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CIE 9231 2023 June Paper 23 Q4

A Level / CIE / FP2

CIE 9231 2023 June Paper 23 Paper · Question 4

题目

Problem

The curve CC has equation

4y3+(x+y)6=109.4y^3+(x+y)^6=109.

(a) Show that, at the point (−4,3)(-4,3) on CC, dydx=117\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{17}.

(3)

(b) Find the value of d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} at the point (−4,3)(-4,3).

(5)
题目中文翻译

曲线 CC 的方程为

4y3+(x+y)6=109.4y^3+(x+y)^6=109.

(a) 证明在曲线 CC 上的点 (−4,3)(-4,3) 处,dydx=117\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{17}。

(3)

(b) 求点 (−4,3)(-4,3) 处 d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} 的值。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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对两个幂隐式求导,特别注意 x+yx+y 的导数是 1+y′1+y'。

答题过程

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Differentiating gives

12y2y′+6(x+y)5(1+y′)=0.12y^2y'+6(x+y)^5(1+y')=0.

At (−4,3)(-4,3) this is 108y′−6(1+y′)=0108y'-6(1+y')=0, so y′=1/17y'=1/17.

(b)

解法一

思路

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再对 (a) 求导。将含 y′′y'' 的项合在一起,再代入给定点坐标及已求斜率。

答题过程

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The derivatives of the two terms in (a) are

ddx(12y2y′)= 24y(y′)2+12y2y′′,ddx[6(x+y)5(1+y′)]= 30(x+y)4(1+y′)2+6(x+y)5y′′.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}(12y^2y') =&\,24y(y')^2+12y^2y'',\\ \frac{\mathrm{d}}{\mathrm{d}x}\big[6(x+y)^5(1+y')\big] =&\,30(x+y)^4(1+y')^2+6(x+y)^5y''. \end{align*}

Combining the terms containing y′′y'' yields

 [12y2+6(x+y)5]y′′+24y(y′)2 +30(x+y)4(1+y′)2=0.\begin{align*} &\,\big[12y^2+6(x+y)^5\big]y''+24y(y')^2\\ &\,\hspace{2pt}+30(x+y)^4(1+y')^2=0. \end{align*}

At x=−4x=-4, y=3y=3, y′=1/17y'=1/17, this gives

102y′′+72+30(18)2172=0.102y''+\frac{72+30(18)^2}{17^2}=0.

Since 72+30(18)2=102⋅9672+30(18)^2=102\cdot96, we obtain y′′=−96289\boxed{y''=-\frac{96}{289}}.