ベクトルの微分

問題

次の式を証明せよ。ただし$\phi$はスカラーとする。

(1) $\frac{\diff}{\diff t}(\phi\vec{A})=\frac{\diff\phi}{\diff t}\vec{A}+\phi\frac{\diff\vec{A}}{\diff t}$

(2) $\frac{\diff}{\diff t}(\vec{A}\cdot\vec{B})=\frac{\diff\vec{A}}{\diff t}\cdot\vec{B}+\vec{A}\cdot\frac{\diff\vec{B}}{\diff t}$

(3) $\frac{\diff}{\diff t}(\vec{A}\times\vec{B})=\frac{\diff\vec{A}}{\diff t}\times\vec{B}+\vec{A}\times\frac{\diff\vec{B}}{\diff t}$

解答

$\vec{A}=(A_x,A_y,A_z)$, $\vec{B}=(B_x,B_y,B_z)$とすると、

(1)
\begin{align*}
\frac{\diff}{\diff t}(\phi\vec{A})&=\left(\frac{\diff}{\diff t}(\phi A_x),\frac{\diff}{\diff t}(\phi A_y),\frac{\diff}{\diff t}(\phi A_z)\right)\\
&=\left(\frac{\diff\phi}{\diff t}A_x+\phi\frac{\diff A_x}{\diff t},\frac{\diff\phi}{\diff t}A_y+\phi\frac{\diff A_y}{\diff t},\frac{\diff\phi}{\diff t}A_z+\phi\frac{\diff A_z}{\diff t}\right)\\
&=\frac{\diff\phi}{\diff t}\vec{A}+\phi\frac{\diff\vec{A}}{\diff t}
\end{align*}

(2)
\begin{align*}
\frac{\diff}{\diff t}(\vec{A}\cdot\vec{B})&=\frac{\diff}{\diff t}(A_xB_x+A_yB_y+A_zB_z)\\
&=\frac{\diff}{\diff t}(A_xB_x)+\frac{\diff}{\diff t}(A_yB_y)+\frac{\diff}{\diff t}(A_zB_z)\\
&=\frac{\diff A_x}{\diff t}B_x+A_x\frac{\diff B_x}{\diff t}+\frac{\diff A_y}{\diff t}B_y+A_y\frac{\diff B_y}{\diff t}+\frac{\diff A_z}{\diff t}B_z+A_z\frac{\diff B_z}{\diff t}\\
&=\frac{\diff A_x}{\diff t}B_x+\frac{\diff A_y}{\diff t}B_y+\frac{\diff A_z}{\diff t}B_z+A_x\frac{\diff B_x}{\diff t}+A_y\frac{\diff B_y}{\diff t}+A_z\frac{\diff B_z}{\diff t}\\
&=\frac{\diff\vec{A}}{\diff t}\cdot\vec{B}+\vec{A}\cdot\frac{\diff\vec{B}}{\diff t}
\end{align*}

(3)
\begin{align*}
\frac{\diff}{\diff t}(\vec{A}\times\vec{B})&=\frac{\diff}{\diff t}(A_yB_z-A_zB_y,A_zB_x-A_xB_z,A_xB_y-A_yB_x)\\
&=\left(\frac{\diff}{\diff t}(A_yB_z-A_zB_y),\frac{\diff}{\diff t}(A_zB_x-A_xB_z),\frac{\diff}{\diff t}(A_xB_y-A_yB_x)\right)\\
\end{align*}
それぞれの成分について計算すると、
\begin{align*}
\frac{\diff}{\diff t}(A_yB_z-A_zB_y)&=\frac{\diff}{\diff t}(A_yB_z)-\frac{\diff}{\diff t}(A_zB_y)\\
&=\frac{\diff A_y}{\diff t}B_z+A_y\frac{\diff B_z}{\diff t}-\left(\frac{\diff A_z}{\diff t}B_y+A_z\frac{\diff B_y}{\diff t}\right)\\
\frac{\diff}{\diff t}(A_zB_x-A_xB_z)&=\frac{\diff}{\diff t}(A_zB_x)-\frac{\diff}{\diff t}(A_xB_z)\\
&=\frac{\diff A_z}{\diff t}B_x+A_z\frac{\diff B_x}{\diff t}-\left(\frac{\diff A_x}{\diff t}B_z+A_x\frac{\diff B_z}{\diff t}\right)\\
\frac{\diff}{\diff t}(A_xB_y-A_yB_x)&=\frac{\diff}{\diff t}(A_xB_y)-\frac{\diff}{\diff t}(A_yB_x)\\
&=\frac{\diff A_x}{\diff t}B_y+A_x\frac{\diff B_y}{\diff t}-\left(\frac{\diff A_y}{\diff t}B_x+A_y\frac{\diff B_x}{\diff t}\right)
\end{align*}
となる。
従って
\begin{align*}
\frac{\diff}{\diff t}(\vec{A}\times\vec{B})&=\left(\frac{\diff A_y}{\diff t}B_z-A_z\frac{\diff B_y}{\diff t}+A_y\frac{\diff B_z}{\diff t}-\frac{\diff A_z}{\diff t}B_y, \\
\frac{\diff A_z}{\diff t}B_x-A_x\frac{\diff B_z}{\diff t}+A_z\frac{\diff B_x}{\diff t}-\frac{\diff A_x}{\diff t}B_z,\\\frac{\diff A_x}{\diff t}B_y-A_y\frac{\diff B_x}{\diff t}+A_x\frac{\diff B_y}{\diff t}-\frac{\diff A_y}{\diff t}B_x\right)\\
&=\frac{\diff\vec{A}}{\diff t}\times\vec{B}+\vec{A}\times\frac{\diff\vec{B}}{\diff t}
\end{align*}

となる。