Langsung ke konten utama

Turunan Vektor Basis Satuan Azimutal terhadap Koordinat Azimutal

\[ \partial\hat{\phi}/\partial\phi = (\partial/\partial\phi)(\vec{e}_\phi/|\vec{e}_\phi|). \]
\[ \vec{e}_\phi = \partial\vec{r}/\partial\phi. \]
\[ \vec{r} = \hat{x}r\sin\theta\cos\phi + \hat{y}r\sin\theta\sin\phi + \hat{z}r\cos\theta. \]
\[ \vec{e}_\phi = -\hat{x}r\sin\theta\sin\phi + \hat{y}r\sin\theta\cos\phi. \]
\[ |\vec{e}_\phi| = r\sin\theta. \]
\[ \hat{\phi} = \vec{e}_\phi/|\vec{e}_\phi| = -\hat{x}\sin\phi + \hat{y}\cos\phi. \]
\[ \partial\hat{\phi}/\partial\phi = -(\hat{x}\cos\phi + \hat{y}\sin\phi). \]
\[ \hat{r} = \hat{x}\sin\theta\cos\phi + \hat{y}\sin\theta\sin\phi + \hat{z}\cos\theta. \]
\[ \hat{\theta} =  \vec{e}_\theta/|\vec{e}_\theta|. \]
\[ \vec{e}_\theta = \partial\vec{r}/\partial\theta = \hat{x}r\cos\theta\cos\phi + \hat{y}r\cos\theta\sin\phi - \hat{z}r\sin\theta. \]
\[ |\vec{e}_\theta| = r. \]
\[ \hat{\theta} = \hat{x}\cos\theta\cos\phi + \hat{y}\cos\theta\sin\phi - \hat{z}\sin\theta. \]
\[ \begin{pmatrix}\hat{r} \\ \hat{\theta} \\ \hat{\phi}\end{pmatrix} = \begin{pmatrix}\sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\phi & \cos\phi & 0\end{pmatrix} \begin{pmatrix}\hat{x} \\ \hat{y} \\ \hat{z}\end{pmatrix}. \]
\[ \Delta = \begin{vmatrix}\sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\phi & \cos\phi & 0\end{vmatrix} = 1. \]
\[ \hat{x} = \frac{1}{\Delta}\begin{vmatrix}\hat{r} & \sin\theta\sin\phi & \cos\theta \\ \hat{\theta} & \cos\theta\sin\phi & -\sin\theta \\ \hat{\phi} & \cos\phi & 0\end{vmatrix} = \hat{r}\sin\theta\cos\phi + \hat{\theta}\cos\theta\cos\phi - \hat{\phi}\sin\phi. \]
\[ \hat{y} = \frac{1}{\Delta}\begin{vmatrix}\sin\theta\cos\phi & \hat{r} & \cos\theta \\ \cos\theta\cos\phi & \hat{\theta} & -\sin\theta \\ -\sin\phi & \hat{\phi} & 0\end{vmatrix} = \hat{r}\sin\theta\sin\phi + \hat{\theta}\cos\theta\sin\phi + \hat{\phi}\cos\phi. \]
\[ \partial\hat{\phi}/\partial\phi = -((\hat{r}\sin\theta\cos\phi + \hat{\theta}\cos\theta\cos\phi - \hat{\phi}\sin\phi)\cos\phi + (\hat{r}\sin\theta\sin\phi + \hat{\theta}\cos\theta\sin\phi + \hat{\phi}\cos\phi)\sin\phi). \]
\[ \partial\hat{\phi}/\partial\phi = -(\hat{r}\sin\theta + \hat{\theta}\cos\theta). \]

