题目

Problem

The parametric equations of a curve are

x=3\sin2t,\quad y=\tan t+\cot t,

for $0<t<\frac12\pi$ .

(a) Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-2}{3\sin^22t}$ .

(5)

(b) Find the equation of the normal to the curve at the point where $t=\frac14\pi$ . Give your answer in the form $py+qx+r=0$ , where $p$ , $q$ and $r$ are integers.

(3)

题目中文翻译

一条曲线的参数方程为

x=3\sin2t,\quad y=\tan t+\cot t,

其中 $0<t<\frac12\pi$ 。

(a) 证明 $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-2}{3\sin^22t}$ 。

(5)

(b) 求曲线上 $t=\frac14\pi$ 处的法线方程。答案以 $py+qx+r=0$ 的形式表示，其中 $p$ 、 $q$ 和 $r$ 为整数。

(3)

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

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先做参数微分。约去 $\cos2t$ 的计算只适用于它非零处；指定中点处的导数可由消参后的同一曲线确定。

答题过程

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Differentiating with respect to $t$ ,

\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t}=&\,6\cos2t,\\[4mm] \frac{\mathrm{d}y}{\mathrm{d}t} =&\,\sec^2t-\operatorname{cosec}^2t\\[4mm] =&\,\frac{\sin^2t-\cos^2t}{\sin^2t\cos^2t}\\[4mm] =&\,-\frac{4\cos2t}{\sin^22t}. \end{align*}

For $t\ne\pi/4$ ,

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{-4\cos2t}{6\cos2t\sin^22t} =\boxed{-\frac2{3\sin^22t}}.

At $t=\pi/4$ , both parameter derivatives are zero, so their ratio cannot be evaluated directly. However,

y=\frac1{\sin t\cos t}=\frac2{\sin2t}, \qquad x=3\sin2t,

giving $y=6/x$ . This curve has derivative $-6/x^2$ , which at $x=3$ is $-2/3$ , agreeing with the displayed expression. Thus it also gives the tangent at that endpoint of the traced arc.

解法二

思路

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直接消参最简洁，整个轨迹满足 $xy=6$ ，求导后再换回参数即可。

答题过程

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Since

\tan t+\cot t =\frac{\sin^2t+\cos^2t}{\sin t\cos t} =\frac2{\sin2t},

and $x=3\sin2t$ , we have

y=\frac6x.

Therefore

\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac6{x^2} =-\frac6{9\sin^22t} =\boxed{-\frac2{3\sin^22t}}.

This description also determines the one-sided tangent to the traced arc at $t=\pi/4$ .

(b)

解法一

思路

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在指定参数处使用(a)由曲线方程确定的切线斜率，再取负倒数求法线。

答题过程

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At $t=\pi/4$ , the point is $(3,2)$ and the tangent gradient is $-2/3$ .

Hence the normal gradient is $3/2$ , giving

y-2=\frac32(x-3).

Rearranging into the requested integer-coefficient form,

\boxed{2y-3x+5=0}.

CIE 9709 2024 November Paper 33 Q7

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程