de-CH
utf-8
math math-format graphie
Reading Slope Field: Quadratic
dgl-09-01
multiple
30030
randRangeExclude(-8,8,[0]) randRangeExclude(-8,8,[0,-A,A]) randRangeExclude(-8,8,[-1,0,1]) randRangeExclude(-8,8,[-1,0,1,-N,N]) fractionReduce(Z,N) A+B fractionReduce(Z*T,N) A*B fractionReduce(Z*D,N)

The slope field of the ODE y' = C y^2 - TTy + DD is given below.

Determine \color{red}\alpha and \color{blue}\beta .

style({ stroke: "black", strokeWidth: 0.05 }); graphInit({ range: [[-8, 8], [-10,10]], scale: [22, 22], axisArrows: "->", tickStep: 10, }); label([0,A-0.5], "\\color{red}\\alpha", "above left"); label([0,B-0.5], "\\color{blue}\\beta", "above left"); // Vektorfeld for(var i = -8; i <= 8; i+=1.25) { for(var j = -9; j <= 9; j+=1.25) { var dy = Z/N* (j - A)*( j - B)/15; line([i - 0.25, j - dy*0.25], [i + 0.25, j + dy*0.25], { arrows: "", strokeWidth: 1, stroke: "gray" }); line([i - 0.25, A ], [i + 0.25, A], { arrows: "", strokeWidth: 1, stroke: "red" }); line([i - 0.25, B ], [i + 0.25, B], { arrows: "", strokeWidth: 1, stroke: "blue" }); } }

x \color{red}\alpha = A
x \color{blue}\beta = B

At a given point (x_0,y_0), the slope field shows a tiny piece of the tangent of a solution of the ODE, on which (x_0,y_0) lies.

The slope of the tangent is determined by the value of the right-hand side of the ODE at the point (x_0,y_0).

At the points \color{red}\alpha und \color{blue}\beta the tangent slopes are zero.

We search for the zeros on the right-hand side of the ODE y' = C y^2 - TTy + DD. .

With y' = C y^2 - TTy + DD = C\left(y^2 - Ty + D \right) = C (y- A) (y- B) one gets \color{red}\alpha = A and \color{blue}\beta = B.