Matrix dimensions must agree error

10 vues (au cours des 30 derniers jours)
Alexandros Panagi
Alexandros Panagi le 13 Déc 2018
Modifié(e) : madhan ravi le 14 Déc 2018
Im trying to make a simple function in which you input the radius, centre of a circle and the intercept coordinates to output a plot with the circle, radius and tangent showing but I seem to be stuck here any help is appreciated
function circle2
x0=input('centre x-axis');
y0=input('centre y-axis');
r=input('radius');
xp=input('P x-coord');
yp=input('P y-coord');
%Parameters
th=0:0.01:2.*pi;
%Value for x
x=r.*cos(th)+x0;
%Value for y
y=r.*sin(th)+y0;
%Differentiation
dxdth = -r.*sin(th);
dydth = r.*cos(th);
%Gradient
dydx = dydth./dxdth;
%Calculate intercept
intcpt = yp - dydx.*xp;
%x vector for tangent
xt= linspace(xp-1,xp+1,2);
%Calculate tangent
yt = dydx.*xt+intcpt;
if yp~=yt
xt=xp-2<=xt<=xp+2
yt=dydx(xt-xp)+yp
elseif yp==y0
xt=xp
yt=yp-5<=yt<=yp+5
end
%Plot intercept
plot(x,y);
hold on
plot(xp, yp);
%Plot tangent
plot(xt, yt);
%Plot radius
plot ([x0 xp],[y0 yp])
%Axis limit values
axis ([-1.25*r 1.25*r -1.25*r 1.25*r]);
  1 commentaire
madhan ravi
madhan ravi le 14 Déc 2018
Modifié(e) : madhan ravi le 14 Déc 2018
It's truly impolite to close a question (and to curse with vulgar words as you flagged my answer which is why I deleted my answer) after someone has made an effort to answer your question.

Connectez-vous pour commenter.

Réponses (1)

Guillaume
Guillaume le 13 Déc 2018
What are these lines supposed to do?
xt=xp-2<=xt<=xp+2
yt=yp-5<=yt<=yp+5
Whatever you meant for these, I'm fairly certain it's not doing it. a <= b is always going to result in an array of true or false (0 or 1). So you're comparing an array of 0 and 1 with xp + 2.
Typically, when novices write this sort of expressions they mean:
xt = xp-2 <= xt & xt <= xp+2
yt = yp-5 <= yt & yt <= yp+5
but here even that would make no sense.
  1 commentaire
Image Analyst
Image Analyst le 14 Déc 2018
And what are "the intercept coordinates"? What is intercepting what?
A diagram would help.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Mathematics dans Help Center et File Exchange

Produits


Version

R2018b

Community Treasure Hunt

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

Start Hunting!

Translated by