Question

If \(u = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right)\), then the value of \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \) is:

• A. \( \frac{1}{\sqrt{y^2 - x^2}} \)
• B. \( \frac{-1}{y^2 - x^2} \)
• C. \( 0 \)
• D. \( 1 \)

Hint:

When computing \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \), consider using Euler's theorem if the function is homogeneous, or apply the chain rule carefully for each term.

Answer 1

The Correct Option is C

We are given: \[ u = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right) \] Let us compute: \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \] We use the chain rule. First, define: \[ u_1 = \sin^{-1}\left(\frac{x}{y}\right), \quad u_2 = \tan^{-1}\left(\frac{y}{x}\right) \] So, \[ \frac{\partial u_1}{\partial x} = \frac{1}{\sqrt{1 - \left(\frac{x}{y}\right)^2}} \cdot \frac{1}{y} = \frac{1}{\sqrt{y^2 - x^2}}, \quad \frac{\partial u_1}{\partial y} = \frac{-x}{y^2 \sqrt{1 - \left(\frac{x}{y}\right)^2}} = \frac{-x}{y \sqrt{y^2 - x^2}} \] \[ \frac{\partial u_2}{\partial x} = \frac{-y}{x^2 + y^2}, \quad \frac{\partial u_2}{\partial y} = \frac{x}{x^2 + y^2} \] Now summing the total derivatives: \[ \frac{\partial u}{\partial x} = \frac{1}{\sqrt{y^2 - x^2}} - \frac{y}{x^2 + y^2} \] \[ \frac{\partial u}{\partial y} = \frac{-x}{y \sqrt{y^2 - x^2}} + \frac{x}{x^2 + y^2} \] Now compute: \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x\left(\frac{1}{\sqrt{y^2 - x^2}} - \frac{y}{x^2 + y^2}\right) + y\left(\frac{-x}{y \sqrt{y^2 - x^2}} + \frac{x}{x^2 + y^2} \right) \] Simplifying: \[ = \frac{x}{\sqrt{y^2 - x^2}} - \frac{xy}{x^2 + y^2} - \frac{x}{\sqrt{y^2 - x^2}} + \frac{xy}{x^2 + y^2} = 0 \] Hence, \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \]