arc length parametrization help me

2 次查看（过去 30 天）
Hyunji Yu
Hyunji Yu 2020-3-8
回答： David Goodmanson 2020-3-8
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

请先登录，再回答此问题。

回答（1 个）

David Goodmanson
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 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by