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Mathematics for Physics

305 Basic Derivation of Jacobian

Example. Let

\[ \mathbf F(x,y) = \begin{pmatrix} x^2+3y \\ 2x-y^2 \end{pmatrix}. \]

where

  • \(F_x=x^2+3y\) is the first component.
  • \(F_y=2x-y^2\) is the second component.
  • \(x\) and \(y\) are input coordinates.

The Jacobian is

\[ J_{\mathbf F} = \begin{pmatrix} \dfrac{\partial F_x}{\partial x} & \dfrac{\partial F_x}{\partial y} \\ \dfrac{\partial F_y}{\partial x} & \dfrac{\partial F_y}{\partial y} \end{pmatrix}. \]

Calculate each partial derivative:

\[ \frac{\partial F_x}{\partial x} = 2x, \]

\[ \frac{\partial F_x}{\partial y} = 3, \]

\[ \frac{\partial F_y}{\partial x} = 2, \]

\[ \frac{\partial F_y}{\partial y} = -2y. \]

Therefore,

\[ J_{\mathbf F}(x,y) = \begin{pmatrix} 2x & 3 \\ 2 & -2y \end{pmatrix}. \]

Evaluation. Choose the point

\[ (x,y)=(1,2). \]

Substituting these values gives

\[ J_{\mathbf F}(1,2) = \begin{pmatrix} 2 & 3 \\ 2 & -4 \end{pmatrix}. \]

This matrix gives the first-order change in \(\mathbf F\) near \((1,2)\).

Choose a small input change

\[ \Delta\mathbf x = \begin{pmatrix} 0.1 \\ 0.05 \end{pmatrix}. \]

Then

\[ \Delta\mathbf F \approx J_{\mathbf F}(1,2)\Delta\mathbf x. \]

Substituting the values,

\[ \Delta\mathbf F \approx \begin{pmatrix} 2 & 3 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} 0.1 \\ 0.05 \end{pmatrix}. \]

Multiplying gives

\[ \Delta\mathbf F \approx \begin{pmatrix} 2(0.1)+3(0.05) \\ 2(0.1)-4(0.05) \end{pmatrix} = \begin{pmatrix} 0.35 \\ 0 \end{pmatrix}. \]

Thus, the input change

\[ \begin{pmatrix} 0.1 \\ 0.05 \end{pmatrix} \]

produces approximately the output change

\[ \begin{pmatrix} 0.35 \\ 0 \end{pmatrix}. \]

Exact Change. At the original point,

\[ \mathbf F(1,2) = \begin{pmatrix} 7 \\ -2 \end{pmatrix}. \]

At the displaced point,

\[ \mathbf F(1.1,2.05) = \begin{pmatrix} 7.36 \\ -2.0025 \end{pmatrix}. \]

Therefore, the exact change is

\[ \Delta\mathbf F = \begin{pmatrix} 0.36 \\ -0.0025 \end{pmatrix}. \]

The Jacobian approximation was

\[ \Delta\mathbf F \approx \begin{pmatrix} 0.35 \\ 0 \end{pmatrix}. \]

The difference comes from the higher-order terms omitted by the linear approximation.

Jacobian Determinant. At \((1,2)\),

\[ \det J_{\mathbf F} = \begin{vmatrix} 2 & 3 \\ 2 & -4 \end{vmatrix}. \]

Therefore,

\[ \det J_{\mathbf F} = 2(-4)-3(2) = -14. \]

Hence,

\[ |\det J_{\mathbf F}|=14. \]

A sufficiently small area near \((1,2)\) is therefore scaled by approximately a factor of \(14\). The negative sign shows that the transformation reverses orientation.