91影视

Consider the change in coordinates is${}^{\mathbf{f}\mathbf{}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{鈫�}\mathbf{f}\mathbf{}\mathbf{(}{\overline{\mathbf{y}}}_{\mathbf{,}\mathbf{}}\overline{\mathbf{z}}\mathbf{}\mathbf{)}}$.

$$\begin{gathered} \overline{y}=y\cos 蠒+z{\sin }^2蠒 \\ \overline{z}\cos 蠒=-y\sin 蠒+z\cos 蠒 \end{gathered}$$

Here, the variables of the function are y and z. The shifted coordinates are ${}^{\overline{\mathrm{y}}}$and ${}^{\overline{\mathrm{z}}}$are the changed coordinates. The angle of rotation is ${}^{鈭�}$.

Rewrite the equations as,

role="math" localid="1654596205298" $$\begin{gathered} \overline{y}\sin 蠒=y\cos 蠒\sin 蠒+z{\sin }^2蠒 \\ \overline{z}\cos 蠒=-y\sin 蠒\cos 蠒+z{\cos }^2蠒 \end{gathered}$$

Add the equation for the two coordinates as,

$\overline{y}\sin 蠒=y\cos 蠒=(y\cos 蠒\sin 蠒+z{\sin }^2蠒)+(-y\sin 蠒\cos 蠒+z{\cos }^2蠒)=z.......\left(1\right)$

Rewrite the coordinates as,

$$\begin{gathered} \overline{y}\cos 蠒=y{\cos }^2蠒+z\sin 蠒\cos 蠒 \\ \overline{z}\sin 蠒=-y\sin 蠒\cos 蠒+z\cos 蠒\sin 蠒 \end{gathered}$$

Subtract the two equations as,

$\overline{y}\cos 蠒-\overline{z}\sin 蠒=y{\cos }^2蠒+z\sin 蠒\cos 蠒-(-y\sin 蠒\cos 蠒+z\cos 蠒\sin 蠒)=y........\left(2\right)$

Determine the partial derivatives of the equation (1).

$$\begin{gathered} \frac{鈭�z}{鈭�y}=\sin 蠒 \\ \frac{鈭�z}{鈭�z}=\cos 蠒 \end{gathered}$$

Determine the partial derivatives of the equation (2).

$$\begin{gathered} \frac{鈭�y}{鈭�y}=\cos 蠒 \\ \frac{鈭�y}{鈭�z}=-\sin 蠒 \end{gathered}$$

Write the expression for the gradient of the function with respect to y.

${\left(\overline{鈭�f}\right)}_y=\frac{鈭�f}{鈭�y}.\frac{鈭�y}{鈭�\overline{y}}+\frac{鈭�f}{鈭�z}.\frac{鈭�z}{鈭�\overline{y}}$

Write the equation for the gradient in terms of the partial derivative as,

${\left(\overline{鈭�f}\right)}_y={\left(鈭�f\right)}_y\cos 蠒+{\left(鈭�f\right)}_z\sin 蠒$

Write the expression for the gradient in terms of the z.

$$\begin{gathered} {\left(\overline{鈭�f}\right)}_z=\frac{鈭�f}{鈭�z}.\frac{鈭�y}{鈭�\overline{z}}+\frac{鈭�\mathrm{f}}{鈭�z}.\frac{鈭�z}{鈭�\overline{z}} \\ ={\left(\overline{鈭�f}\right)}_y-\sin 蠒+{\left(\overline{鈭�f}\right)}_z\cos 蠒 \end{gathered}$$

Write the expression for the gradient in the matrix form.

$\left(\overline{鈭�f}\right)=\left[\begin{array}{cc}\cos 蠒& \sin 蠒\\ -\sin 蠒& \cos 蠒\end{array}\right]\left(鈭�f\right)$

Thus, the matrix $\left(\overline{鈭�f}\right)=\left[\begin{array}{cc}\cos 蠒& \sin 蠒\\ -\sin 蠒& \cos 蠒\end{array}\right]\left(鈭�f\right)$proves that鈥檚 the $鈭�f$is used to transform the vector under the rotation.