📝 题目
例 3 得
$$ {\left( r\mathrm{\;d}\theta \right) }^{2} + {\left( \mathrm{d}r\right) }^{2} = {\left( r \cdot \frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{r}^{2}}\right) }^{2} + {\left( \frac{x\mathrm{\;d}x + y\mathrm{\;d}y}{r}\right) }^{2} $$
$$ = \frac{1}{{r}^{2}}\left\lbrack {{\left( x\mathrm{\;d}y - y\mathrm{\;d}x\right) }^{2} + {\left( x\mathrm{\;d}x + y\mathrm{\;d}y\right) }^{2}}\right\rbrack $$
$$ = {\left( \mathrm{d}x\right) }^{2} + {\left( \mathrm{d}y\right) }^{2}. $$
💡 答案与解析
证法 2 由 $x = r\left( \theta \right) \cos \theta ,y = r\left( \theta \right) \sin \theta \left( {\theta = \theta \left( t\right) }\right)$ ,从而
$$ {\left( \mathrm{d}x\right) }^{2} + {\left( \mathrm{d}y\right) }^{2} = {\left( \cos \theta \mathrm{d}r - r\sin \theta \mathrm{d}\theta \right) }^{2} + {\left( \sin \theta \mathrm{d}r + r\cos \theta \mathrm{d}\theta \right) }^{2} $$
$$ = {\left( r\mathrm{\;d}\theta \right) }^{2} + {\left( \mathrm{d}r\right) }^{2}. $$