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CIE 9709 2025 November Paper 31 Q7

A Level / CIE / P3

CIE 9709 2025 Nov Paper 31 Paper · Question 7

题目

Problem

The parametric equations of a curve are

x=t2−ln⁡(2t+1),y=t2t+1.x=t^2-\ln(2t+1),\qquad y=\frac{t}{2t+1}.

Obtain a simplified expression for dydx\frac{\mathrm{d}y}{\mathrm{d}x} in terms of tt.

(5)
题目中文翻译

一条曲线的参数方程为

x=t2−ln⁡(2t+1),y=t2t+1.x=t^2-\ln(2t+1),\qquad y=\frac{t}{2t+1}.

求用 tt 表示的 dydx\frac{\mathrm{d}y}{\mathrm{d}x} 的最简表达式。

(5)
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

解法一

思路

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分别对参数求导，再相除。分子先约成一，分母的二次式继续分解。

答题过程

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Differentiating with respect to tt,

dxdt=2t−22t+1=2(2t−1)(t+1)2t+1,\frac{\mathrm{d}x}{\mathrm{d}t} =2t-\frac2{2t+1} =\frac{2(2t-1)(t+1)}{2t+1},

and

dydt=(2t+1)−2t(2t+1)2=1(2t+1)2.\frac{\mathrm{d}y}{\mathrm{d}t} =\frac{(2t+1)-2t}{(2t+1)^2} =\frac1{(2t+1)^2}.

Hence

dydx= dy/dtdx/dt= 1(2t+1)22t+12(2t−1)(t+1)= 12(2t+1)(2t−1)(t+1).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\\ =&\,\frac1{(2t+1)^2} \frac{2t+1}{2(2t-1)(t+1)}\\ =&\,\boxed{\frac1{2(2t+1)(2t-1)(t+1)}}. \end{align*}

The parameter domain is t>−1/2t>-1/2; the expression gives a finite gradient for t≠1/2t\ne1/2.

解法二

思路

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先从 yy 解出参数，把 xx 写成 yy 的函数。求出 dx/dy\mathrm{d}x/\mathrm{d}y 后取倒数，最后换回参数。

答题过程

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From y=t/(2t+1)y=t/(2t+1),

t=y1−2y,2t+1=11−2y.t=\frac{y}{1-2y},\qquad 2t+1=\frac1{1-2y}.

Thus

x=y2(1−2y)2+ln⁡(1−2y).x=\frac{y^2}{(1-2y)^2}+\ln(1-2y).

Differentiating,

dxdy= 2y(1−2y)3−21−2y= 2y−2(1−2y)2(1−2y)3= −2(1−4y)(1−y)(1−2y)3.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}y} =&\,\frac{2y}{(1-2y)^3}-\frac2{1-2y}\\ =&\,\frac{2y-2(1-2y)^2}{(1-2y)^3}\\ =&\,\frac{-2(1-4y)(1-y)}{(1-2y)^3}. \end{align*}

Therefore

dydx=(1−2y)3−2(1−4y)(1−y).\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{(1-2y)^3}{-2(1-4y)(1-y)}.

Substituting

1−2y= 12t+1,1−4y= 1−2t2t+1,1−y= t+12t+1\begin{align*} 1-2y=&\,\frac1{2t+1},\\[4mm] 1-4y=&\,\frac{1-2t}{2t+1},\\[4mm] 1-y=&\,\frac{t+1}{2t+1} \end{align*}

gives

dydx=12(2t+1)(2t−1)(t+1).\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =\frac1{2(2t+1)(2t-1)(t+1)}}.

Here t>−1/2t>-1/2, with t≠1/2t\ne1/2 for a finite gradient.