📝 题目
3．设 $T=2 \pi \sqrt{\frac{l}{g}}$ ，求证 $l \frac{\partial T}{\partial l}+g \frac{\partial T}{\partial g}=0$ ．
💡 答案与解析
[AI解答]
[AI解答]
已知 $$ T = 2\pi \sqrt{\frac{l}{g}} = 2\pi l^{\frac12} g^{-\frac12}. $$
首先求偏导数：
$$ \frac{\partial T}{\partial l} = 2\pi \cdot \frac12 l^{-\frac12} g^{-\frac12} = \pi l^{-\frac12} g^{-\frac12}. $$
$$ \frac{\partial T}{\partial g} = 2\pi l^{\frac12} \cdot \left(-\frac12\right) g^{-\frac32} = -\pi l^{\frac12} g^{-\frac32}. $$
现在计算表达式：
$$ l \frac{\partial T}{\partial l} + g \frac{\partial T}{\partial g} = l \cdot \left( \pi l^{-\frac12} g^{-\frac12} \right) + g \cdot \left( -\pi l^{\frac12} g^{-\frac32} \right). $$
第一项： $$ l \cdot \pi l^{-\frac12} g^{-\frac12} = \pi l^{\frac12} g^{-\frac12}. $$
第二项： $$ g \cdot \left( -\pi l^{\frac12} g^{-\frac32} \right) = -\pi l^{\frac12} g^{-\frac12}. $$
相加得： $$ \pi l^{\frac12} g^{-\frac12} - \pi l^{\frac12} g^{-\frac12} = 0. $$
因此原式成立。
难度：★☆☆☆☆