arc length parametrization help me

4 vues (au cours des 30 derniers jours)
Hyunji Yu
Hyunji Yu le 8 Mar 2020
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 commentaires
darova
darova le 8 Mar 2020
What is "arc length parametrization"? Is it length of a curve?
Star Strider
Star Strider le 8 Mar 2020
Some necessary context: nothing appears with this code

Connectez-vous pour commenter.

Réponses (1)

David Goodmanson
David Goodmanson le 8 Mar 2020
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.

Catégories

En savoir plus sur Symbolic Math Toolbox dans Help Center et File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by