拉格朗日乘数法问题请教
请教各位大侠，划线句子怎么理解比较好Theorem 12 is the key to the method of Lagrange multipliers. Suppose that ƒ(x, y, z) and g(x, y, z) are differentiable and that P0 is a point on the surface g(x, y, z) = 0 where ƒ has a local maximum or minimum value relative to its other values on the surface. We assume also that ∇g ≠ 0 at points on the surface g(x, y, z) = 0. Then ƒ takes on a local maximum or minimum at P0 relative to its values on every differentiable curve through P0 on the surface g(x, y, z) = 0. Therefore, ∇ƒ is orthogonal to the tangent vector of every such differentiable curve through P0. So is ∇g, moreover (because ∇g is orthogonal to the level surface g = 0, as we saw in Section 14.5). Therefore, at P0, ∇ƒ is some scalar multiple l of ∇g.
附：theorem 12—the orthogonal Gradient theorem Suppose that ƒ(x, y, z) is differentiable in a region whose interior contains a smooth curve C: r(t) = x(t)i + y(t)j + z(t)k. If P0 is a point on C where ƒ has a local maximum or minimum relative to its values on C, then ∇ƒ is orthogonal to C at P0.
文章出自Thomas' Calculus 13th，第13.8章节
简单点说，就是梯度会垂直于等势面，也就是曲面的法线方向，
当然与经过这个点的所有曲面的切线正交，
其实就是，标量场f 的等势面扫过约束曲面时，会当与曲面相切，
此时的等势面就是一个局部极值，在相切的点，两个曲面的法线方向一致
也就是两个梯度会差一个常数 茉莉素馨 发表于 2019-6-24 17:42
简单点说，就是梯度会垂直于等势面，也就是曲面的法线方向，
当然与经过这个点的所有曲面的切线正交，
Then ƒ takes on a local maximum or minimum at P0 relative to its values on every differentiable curve through P0 on the surface g(x, y, z) = 0。这句话中relative to its values on every differentiable curve through P0 on the surface g(x, y, z) = 0是不是翻译为“在曲面g上，经过p0点的，所有值都等于p0点的值的每个可微分曲线”，大侠能否帮我看看，谢谢
茉莉素馨 发表于 2019-6-24 17:42
简单点说，就是梯度会垂直于等势面，也就是曲面的法线方向，
当然与经过这个点的所有曲面的切线正交，
大侠，我觉得书中给出的应该是“梯度f垂直于等位线，梯度垂直于等位面”，不知道我又没有理解错 惯性矩 发表于 2019-6-24 22:04
大侠，我觉得书中给出的应该是“梯度f垂直于等位线，梯度垂直于等位面”，不知道我又没有理解错 ...
从百度百科取张图，f=C1,C2,...,Cn 都是等势线，三维就是等势面
可以看成f 是一个标量场，梯度方向是垂直于等势线（面）的
而g=0，就是其中一个等势面，在图上表示出来就是L围成的区域
或者说是一个曲面组成的区域，那些相切的点就是满足约束的极值点
本帖最后由 惯性矩 于 2019-6-25 12:59 编辑
茉莉素馨 发表于 2019-6-24 23:13
从百度百科取张图，f=C1,C2,...,Cn 都是等势线，三维就是等势面
可以看成f 是一个标量场，梯度方向是垂 ...
根据定理12，梯度f是垂直于通过P0出的曲线C，梯度g垂直于曲面，推导出梯度g垂直于通过P0出的曲线C，然后P0位于曲面g(x,y,z)=0上，根据这个就可以得出拉格朗日乘数法。
附：The Method of Lagrange Multipliers
Suppose that ƒ(x, y, z) and g(x, y, z) are differentiable and ∇g ≠ 0 when g(x, y, z) = 0. To find
the local maximum and minimum values of ƒ subject to the constraint g(x, y, z) = 0 (if these exist),
find the values of x, y, z, and l that simultaneously satisfy the equations
∇ƒ = a∇gandg(x, y, z) = 0.
For functions of two independent variables, the condition is similar, but without the variable z.
这些就是我理解后所总结的，我刚开始接触这个，先做个标记，以后学习场论再来瞧瞧
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