Question

The family of surfaces given by \[ u = xy + f(x^2 - y^2), \text{where} \, f : \mathbb{R} \to \mathbb{R} \, \text{is a differentiable function, satisfies:} \]

• A. \( y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = x^2 + y^2 \)
• B. \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x^2 + y^2 \)
• C. \( y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = x^2 - y^2 \)
• D. \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x^2 - y^2 \)

Hint:

When calculating partial derivatives of a composite function, apply the chain rule to handle the function inside the function.

Answer 1

The Correct Option is A

The given function is: \[ u = xy + f(x^2 - y^2). \] We need to compute the partial derivatives with respect to \(x\) and \(y\): \[ \frac{\partial u}{\partial x} = y + f'(x^2 - y^2) \cdot 2x, \] \[ \frac{\partial u}{\partial y} = x - f'(x^2 - y^2) \cdot 2y. \] Now, multiplying \(y\) and \(x\) with these derivatives and adding them: \[ y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = y \left( y + f'(x^2 - y^2) \cdot 2x \right) + x \left( x - f'(x^2 - y^2) \cdot 2y \right). \] Simplifying this expression: \[ y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = y^2 + x^2 = x^2 + y^2. \] Thus, the correct answer is (A). Final Answer: (A) \( y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = x^2 + y^2 \).