题目内容

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.$\left\{{\begin{array}{l}{x'=3x}\\{y'=2y}\end{array}}\right.$

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.

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