问题
I have the following function (Viviani's curve):
Phi = @(t)[ cos(t)^2, cos(t)*sin(t), sin(t) ]
Just a check that it's valid:
s = linspace(0,T,1000);
plot3(cos(s).^2, cos(s).*sin(s), sin(s));
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.
If the second derivative would be Phi_d2, I need to get it's value (for example in t = 0).
How can I achieve this?
回答1:
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)
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.
回答2:
For the sake of completeness, the numerical solution without using any additional toolboxes:
N = 999;
t = linspace(0,2*pi,N+1);
Phi = [cos(t); cos(t).*sin(t); sin(t)];
dPhi = gradient(Phi,2*pi/N)
For non-uniformly spaced argument vectors, the second parameter of gradient is defined by a spacing vector instead of a scalar. (time or angle vector is appropriate) - in this case it is obviously necessary to split up the dimensions. (Though I don't know why.)
So alternatively, though not necessary:
dX = gradient(Phi(1,:),t);
dY = gradient(Phi(2,:),t);
dZ = gradient(Phi(3,:),t);
dPhi = [dX; dY; dZ];
来源：https://stackoverflow.com/questions/20613859/calculate-the-derivative-of-a-vector