arc length parametrization help me
显示 更早的评论
syms t;
x(t) = sin(3*t^2)*(12*t + (10*13^(1/2))/13);
y(t) = t*(6*13^(1/2)*t + 5);
z(t) = cos(3*t^2)*(12*t + (10*13^(1/2))/13);
%Arc-Length Parametrization
syms tau;
L(t) = int(speed(tau), tau, 0, t);
syms s;
solve(s == L(t), t);
assume(t, 'positive');
g(s) = subs(finverse(L(t)), t, s);
x2(s) = x(g(s))
y2(s) = y(g(s))
z2(s) = z(g(s))
I have no idea how to make code for arc length parametrization. please help me
2 个评论
darova
2020-3-8
What is "arc length parametrization"? Is it length of a curve?
Star Strider
2020-3-8
回答（1 个）
David Goodmanson
2020-3-8
Hello HY,
You just need to go back to the basics. There is a vector x,y,z defined by a parameter (t in this case), so the rate of change of the arc length is
ds/dt = sqrt( (dx/dt)^2 + (dy/dt)^2 (dz/dt)^2 )
You can get that quantity in your code with
dsdt = sqrt(diff(x)^2 + diff(y)^2+diff(z)^2)
but actually being able to integrate the result algebraically to find s(t) is a whole different matter. In this case probably not, but once you have the expression you can integrate it numerically.
类别
在 帮助中心 和 File Exchange 中查找有关 Symbolic Math Toolbox 的更多信息
2020-3-8
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
