📝 题目
2.1.7 设曲线由隐式方程 $\sqrt[3]{{x}^{2}} + \sqrt[3]{{y}^{2}} = \sqrt[3]{{a}^{2}}\left( {a > 0}\right)$ 给出.
(1)求证:曲线的切线被坐标轴所截的长度为一常数；
(2)写出曲线的参数式,利用参数式求导给出上一小题的另一证法.
💡 答案与解析
2.1.7 (1) 用隐函数求导,
$$ \frac{2}{3{x}^{1/3}} + \frac{2}{3{y}^{1/3}}{y}^{\prime } = 0 \Rightarrow {y}^{\prime } = - \frac{{y}^{1/3}}{{x}^{1/3}}. $$
切线在 $x$ 轴上的截距为
$$ u = x - \frac{y}{{y}^{\prime }} = x + {y}^{2/3}{x}^{1/3} = x + \left( {{a}^{2/3} - {x}^{2/3}}\right) {x}^{1/3} = {a}^{2/3}{x}^{1/3}; $$
切线在 $y$ 轴上的截距为
$$ v = y - x{y}^{\prime } = y + {x}^{2/3}{y}^{1/3} = {y}^{1/3}\left( {{x}^{2/3} + {y}^{2/3}}\right) = {a}^{2/3}{y}^{1/3}. $$
于是切线夹在两坐标轴之间的长度为
$$ \sqrt{{u}^{2} + {v}^{2}} = {a}^{2/3}\sqrt{{x}^{2/3} + {y}^{2/3}} = {a}^{2/3}\sqrt{{a}^{2/3}} = a. $$
(2) $x = a{\cos }^{3}t,y = a{\sin }^{3}t$ .