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P4 2023 Oct Q5

A Level / Edexcel / P4

IAL 2023 Oct Paper · Question 5

题目

Problem

Figure 2 shows a sketch of the curve CC with equation

y3−x2+4x2y=ky^3-x^2+4x^2y=k

where kk is a positive constant greater than 11

(a) Find

dydx\frac{\mathrm{d}y}{\mathrm{d}x}

in terms of xx and yy.

(5)

The point PP lies on CC.

Given that the normal to CC at PP has equation y=xy=x, as shown in Figure 2,

(b) find the value of kk.

(5)
题目中文翻译

图 2 给出了曲线 CC 的示意图,其方程为

y3−x2+4x2y=ky^3-x^2+4x^2y=k

其中 kk 是一个大于 11 的正常数。

(a) 求 dydx\frac{\mathrm{d}y}{\mathrm{d}x},用 xx 和 yy 表示。

(5)

点 PP 在曲线 CC 上。

已知曲线 CC 在点 PP 处的法线方程为 y=xy=x,如图 2 所示,

(b) 求 kk 的值。

(5)

解答

(a)

解法一

思路

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对曲线方程关于 xx 隐式求导,其中 4x2y4x^2y 要用乘积法则。用 g=dy/dxg=\mathrm{d}y/\mathrm{d}x 暂记导数,先收集含 gg 的项,最后恢复导数记号。

答题过程

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Let

g=dydx.g=\frac{\mathrm{d}y}{\mathrm{d}x}.

Differentiate each term with respect to xx:

ddx(y3)=3y2g,\frac{\mathrm{d}}{\mathrm{d}x}(y^3)=3y^2g, ddx(−x2)=−2x,\frac{\mathrm{d}}{\mathrm{d}x}(-x^2)=-2x,

and, by the product rule,

ddx(4x2y)=8xy+4x2g.\frac{\mathrm{d}}{\mathrm{d}x}(4x^2y)=8xy+4x^2g.

Since kk is constant, differentiating the curve equation gives

3y2g−2x+8xy+4x2g=0.3y^2g-2x+8xy+4x^2g=0.

Collecting the terms in gg,

(3y2+4x2)g=2x−8xy.(3y^2+4x^2)g=2x-8xy.

Hence

dydx=2x−8xy3y2+4x2.\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{2x-8xy}{3y^2+4x^2}}.

(b)

解法一

思路

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法线 y=xy=x 的斜率是 11,因此切线斜率为 −1-1。点 PP 也在该法线上,将 y=xy=x 代入导数式,再以切线斜率为 −1-1 求坐标,最后代回曲线求 kk。

答题过程

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The normal y=xy=x has gradient 11, so the tangent at PP has gradient

dydx=−1.\frac{\mathrm{d}y}{\mathrm{d}x}=-1.

Since PP lies on the normal, y=xy=x at PP. Substituting in part (a),

−1=2x−8x27x2.-1=\frac{2x-8x^2}{7x^2}.

As k>1k>1, the origin is not on CC. Since y=xy=x, we must have x≠0x\ne0 at PP.

Therefore

−7x2= 2x−8x2,x2−2x= 0,x(x−2)= 0.\begin{align*} -7x^2=&\,2x-8x^2,\\ x^2-2x=&\,0,\\ x(x-2)=&\,0. \end{align*}

Rejecting x=0x=0 gives x=2x=2. Hence y=2y=2 and P=(2,2)P=(2,2).

Substituting in the curve equation,

k= 23−22+4(22)(2)= 8−4+32= 36.\begin{align*} k=&\,2^3-2^2+4(2^2)(2)\\ =&\,8-4+32\\ =&\,\boxed{36}. \end{align*}