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P3 2021 Jan Q10

A Level / Edexcel / P3

IAL 2021 Jan Paper · Question 10

题目

Problem

The curve CC has equation

x=3sec⁡22yx>3, 0<y<π4x=3\sec^2 2y \qquad x>3,\ 0<y<\frac{\pi}{4}

(a) Find dxdy\dfrac{\mathrm{d}x}{\mathrm{d}y} in terms of yy.

(2)

(b) Hence show that

dydx=pqxx−3\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{p}{qx\sqrt{x-3}}

where pp is irrational and qq is an integer, stating the values of pp and qq.

(3)

(c) Find the equation of the normal to CC at the point where y=π12y=\dfrac{\pi}{12}, giving your answer in the form y=mx+cy=mx+c, giving mm and cc as exact irrational numbers.

(5)
题目中文翻译

曲线 CC 的方程为

x=3sec⁡22yx>3, 0<y<π4x=3\sec^2 2y \qquad x>3,\ 0<y<\frac{\pi}{4}

(a) 用 yy 表示 dxdy\dfrac{\mathrm{d}x}{\mathrm{d}y}。

(b) 由此证明

dydx=pqxx−3\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{p}{qx\sqrt{x-3}}

其中 pp 为无理数，qq 为整数，并写出 pp 与 qq 的值。

(c) 求曲线 CC 在 y=π12y=\dfrac{\pi}{12} 处的法线方程。答案写成 y=mx+cy=mx+c 的形式，其中 mm 与 cc 都要写成精确无理数。

解答

(a)

解法一

思路

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把 sec⁡22y\sec^22y 看成 (sec⁡2y)2(\sec2y)^2，连续使用链式法则求 dxdy\frac{\mathrm{d}x}{\mathrm{d}y}。

答题过程

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We have

x=3sec⁡22yx=3\sec^2 2y

Differentiate with respect to yy:

dxdy=3⋅2sec⁡2y⋅ddy(sec⁡2y)\frac{\mathrm{d}x}{\mathrm{d}y} =3\cdot2\sec2y\cdot \frac{\mathrm{d}}{\mathrm{d}y}(\sec2y)

Since

ddy(sec⁡2y)=2sec⁡2ytan⁡2y\frac{\mathrm{d}}{\mathrm{d}y}(\sec2y)=2\sec2y\tan2y

we get

dxdy=3⋅2sec⁡2y(2sec⁡2ytan⁡2y)\frac{\mathrm{d}x}{\mathrm{d}y} =3\cdot2\sec2y(2\sec2y\tan2y)

Therefore

dxdy=12sec⁡22ytan⁡2y\boxed{\frac{\mathrm{d}x}{\mathrm{d}y}=12\sec^2 2y\tan2y}

解法二

思路

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也可以先把正割平方写成余弦的负二次幂，再使用链式法则求导。

答题过程

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Write

x=3(cos⁡2y)−2.x=3(\cos2y)^{-2}.

Then

dxdy= 3(−2)(cos⁡2y)−3(−2sin⁡2y)= 12(cos⁡2y)−3sin⁡2y= 12sec⁡22ytan⁡2y.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}y} =&\,3(-2)(\cos2y)^{-3}(-2\sin2y)\\[4mm] =&\,12(\cos2y)^{-3}\sin2y\\[4mm] =&\,\boxed{12\sec^22y\tan2y}. \end{align*}

(b)

解法一

思路

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由原方程把 sec⁡22y\sec^22y 和 tan⁡2y\tan2y 都写成 xx 的式子，再代入 (a) 并取倒数。角度范围保证平方根取正值。

