Question

11．在同一坐标系中，将曲线y=$\frac{1}{2}$sin3x变为曲线y'=sinx′的伸缩变换是（　　）

• A． $\left\{{\begin{array}{l}{x=3x'}\\{y=\frac{1}{2}y'}\end{array}}\right.$
• B． $\left\{{\begin{array}{l}{x'=3x}\\{y'=\frac{1}{2}y}\end{array}}\right.$
• C． $\left\{{\begin{array}{l}{x=3x'}\\{y=2y'}\end{array}}\right.$
• D． $\left\{{\begin{array}{l}{x'=3x}\\{y'=2y}\end{array}}\right.$

Answer 1

分析 把伸缩变换公式代入变换后的方程，整理化简与变换前方程比较即可得出变换公式．

解答 解：设伸缩变换为$\left\{\begin{array}{l}{x′=λx}\\{y′=μy}\end{array}\right.$，则μy=sinλx，即y=$\frac{1}{μ}$sinλx，
∴$\left\{\begin{array}{l}{λ=3}\\{μ=2}\end{array}\right.$，
∴伸缩变换公式为$\left\{\begin{array}{l}{x′=3x}\\{y′=2y}\end{array}\right.$．
故选D．

点评 本题考查了伸缩变换，属于基础题．