To calculate the Jacobian $\frac{\partial(x, y)}{\partial(r, \theta)}$, we need to compute the partial derivatives of $x$ and $y$ with respect to $r$ and $\theta$. The given relations are:

1. $x = a(u + v)$
2. $y = b(u - v)$
3. $u = R^2 \cos 2\theta$
4. $v = R^2 \sin 2\theta$

Steps to Compute the Jacobian:

The Jacobian is defined as:

\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix}.$$ #### 1. Express $$x$$ and $$y$$ in terms of $$r$$ and $$\theta$$: From the given equations: $$x = a(R^2 \cos 2\theta + R^2 \sin 2\theta),$$ $$y = b(R^2 \cos 2\theta - R^2 \sin 2\theta).$$ #### 2. Differentiate $$x$$ and $$y$$: First, let $$R = r$$ (to simplify notation). $$\frac{\partial x}{\partial r} = a \frac{\partial}{\partial r} \left( r^2 \cos 2\theta + r^2 \sin 2\theta \right) = a(2r \cos 2\theta + 2r \sin 2\theta),$$ $$\frac{\partial x}{\partial \theta} = a \frac{\partial}{\partial \theta} \left( r^2 \cos 2\theta + r^2 \sin 2\theta \right) = a r^2 (-2\sin 2\theta + 2\cos 2\theta).$$ Similarly for $$y$$: $$\frac{\partial y}{\partial r} = b \frac{\partial}{\partial r} \left( r^2 \cos 2\theta - r^2 \sin 2\theta \right) = b(2r \cos 2\theta - 2r \sin 2\theta),$$ $$\frac{\partial y}{\partial \theta} = b \frac{\partial}{\partial \theta} \left( r^2 \cos 2\theta - r^2 \sin 2\theta \right) = b r^2 (-2\sin 2\theta - 2\cos 2\theta).$$ #### 3. Construct the Jacobian matrix: $$J = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{bmatrix} = \begin{bmatrix} a(2r \cos 2\theta + 2r \sin 2\theta) & a r^2 (-2\sin 2\theta + 2\cos 2\theta) \\ b(2r \cos 2\theta - 2r \sin 2\theta) & b r^2 (-2\sin 2\theta - 2\cos 2\theta) \end{bmatrix}.$$ #### 4. Compute the determinant of $$J$$: The determinant is: $$J = \begin{vmatrix} a(2r \cos 2\theta + 2r \sin 2\theta) & a r^2 (-2\sin 2\theta + 2\cos 2\theta) \\ b(2r \cos 2\theta - 2r \sin 2\theta) & b r^2 (-2\sin 2\theta - 2\cos 2\theta) \end{vmatrix}.$$ Expand the determinant: $$J = a b \left[ (2r \cos 2\theta + 2r \sin 2\theta)(-2r^2 \sin 2\theta - 2r^2 \cos 2\theta) - (2r \cos 2\theta - 2r \sin 2\theta)(-2r^2 \sin 2\theta + 2r^2 \cos 2\theta) \right].$$ Simplify: $$J = a b \cdot (-4r^3 (\cos 2\theta + \sin 2\theta)(\sin 2\theta + \cos 2\theta) + 4r^3 (\cos 2\theta - \sin 2\theta)(\sin 2\theta - \cos 2\theta)).$$ Using $$(x + y)(x - y) = x^2 - y^2$$: $$J = a b \cdot (-4r^3 (\cos^2 2\theta - \sin^2 2\theta) + 4r^3 (\cos^2 2\theta - \sin^2 2\theta)).$$ Thus: $$J = 0.$$ ### Final Answer: $$\boxed{J = 0.}$$ Would you like a step-by-step clarification or further details? ### Related Questions: 1. How is the Jacobian determinant useful in coordinate transformations? 2. What happens if $$a \neq b$$ in the given equations? 3. Can you compute the Jacobian determinant if $$u$$ and $$v$$ have different functional forms? 4. How do trigonometric identities simplify the determinant calculation? 5. How does the Jacobian behave under singular transformations? ### Tip: Always verify the functional independence of variables before calculating the Jacobian to ensure the transformation is valid.