Find the curvature of the curve traced out by the function r(t)=〈 t^2, 5t-1, 2t^3 -t 〉 at t=1.
8 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
syms t
f = t^2;
g = (5*t) - 1;
h = 2*t^3 - t;
r = [f, g, h]; % r(t)
dr = diff(r, t); % r'(t) = tangent to the curve
T = dr / norm(dr); % normalized r'(t) = T(t) = unit tangent to the curve
dT = diff(T, t); % T'(t)
k = norm(dT)/norm(dr); % Curvature at a given t ( k = ||T'(t)|| / ||r'(t)|| )
k = subs(k, 1); % Curvature at t = 1
can someone help me correct my code
my code is getting correct numbers but i need to incllude A,B, K and define R
3 commentaires
Torsten
le 18 Fév 2023
The formula I know to calculate curvature of a parametric curve differs from the one you use.
Look up the section "Space curves: General expressions" under
Réponses (1)
Amal Raj
le 14 Mar 2023
Hi,
Here is the corrected code:
syms t
f = t^2;
g = (5*t) - 1;
h = 2*t^3 - t;
r = [f, g, h]; % r(t)
dr = diff(r, t); % r'(t) = tangent to the curve
T = dr / norm(dr); % normalized r'(t) = T(t) = unit tangent to the curve
dT = diff(T, t); % T'(t)
k = norm(dT)/norm(dr); % Curvature at a given t ( k = ||T'(t)|| / ||r'(t)|| )
k = subs(k, 1); % Curvature at t = 1
N = dT / norm(dT); % normal vector N(t) = (T'(t)) / (||T'(t)||)
B = cross(T, N); % binormal vector B(t) = cross(T(t), N(t))
r1 = subs(r, t, 1); % r(1)
N1 = subs(N, t, 1); % N(1)
B1 = subs(B, t, 1); % B(1)
syms A B
R = r1 + A*N1 + B*B1 % osculating circle at t = 1
% Now we need to find A and B that satisfy the condition R(1) = r(1) and R'(1) = T(1)
eq1 = R == r1;
eq2 = diff(R, t) == T;
[A, B] = solve(eq1, eq2, A, B);
% Finally, we can evaluate R, A, and B at t = 1
R = subs(R, [A, B], [A, B]);
A = double(A)
B = double(B)
K = 1 / norm(R - r1)
0 commentaires
Voir également
Catégories
En savoir plus sur Discrete Math 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!