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IAL 2022 June Q4

A Level / Edexcel / P4

IAL 2022 June Paper · Question 4

题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

A curve has equation

16x3−9kx2y+8y3=87516x^3-9kx^2y+8y^3=875

where kk is a constant.

(a) Show that

dydx=6kxy−16x28y2−3kx2\frac{dy}{dx}=\frac{6kxy-16x^2}{8y^2-3kx^2}
(4)

Given that the curve has a turning point at x=52x=\dfrac52

(b) find the value of kk

(4)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

依赖计算器技术的解法不被接受。

一条曲线的方程为

16x3−9kx2y+8y3=87516x^3-9kx^2y+8y^3=875

其中 kk 为常数。

(a) 证明

dydx=6kxy−16x28y2−3kx2\frac{dy}{dx}=\frac{6kxy-16x^2}{8y^2-3kx^2}

。

已知该曲线在 x=52x=\dfrac52 处有一个转折点，

(b) 求 kk 的值。

解答

(a)

解法一

思路

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对曲线方程两边关于 xx 作隐函数求导。处理 x2yx^2y 时使用乘积法则，然后把所有含 dydx\frac{\mathrm{d}y}{\mathrm{d}x} 的项移到同一边并提取公因式。

答题过程

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Differentiating implicitly with respect to xx gives

 48x2−9k(2xy+x2dydx) +24y2dydx=0.\begin{align*} &\,48x^2 -9k\bigg( 2xy+x^2\frac{\mathrm{d}y}{\mathrm{d}x} \bigg)\\ &\,\hspace{2pt} +24y^2\frac{\mathrm{d}y}{\mathrm{d}x} =0. \end{align*}

Therefore,

 48x2−18kxy−9kx2dydx +24y2dydx=0(24y2−9kx2)dydx= 18kxy−48x2.\begin{align*} &\,48x^2-18kxy -9kx^2\frac{\mathrm{d}y}{\mathrm{d}x}\\ &\,\hspace{2pt} +24y^2\frac{\mathrm{d}y}{\mathrm{d}x} =0\\ \big(24y^2-9kx^2\big) \frac{\mathrm{d}y}{\mathrm{d}x} =&\,18kxy-48x^2. \end{align*}

Hence

dydx= 18kxy−48x224y2−9kx2= 6kxy−16x28y2−3kx2,\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{18kxy-48x^2} {24y^2-9kx^2}\\ =&\,\boxed{ \frac{6kxy-16x^2}{8y^2-3kx^2} }, \end{align*}

as required.

(b)

解法一

思路

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转折点处切线水平，因此令 (a) 中导数的分子为零，并代入 x=52x=\frac52，得到 kyky 的值。再把这一关系和 x=52x=\frac52 一同代回原曲线方程，依次求出 yy 与 kk。

答题过程

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At a turning point,

dydx=0.\frac{\mathrm{d}y}{\mathrm{d}x}=0.

Using the numerator of the derivative from part (a), with x=52x=\frac52,

6k(52)y−16(52)2= 015ky−100= 0.\begin{align*} 6k\bigg(\frac52\bigg)y -16\bigg(\frac52\bigg)^2 =&\,0\\ 15ky-100=&\,0. \end{align*}

Thus

ky=203.ky=\frac{20}{3}.

Substituting x=52x=\frac52 and ky=203ky=\frac{20}{3} into the equation of the curve gives

16(52)3−9(52)2(ky)+8y3= 875250−375+8y3= 8758y3= 1000y= 5.\begin{align*} 16\bigg(\frac52\bigg)^3 -9\bigg(\frac52\bigg)^2(ky) +8y^3 =&\,875\\ 250-375+8y^3=&\,875\\ 8y^3=&\,1000\\ y=&\,5. \end{align*}

Therefore,

k=203y=43.k=\frac{20}{3y} =\boxed{\frac43}.

For x=52x=\frac52, y=5y=5 and k=43k=\frac43, the denominator of the derivative is

8y2−3kx2=175≠0,8y^2-3kx^2=175\ne0,

so this value does give a horizontal tangent.