Radius and centre of curvature
67 views (last 30 days)
Subramani Nagaraj on 18 Jan 2021
Can anyone say
why it is showing inf as output as shown in the screenshot attached, when I ask radius and centre of curvature for sqrt(X)+sqrt(y)==sqrt(a),when x = 3???
Why it is not showing a numerical value???
Kindly solve this problem.
Divija Aleti on 25 Jan 2021
You will have to first change your equation to the form of y = f(x) and then use the formulae to find the radius and center of curvature. The code is given below for your reference:
syms y(x) a
y(x) = (sqrt(a)-sqrt(x))^2; % Re-arranged the equation
x1 = 3;
c1 = x - (diff(y,x)*(1+(diff(y,x))^2))/diff(y,x,2);
c1 = c1(x1) % Found x-coordinate of center of curvature
y1 = y(x1);
c2 = y + (1+(diff(y,x))^2)/diff(y,x,2);
c2 = c2(y1) % Found y-coordinate of center of curvature
Center = [c1,c2];
Point = [x1,y1];
Radius = ((x1-c1)^2 + (y1-c2)^2) % Distance between the point and the center gives radius of curvature
For additional information on the 'diff' function, refer to the following link: