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P4 2022 Oct Q11

A Level / Edexcel / P4

IAL 2022 Oct Paper · Question 11

题目

Problem

Figure 4 shows a sketch of the closed curve with equation

 (x+y)3 +10y2=108x\begin{align*} &\,(x+y)^3\\ &\,\hspace{2pt}+10y^2=108x \end{align*}

(a) Show that

dydx=  108−3(x+y)220y+3(x+y)2\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=&\,\\ &\,\frac{108-3(x+y)^2}{20y+3(x+y)^2} \end{align*}
(5)

The curve is used to model the shape of a cycle track with both xx and yy measured in km.

The points PP and QQ represent points that are furthest north and furthest south of the origin OO, as shown in Figure 4.

Using the result given in part (a),

(b) find how far the point QQ is south of OO. Give your answer to the nearest 100100 m.

(4)
题目中文翻译

图 4 给出了封闭曲线的示意图，其方程为

(x+y)3+10y2=108x(x+y)^3+10y^2=108x

(a) 证明

dydx=  108−3(x+y)220y+3(x+y)2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=&\,\\ &\,\frac{108-3(x+y)^2}{20y+3(x+y)^2}. \end{align*}
(5)

该曲线用来模拟一条自行车赛道的形状，其中 xx 和 yy 的单位都是 km。

如图 4 所示，点 PP 和点 QQ 分别表示相对于原点 OO 最北和最南的位置。

利用 (a) 中给出的结果，

(b) 求点 QQ 在 OO 的正南方多远。答案精确到最近的 100100 m。

(4)

解答

(a)

解法一

思路

展开

对曲线方程关于 xx 隐式求导。(x+y)3(x+y)^3 要同时使用链式法则，并注意 ddx(x+y)=1+dydx\frac{\mathrm{d}}{\mathrm{d}x}(x+y)=1+\frac{\mathrm{d}y}{\mathrm{d}x}。展开后收集所有含 dydx\frac{\mathrm{d}y}{\mathrm{d}x} 的项，即可自然整理出目标式。

答题过程

展开

Write y′=dydxy'=\frac{\mathrm{d}y}{\mathrm{d}x}. Differentiating

(x+y)3+10y2=108x(x+y)^3+10y^2=108x

implicitly with respect to xx gives

 3(x+y)2(1+y′) +20yy′=108.\begin{align*} &\,3(x+y)^2(1+y')\\ &\,\hspace{2pt}+20yy'=108. \end{align*}

Therefore,

 3(x+y)2+3(x+y)2y′ +20yy′=108,\begin{align*} &\,3(x+y)^2 +3(x+y)^2y'\\ &\,\hspace{2pt} +20yy' =108, \end{align*}

Collecting the terms in y′y',

 [20y+3(x+y)2]y′= 108−3(x+y)2.\begin{align*} &\,\big[20y+3(x+y)^2\big]y'\\ =&\,108-3(x+y)^2. \end{align*}

Hence

y′=108−3(x+y)220y+3(x+y)2.\boxed{ y' =\frac{108-3(x+y)^2} {20y+3(x+y)^2} }.

This is the required expression for dydx\frac{\mathrm{d}y}{\mathrm{d}x}.

解法二：展开后逐项求导

思路

展开

先把 (x+y)3(x+y)^3 展开，再对每一项求导。含 xx 和 yy 的乘积要用乘积法则，含 yy 的幂要乘上 y′y'。收集含 y′y' 的项后，把 x2+2xy+y2x^2+2xy+y^2 重新写成 (x+y)2(x+y)^2，即可推出目标式。

答题过程

展开

Starting from the curve equation,

(x+y)3+10y2=108x,(x+y)^3+10y^2=108x,

expand the cubic:

 x3+3x2y +3xy2+y3 +10y2=108x.\begin{align*} &\,x^3+3x^2y\\ &\,\hspace{2pt}+3xy^2+y^3\\ &\,\hspace{4pt}+10y^2=108x. \end{align*}

Write y′=dydxy'=\frac{\mathrm{d}y}{\mathrm{d}x}. Applying the power and product rules,

