微积分/弧长/解答

1. 求曲线 $y=x^{\frac {3}{2}}$ 从 $x=0$ 到 $x=1$ 的长度。

:${\begin{aligned}s&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {d}{dx}}{\big (}x^{\frac {3}{2}}{\big )}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+\left({\tfrac {3}{2}}{\sqrt {x}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+{\tfrac {9}{4}}x}}\,dx\end{aligned}}$

令

$u=1+{\frac {9}{4}}x;\quad du={\frac {9}{4}}dx;\quad dx={\frac {4}{9}}du$

则

${\begin{aligned}s&=\int \limits _{u(0)}^{u(1)}{\frac {4}{9}}{\sqrt {u}}\,du\\&={\frac {4}{9}}\cdot {\frac {2}{3}}{\Big [}u^{\frac {3}{2}}{\Big ]}_{u(0)}^{u(1)}\\&={\frac {8}{27}}\left[\left(1+{\frac {9}{4}}x\right)^{\frac {3}{2}}\right]_{0}^{1}\\&={\frac {8}{27}}\left[\left(1+{\frac {9}{4}}\right)^{\frac {3}{2}}-1\right]\\&=\mathbf {\frac {13{\sqrt {13}}-8}{27}} \end{aligned}}$

2. 求曲线 $y={\frac {e^{x}+e^{-x}}{2}}$ 从 $x=0$ 到 $x=1$ 的长度。

:${\begin{aligned}s&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {d}{dx}}\left({\frac {e^{x}+e^{-x}}{2}}\right)\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {e^{x}-e^{-x}}{2}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+{\frac {e^{2x}-2+e^{-2x}}{4}}}}dx\\&=\int \limits _{0}^{1}{\sqrt {\frac {e^{2x}+2+e^{-2x}}{4}}}dx\\&=\int \limits _{0}^{1}{\sqrt {\left({\frac {e^{x}+e^{-x}}{2}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\frac {e^{x}+e^{-x}}{2}}dx\\&={\frac {e^{x}-e^{-x}}{2}}{\bigg |}_{0}^{1}\\&=\mathbf {\frac {e-{\frac {1}{e}}}{2}} \end{aligned}}$

3. 求由参数方程 $x(t)=R\cos(t),y(t)=R\sin(t)$ 给出的圆的周长，其中 $t\in [0,2\pi ]$ 。

:${\begin{aligned}s&=\int \limits _{0}^{2\pi }{\sqrt {\left({\tfrac {d}{dt}}{\big (}R\cos(t){\big )}\right)^{2}+\left({\tfrac {d}{dt}}{\big (}R\sin(t){\big )}\right)^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {{\big (}-R\sin(t){\big )}^{2}+{\big (}R\cos(t){\big )}^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {R^{2}{\big (}\sin ^{2}(t)+\cos ^{2}(t){\big )}}}dt\\&=\int \limits _{0}^{2\pi }Rdt\\&=R\cdot t{\Big |}_{0}^{2\pi }\\&=\mathbf {2\pi R} \end{aligned}}$

4. 求由参数方程 $x(t)=R{\big (}t-\sin(t){\big )},y(t)=R{\big (}1-\cos(t){\big )}$ 给出的摆线的一段弧的长度，其中 $t\in [0,2\pi ]$ 。

:${\begin{aligned}s&=\int \limits _{0}^{2\pi }{\sqrt {\left({\tfrac {d}{dt}}R(t-\sin(t))\right)^{2}+\left({\tfrac {d}{dt}}R(1-\cos(t))\right)^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {{\big (}R(1-\cos(t)){\big )}^{2}+{\big (}R\sin(t){\big )}^{2}}}dt\\&=\int \limits _{0}^{2\pi }R{\sqrt {{\big (}1-\cos(t){\big )}^{2}+\sin ^{2}(t)}}dt\\&=\int \limits _{0}^{2\pi }R{\sqrt {1-2\cos(t)+\cos ^{2}(t)+\sin ^{2}(t)}}dt\\&=\int \limits _{0}^{2\pi }R{\sqrt {2-2\cos(t)}}dt\end{aligned}}$

使用 三角恒等式

$\sin ^{2}\left({\frac {t}{2}}\right)={\frac {1-\cos(t)}{2}}$

，我们有

${\begin{aligned}s&=\int \limits _{0}^{2\pi }R{\sqrt {4\sin ^{2}\left({\frac {t}{2}}\right)}}dt\\&=\int \limits _{0}^{2\pi }2R\sin \left({\frac {t}{2}}\right)dt\\&=-4R\cos \left({\frac {t}{2}}\right){\Bigg |}_{0}^{2\pi }\\&=-4R(-1-1)\\&=\mathbf {8R} \end{aligned}}$