方企勤 第二章 一元函数微分学 第9题
📝 题目
例 9 设函数 $y = y\left( x\right)$ 由 $\frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}} = 1$ 确定,求 ${y}^{\prime \prime }$ .
💡 答案解析
解法 1 对隐函数方程两边求一次导, 得
$$ 2\frac{x}{{a}^{2}} + \frac{2}{{b}^{2}}y{y}^{\prime } = 0. \tag{1.6} $$
由此求出 ${y}^{\prime } = - \frac{{b}^{2}x}{{a}^{2}y}$ . 对方程 (1.6) 两边再求一次导,得
$$ \frac{2}{{a}^{2}} + \frac{2}{{b}^{2}}y{y}^{\prime \prime } + \frac{2}{{b}^{2}}{\left( {y}^{\prime }\right) }^{2} = 0. \tag{1.7} $$
用 ${y}^{\prime }$ 代入 (1.7) 式,即可解出
$$ {y}^{\prime \prime } = \frac{1}{2}\frac{{b}^{2}}{y}\left( {-\frac{2}{{a}^{2}} - \frac{2{b}^{2}{x}^{2}}{{a}^{4}{y}^{2}}}\right) = - \frac{{b}^{4}}{{a}^{2}{y}^{3}}\left( {\frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}}}\right) = - \frac{{b}^{4}}{{a}^{2}{y}^{3}}. $$
解法 2 由椭圆的参数方程 $x = a\cos t,y = b\sin t$ 得
$$ \frac{\mathrm{d}y}{\mathrm{\;d}x} = - \frac{b\cos t}{a\sin t}, $$
$$ \frac{{\mathrm{d}}^{2}y}{\mathrm{\;d}{x}^{2}} = \frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\mathrm{d}y}{\mathrm{\;d}x}\right) = \frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\mathrm{d}y}{\mathrm{\;d}x}\right) \cdot \frac{\mathrm{d}t}{\mathrm{\;d}x} = \frac{\frac{\mathrm{d}}{\mathrm{d}t}\left( {-\frac{b\cos t}{a\sin t}}\right) }{\frac{\mathrm{d}x}{\mathrm{\;d}t}} $$
$$ = - \frac{b}{{a}^{2}{\sin }^{3}t} = - \frac{{b}^{4}}{{a}^{2}{y}^{3}}. $$