题目
Problem
The curve C has equation
4y2+4ln(xy)=1.
(a) Show that, at the point (2,21) on C,
dxdy=−61.
[3]
(b) Find the value of dx2d2y at the point (2,21).
[4]
题目中文翻译
曲线 C 的方程为
4y2+4ln(xy)=1
(a) 证明在 C 上的点 (2,21) 处,
dxdy=−61
(3)
(b) 求点 (2,21) 处 dx2d2y 的值。
(4)
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 ln(xy) 用链式法则。把导数写成 x1+yy′,再代入给定点。
答题过程
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Since xy>0, the logarithm is defined, and
xyy+xy′=x1+yy′.
Differentiating gives
8yy′+4xyy+xy′=(8y+y4)y′+x4=0.
At (2,21), this becomes 12y′+2=0, so y′=−61.
(b)
解法一
思路
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对 (a) 的隐式导数等式再求导。把 y′ 前的括号作为一个整体,用乘积法则。
答题过程
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Differentiating the equation from (a) yields
(8y+y4)y′′+(8−y24)(y′)2−x24=0.
Substituting x=2, y=21 and y′=−61 gives
12y′′=y′′=1+368=911,10811.
解法二
思路
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也可先把 y′ 单独表示成 x,y 的函数,再用乘积法则和链式法则求导。
答题过程
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From (a), y′=−x−1(2y+y−1)−1. Let H=2y+y−1. Then
y′′=x−2H−1+x−1H−2(2−y−2)y′.
At the point, H=3, x=2, 2−y−2=−2 and y′=−61. Therefore
y′′=121+541=10811.