计算一个矢量的衍生物 [英] Calculate the derivative of a vector

问题描述

我有以下功能（维维安尼的曲线）：

I have the following function (Viviani's curve):

Phi     = @(t)[ cos(t)^2, cos(t)*sin(t), sin(t) ]

只是一个检查，它是有效的：

Just a check that it's valid:

s = linspace(0,T,1000);
plot3(cos(s).^2, cos(s).*sin(s), sin(s));

如何衍生功能披（也许多次），从而重新presents维维安尼的曲线中的一个点 T ，其中 T 从 0 进入 2 * PI ？难道我定义的披适用于这样的衍生品？我试过差异，但它并没有保持披因为我需要它。

How to derivate the function Phi (maybe multiple times), which represents Viviani's curve in a point t where t goes from 0 to 2*pi? Did I defined Phi suitable for such a derivative? I've tried diff, but it did not keep the Phi as I would need it.

如果二阶导数是 Phi_d2 ，我需要得到它的价值（例如 T = 0 ）。

If the second derivative would be Phi_d2, I need to get it's value (for example in t = 0).

我怎样才能做到这一点？

How can I achieve this?

推荐答案

下面是您可以用三种方式实现。第一种使用 潜艇 ，第二次使用 symfun ，第三个使用复杂的一步分化

Here are three ways you can accomplish this. The first uses subs, the second uses a symfun, and the third uses complex step differentiation:

% Using subs
syms t
Phi = [cos(t) cos(t).*sin(t) sin(t)];
Phi_d2 = diff(Phi,t)
double(subs(Phi_d2,t,0))

% Using symfun
syms t
Phi(t) = [cos(t) cos(t).*sin(t) sin(t)];
Phi_d2 = diff(Phi,t)
double(Phi_d2(0))

% Using complex step differentiation
Phi = @(t)[cos(t) cos(t).*sin(t) sin(t)];
h = 2^-28;
cdiff = @(f,x)imag(f(x(:)+1i*h))/h;
Phi_d2 = cdiff(Phi,0)

您可以找到一个函数来执行第一级和第二级复杂的一步分化在我的GitHub： CDIFF 。需要注意的是复杂的一步分化将不会为高阶导工作。这是最好的时候，人们只要一不可微函数或需要快速数值一阶导数。

You can find a function for performing first- and second-order complex step differentiation on my GitHub: cdiff. Note that complex step differentiation won't work well for higher order derivatives. It's best when one only has a non-differentiable function or needs fast numerical first derivatives.

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