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用Microsoft Mathematics求曲线的曲率

2012-01-10 16:33:09| 分类：微软数学软件 | 标签： |举报 |字号大中小订阅

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本文转载自calculus《用Microsoft Mathematics求曲线的凹凸区间和拐点》

用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

输入：abs(derivN(ln(sec(x)),x,2))/(1+(deriv(ln(sec(x)),x))^2)^(3/2) 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

先求曲率表达式：abs(derivN(1/x,x,2))/(1+(deriv(1/x,x))^2)^(3/2)

用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

再求x=1处的曲率k：solve({k=2/((1+1/x^4)^3/2(abs(x))^3),x=1}) 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

先求一阶导数：deriv(2sin(t), t)/deriv(cos(t), t) 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

再求二阶导数：deriv((-2)cot(t), t)/deriv(cos(t), t) 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

求曲率表达式：abs((-2)(csc(t))^3)/(1+((-2)cot(t))^2)^3/2 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

求t=pi/2时的曲率：solve({K=2/((abs(sin(t)))^3(4(cot(t))^2+1)^3/2),t=pi/2}) 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

先求一阶导数：deriv(a*sin(t)^3, t)/deriv(a*cos(t)^3, t) 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

再求二阶导数：deriv(-tan(t), t)/deriv(a(cos(t))^3, t) 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

求曲率表达式：abs(csc(t)(sec(t))^4/(3a))/(1+(-tan(t))^2)^3/2 用Microsoft Mathematics求曲线的曲率 - calculus - 徐小湛的博客

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