If $x=u+e^{-v}sin~u, y=v+e^{-v}cos~u$, then

Question

Accepted Answer

To determine why Option 2 is the correct answer, we need to analyze the given functions and their partial derivatives. The functions are:

$$ x = u + e^{-v} \sin u $$
$$ y = v + e^{-v} \cos u $$

We are tasked with finding the relationship between the partial derivatives of $ u $ and $ v $ with respect to $ x $ and $ y $.

### Explanation for Option 2: $\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}$

To verify this, we need to consider the inverse relationship between the variables. The given functions can be seen as a transformation from $(u, v)$ to $(x, y)$. To find the partial derivatives, we can use the inverse function theorem, which states that if a transformation is invertible and differentiable, the partial derivatives of the inverse functions are related to the partial derivatives of the original functions.

First, we calculate the Jacobian determinant of the transformation from $(u, v)$ to $(x, y)$:

$$
J = \begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{vmatrix}
$$

Calculating each component:

$$
\frac{\partial x}{\partial u} = 1 + e^{-v} \cos u
$$
$$
\frac{\partial x}{\partial v} = -e^{-v} \sin u
$$
$$
\frac{\partial y}{\partial u} = -e^{-v} \sin u
$$
$$
\frac{\partial y}{\partial v} = 1 - e^{-v} \cos u
$$

The Jacobian determinant $ J $ is:

$$
J = \left(1 + e^{-v} \cos u\right)\left(1 - e^{-v} \cos u\right) - \left(-e^{-v} \sin u\right)\left(-e^{-v} \sin u\right)
$$

Simplifying, we find:

$$
J = 1 - e^{-2v} \cos^2 u - e^{-2v} \sin^2 u = 1 - e^{-2v} (\cos^2 u + \sin^2 u) = 1 - e^{-2v}
$$

Assuming $ J \neq 0 $, the inverse function theorem applies, and we can find the partial derivatives of the inverse transformation:

$$
\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}
$$

This confirms that Option 2 is correct.

### Explanation for why other options are incorrect:

1. **Option 1: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$**

This option suggests a symmetry that does not hold in the context of the given transformation. The inverse function theorem does not support this relationship based on the calculated Jacobian.

3. **Option 3: $\frac{\partial u}{\partial x} = -\frac{\partial v}{\partial y}$**

This option implies an inverse relationship that is not supported by the structure of the transformation and the inverse function theorem.

4. **Option 4: $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$**

This option suggests a negative correlation between the derivatives, which is not consistent with the positive correlation derived from the inverse function theorem.

In conclusion, the correct relationship between the partial derivatives, as supported by the inverse function theorem and the structure of the transformation, is given by Option 2: $\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}$.