答题过程

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From

x=3sec⁡22yx=3\sec^2 2y

we have

sec⁡22y=x3\sec^2 2y=\frac{x}{3}

Using

tan⁡22y=sec⁡22y−1\tan^2 2y=\sec^2 2y-1

gives

tan⁡22y=x3−1=x−33\tan^2 2y=\frac{x}{3}-1=\frac{x-3}{3}

Since

0<y<π40<y<\frac{\pi}{4}

we have 0<2y<π20<2y<\dfrac{\pi}{2}, so tan⁡2y>0\tan2y>0. Hence

tan⁡2y=x−33\tan2y=\sqrt{\frac{x-3}{3}}

Using part (a),

dxdy=12sec⁡22ytan⁡2y\frac{\mathrm{d}x}{\mathrm{d}y}=12\sec^2 2y\tan2y

Substitute the expressions in xx:

dxdy= 12(x3)x−33= 4x⋅x−33= 4xx−33\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}y} =&\,12\big(\frac{x}{3}\big)\sqrt{\frac{x-3}{3}}\\[4mm] =&\,4x\cdot\frac{\sqrt{x-3}}{\sqrt3}\\[4mm] =&\,\frac{4x\sqrt{x-3}}{\sqrt3} \end{align*}

Therefore

dydx=34xx−3\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\sqrt3}{4x\sqrt{x-3}}

So

p=3,q=4\boxed{p=\sqrt3,\qquad q=4}

解法二

思路

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先把原式反解成 y=12arccos⁡3/xy=\frac12\arccos\sqrt{3/x}，然后直接对 xx 求导。

答题过程

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Since 0<2y<π20<2y<\frac{\pi}{2},

cos⁡2y=3x,\cos2y=\sqrt{\frac{3}{x}},

and therefore

y=12arccos⁡(3 x−1/2).y=\frac12\arccos\big(\sqrt3\,x^{-1/2}\big).

Differentiating gives

dydx= 12(−11−3/x)(−32x−3/2)= 34xx−3.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac12\left( -\frac{1}{\sqrt{1-3/x}} \right) \left(-\frac{\sqrt3}{2}x^{-3/2}\right)\\[4mm] =&\,\frac{\sqrt3}{4x\sqrt{x-3}}. \end{align*}

Hence

p=3,q=4.\boxed{p=\sqrt3,\qquad q=4}.

(c)

解法一

思路

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先由给定的 yy 求曲线上的 xx 坐标，再用 (b) 求切线斜率；法线斜率是其负倒数。

答题过程

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At

y=π12y=\frac{\pi}{12}

we have

2y=π62y=\frac{\pi}{6}

Therefore

x=3sec⁡2π6x=3\sec^2\frac{\pi}{6}

Since

cos⁡π6=32\cos\frac{\pi}{6}=\frac{\sqrt3}{2}

we get

sec⁡2π6=43\sec^2\frac{\pi}{6}=\frac{4}{3}

Thus

x=3⋅43=4x=3\cdot \frac43=4

So the point is

(4,π12)\left(4,\frac{\pi}{12}\right)

Now find the gradient of the tangent:

dydx=34xx−3\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sqrt3}{4x\sqrt{x-3}}

At x=4x=4,

dydx=34(4)1=316\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\sqrt3}{4(4)\sqrt{1}} =\frac{\sqrt3}{16}

So the gradient of the normal is the negative reciprocal:

m=−163=−1633m=-\frac{16}{\sqrt3}=-\frac{16\sqrt3}{3}

Use

y−y1=m(x−x1)y-y_1=m(x-x_1)

with

(x1,y1)=(4,π12)\left(x_1,y_1\right)=\left(4,\frac{\pi}{12}\right)

So

y−π12=−1633(x−4)y-\frac{\pi}{12}=-\frac{16\sqrt3}{3}(x-4)

Expand:

y=−1633x+6433+π12y=-\frac{16\sqrt3}{3}x+\frac{64\sqrt3}{3}+\frac{\pi}{12}

Hence

y=−1633x+6433+π12\boxed{y=-\frac{16\sqrt3}{3}x+\frac{64\sqrt3}{3}+\frac{\pi}{12}}