Considering \( x \) and \( y \) as independent variables, find \( \frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y} \) when \( x=e^{2 r} \cos \theta \) \( y=e^{3 r} \sin \theta \)

To find the partial derivatives \( \frac{\partial r}{\partial x} \), \( \frac{\partial r}{\partial y} \), \( \frac{\partial \theta}{\partial x} \), and \( \frac{\partial \theta}{\partial y} \) given the equations \( x = e^{2r} \cos \theta \) and \( y = e^{3r} \sin \theta \), we will use implicit differentiation and the chain rule. ### Step 1: Express \( r \) and \( \theta \) in terms of \( x \) and \( y \) We have the following equations: \[ x = e^{2r} \cos \theta \] \[ y = e^{3r} \sin \theta \] ### Step 2: Differentiate with respect to \( x \) and \( y \) #### Finding \( \frac{\partial r}{\partial x} \) and \( \frac{\partial \theta}{\partial x} \) 1. Differentiate \( x \) with respect to \( r \) and \( \theta \): \[ \frac{\partial x}{\partial r} = 2 e^{2r} \cos \theta \] \[ \frac{\partial x}{\partial \theta} = -e^{2r} \sin \theta \] 2. Differentiate \( y \) with respect to \( r \) and \( \theta \): \[ \frac{\partial y}{\partial r} = 3 e^{3r} \sin \theta \] \[ \frac{\partial y}{\partial \theta} = e^{3r} \cos \theta \] #### Using the chain rule Using the chain rule, we can express the derivatives of \( r \) and \( \theta \) in terms of \( x \) and \( y \): \[ \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{pmatrix} \begin{pmatrix} \frac{\partial r}{\partial x} \\ \frac{\partial \theta}{\partial x} \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \] This gives us the system of equations: \[ \begin{pmatrix} 2 e^{2r} \cos \theta & -e^{2r} \sin \theta \\ 3 e^{3r} \sin \theta & e^{3r} \cos \theta \end{pmatrix} \begin{pmatrix} \frac{\partial r}{\partial x} \\ \frac{\partial \theta}{\partial x} \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \] #### Finding the inverse To solve for \( \frac{\partial r}{\partial x} \) and \( \frac{\partial \theta}{\partial x} \), we need the inverse of the Jacobian matrix: \[ J = \begin{pmatrix} 2 e^{2r} \cos \theta & -e^{2r} \sin \theta \\ 3 e^{3r} \sin \theta & e^{3r} \cos \theta \end{pmatrix} \] The determinant of \( J \) is: \[ \text{det}(J) = (2 e^{2r} \cos \theta)(e^{3r} \cos \theta) - (-e^{2r} \sin \theta)(3 e^{3r} \sin \theta) \] \[ = 2 e^{5r} \cos^2 \theta + 3 e^{5r} \sin^2 \theta = e^{5r} (2 \cos^2 \theta + 3 \sin^2 \theta) \] The inverse of \( J \) is given by: \[ J^{-1} = \frac{1}{\text{det}(J)} \begin{pmatrix} e^{3r} \cos \theta & e^{2r} \sin \theta \\ -3 e^{3r} \sin \theta & 2 e^{2r} \cos \theta \end{pmatrix} \] ### Step 3: Solve for \( \frac{\partial r}{\partial x} \) and \( \frac{\partial \theta