Finding tangent plane with the max range value and z value being equal
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Rahal Rodrigo
le 16 Avr 2021
Modifié(e) : Andrew Newell
le 16 Avr 2021
Hi there. I'm plotting a tangent plane to a particular surface. However, the range for both x and y if +- sqrt(2), which is +-1.41. The expected value of z for the tangent plane is sqrt(2) as well. However, when i run the command, MATLAB produces an error for that particulat range (produces sucessful output on ranges such as +-2). Would be grateful if someone provided insight what I could do to solve this issue. The code is shown below. Any advice would be greatly apreciated.
0 commentaires
Réponse acceptée
Andrew Newell
le 16 Avr 2021
Modifié(e) : Andrew Newell
le 16 Avr 2021
Notice that j is empty, then try this:
X = -1.41:0.01:1.41;
[~,idx] = min(abs(X-1));
disp(X(idx)-1)
This being floating point, you're not getting exact increments of . You could scale everything so x, y are and then find and . Or you could do it like this:
z = @(x,y) sqrt(4-x.^2-y.^2);
X = -1.41:0.01:1.41;
[x,y] = meshgrid(X);
[fx,fy] = gradient(z(x,y),0.01);
[~,j] = min(abs(X-1));
fx0 = fx(j,j);
fy0 = fy(j,j);
z1 = @(x,y) z(1,1) + fx0*(x-1) + fy0*(y-1);
As a side note, you don't need . You can use j directly as logical indexing.
0 commentaires
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Surfaces, Volumes, and Polygons 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!