Effacer les filtres
Effacer les filtres

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)
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
mnera almansoorie
mnera almansoorie le 18 Fév 2023
Déplacé(e) : Walter Roberson le 18 Fév 2023
R is the r(t) K is the formula A derivative B derivative
It's a Mathlab assignment and I keep getting 20/100 so I don't know what's wrong
Torsten
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

Connectez-vous pour commenter.

Réponses (1)

Amal Raj
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)

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!

Translated by