To show that $J \cdot J' = 1$, where $J = \frac{D(u, v)}{D(x, y)}$, we first need to calculate the Jacobian determinant $J$ and its inverse $J'$.

Given: $u = x + \frac{y^2}{x}$ $v = \frac{y^2}{x}$

Step 1: Calculate the Partial Derivatives

Partial derivatives of $u$:

• $\frac{\partial u}{\partial x} = 1 - \frac{y^2}{x^2}$
• $\frac{\partial u}{\partial y} = \frac{2y}{x}$

Partial derivatives of $v$:

• $\frac{\partial v}{\partial x} = -\frac{y^2}{x^2}$
• $\frac{\partial v}{\partial y} = \frac{2y}{x}$

Step 2: Construct the Jacobian Matrix

The Jacobian matrix $J$ is given by:

\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 - \frac{y^2}{x^2} & \frac{2y}{x} \\ -\frac{y^2}{x^2} & \frac{2y}{x} \end{bmatrix}$$ ### Step 3: Calculate the Determinant $$D(u, v)$$ The determinant $$D(u, v)$$ is calculated as follows: $$D(u, v) = \left( 1 - \frac{y^2}{x^2} \right) \left( \frac{2y}{x} \right) - \left( -\frac{y^2}{x^2} \right) \left( \frac{2y}{x} \right)$$ $$= \frac{2y}{x} \left( 1 - \frac{y^2}{x^2} + \frac{y^2}{x^2} \right)$$ $$= \frac{2y}{x}$$ ### Step 4: Calculate the Jacobian of $$D(x, y)$$ The Jacobian $$D(x, y)$$ is: $$D(x, y) = 1 \cdot 1 - 0 \cdot 0 = 1$$ ### Step 5: Calculate $$J$$ Thus, $$J = \frac{D(u, v)}{D(x, y)} = \frac{\frac{2y}{x}}{1} = \frac{2y}{x}$$ ### Step 6: Calculate the Inverse $$J'$$ The inverse Jacobian $$J'$$ will be the reciprocal: $$J' = \frac{x}{2y}$$ ### Step 7: Verify $$J \cdot J' = 1$$ Now, we check: $$J \cdot J' = \left( \frac{2y}{x} \right) \left( \frac{x}{2y} \right) = 1$$ Thus, we have shown that $$J \cdot J' = 1$$. ### Conclusion The result $$J \cdot J' = 1$$ is confirmed. Do you need further details or explanations? Here are some questions for you: 1. What are the applications of Jacobians in multivariable calculus? 2. How do Jacobians relate to transformations in coordinate systems? 3. Can you provide examples of other functions where you calculate the Jacobian? 4. What are the implications of the determinant of the Jacobian being non-zero? 5. How would you find the Jacobian for a more complex function? **Tip:** Always check if the determinant is zero to ensure that the transformation is valid.