题目
Problem
The curve C has equation
4y3+(x+y)6=109.
(a) Show that, at the point (−4,3) on C, dxdy=171.
(3)
(b) Find the value of dx2d2y at the point (−4,3).
(5)
题目中文翻译
曲线 C 的方程为
4y3+(x+y)6=109.
(a) 证明在曲线 C 上的点 (−4,3) 处,dxdy=171。
(3)
(b) 求点 (−4,3) 处 dx2d2y 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+y 的导数是 1+y′。
答题过程
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Differentiating gives
12y2y′+6(x+y)5(1+y′)=0.
At (−4,3) this is 108y′−6(1+y′)=0, so y′=1/17.
(b)
解法一
思路
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再对 (a) 求导。将含 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
dxd(12y2y′)=dxd[6(x+y)5(1+y′)]=24y(y′)2+12y2y′′,30(x+y)4(1+y′)2+6(x+y)5y′′.
Combining the terms containing y′′ yields
[12y2+6(x+y)5]y′′+24y(y′)2+30(x+y)4(1+y′)2=0.
At x=−4, y=3, y′=1/17, this gives
102y′′+17272+30(18)2=0.
Since 72+30(18)2=102⋅96, we obtain y′′=−28996.