Question

In Cartesian coordinates, consider the functions \(u(x, y) = \frac{1{2}(x^2 - y^2)\) and \(v(x,y) = xy\). If \((r, \theta)\) are the polar coordinates, the Jacobian determinant \(|\frac{\partial(u,v)}{\partial(r,\theta)}|\) is}

• A. r
• B. \(\frac{1}{r}\)
• C. \(r^2\)
• D. \(r^3\)

Hint:

For transformations involving standard coordinate systems, remembering their Jacobians (e.g., \(\frac{\partial(x,y)}{\partial(r,\theta)} = r\)) can save significant time. The chain rule for Jacobians is a powerful tool to relate transformations indirectly.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem requires calculating the Jacobian determinant of a coordinate transformation from polar coordinates \((r, \theta)\) to a new coordinate system \((u, v)\). We can use the chain rule for Jacobians, which states that \(\frac{\partial(u,v)}{\partial(r,\theta)} = \frac{\partial(u,v)}{\partial(x,y)} \frac{\partial(x,y)}{\partial(r,\theta)}\).
Step 2: Key Formula or Approach:
1. The transformation from polar to Cartesian coordinates is \(x = r\cos\theta\), \(y = r\sin\theta\).
2. The Jacobian determinant for this transformation is \(\frac{\partial(x,y)}{\partial(r,\theta)} = r\).
3. The Jacobian determinant for the transformation from \((x,y)\) to \((u,v)\) is \(\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}\).
Step 3: Detailed Explanation:
First, we calculate the Jacobian determinant \(\frac{\partial(u,v)}{\partial(x,y)}\). The partial derivatives of \(u(x, y) = \frac{1}{2}(x^2 - y^2)\) and \(v(x, y) = xy\) are: \[ \frac{\partial u}{\partial x} = \frac{1}{2}(2x) = x \] \[ \frac{\partial u}{\partial y} = \frac{1}{2}(-2y) = -y \] \[ \frac{\partial v}{\partial x} = y \] \[ \frac{\partial v}{\partial y} = x \] So, the Jacobian determinant is: \[ \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} x & -y \\ y & x \end{vmatrix} = (x)(x) - (-y)(y) = x^2 + y^2 \] Next, we find the Jacobian determinant \(\frac{\partial(x,y)}{\partial(r,\theta)}\). This is a standard result for polar coordinates: \[ \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta - (-r\sin^2\theta) = r(\cos^2\theta + \sin^2\theta) = r \] Now, we apply the chain rule for Jacobians: \[ \frac{\partial(u,v)}{\partial(r,\theta)} = \frac{\partial(u,v)}{\partial(x,y)} \frac{\partial(x,y)}{\partial(r,\theta)} = (x^2 + y^2)(r) \] Finally, we express the result in terms of polar coordinates. We know that \(x^2 + y^2 = r^2\). \[ \frac{\partial(u,v)}{\partial(r,\theta)} = (r^2)(r) = r^3 \] The question asks for the absolute value of this determinant. Since \(r\) (radius) is non-negative, \(r^3\) is also non-negative. \[ \left|\frac{\partial(u,v)}{\partial(r,\theta)}\right| = r^3 \] Step 4: Final Answer:
The Jacobian determinant is \(r^3\). Therefore, option (D) is the correct answer.