📝 题目
19．设 $\left\{\begin{array}{l}x=\ln t+\mathrm{e}^{2}, \\ y=\frac{1}{1-t},\end{array}\right.$ 求 $\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{t=\frac{1}{2}}$ ．
💡 答案与解析
[AI解答]
[AI解答] 已知参数方程： $$ \begin{cases} x = \ln t + e^{2}, \\ y = \dfrac{1}{1 - t}, \end{cases} $$ 要求 $\left.\dfrac{\mathrm{d} y}{\mathrm{d} x}\right|_{t=\frac{1}{2}}$。
首先，由参数方程求导公式： $$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\displaystyle\frac{\mathrm{d} y}{\mathrm{d} t}}{\displaystyle\frac{\mathrm{d} x}{\mathrm{d} t}}. $$
分别计算： $$ \frac{\mathrm{d} x}{\mathrm{d} t} = \frac{1}{t}, \quad \frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\mathrm{d}}{\mathrm{d} t} (1-t)^{-1} = (-1)\cdot(-1)(1-t)^{-2} = \frac{1}{(1-t)^{2}}. $$
因此： $$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\displaystyle\frac{1}{(1-t)^{2}}}{\displaystyle\frac{1}{t}} = \frac{t}{(1-t)^{2}}. $$
代入 $t = \frac{1}{2}$： $$ \left.\frac{\mathrm{d} y}{\mathrm{d} x}\right|_{t=\frac{1}{2}} = \frac{\frac{1}{2}}{\left(1 - \frac{1}{2}\right)^{2}} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^{2}} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2. $$
最终结果： $$ \boxed{2} $$
难度：★☆☆☆☆