Polynomial angle definition in boundary conditions help

3 views (last 30 days)
Hello, I need help figuring out how to write a piece of code to define a polinomial. I have 2 points (x1,y1 and x2,y2), 2 angles (a1 and a2) and an area (Area) as boundary conditions. The shape I need the polinomial to make is a curve leaving from x1 at an angle of 0 degrees and arriving at x2 with an angle of 90 degrees. The help I need is to bypass the tan(90) problem. I know that matlab tan works with radians. The points, angles and area are user defined and will change at each iteration, but the angles will reach 90 degrees.
I made the code like this:
A = zeros(5,5); B = zeros(5,1);
A(1,:)=[x1^4 x1^3 x1^2 x1 1] B(1)=y1
A(2,:)=[4*x1^3 3*x1^2 2*x1 1 0] B(2)=a1
A(3,:)=[x2^4 x2^3 x2^2 x2 1] B(3)=y2
A(4,:)=[4*x2^3 3*x2^2 2*x2 1 0] B(4)=a2
A(5,:)=[integral between x1 and x2] B(5)=Area
and then I proceed to calculate the coefficients and plot the function for different values.
When a2 is 90 degrees or pi/2, how do I make it run smoothly without crashing?
Thanks guys
  1 Comment
Roger Stafford
Roger Stafford on 6 Dec 2017
Edited: Roger Stafford on 6 Dec 2017
I am guessing that you might solve your problem by creating a parametric curve that satisfies your boundary conditions. That way you would have no trouble with infinite slopes. If x and y are each given as cubic, or possibly fourth-order, polynomials of a common parameter, t, that should suffice for the conditions you describe.

Sign in to comment.

Answers (1)

Torsten on 5 Dec 2017
A polynomial cannot have an infinite slope.
Best wishes
Torsten on 5 Dec 2017
You can't find a polynomial that is tangent to x=x2.
Of yourse, you can try to make the slope arbitrary large by setting a2 to a large number, but I doubt that the polynomial will look fine over the interval.
Best wishes

Sign in to comment.


Find more on Polynomials in Help Center and File Exchange


Community Treasure Hunt

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

Start Hunting!