# Polynomial angle definition in boundary conditions help

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João on 5 Dec 2017
Edited: Roger Stafford on 6 Dec 2017
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
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.

Torsten on 5 Dec 2017
A polynomial cannot have an infinite slope.
Best wishes
Torsten.
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
Torsten.

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