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青青子衿
发表于 2020-9-24 01:35
本帖最后由 青青子衿 于 2023-9-6 21:59 编辑 回复 7# wwdwwd117
不能叫定积分解法，只能说是知道答案后，定积分验证。
而且呢，只能被动验证，而不能编题 ...
wwdwwd117 发表于 2018-9-5 08:55
利用“椭圆积分的加法定理”，就可以编题了。
椭圆积分的加法定理：
\begin{align*}
\boxed{\quad\begin{gather*}
\int_0^u\sqrt{\frac{1-k^2t^2}{1-t^2}}{\mathrm{d}}t+\int_0^v\sqrt{\frac{1-k^{2}t^{2}}{1-t^2}}{\mathrm{d}}t=\int_0^w\sqrt{\frac{1-k^2t^2}{1-t^2}}{\mathrm{d}}t+k^2uvw\\
\\
w\,\,\colon\!=\frac{u\sqrt{\big(1-v^2\big)\big(1-k^2v^2\big)}+v\sqrt{\big(1-u^2\big)\big(1-k^2u^2\big)}}{1-k^2u^2v^2}
\end{gather*}\quad}
\end{align*}
椭圆积分的倍加定理：
\[\boxed{\quad\begin{gather*}
\int_0^{\color{purple}{u}}\sqrt{\frac{1-k^2t^2}{1-t^2}}{\mathrm{d}}t=\int_{\color{purple}{u}}^w\sqrt{\frac{1-k^2t^2}{1-t^2}}{\mathrm{d}}t+k^2u^2w\\
\\
w\,\,\colon\!=\frac{{\color{blue}{2}}u\sqrt{\big(1-u^2\big)\big(1-k^2u^2\big)}}{1-k^2u^4}
\end{gather*}\quad}\]
椭圆\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)介于点\(B(0,b)\)与点\(X_0\left(x_0,b\sqrt{1-\tfrac{{x_0}^2}{a^2}}\,\right)\)之间的弧长\(\operatorname{arc}\left(BX_0\right)\)为
\begin{align*}
\int_0^{x_0}\sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}{\mathrm{d}}x
&=a\int_0^{x_0}\sqrt{\frac{1-\tfrac{e^2x^2}{a^2}}{1-\tfrac{x^2}{a^2}}}{\mathrm{d}}\left(\frac{x}{a}\right)\\
&=a\int_0^\tfrac{x_0}{a}\sqrt{\frac{1-e^2t^2}{1-t^2}}{\mathrm{d}}t
\end{align*}
同理，椭圆\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)介于点\(X_0\left(x_0,b\sqrt{1-\tfrac{{x_0}^2}{a^2}}\,\right)\)与点\(X_1(x_1,b\sqrt{1-\tfrac{{x_1}^2}{a^2}})\)到之间的弧长\(\operatorname{arc}\left(X_0X_1\right)\)为
\begin{align*}
\int_{x_0}^{x_1}\sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}{\mathrm{d}}x
&=a\int_{\tfrac{x_0}{a}}^{\tfrac{x_1}{a}}\sqrt{\frac{1-e^2t^2}{1-t^2}}{\mathrm{d}}t
\end{align*}
其中\(\dfrac{x_1}{a}\,\,\colon\!=\frac{\tfrac{2x_0}{a}\sqrt{\left(1-\tfrac{{x_0}^2}{a^2}\right)\left(1-\tfrac{e^2{x_0}^2}{a^2}\right)}}{1-\tfrac{e^2{x_0}^4}{a^4}}\)
于是
\begin{align*}
\operatorname{arc}\left(BX_0\right)-\operatorname{arc}\left(X_0X_1\right)&=\int_0^{x_0}\sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}{\mathrm{d}}x-\int_{x_0}^{x_1}\sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}{\mathrm{d}}x\\
&=a\int_{0}^{\tfrac{x_0}{a}}\sqrt{\frac{1-e^2t^2}{1-t^2}}{\mathrm{d}}t-a\int_{\tfrac{x_0}{a}}^{\tfrac{x_1}{a}}\sqrt{\frac{1-e^2t^2}{1-t^2}}{\mathrm{d}}t\\
&=ae^2\left(\dfrac{x_0}{a}\right)^2\left(\dfrac{x_1}{a}\right)
\end{align*}
代入\(a=2\),\(e=\dfrac{\sqrt{3}}{2}\),\(x_0=\dfrac{2\sqrt{6}}{3}\)，便可知弧长差为1 |
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kuing
发表于 2018-9-4 22:57
wwdwwd117
发表于 2018-9-5 08:55
