题目
Problem
The equation of a curve is x3+y3+2xy+8=0.
(a) Express dxdy in terms of x and y.
(4)
The tangent to the curve at the point where x=0 and the tangent at the point where y=0 intersect at the acute angle α.
(b) Find the exact value of tanα.
(5)
题目中文翻译
一条曲线的方程为 x3+y3+2xy+8=0。
(a) 用 x 和 y 表示 dxdy。
(4)
曲线在 x=0 处的切线与在 y=0 处的切线相交,所成的锐角为 α。
(b) 求 tanα 的精确值。
(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
对含 y 的项用链式法则,对 2xy 用乘积法则,最后收集导数项。
答题过程
展开
Differentiating implicitly,
3x2+3y2dxdy+2y+2xdxdy=0.
Hence
(3y2+2x)dxdy=−(3x2+2y).
Therefore
dxdy=−3y2+2x3x2+2y.
(b)
解法一
思路
展开
先分别求两个坐标轴交点及切线斜率。两条切线的倾角作差,使用正切差角公式得到锐角的正切。
答题过程
展开
At x=0, the curve gives y3+8=0, so the point is (0,−2). From part (a), its tangent gradient is
m1=−12−4=31.
At y=0, x3+8=0, so the point is (−2,0). Its tangent gradient is
m2=−−412=3.
If their inclination angles are β1,β2, then tanβ1=1/3 and tanβ2=3. The acute angle is α=β2−β1. Thus
tanα=1+3(1/3)3−1/3=34.