問題
極座標の平面を考える。
速度$\vec{v}$において$r$方向の速度$v_r$と$\theta$方向の速度$v_\theta$を求めよ。
解答
$v_r,v_\theta$を$v_x,v_y$を用いて表すと、
\begin{align*}
v_r&=v_x\cos\theta+v_y\sin\theta\\
v_\theta&=-v_x\sin\theta+v_y\cos\theta
\end{align*}
となる。
ここで
\begin{align*}
\begin{cases}
x=r\cos\theta\\
y=r\sin\theta
\end{cases}
\end{align*}
であるので
\begin{align*}
v_x=\frac{\diff x}{\diff t}=\frac{\diff}{\diff t}(r\cos\theta)&=\frac{\diff r}{\diff t}\cos\theta+r\frac{\diff}{\diff t}(\cos\theta)\\
&=\frac{\diff r}{\diff t}\cos\theta+r(-\sin\theta)\frac{\diff \theta}{\diff t}\\
&=\frac{\diff r}{\diff t}\cos\theta-r\sin\theta\frac{\diff\theta}{\diff t}\\
v_y=\frac{\diff y}{\diff t}=\frac{\diff}{\diff t}(r\sin\theta)&=\frac{\diff r}{\diff t}\sin\theta+r\frac{\diff}{\diff t}(\sin\theta)\\
&=\frac{\diff r}{\diff t}\sin\theta+r\cos\theta\frac{\diff\theta}{\diff t}
\end{align*}
これらを$v_r,v_\theta$の式に代入すると、
\begin{align*}
v_r&=\left(\frac{\diff r}{\diff t}\cos\theta-r\sin\theta\frac{\diff\theta}{\diff t}\right)\cos\theta+\left(\frac{\diff r}{\diff t}\sin\theta+r\cos\theta\frac{\diff\theta}{\diff t}\right)\sin\theta\\
&=\frac{\diff r}{\diff t}\cos^2\theta-r\cos\theta\sin\theta\frac{\diff\theta}{\diff t}+\frac{\diff r}{\diff t}\sin^2\theta+r\cos\theta\sin\theta\frac{\diff\theta}{\diff t}\\
&=\frac{\diff r}{\diff t}(\cos^2\theta+\sin^2\theta)\\
&=\frac{\diff r}{\diff t}\\
v_\theta&=-\left(\frac{\diff r}{\diff t}\cos\theta-r\sin\theta\frac{\diff\theta}{\diff t}\right)\sin\theta+\left(\frac{\diff r}{\diff t}\sin\theta+r\cos\theta\frac{\diff\theta}{\diff t}\right)\cos\theta\\
&=-\frac{\diff r}{\diff t}\cos\theta\sin\theta-r\sin^2\theta\frac{\diff\theta}{\diff t}+\frac{\diff r}{\diff t}\sin\theta\cos\theta+r\cos^2\theta\frac{\diff\theta}{\diff t}\\
&=r\frac{\diff\theta}{\diff t}(\sin^2\theta+\cos^2\theta)\\
&=r\frac{\diff\theta}{\diff t}
\end{align*}
となる。