Question

Let \( D = \mathbb{R}^2 \setminus \{(0,0)\} \). Consider the two functions \( u,v : D \to \mathbb{R} \) defined by \[ u(x,y) = x^2 - y^2 \quad \text{and} \quad v(x,y) = xy. \] Consider the gradients \(\nabla u\) and \(\nabla v\) of the functions \(u\) and \(v\), respectively. Then

• A. \(\nabla u\) and \(\nabla v\) are parallel at each point \((x,y)\) of \(D.\)
• B. \(\nabla u\) and \(\nabla v\) are perpendicular at each point \((x,y)\) of \(D.\)
• C. \(\nabla u\) and \(\nabla v\) do not exist at some points of \(D.\)
• D. \(\nabla u\) and \(\nabla v\) at each point \((x,y)\) of \(D\) span \(\mathbb{R}^2.\)

Hint:

For functions \(u,v\) of two variables, if \(\nabla u \cdot \nabla v = 0\) everywhere, their level curves intersect orthogonally.

Answer 1

The Correct Option is B, D

Step 1: Compute gradients.
\[ \nabla u = (2x, -2y), \quad \nabla v = (y, x). \]
Step 2: Check dot product.
\[ \nabla u \cdot \nabla v = 2x \cdot y + (-2y) \cdot x = 2xy - 2xy = 0. \] Hence, \(\nabla u\) and \(\nabla v\) are perpendicular at every \((x,y) \ne (0,0).\)
Step 3: Conclusion.
Therefore, the correct option is (B).