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The line with equation ax + by + c = 0 has slope -a/b, so any line perpendicular to it will have slope b/a (the negative reciprocal). Let (m, n) be the point of intersection of the line ax + by + c = 0 and the line perpendicular to it which passes through the point (x 0, y 0). The line through these two points is perpendicular to the original ...
All such algorithms proceed in two steps: The initial, "prediction" step, starts from a function fitted to the function-values and derivative-values at a preceding set of points to extrapolate ("anticipate") this function's value at a subsequent, new point.
the slope field is an array of slope marks in the phase space (in any number of dimensions depending on the number of relevant variables; for example, two in the case of a first-order linear ODE, as seen to the right). Each slope mark is centered at a point (,,, …,) and is parallel to the vector
Slope illustrated for y = (3/2)x − 1.Click on to enlarge Slope of a line in coordinates system, from f(x) = −12x + 2 to f(x) = 12x + 2. The slope of a line in the plane containing the x and y axes is generally represented by the letter m, [5] and is defined as the change in the y coordinate divided by the corresponding change in the x coordinate, between two distinct points on the line.
First we consider the intersection of two lines L 1 and L 2 in two-dimensional space, with line L 1 being defined by two distinct points (x 1, y 1) and (x 2, y 2), and line L 2 being defined by two distinct points (x 3, y 3) and (x 4, y 4). [2] The intersection P of line L 1 and L 2 can be defined using determinants.
The above procedure now is reversed to find the form of the function F(x) using its (assumed) known log–log plot. To find the function F, pick some fixed point (x 0, F 0), where F 0 is shorthand for F(x 0), somewhere on the straight line in the above graph, and further some other arbitrary point (x 1, F 1) on the same graph.
Two-stage 2nd order Diagonally Implicit Runge–Kutta method: x x 0 1 1 − x x 1 − x x {\displaystyle {\begin{array}{c|cc}x&x&0\\1&1-x&x\\\hline &1-x&x\\\end{array}}} Again, this Diagonally Implicit Runge–Kutta method is A-stable if and only if x ≥ 1 4 {\textstyle x\geq {\frac {1}{4}}} .
The slope number of a graph of maximum degree d is clearly at least ⌈ / ⌉, because at most two of the incident edges at a degree-d vertex can share a slope. More precisely, the slope number is at least equal to the linear arboricity of the graph, since the edges of a single slope must form a linear forest, and the linear arboricity in turn is at least ⌈ / ⌉.