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IAL 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{dx}{dy} in terms of yy.

(2)

(b) Hence show that

dydx=pqxx−3\frac{dy}{dx}=\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{dx}{dy}。

(b) 由此证明

dydx=pqxx−3\frac{dy}{dx}=\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)

We have

x=3sec⁡22yx=3\sec^2 2y

Differentiate with respect to yy:

dxdy=3⋅2sec⁡2y⋅ddy(sec⁡2y)\frac{dx}{dy}=3\cdot 2\sec2y\cdot \frac{d}{dy}(\sec2y)

Since

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

we get

dxdy=3⋅2sec⁡2y(2sec⁡2ytan⁡2y)\frac{dx}{dy}=3\cdot 2\sec2y(2\sec2y\tan2y)

Therefore

dxdy=12sec⁡22ytan⁡2y\boxed{\frac{dx}{dy}=12\sec^2 2y\tan2y}

(b)

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{dx}{dy}=12\sec^2 2y\tan2y

Substitute the expressions in xx:

dxdy=12(x3)x−33=4x⋅x−33=4xx−33\begin{aligned} \frac{dx}{dy} &=12\left(\frac{x}{3}\right)\sqrt{\frac{x-3}{3}}\\ &=4x\cdot \frac{\sqrt{x-3}}{\sqrt3}\\ &=\frac{4x\sqrt{x-3}}{\sqrt3} \end{aligned}

Therefore

dydx=34xx−3\frac{dy}{dx} =\frac{\sqrt3}{4x\sqrt{x-3}}

So

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

(c)

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{dy}{dx}=\frac{\sqrt3}{4x\sqrt{x-3}}

At x=4x=4,

dydx=34(4)1=316\frac{dy}{dx}=\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}}