Question

Let \(\theta \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)\). Consider the functions
\[ u : \mathbb{R}^2 - \{ (0, 0) \} \to \mathbb{R} \quad \text{and} \quad v : \mathbb{R}^2 - \{ (0, 0) \} \to \mathbb{R} \]
given by
\[ u(x,y) = x - \frac{x}{x^2 + y^2} \quad \text{and} \quad v(x,y) = y + \frac{y}{x^2 + y^2}. \]
The value of the determinant \(\begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}\) at the point \((\cos \theta, \sin \theta)\) is equal to

• A. \(4 \sin \theta.\)
• B. \(4 \cos \theta.\)
• C. \(4 \sin^2 \theta.\)
• D. \(4 \cos^2 \theta.\)

Answer 1

The Correct Option is D

To find the value of the determinant \( \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \) at the point \((\cos \theta, \sin \theta)\), we must first calculate the partial derivatives of the functions \( u \) and \( v \).

The function \( u(x, y) \) is given by:

\(u(x, y) = x - \frac{x}{x^2 + y^2}\)

The partial derivatives of \( u \) with respect to \( x \) and \( y \) are:

1. \(\frac{\partial u}{\partial x} = 1 - \left( \frac{(x^2 + y^2) - x(2x)}{(x^2 + y^2)^2} \right) = 1 - \frac{y^2 - x^2}{(x^2 + y^2)^2}\)
2. \(\frac{\partial u}{\partial y} = - \frac{x(2y)}{(x^2 + y^2)^2} = - \frac{2xy}{(x^2 + y^2)^2}\)

The function \( v(x, y) \) is given by:

\(v(x, y) = y + \frac{y}{x^2 + y^2}\)

The partial derivatives of \( v \) with respect to \( x \) and \( y \) are:

1. \(\frac{\partial v}{\partial x} = - \frac{y(2x)}{(x^2 + y^2)^2} = - \frac{2xy}{(x^2 + y^2)^2}\)
2. \(\frac{\partial v}{\partial y} = 1 + \frac{(x^2 + y^2) - y(2y)}{(x^2 + y^2)^2} = 1 + \frac{x^2 - y^2}{(x^2 + y^2)^2}\)

The determinant of the Jacobian matrix at the point \((\cos \theta, \sin \theta)\) is:

\(\begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} 1 - \frac{\sin^2 \theta - \cos^2 \theta}{1^2} & - \frac{2 \cos \theta \sin \theta}{1^2} \\ - \frac{2 \cos \theta \sin \theta}{1^2} & 1 + \frac{\cos^2 \theta - \sin^2 \theta}{1^2} \end{vmatrix}\)

Simplifying this determinant:

\(= \begin{vmatrix} \cos 2\theta & -\sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{vmatrix}\)

The determinant of a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is calculated as \(ad - bc\).

Thus, the determinant becomes:

\((\cos 2\theta)(\cos 2\theta) - (-\sin 2\theta)(-\sin 2\theta) = \cos^2 2\theta - \sin^2 2\theta = \cos 4\theta\)

At \((\cos \theta, \sin \theta)\), where \(\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), we have:

\(\cos 4\theta = \cos(2 \times 2 \theta) = \cos(2 \times (\pi/2 - \theta)) = \cos(\pi - 2\theta)\)

Since \(\cos(\pi - 2\theta) = -\cos(2\theta)\), substituting back will yield:

\(4\cos^2 \theta\)

Therefore, the value of the determinant at the given point is \(4 \cos^2 \theta\).