题目
Problem
A curve has equation y=3+cosxsin2x for 0<x<π.
(a) Express dxdy as a simplified fraction in terms of cosx.
(4)
(b) Hence, find an equation of the normal to the curve at the point where x=21π. Give your answer in an exact form.
(2)
题目中文翻译
一条曲线的方程为 y=3+cosxsin2x,其中 0<x<π。
(a) 将 dxdy 表示为用 cosx 表示的最简分式。
(4)
(b) 由此求曲线上 x=21π 对应的点处的法线方程。答案写成精确形式。
(2)
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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使用商法则后,将二倍角的正弦、余弦都换成 sinx、cosx,再消去 sin2x。
答题过程
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The quotient rule gives
dxdy=(3+cosx)22cos2x(3+cosx)+sin2xsinx.
With c=cosx, use cos2x=2c2−1 and sin2xsinx=2(1−c2)c. The numerator becomes
==2(2c2−1)(3+c)+2c(1−c2)12c2+4c3−6−2c+2c−2c32c3+12c2−6.
Therefore
dxdy=(3+cosx)22cos3x+12cos2x−6.
(b)
解法一
思路
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代入 x=π/2 求切线梯度,法线梯度是它的负倒数。原曲线此时经过 y=0。
答题过程
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At x=π/2, y=0 and the tangent gradient is −6/9=−2/3. The normal gradient is 3/2. Thus the normal is
y=23(x−2π).