r=sin^2 5πθ 求dr/dθ (求导)
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解决时间 2021-02-09 15:22
- 提问者网友:轻浮
- 2021-02-09 05:46
r=sin^2 5πθ 求dr/dθ (求导)
最佳答案
- 五星知识达人网友:风格不统一
- 2021-02-09 06:50
r'=2 sin(5πθ)cos(5πθ)*5π=5π sin(10πθ)
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- 1楼网友:孤老序
- 2021-02-09 07:43
∫[π/6~π/3]dθ∫[2sinθ~4sinθ]r^3dr
=(1/4)∫[π/6→π/3] [64(sinθ)^4-16(sinθ)^4] dθ
=12∫[π/6→π/3] (sinθ)^4 dθ
=3∫[π/6→π/3] (1-cos2θ)^2 dθ
=3∫[π/6→π/3] (1-2cos2θ+cos²2θ) dθ
=3∫[π/6→π/3] [1-2cos2θ+(1/2)(1+cos4θ)] dθ
=3∫[π/6→π/3] [(3/2)-2cos2θ+(1/2)cos4θ] dθ
=3[(3/2)θ - sin2θ + (1/8)sin4θ] |[π/6→π/3]
=3π/4 - 3√3/8
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