“mathematica”计算方程组是什么？
1个回答
展开全部
用t表示θ,
方程为Solve[{a2 Cos[t1 + t2] + a1 Cos[t1] == x0, a2 Sin[t1 + t2] + a1 Sin[t1] == y0}, {t1, t2}]
解得:
{{t2 -> -ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> -ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 - \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 -> -ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(
2 a1 a2)],
t1 -> ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 - \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 -> -ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(
2 a1 a2)],
t1 -> -ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 + \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 -> -ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(
2 a1 a2)],
t1 -> ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 + \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 ->
ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> -ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 - \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 ->
ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 - \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 ->
ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> -ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 + \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 ->
ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 + \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}}
方程为Solve[{a2 Cos[t1 + t2] + a1 Cos[t1] == x0, a2 Sin[t1 + t2] + a1 Sin[t1] == y0}, {t1, t2}]
解得:
{{t2 -> -ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> -ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 - \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 -> -ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(
2 a1 a2)],
t1 -> ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 - \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 -> -ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(
2 a1 a2)],
t1 -> -ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 + \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 -> -ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(
2 a1 a2)],
t1 -> ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 + \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 ->
ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> -ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 - \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 ->
ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 - \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 ->
ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> -ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 + \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}, {t2 ->
ArcCos[(-a1^2 - a2^2 + x0^2 + y0^2)/(2 a1 a2)],
t1 -> ArcCos[(a1^3 x0 - a1 a2^2 x0 + a1 x0^3 +
a1 x0 y0^2 + \[Sqrt](-a1^6 y0^2 + 2 a1^4 a2^2 y0^2 -
a1^2 a2^4 y0^2 + 2 a1^4 x0^2 y0^2 + 2 a1^2 a2^2 x0^2 y0^2 -
a1^2 x0^4 y0^2 + 2 a1^4 y0^4 + 2 a1^2 a2^2 y0^4 -
2 a1^2 x0^2 y0^4 - a1^2 y0^6))/(2 (a1^2 x0^2 +
a1^2 y0^2))]}}
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