Komentar

Posting Komentar

Postingan populer dari blog ini

Fungsi Homogen Berderajat Sebarang

Diketahui ada kuantitas  $f:=\sum_{i_1,\dots,i_n=1}^r{M}_{i_1\cdots{i}_n}x_{i_1}\cdots{x}_{i_n}$, sehingga \[ \frac{\partial{f}}{\partial{x_i}}=\sum_{i_1,\dots,i_n}^r\sum_{k=1}^n{M}_{i_1\cdots{i}_n}{x}_{i_1}\cdots{x}_{i_{k-1}}\delta_{ii_k}{x}_{i_{k+1}}\cdots{x}_{i_n}, \] \[ \frac{\partial{f}}{\partial{x_i}}=\sum_{k=1}^n\sum_{i_1,\dots,i_{k-1},i_{k+1},\dots,i_n=1}^r{M}_{i_1\cdots{i}_{k-1}ii_{k+1}\cdots{i}_n}{x}_{i_1}\cdots{x}_{i_{k-1}}{x}_{i_{k+1}}\cdots{x}_{i_n}, \] \[ \sum_{i=1}^r{x_i}\frac{\partial{f}}{\partial{x_i}}=\sum_{k=1}^n\sum_{i,i_1,\dots,i_{k-1},i_{k+1},\dots,i_n=1}^r{M}_{i_1\cdots{i}_{k-1}ii_{k+1}\cdots{i}_n}{x}_{i_1}\cdots{x}_{i_{k-1}}x_i{x}_{i_{k+1}}\cdots{x}_{i_n}. \] Karena ${M}_{i_1\cdots{i}_{k-1}ii_{k+1}\cdots{i}_n}x_i=\sum_{i_k=1}^r\delta_{ii_k}{M}_{i_1\cdots{i}_{k-1}i_ki_{k+1}\cdots{i}_n}x_{i_k}$, maka \[ \sum_{i=1}^r{x_i}\frac{\partial{f}}{\partial{x_i}}=\sum_{k=1}^n\sum_{i,i_1,\dots,i_n=1}^r{M}_{i_1\cdots{i}_n}{x}_{i_1}\cdots{x}_{i_n}\delta_{ii_k}. \] Ka...
Posting Komentar
Baca selengkapnya

Sebuah Kulit Bola dalam Sistem Koordinat Kulit Bola

Sebuah kulit bola memiliki tempat kedudukan \[ S^2(\vec{r}_0, R) := \{\vec{r} \in \mathbb{R}^3 ~|~ |\vec{r} - \vec{r}_0| = R\} \] di mana $\vec{r}_0 \in \mathbb{R}^3$ adalah pusat kulit bola tersebut, dan $R \in \mathbb{R}^+$ adalah jari-jarinya. Tentu saja, \[ \vec{r} := r(\hat{x}\sin\theta\cos\phi + \hat{y}\sin\theta\sin\phi + \hat{z}\cos\theta) \] dan \[ \vec{r}_0 := r_0(\hat{x}\sin\theta_0\cos\phi_0 + \hat{y}\sin\theta_0\sin\phi_0 + \hat{z}\cos\theta_0) \] di mana $r, r_0 \in \mathbb{R}^+\cup\{0\}$, $\theta, \theta_0 \in [0, \pi]$, dan $\phi, \phi_0 \in \{0\}\cup(0, 2\pi)$, serta $\hat{x} := (1, 0, 0)$, $\hat{y} := (0, 1, 0)$, dan $\hat{z} := (0, 0, 1)$. Tentu saja, \[ \vec{r} - \vec{r}_0 = \hat{x}(r\sin\theta\cos\phi - r_0\sin\theta_0\cos\phi_0) \] \[ + \hat{y}(r\sin\theta\sin\phi - r_0\sin\theta_0\sin\phi_0) + \hat{z}(r\cos\theta - r_0\cos\theta_0) \] sehingga \[ |\vec{r} - \vec{r}_0|^2 = r^2\sin^2\theta\cos^2\phi + r_0^2\sin^2\theta_0\cos^2\phi_0 - 2r_0r\sin\theta_...
Posting Komentar
Baca selengkapnya

Menurunkan Hukum Arus Kirchhoff dari Hukum Ampere

Hukum arus Kirchhoff dapat diperoleh dari persamaan kontinyuitas yang diperoleh dari hukum Ampere. Hukum Ampere adalah \[ \nabla\times\vec{H} = \vec{J} + \partial\vec{D}/\partial t. \] Pengambilan divergensi pada kedua ruas persamaan terakhir menghasilkan \[ 0 = \nabla\cdot\vec{J} + \partial\rho/\partial t, \] di mana $\nabla\cdot\vec{D} = \rho$ merupakan hukum Gauss. Persamaan terakhir ini merupakan persamaan kontinyuitas untuk elektromagnetisme. Pengintegralan kedua ruas persamaan kontinyuitas tersebut ke seluruh volume $V$, dengan menerapkan teorema divergensi Gauss, menghasilkan \[ 0 = \oint_{\partial V}\vec{J}\cdot d^2\vec{r} + \frac{dq}{dt}, \] di mana $q = \int_V \rho d^3\vec{r}$ adalah muatan listrik pada $V$. Untuk volume $V$ yang mendekati titik matematis yang berupa simpul, maka $\oint_{\partial V}\vec{J}\cdot d^2\vec{r} = 0$.  Sementara itu $I := dq/dt = \sum_{j=1}^n I_j$ adalah total arus listrik yang melewati titik simpul tersebut, sehingga \[ \sum_{j=1...
Posting Komentar
Baca selengkapnya