Polynomial angle definition in boundary conditions help

Question

João 2017-12-5

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编辑： Roger Stafford 2017-12-6

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

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Roger Stafford 2017-12-6

编辑：Roger Stafford 2017-12-6

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.

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Answer 1

Torsten 2017-12-5

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A polynomial cannot have an infinite slope.

Best wishes

Torsten.

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João 2017-12-5

The goal is that the polynomial will be tangent at y2.

Torsten 2017-12-5

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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