Question

11．在同一坐標(biāo)系中，將曲線y=$\frac{1}{2}$sin3x變?yōu)榍�線y'=sinx′的伸縮變換是（　�。�

• 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 把伸縮變換公式代入變換后的方程，整理化簡與變換前方程比較即可得出變換公式．

解答 解：設(shè)伸縮變換為$\left\{\begin{array}{l}{x′=λx}\\{y′=μy}\end{array}\right.$，則μy=sinλx，即y=$\frac{1}{μ}$sinλx，
∴$\left\{\begin{array}{l}{λ=3}\\{μ=2}\end{array}\right.$，
∴伸縮變換公式為$\left\{\begin{array}{l}{x′=3x}\\{y′=2y}\end{array}\right.$．
故選D．

點(diǎn)評(píng) 本題考查了伸縮變換，屬于基礎(chǔ)題．