题目
Problem
The curve C has equation
x3+2xy+8y3=−12.
(a) Show that, at the point (−2,−1) on C,
dxdy=−21.
[3]
(b) Find the value of dx2d2y at the point (−2,−1).
[5]
题目中文翻译
曲线 C 的方程为
x3+2xy+8y3=−12
(a) 证明在曲线 C 上的点 (−2,−1) 处,
dxdy=−21
(3)
(b) 求点 (−2,−1) 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
展开
隐式求导,将含 y′ 的项收集在一起,再代入 (−2,−1)。
答题过程
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Differentiating with respect to x gives
3x2+2y+(2x+24y2)y′=0.
At (−2,−1) this is 10+20y′=0, so y′=−21.
(b)
解法一
思路
展开
对 (a) 的整个等式再求一次导数。用乘积法则展开后,把 x,y,y′ 同时代入。
答题过程
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Differentiating the equation from (a), using the product rule on (2x+24y2)y′, gives
6x+2y′+(2+48yy′)y′+(2x+24y2)y′′=0.
Collecting terms gives
6x+4y′+48y(y′)2+(2x+24y2)y′′=0.
At (−2,−1), with y′=−21, this becomes
−12−2−12+20y′′=0.
Hence y′′=1013.