(x3)′= 3x2,(3x2y)′= 6xy+3x2y′,(3xy2)′= 3y2+6xyy′,(y3)′= 3y2y′,(10y2)′= 20yy′.\begin{align*} (x^3)'=&\,3x^2,\\ (3x^2y)'=&\,6xy+3x^2y',\\ (3xy^2)'=&\,3y^2+6xyy',\\ (y^3)'=&\,3y^2y',\\ (10y^2)'=&\,20yy'. \end{align*}

Therefore,

 3x2+6xy+3x2y′ +3y2+6xyy′ +3y2y′+20yy′=108.\begin{align*} &\,3x^2+6xy+3x^2y'\\ &\,\hspace{2pt}+3y^2+6xyy'\\ &\,\hspace{4pt}+3y^2y'+20yy'=108. \end{align*}

Let

A=3x2+6xy+3y2.A=3x^2+6xy+3y^2.

Collecting the terms containing y′y' gives

(A+20y)y′=108−A.(A+20y)y'=108-A.

Since A=3(x+y)2A=3(x+y)^2,

 [20y+3(x+y)2]y′= 108−3(x+y)2.\begin{align*} &\,\big[20y+3(x+y)^2\big]y'\\ =&\,108-3(x+y)^2. \end{align*}

Hence

y′=108−3(x+y)220y+3(x+y)2.\boxed{y'=\frac{108-3(x+y)^2}{20y+3(x+y)^2}}.

This is the required expression for dydx\frac{\mathrm{d}y}{\mathrm{d}x}.

(b)

解法一

思路

展开

最北点和最南点的切线水平，所以令 dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0，即令 (a) 中分子为零，得到 x+y=±6x+y=\pm6。需要分别检查两个符号是否能在原曲线上产生实点；然后选取负的 yy 坐标，并把距离换算成题目要求的精度。

答题过程

展开

At the northernmost and southernmost points, the tangent is horizontal. Hence

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

so

108−3(x+y)2= 0,(x+y)2= 36.\begin{align*} 108-3(x+y)^2=&\,0,\\ (x+y)^2=&\,36. \end{align*}

Thus

x+y=±6.x+y=\pm6.

If x+y=−6x+y=-6, then x=−6−yx=-6-y. Substitution into the curve gives

5y2+54y+216=0.5y^2+54y+216=0.

Its discriminant is

Δ= 542−4(5)(216)= −1404<0,\begin{align*} \Delta=&\,54^2-4(5)(216)\\ =&\,-1404<0, \end{align*}

so this case has no real points.

Therefore, x+y=6x+y=6, so x=6−yx=6-y. Substituting into the curve,

 63+10y2= 108(6−y).\begin{align*} &\,6^3+10y^2\\ =&\,108(6-y). \end{align*}

This simplifies to

5y2+54y−216=0.5y^2+54y-216=0.

Solving,

Δ= 542+4(5)(216)= 7236.\begin{align*} \Delta=&\,54^2+4(5)(216)\\ =&\,7236. \end{align*}

Thus

y= −54±723610= −27±32015.\begin{align*} y=&\,\frac{-54\pm\sqrt{7236}}{10}\\ =&\,\frac{-27\pm3\sqrt{201}}{5}. \end{align*}

The southernmost point has the negative value

y= −27−32015= −13.906… km.\begin{align*} y=&\,\frac{-27-3\sqrt{201}}{5}\\ =&\,-13.906\ldots\text{ km}. \end{align*}

Therefore, QQ is, to the nearest 100100 m,

13 900 m\boxed{13\,900\text{ m}}

south of OO.

The same substitution can be carried out in terms of xx. With y=6−xy=6-x,

216+10(6−x)2= 108x,5x2−114x+288= 0.\begin{align*} 216+10(6-x)^2=&\,108x,\\ 5x^2-114x+288=&\,0. \end{align*}

Its discriminant is

Δ= 1142−4(5)(288)= 7236.\begin{align*} \Delta=&\,114^2-4(5)(288)\\ =&\,7236. \end{align*}

Since y=6−xy=6-x, the southernmost point corresponds to the larger root:

x= 114+723610= 57+32015.\begin{align*} x=&\,\frac{114+\sqrt{7236}}{10}\\ =&\,\frac{57+3\sqrt{201}}{5}. \end{align*}

Then

y= 6−x= −27−32015,\begin{align*} y=&\,6-x\\ =&\,\frac{-27-3\sqrt{201}}{5}, \end{align*}

giving the same southward distance, 13 90013\,900 m to the nearest 100100 m.