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FP3 2026 Jan Q5

A Level / Edexcel / FP3

IAL 2026 Jan Paper · Question 5

题目

Problem

5.

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Figure 2 shows a sketch of the curve C defined by the parametric equations

x= cosh⁡2ty= 4sinh⁡t 0⩽t⩽32\begin{align*} x =&\,\cosh2t\\[2mm] y =&\,4\sinh t\\[2mm] &\,0\leqslant t\leqslant\frac32 \end{align*}

(a) Show that

(dxdt)2+(dydt)2= mcosh⁡nt\begin{align*} \Big(\frac{\mathrm{d}x}{\mathrm{d}t}\Big)^2 + \Big(\frac{\mathrm{d}y}{\mathrm{d}t}\Big)^2 =&\, m\cosh^n t \end{align*}

where mm and nn are integers to be determined.

(5)

(b) Use algebraic integration to determine the exact length of C.

Give the answer in the form a+sinh⁡ba + \sinh b where aa and bb are integers to be found.

(3)

(c) Use algebraic integration to determine the exact area of the surface formed when C is rotated through 360° about the x-axis.

Give the answer in simplest form in terms of hyperbolic functions.

(4)
题目中文翻译

在本题中，你必须写出所有解题步骤。

不能依赖计算器技术求解。

图 2 展示了曲线 C 的草图，其参数方程为

x= cosh⁡2ty= 4sinh⁡t 0⩽t⩽32\begin{align*} x =&\,\cosh2t\\[2mm] y =&\,4\sinh t\\[2mm] &\,0\leqslant t\leqslant\frac32 \end{align*}

(a) 证明

(dxdt)2+(dydt)2= mcosh⁡nt\begin{align*} \Big(\frac{\mathrm{d}x}{\mathrm{d}t}\Big)^2 + \Big(\frac{\mathrm{d}y}{\mathrm{d}t}\Big)^2 =&\, m\cosh^n t \end{align*}

其中 mm 和 nn 为待定整数。

(5)

(b) 用代数积分求曲线 C 的精确长度。

答案写成 a+sinh⁡ba + \sinh b 的形式，其中 aa 和 bb 为待求整数。

(3)

(c) 用代数积分求曲线 C 绕 x 轴旋转 360° 所形成曲面的精确面积。

答案用双曲函数表示，并化到最简形式。

(4)

解答

(a)

解法一

思路

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先分别求参数方程关于 tt 的导数。平方相加后，用 sinh⁡2t=2sinh⁡tcosh⁡t\sinh 2t=2\sinh t\cosh t，再以 sinh⁡2t+1=cosh⁡2t\sinh^2t+1=\cosh^2t 合并成单项式，由此读出 m,nm,n。

答题过程

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Differentiating with respect to tt,

dxdt=2sinh⁡2t\frac{\mathrm{d}x}{\mathrm{d}t}=2\sinh 2t

and

dydt=4cosh⁡t.\frac{\mathrm{d}y}{\mathrm{d}t}=4\cosh t.

Hence

(dxdt)2+(dydt)2= 4sinh⁡22t+16cosh⁡2t= 16sinh⁡2tcosh⁡2t+16cosh⁡2t= 16cosh⁡2t(sinh⁡2t+1)= 16cosh⁡4t.\begin{align*} \Big(\frac{\mathrm{d}x}{\mathrm{d}t}\Big)^2 +\Big(\frac{\mathrm{d}y}{\mathrm{d}t}\Big)^2 =&\,4\sinh^2 2t+16\cosh^2t\\ =&\,16\sinh^2t\cosh^2t+16\cosh^2t\\ =&\,16\cosh^2t(\sinh^2t+1)\\ =&\,16\cosh^4t. \end{align*}

Therefore,

m=16,n=4.\boxed{m=16,\qquad n=4}.

(b)

解法一

思路

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参数曲线弧长为 ∫(dx/dt)2+(dy/dt)2 dt\int\sqrt{(\mathrm{d}x/\mathrm{d}t)^2+(\mathrm{d}y/\mathrm{d}t)^2}\,\mathrm{d}t。由 (a)，平方根在给定区间上是 4cosh⁡2t4\cosh^2t；再用二倍角公式 2cosh⁡2t=1+cosh⁡2t2\cosh^2t=1+\cosh2t 化成可直接积分的形式。

答题过程

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Using the result from part (a), the length of CC is

L= ∫03/2(dxdt)2+(dydt)2 dt= ∫03/24cosh⁡2t dt= ∫03/22(1+cosh⁡2t) dt= [2t+sinh⁡2t]03/2= 3+sinh⁡3.\begin{align*} L =&\,\int_0^{3/2} \sqrt{\Big(\frac{\mathrm{d}x}{\mathrm{d}t}\Big)^2 +\Big(\frac{\mathrm{d}y}{\mathrm{d}t}\Big)^2}\,\mathrm{d}t\\ =&\,\int_0^{3/2}4\cosh^2t\,\mathrm{d}t\\ =&\,\int_0^{3/2}2(1+\cosh2t)\,\mathrm{d}t\\ =&\,\Big[2t+\sinh2t\Big]_0^{3/2}\\ =&\,\boxed{3+\sinh3}. \end{align*}

Thus a=3a=3 and b=3b=3.

