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FP3 2022 Jan Q2

A Level / Edexcel / FP3

IAL 2022 Jan Paper · Question 2

题目

Problem

2.

Figure 1 shows a sketch of the curve CC with parametric equations

x=ln⁡(sec⁡θ+tan⁡θ)−sin⁡θ,y=cos⁡θ,0≤θ≤π4x=\ln(\sec\theta+\tan\theta)-\sin\theta, \qquad y=\cos\theta, \qquad 0\le\theta\le\frac{\pi}{4}

The curve CC is rotated through 2π2\pi radians about the xx-axis and is used to form a solid of revolution SS.

Using calculus, show that the total surface area of SS is given by

π2(p+q2)\frac{\pi}{2}(p+q\sqrt2)

where pp and qq are integers to be determined.

(8)
题目中文翻译

2.

图 1 给出了曲线 CC 的示意图，其参数方程为

x=ln⁡(sec⁡θ+tan⁡θ)−sin⁡θ,y=cos⁡θ,0≤θ≤π4x=\ln(\sec\theta+\tan\theta)-\sin\theta, \qquad y=\cos\theta, \qquad 0\le\theta\le\frac{\pi}{4}

曲线 CC 绕 xx 轴旋转 2π2\pi 弧度，形成旋转体 SS。

利用微积分证明，SS 的总表面积为

π2(p+q2)\frac{\pi}{2}(p+q\sqrt2)

其中 pp 和 qq 为待定整数。

(8)

解答

解法一：使用参数曲线的弧长元素

思路

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先分别求 dxdθ\frac{\mathrm{d}x}{\mathrm{d}\theta} 和 dydθ\frac{\mathrm{d}y}{\mathrm{d}\theta}，化简参数曲线的弧长因子。曲面侧面积由 2π∫y ds2\pi\int y\,\mathrm{d}s 求得；题目问的是总表面积，所以还必须加上两端半径分别为 11 和 12\frac1{\sqrt2} 的圆面面积。

答题过程

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First,

dxdθ= sec⁡θtan⁡θ+sec⁡2θsec⁡θ+tan⁡θ−cos⁡θ= sec⁡θ−cos⁡θ= sin⁡2θcos⁡θ,\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta} =&\,\frac{\sec\theta\tan\theta+\sec^2\theta} {\sec\theta+\tan\theta}-\cos\theta \\ =&\,\sec\theta-\cos\theta \\ =&\,\frac{\sin^2\theta}{\cos\theta}, \end{align*}

and

dydθ=−sin⁡θ.\frac{\mathrm{d}y}{\mathrm{d}\theta}=-\sin\theta.

Since 0≤θ≤π40\le\theta\le\frac\pi4,

(dxdθ)2+(dydθ)2= sin⁡4θcos⁡2θ+sin⁡2θ= sin⁡2θcos⁡2θ= tan⁡θ.\begin{align*} \sqrt{\bigg(\frac{\mathrm{d}x}{\mathrm{d}\theta}\bigg)^2 +\bigg(\frac{\mathrm{d}y}{\mathrm{d}\theta}\bigg)^2} =&\,\sqrt{\frac{\sin^4\theta}{\cos^2\theta} +\sin^2\theta} \\ =&\,\sqrt{\frac{\sin^2\theta}{\cos^2\theta}} \\ =&\,\tan\theta. \end{align*}

Therefore, the curved surface area is

AC= 2π∫0π/4y(dxdθ)2+(dydθ)2 dθ= 2π∫0π/4cos⁡θtan⁡θ dθ= 2π∫0π/4sin⁡θ dθ= 2π[−cos⁡θ]0π/4= 2π(1−12).\begin{align*} A_C =&\,2\pi\int_0^{\pi/4} y\sqrt{\bigg(\frac{\mathrm{d}x}{\mathrm{d}\theta}\bigg)^2 +\bigg(\frac{\mathrm{d}y}{\mathrm{d}\theta}\bigg)^2} \,\mathrm{d}\theta \\ =&\,2\pi\int_0^{\pi/4}\cos\theta\tan\theta \,\mathrm{d}\theta \\ =&\,2\pi\int_0^{\pi/4}\sin\theta \,\mathrm{d}\theta \\ =&\,2\pi\bigl[-\cos\theta\bigr]_0^{\pi/4} \\ =&\,2\pi\bigg(1-\frac1{\sqrt2}\bigg). \end{align*}

The radii of the two circular ends are

y(0)=1,y(π4)=12.y(0)=1, \qquad y\bigg(\frac\pi4\bigg)=\frac1{\sqrt2}.

Hence the total surface area is

A= 2π(1−12)+π(1)2+π(12)2= 7π2−π2= π2(7−22).\begin{align*} A =&\,2\pi\bigg(1-\frac1{\sqrt2}\bigg) +\pi(1)^2+\pi\bigg(\frac1{\sqrt2}\bigg)^2 \\ =&\,\frac{7\pi}{2}-\pi\sqrt2 \\ =&\,\boxed{\frac\pi2\bigl(7-2\sqrt2\bigr)}. \end{align*}

Thus p=7p=7 and q=−2q=-2.

解法二：使用 dydx\frac{\mathrm{d}y}{\mathrm{d}x} 形式

思路

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也可以使用 2π∫y1+(dydx)2 dx2\pi\int y\sqrt{1+(\frac{\mathrm{d}y}{\mathrm{d}x})^2}\,\mathrm{d}x。把 dx\mathrm{d}x 换回参数 θ\theta 后，根式与 dxdθ\frac{\mathrm{d}x}{\mathrm{d}\theta} 的乘积仍化成 tan⁡θ\tan\theta，再补上两个端面。

答题过程

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From

dxdθ=sin⁡2θcos⁡θ\frac{\mathrm{d}x}{\mathrm{d}\theta} =\frac{\sin^2\theta}{\cos\theta}

and

dydθ=−sin⁡θ,\frac{\mathrm{d}y}{\mathrm{d}\theta}=-\sin\theta,

we have

dydx=−cos⁡θsin⁡θ.\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{\cos\theta}{\sin\theta}.

Therefore,

 1+(dydx)2dxdθ= 1+cos⁡2θsin⁡2θsin⁡2θcos⁡θ= 1sin⁡θsin⁡2θcos⁡θ= tan⁡θ.\begin{align*} &\,\sqrt{1+\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2} \frac{\mathrm{d}x}{\mathrm{d}\theta} \\ =&\,\sqrt{1+\frac{\cos^2\theta}{\sin^2\theta}} \frac{\sin^2\theta}{\cos\theta} \\ =&\,\frac1{\sin\theta} \frac{\sin^2\theta}{\cos\theta} \\ =&\,\tan\theta. \end{align*}

Thus the curved surface area is

AC= 2π∫0π/4y1+(dydx)2dxdθ dθ= 2π∫0π/4sin⁡θ dθ= 2π(1−12).\begin{align*} A_C =&\,2\pi\int_0^{\pi/4} y\sqrt{1+\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2} \frac{\mathrm{d}x}{\mathrm{d}\theta}\,\mathrm{d}\theta \\ =&\,2\pi\int_0^{\pi/4}\sin\theta\,\mathrm{d}\theta \\ =&\,2\pi\bigg(1-\frac1{\sqrt2}\bigg). \end{align*}

Adding the two circular ends gives

A= AC+π+π2= π2(7−22).\begin{align*} A =&\,A_C+\pi+\frac\pi2 \\ =&\,\boxed{\frac\pi2\bigl(7-2\sqrt2\bigr)}. \end{align*}

Therefore, p=7p=7 and q=−2q=-2.