题目

Problem

The curve $C$ has equation

y=\frac12\operatorname{arcosh}(2x) \qquad \frac72\le x\le13.

Using calculus, determine the exact length of the curve $C$ . Give your answer in the form $p\sqrt{q}$ , where $p$ and $q$ are constants to be found.

(6)

题目中文翻译

曲线 $C$ 的方程为

y=\frac12\operatorname{arcosh}(2x) \qquad \frac72\le x\le13

利用微积分求曲线 $C$ 的精确长度。答案应写成 $p\sqrt{q}$ 的形式，其中 $p$ 和 $q$ 为待求常数。

(6)

解答

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先求导并代入弧长公式。根式可化简成 $\frac{2x}{\sqrt{4x^2-1}}$ ；由于整个区间内 $x>0$ ，不需要额外的绝对值。最后用代换 $u=4x^2-1$ 积分并代入精确上下限。

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Differentiating,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac12\cdot \frac{2}{\sqrt{(2x)^2-1}} \\ =&\,\frac1{\sqrt{4x^2-1}}. \end{align*}

Since $x>0$ throughout the given interval,

\begin{align*} \sqrt{1+\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2} =&\,\sqrt{1+\frac1{4x^2-1}} \\ =&\,\frac{2x}{\sqrt{4x^2-1}}. \end{align*}

Therefore, the length is

s=\int_{7/2}^{13} \frac{2x}{\sqrt{4x^2-1}}\,\mathrm{d}x.

Let $u=4x^2-1$ , so $\mathrm{d}u=8x\,\mathrm{d}x$ . Then

\begin{align*} s =&\,\frac14\int_{48}^{675}u^{-1/2}\,\mathrm{d}u \\ =&\,\frac12\bigl[u^{1/2}\bigr]_{48}^{675} \\ =&\,\frac12\bigl(15\sqrt3-4\sqrt3\bigr) \\ =&\,\boxed{\frac{11\sqrt3}{2}}. \end{align*}

Thus $p=\frac{11}{2}$ and $q=3$ .

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求出相同的弧长积分后，令 $2x=\cosh u$ 。这样分母变成 $\sinh u$ ，与 $\mathrm{d}x$ 中的因子约去，积分化成 $\frac12\int\cosh u\,\mathrm{d}u$ ；上下限也相应改为反双曲余弦形式。

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As in the first method,

s=\int_{7/2}^{13} \frac{2x}{\sqrt{4x^2-1}}\,\mathrm{d}x.

Let

2x=\cosh u.

Then

\mathrm{d}x=\frac12\sinh u\,\mathrm{d}u, \qquad \sqrt{4x^2-1}=\sinh u.

The new limits are

u=\operatorname{arcosh}7 \quad\text{and}\quad u=\operatorname{arcosh}26.

Hence

\begin{align*} s =&\,\frac12 \int_{\operatorname{arcosh}7}^{\operatorname{arcosh}26} \cosh u\,\mathrm{d}u \\ =&\,\frac12 \bigl[\sinh u\bigr]_{\operatorname{arcosh}7}^{ \operatorname{arcosh}26} \\ =&\,\frac12\bigl(\sqrt{26^2-1}-\sqrt{7^2-1}\bigr) \\ =&\,\boxed{\frac{11\sqrt3}{2}}. \end{align*}