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CIE 9231 2025 Nov Paper 21 Q2

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 21 Paper · Question 2

题目

Problem

The curve CC has parametric equations

x=sinh⁡tandy=t+cosh⁡t.x = \sinh t \quad\text{and}\quad y = t + \cosh t.

(a) Find dydx\frac{\mathrm{d}y}{\mathrm{d}x} in terms of tt.

[2]

(b) Show that

d2ydx2=1−sinh⁡tcosh⁡3t.\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{1 - \sinh t}{\cosh^3 t}.
[4]

(c) Find the Maclaurin’s series for yy in terms of xx up to and including the term in x2x^2.

[2]
题目中文翻译

曲线 CC 的参数方程为

x=sinh⁡t且y=t+cosh⁡t.x = \sinh t \quad\text{且}\quad y = t + \cosh t.

(a) 求用 tt 表示的 dydx\frac{\mathrm{d}y}{\mathrm{d}x}。

(2)

(b) 证明

d2ydx2=1−sinh⁡tcosh⁡3t.\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{1 - \sinh t}{\cosh^3 t}.
(4)

(c) 求 yy 关于 xx 的 Maclaurin 级数,直到并包括 x2x^2 项。

(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

分别对参数 tt 求导,再用链式法则求 dydx\frac{\mathrm{d}y}{\mathrm{d}x}。

答题过程

展开

Since dxdt=cosh⁡t\frac{\mathrm{d}x}{\mathrm{d}t}=\cosh t and dydt=1+sinh⁡t\frac{\mathrm{d}y}{\mathrm{d}t}=1+\sinh t,

dydx=1+sinh⁡tcosh⁡t.\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1+\sinh t}{\cosh t}.

(b)

解法一

思路

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先对 (a) 的商关于 tt 求导,再乘 dtdx\frac{\mathrm{d}t}{\mathrm{d}x}。用 cosh⁡2t−sinh⁡2t=1\cosh^2t-\sinh^2t=1 化简分子。

答题过程

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The quotient rule gives

ddt(dydx)= cosh⁡2t−(1+sinh⁡t)sinh⁡tcosh⁡2t= 1−sinh⁡tcosh⁡2t.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)=&\,\frac{\cosh^2t-(1+\sinh t)\sinh t}{\cosh^2t}\\ =&\,\frac{1-\sinh t}{\cosh^2t}. \end{align*}

Multiplying by dtdx=1cosh⁡t\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{\cosh t} gives

d2ydx2=1−sinh⁡tcosh⁡3t.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{1-\sinh t}{\cosh^3t}.

(c)

解法一

思路

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x=0x=0 对应 t=0t=0。在这里求 yy、y′y' 和 y′′y'',再代入 Maclaurin 展开。

答题过程

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At x=0x=0 we have t=0t=0, so y(0)=1y(0)=1, y′(0)=1y'(0)=1 and y′′(0)=1y''(0)=1. Therefore

y= y(0)+xy′(0)+x22!y′′(0)+O(x3)= 1+x+x22+O(x3).\begin{align*} y=&\,y(0)+xy'(0)+\frac{x^2}{2!}y''(0)+O(x^3)\\ =&\,1+x+\frac{x^2}{2}+O(x^3). \end{align*}