Question

If \( x = r\cos\theta, y = r\sin\theta \) then Match the LIST-I with LIST-II

LIST-I		LIST-II
A.	\( \frac{\partial r}{\partial x} \)	I.	\( \frac{1}{r} \)
B.	\( \frac{\partial r}{\partial y} \)	II.	\( \frac{y}{r} \)
C.	\( \frac{\partial(x,y)}{\partial(r,\theta)} \)	III.	\( \frac{x}{r} \)
D.	\( \frac{\partial(r,\theta)}{\partial(x,y)} \)	IV.	\( r \)

(Note: There is a typo in the question; it should be \( y = r \sin\theta \))

• A. A - III, B - II, C - I, D - IV
• B. A - III, B - II, C - IV, D - I
• C. A - II, B - III, C - IV, D - I
• D. A - II, B - III, C - I, D - IV

Hint:

The Jacobian \(\frac{\partial(x,y)}{\partial(r,\theta)} = r\) is a crucial result used for changing variables in double integrals from Cartesian to polar coordinates: \(dx dy = r dr d\theta\).

Answer 1

The Correct Option is B

Step 1: Establish the relationships between Cartesian and polar coordinates. We are given \(x = r\cos\theta\) and \(y = r\sin\theta\). The inverse relationships are \(r^2 = x^2 + y^2\) and \(\theta = \tan^{-1}(y/x)\).
Step 2: Calculate the partial derivatives. - A. \(\frac{\partial r}{\partial x}\): Differentiate \(r^2 = x^2 + y^2\) with respect to \(x\): \(2r \frac{\partial r}{\partial x} = 2x \implies \frac{\partial r}{\partial x} = \frac{x}{r}\). This matches III. - B. \(\frac{\partial r}{\partial y}\): Differentiate \(r^2 = x^2 + y^2\) with respect to \(y\): \(2r \frac{\partial r}{\partial y} = 2y \implies \frac{\partial r}{\partial y} = \frac{y}{r}\). This matches II.
Step 3: Calculate the Jacobians. - C. \(\frac{\partial(x,y){\partial(r,\theta)}\):} This is the determinant of the Jacobian matrix for the transformation from polar to Cartesian coordinates. \[ J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta - (-r\sin^2\theta) = r(\cos^2\theta + \sin^2\theta) = r \] This matches IV. - D. \(\frac{\partial(r,\theta){\partial(x,y)}\):} This is the inverse of the Jacobian calculated in C. \[ \frac{\partial(r,\theta)}{\partial(x,y)} = \left(\frac{\partial(x,y)}{\partial(r,\theta)}\right)^{-1} = \frac{1}{r} \] This matches I.
Step 4: Formulate the correct matching sequence. The matches are: A\(\rightarrow\)III, B\(\rightarrow\)II, C\(\rightarrow\)IV, D\(\rightarrow\)I. This corresponds to option (2).