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FP3 2021 Jan Q8

A Level / Edexcel / FP3

IAL 2021 Jan Paper · Question 8

题目

Problem

8.

The curve CC has equation

y=2+ln⁡(1−x2),12≤x≤34.y=2+\ln(1-x^2), \qquad \frac12\le x\le\frac34.

(a) Show that the length of the curve CC is given by

∫1/23/41+x21−x2 dx.\int_{1/2}^{3/4}\frac{1+x^2}{1-x^2}\,\mathrm{d}x.
(4)

(b) Hence, using algebraic integration, show that the length of the curve CC is p+ln⁡qp+\ln q, where pp and qq are rational numbers to be determined.

(5)
题目中文翻译

曲线 CC 的方程为

y=2+ln⁡(1−x2),12≤x≤34.y=2+\ln(1-x^2), \qquad \frac12\le x\le\frac34.

(a) 证明曲线 CC 的长度由下式给出：

∫1/23/41+x21−x2 dx.\int_{1/2}^{3/4}\frac{1+x^2}{1-x^2}\,\mathrm{d}x.
(4)

(b) 因此，使用代数积分证明曲线 CC 的长度为 p+ln⁡qp+\ln q，其中 pp 和 qq 为待定有理数。

(5)

解答

(a)

解法一

思路

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使用 Cartesian 曲线弧长公式。求导后把根号内通分，分子恰好成为完全平方；同时利用给定区间判断 1−x2>01-x^2>0，从而正确处理开方后的绝对值。

答题过程

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Differentiating,

dydx=−2x1−x2.\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{2x}{1-x^2}.

Therefore

1+(dydx)2= 1+4x2(1−x2)2= (1−x2)2+4x2(1−x2)2= 1+2x2+x4(1−x2)2= (1+x2)2(1−x2)2.\begin{align*} 1+\big(\frac{\mathrm{d}y}{\mathrm{d}x}\big)^2 =&\,1+\frac{4x^2}{(1-x^2)^2}\\[4mm] =&\,\frac{(1-x^2)^2+4x^2}{(1-x^2)^2}\\[4mm] =&\,\frac{1+2x^2+x^4}{(1-x^2)^2}\\[4mm] =&\,\frac{(1+x^2)^2}{(1-x^2)^2}. \end{align*}

On 12≤x≤34\frac12\le x\le\frac34, both 1+x21+x^2 and 1−x21-x^2 are positive. Hence

1+(dydx)2=1+x21−x2.\sqrt{1+\big(\frac{\mathrm{d}y}{\mathrm{d}x}\big)^2} =\frac{1+x^2}{1-x^2}.

Thus the length of CC is

∫1/23/41+x21−x2 dx.\boxed{ \int_{1/2}^{3/4}\frac{1+x^2}{1-x^2}\,\mathrm{d}x}.

(b)

解法一

思路

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承接 (a) 的弧长积分，先将 improper fraction 写成 −1+2/(1−x2)-1+2/(1-x^2)。后者可用部分分式或 artanh⁡x\operatorname{artanh}x 积分，再精确代入上下限。

答题过程

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Using the integral from part (a), first write

1+x21−x2=−1+21−x2.\frac{1+x^2}{1-x^2} =-1+\frac{2}{1-x^2}.

Since

21−x2=11−x+11+x,\frac{2}{1-x^2} =\frac{1}{1-x}+\frac{1}{1+x},

an antiderivative is

−x+ln⁡ ⁣(1+x1−x).-x+\ln\!\big(\frac{1+x}{1-x}\big).

Therefore the length is

L= [−x+ln⁡ ⁣(1+x1−x)]1/23/4= (−34+ln⁡7)−(−12+ln⁡3)= −14+ln⁡73.\begin{align*} L =&\,\big[ -x+\ln\!\big(\frac{1+x}{1-x}\big) \big]_{1/2}^{3/4}\\[4mm] =&\,\big(-\frac34+\ln7\big) -\big(-\frac12+\ln3\big)\\[4mm] =&\,-\frac14+\ln\frac73. \end{align*}

Hence

p=−14,q=73.\boxed{p=-\frac14, \qquad q=\frac73}.

解法二

思路

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令 x=tanh⁡θx=\tanh\theta。这样 1−x2=sech⁡2θ1-x^2=\operatorname{sech}^2\theta，积分约去后变成 2−sech⁡2θ2-\operatorname{sech}^2\theta，可直接积分；最后用 artanh⁡x\operatorname{artanh}x 表示新上下限。

答题过程

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Let

x=tanh⁡θ,dx=sech⁡2θ dθ.x=\tanh\theta, \qquad \mathrm{d}x=\operatorname{sech}^2\theta\,\mathrm{d}\theta.

Then

1+x21−x2 dx= 1+tanh⁡2θ1−tanh⁡2θsech⁡2θ dθ= (1+tanh⁡2θ) dθ= (2−sech⁡2θ) dθ.\begin{align*} \frac{1+x^2}{1-x^2}\,\mathrm{d}x =&\,\frac{1+\tanh^2\theta} {1-\tanh^2\theta} \operatorname{sech}^2\theta\,\mathrm{d}\theta\\[4mm] =&\,(1+\tanh^2\theta)\,\mathrm{d}\theta\\[4mm] =&\,(2-\operatorname{sech}^2\theta)\,\mathrm{d}\theta. \end{align*}

Hence an antiderivative is

2θ−tanh⁡θ=2artanh⁡x−x.2\theta-\tanh\theta =2\operatorname{artanh}x-x.

Therefore

L= [2artanh⁡x−x]1/23/4= (ln⁡7−34)−(ln⁡3−12)= −14+ln⁡73.\begin{align*} L =&\,\bigl[2\operatorname{artanh}x-x\bigr]_{1/2}^{3/4}\\[4mm] =&\,\big(\ln7-\frac34\big) -\big(\ln3-\frac12\big)\\[4mm] =&\,\boxed{-\frac14+\ln\frac73}. \end{align*}

Thus p=−14p=-\frac14 and q=73q=\frac73.