Find the curvature of the curve traced out by the function r(t)=〈 t^2, 5t-1, 2t^3 -t 〉 at t=1.

Question

mnera almansoorie 2023-2-18

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1915010-find-the-curvature-of-the-curve-traced-out-by-the-function-r-t-t-2-5t-1-2t-3-t-at-t-1

回答： Amal Raj 2023-3-14

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 个评论
显示 1更早的评论隐藏 1更早的评论

mnera almansoorie 2023-2-18

移动：Walter Roberson 2023-2-18

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 2023-2-18

@mnera almansoorie

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

https://en.wikipedia.org/wiki/Curvature

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Amal Raj 2023-3-14

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1915010-find-the-curvature-of-the-curve-traced-out-by-the-function-r-t-t-2-5t-1-2t-3-t-at-t-1#answer_1192475

在 MATLAB Online 中打开

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)