The Jacobian of x, y is defined as follows: x = r cos θ, y = r sin θ. What is the Jacobian?
Understand the Problem
The question is asking to find the Jacobian of the functions x and y defined in terms of the independent variables r and θ, specifically x = r cos(θ) and y = r sin(θ).
Answer
The Jacobian is \( r \).
Answer for screen readers
The Jacobian determinant of the functions is ( r ).
Steps to Solve
- Define the Jacobian Matrix
The Jacobian matrix for the functions $x$ and $y$ with respect to the variables $r$ and $\theta$ is defined as:
$$ J = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{bmatrix} $$
- Calculate Partial Derivatives of x
Given $x = r \cos(\theta)$:
- The partial derivative with respect to $r$ is:
$$ \frac{\partial x}{\partial r} = \cos(\theta) $$
- The partial derivative with respect to $\theta$ is:
$$ \frac{\partial x}{\partial \theta} = -r \sin(\theta) $$
- Calculate Partial Derivatives of y
Given $y = r \sin(\theta)$:
- The partial derivative with respect to $r$ is:
$$ \frac{\partial y}{\partial r} = \sin(\theta) $$
- The partial derivative with respect to $\theta$ is:
$$ \frac{\partial y}{\partial \theta} = r \cos(\theta) $$
- Construct the Jacobian Matrix
Substituting the partial derivatives into the Jacobian matrix, we get:
$$ J = \begin{bmatrix} \cos(\theta) & -r \sin(\theta) \ \sin(\theta) & r \cos(\theta) \end{bmatrix} $$
- Calculate the Determinant of the Jacobian
The determinant of the Jacobian matrix $J$ is calculated as follows:
$$ \text{det}(J) = \left(\cos(\theta) \cdot r \cos(\theta)\right) - \left(-r \sin(\theta) \cdot \sin(\theta)\right) $$
This simplifies to:
$$ \text{det}(J) = r \cos^2(\theta) + r \sin^2(\theta) = r(\cos^2(\theta) + \sin^2(\theta)) $$
Using the Pythagorean identity $\cos^2(\theta) + \sin^2(\theta) = 1$, we find:
$$ \text{det}(J) = r $$
The Jacobian determinant of the functions is ( r ).
More Information
The Jacobian determinant indicates how the area changes when transforming coordinates from $(r, \theta)$ to $(x, y)$. In polar coordinates, this determinant being equal to ( r ) shows that the area element in Cartesian coordinates is proportional to ( r , dr , d\theta ).
Tips
- Confusing partial derivatives: Make sure to apply the correct function when taking derivatives, especially with respect to different variables.
- Forgetting the Pythagorean identity: Remember that $\cos^2(\theta) + \sin^2(\theta) = 1$ is fundamental in simplifying the determinant.
