61512036_746096_APM2616_DD_Wilson_Assignment02_Semester1.pdf...

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61512036 DD Wilson 746096 Assignment 2 APM2616 Semester 1 Question 1 f := plot::Function2d(exp(sin(t)), t = -1..5 , Color = RGB::Blue, Legend = "f(t) = exp(sin(t))")g := plot::Function2d(t^2/(1+t^2), t = -1..5 , Color = RGB::Green, Legend = "f(t) = t^2/(1+t^2)" )plot(f,g)
Question 2 spiral := plot::Curve2d([u*cos(u),u*sin(u)],u = 0..3*PI, Color = RGB::Blue) plot(spiral) Question 3 plot(plot::Implicit2d(u^2 - 3*v, u = -1..1, v = -1..1, Legend = "Contours of f(u,v) = u^2 -3*v", Contours = [0, 1, 2, 3, 4]),XAxisTitle = "v",YAxisTitle = "u")
Question 4 plot(plot::Cylinder(50,[0,0,0],[0,0,2]),Color = RGB::Brown, plot::Function3d(sqrt(800-(1.05*x)^2-(1.05*y)^2), x=-100..100, y=-100..100, Color = RGB::Brown, FillColorType = Flat, XLinesVisible = FALSE, YLinesVisible = FALSE), Mesh = 100)
Question 5 A := matrix([[5, 2, 1], [2, 3, 2], [1, 2, 6]]) Ev:= linalg::eigenvectors(A) EigenVector := matrix(1,3) for v from 1 to 3 do EigenVector[v] := Ev[v][3][1]; v = v+1; end_for //The Eigenvalues of the given matrix are complex and thus 3 dimensions is not sufficient in order to plot these vectors, therefor the process of determining and graphing the eigenvectors was repeated for a matrix with real eigenvalues// A := matrix([[1, -3, 3], [6, -10, 6], [6, 6, 4]])
Ev:= linalg::eigenvectors(A) EigenVector := matrix(1,3) for v from 1 to 3 do EigenVector[v] := Ev[v][3][1]; v = v+1; end_for Vector1 := plot::Arrow3d([EigenVector[1][1],EigenVector[1][2],EigenVector[1][3]], Color = RGB::Green, Legend = "EigenVector1") Vector2 := plot::Arrow3d([EigenVector[2][1],EigenVector[2][2],EigenVector[2][3]], Color = RGB::Red, Legend = "EigenVector2") Vector3 := plot::Arrow3d([EigenVector[3][1],EigenVector[3][2],EigenVector[3][3]], Legend = "EigenVector3")

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Term
Winter
Professor
N/A
Tags
Equations, Munganga J M W, Stability of Solutions

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