(c)

解法一

思路

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曲线绕 xx 轴旋转的曲面面积为 2π∫y ds2\pi\int y\,\mathrm{d}s。代入 y=4sinh⁡ty=4\sinh t 和 (a) 得到的 ds/dt=4cosh⁡2t\mathrm{d}s/\mathrm{d}t=4\cosh^2t，被积函数含 sinh⁡tcosh⁡2t\sinh t\cosh^2t，令 u=cosh⁡tu=\cosh t 即可直接积分。

答题过程

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The surface area formed by rotation about the xx-axis is

S=2π∫03/2y(dxdt)2+(dydt)2 dt.S=2\pi\int_0^{3/2} y\sqrt{\Big(\frac{\mathrm{d}x}{\mathrm{d}t}\Big)^2 +\Big(\frac{\mathrm{d}y}{\mathrm{d}t}\Big)^2}\,\mathrm{d}t.

Therefore,

S= 2π∫03/2(4sinh⁡t)(4cosh⁡2t) dt= 32π∫03/2sinh⁡tcosh⁡2t dt= 32π3[cosh⁡3t]03/2= 32π3(cosh⁡332−1).\begin{align*} S =&\,2\pi\int_0^{3/2} (4\sinh t)(4\cosh^2t)\,\mathrm{d}t\\ =&\,32\pi\int_0^{3/2} \sinh t\cosh^2t\,\mathrm{d}t\\ =&\,\frac{32\pi}{3} \Big[\cosh^3t\Big]_0^{3/2}\\ =&\,\boxed{\frac{32\pi}{3} \Big(\cosh^3\frac32-1\Big)}. \end{align*}

解法二

思路

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也可以用三倍角公式积分。先将 sinh⁡tcosh⁡2t\sinh t\cosh^2t 写成 sinh⁡t+sinh⁡3t\sinh t+\sinh^3t，再用 sinh⁡3t=3sinh⁡t+4sinh⁡3t\sinh3t=3\sinh t+4\sinh^3t，把被积函数变成两个可直接积分的双曲函数。

答题过程

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Using y=4sinh⁡ty=4\sinh t and the result from part (a),

S= 32π∫03/2sinh⁡tcosh⁡2t dt= 32π∫03/2(sinh⁡t+sinh⁡3t) dt.\begin{align*} S=&\,32\pi\int_0^{3/2}\sinh t\cosh^2t\,\mathrm{d}t\\[2mm] =&\,32\pi\int_0^{3/2}\big(\sinh t+\sinh^3t\big)\,\mathrm{d}t. \end{align*}

From the triple-angle identity,

sinh⁡3t= sinh⁡3t−3sinh⁡t4,sinh⁡t+sinh⁡3t= 14(sinh⁡t+sinh⁡3t).\begin{align*} \sinh^3t=&\,\frac{\sinh3t-3\sinh t}{4},\\[2mm] \sinh t+\sinh^3t=&\,\frac14\big(\sinh t+\sinh3t\big). \end{align*}

Therefore,

S= 8π∫03/2(sinh⁡t+sinh⁡3t) dt= 8π[cosh⁡t+13cosh⁡3t]03/2= 8π(cosh⁡32+13cosh⁡92−43)= 8π3(3cosh⁡32+cosh⁡92−4).\begin{align*} S=&\,8\pi\int_0^{3/2}\big(\sinh t+\sinh3t\big)\,\mathrm{d}t\\[2mm] =&\,8\pi\Big[\cosh t+\frac13\cosh3t\Big]_0^{3/2}\\[2mm] =&\,8\pi\Big(\cosh\frac32+\frac13\cosh\frac92-\frac43\Big)\\[2mm] =&\,\boxed{\frac{8\pi}{3}\Big(3\cosh\frac32+\cosh\frac92-4\Big)}. \end{align*}

Using cosh⁡3t=4cosh⁡3t−3cosh⁡t\cosh3t=4\cosh^3t-3\cosh t, this is equivalently

S= 8π3(4cosh⁡332−4)= 32π3(cosh⁡332−1).\begin{align*} S=&\,\frac{8\pi}{3}\Big(4\cosh^3\frac32-4\Big)\\[2mm] =&\,\frac{32\pi}{3}\Big(\cosh^3\frac32-1\Big). \end{align*}