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谁教我画一下:r=a(1-sinA) 这个函数图像!
人气:147 ℃ 时间:2020-02-05 19:33:39
解答
a = 6;
PolarPlot[a (1 - Sin[A]),{A,0,2*Pi}]能不能叫我下这个:ContourPlot3D[(x^2 + 9 /4 y^2 + z^2 - 1)^3 - x^2 z^3 - 9/80y^2 z^3, {x, -3, 3}, {y, -3, 3}, {z, -3, 3},PlotPoints \[RightArrow] (1, 1), Axes \[RightArrow] True,Lighting \[RightArrow] False ContourStyle \[RightArrow] {RGBColor[1, 0.4, 0.7]}]ContourPlot3D[(x^2 + 9 /4 y^2 + z^2 - 1)^3 - x^2 z^3 - 9/80y^2 z^3, {x, -3, 3}, {y, -3, 3}, {z, -3, 3},PlotPoints->10, Axes->True, Lighting->Automatic,ContourStyle->RGBColor[1, 0.4, 0.7]]
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