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111學測數A第18題

戴忠淵 於 Sunday, January 23, 2022 2:41 PM 發表
利用三角板(30°-60°-90°)轉半圈即可。
pt[x0_]:={{0,0},{x0,Sqrt[4-x0^2]},{x,Sqrt[3-x^2]},{0,0}}/.
Quiet@Solve[{x^2+y^2==3,(x-x0)^2+(y-Sqrt[4-x0^2])^2==1},{x,y}][[2]]

Manipulate[
Plot[{Sqrt[4-x^2],Sqrt[3-x^2]},{x,-2,2},AspectRatio->1/2,
Epilog->{Green,Thickness[0.005],
Line[pt[Sqrt[3+10^(-7)]][[1;;3]]],
Red,PointSize[0.02],Point[pt[Sqrt[3+10^(-7)]]],
Point[pt[z][[1;;3]]],
Blue,Line[pt[z][[1;;2]]],Line[pt[z][[2;;3]]],
Line[pt[z][[3;;4]]]}
],{z,Sqrt[3]+10^(-7),-2}]

Integrate[Sqrt[4-x^2]-Sqrt[3-x^2],{x,0,Sqrt[3]}]

Integrate[Sqrt[4-x^2]-Sqrt[3](2+x),{x,-2,-(3/2)}]+
Integrate[Sqrt[4-x^2]-Sqrt[3-x^2],{x,-(3/2),Sqrt[3]}]//FullSimplify
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讀者回應 ( 1 意見 )

戴老師好久不見~ 看到您針對學測題目寫的程式還真是挺有趣的~ 讚~

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