-1}\left(\frac{y}{x}\right) \] Recall the derivative rule: \( \frac{d}{dt}(\tan^{-1}(t)) = \frac{1}{1+t^2} \). Applying the chain rule for partial differentiation with respect to \( x \): \[ \frac{\partial v}{\partial x} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial x}\left(\frac{y}{x}\right) = \frac{1}{\frac{x^2 + y^2}{x^2}} \cdot \left( -\frac{y}{x^2} \right) \] Simplifying this expression: \[ \frac{\partial v}{\partial x} = \frac{x^2}{x^2 + y^2} \cdot \left( -\frac{y}{x^2} \right) = -\frac{y}{x^2 + y^2} \] Now, performing partial differentiation with respect to \( y \): \[ \frac{\partial v}{\partial y} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial y}\left(\frac{y}{x}\right) = \frac{x^2}{x^2 + y^2} \cdot \left( \frac{1}{x} \right) \] Simplifying this expression: \[ \frac{\partial v}{\partial y} = \frac{x}{x^2 + y^2} \]

Step 3: Setting up and computing the Jacobian determinant \( \frac{\partial(u,v)}{\partial(x,y)} \).
We construct the Jacobian matrix determinant using our calculated partial derivative expressions: \[ \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 2x & 2y \\ -\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2} \end{vmatrix} \] Evaluating the \( 2 \times 2 \) determinant product: \[ \frac{\partial(u,v)}{\partial(x,y)} = (2x) \left( \frac{x}{x^2 + y^2} \right) - (2y) \left( -\frac{y}{x^2 + y^2} \right) \] Combining the terms over their common denominator: \[ \frac{\partial(u,v)}{\partial(x,y)} = \frac{2x^2}{x^2 + y^2} + \frac{2y^2}{x^2 + y^2} = \frac{2(x^2 + y^2)}{x^2 + y^2} \] Canceling out the common term \( (x^2 + y^2) \) (assuming \( x, y \neq 0 \)): \[ \frac{\partial(u,v)}{\partial(x,y)} = 2 \]

Step 4: Evaluating the truth value of both statements.
Since our calculation reveals that the Jacobian value is exactly \( 2 \), it is clearly non-zero (\( 2 \neq 0 \)).

• This directly confirms that

Statement-II is true because the Jacobian is non-zero.

• Because a non-zero Jacobian structurally guarantees functional independence,

Statement-I is also true.
Thus, both statements are mathematically true, matching Option (C).