Consider the transformation T(x)=[(1,-2),(3,4)]. (2x2 matrix) Sketch the image o

Question

Consider the transformation T(x)=[(1,-2),(3,4)]. (2x2 matrix) Sketch the image of the unit square under this transformation. Find a vector u so that T(u) = [(1),(0)]. (2x1 matrix) Find a vector v so that T(v) = [(0),(1)]. (2x1 matrix) Find a vector w so that T(w) = [(5),(7)]. Do this without solving the matrix equation; instead use your results from the previous two questions. Define a transformation Q(x) = [u v]. Find a matrix for the composed transformation T(Q(x)). What does this tell you about Q?

Explanation / Answer

The Jacobian of a Transformation In this section, we explore the concept of a "derivative" of a coordinate transfor- mation, which is known as the Jacobian of the transformation. However, in this course, it is the determinant of the Jacobian that will be used most frequently. If we let u = hu; vi ; p = hp; qi, and x = hx; yi, then (x; y) = T (u; v) is given in vector notation by x = T (u) This notation allows us to extend the concept of a total derivative to the total derivative of a coordinate transformation. De?nition 5.1: A coordinate transformation T (u) is di¤erentiable at a point p if there exists a matrix J (p) for which lim u!p jjT (u) ?? T (p) ?? J (p) (u ?? p)jj jju ?? pjj = 0 (1) When it exists, J (p) is the total derivative of T (u) at p. In non-vector notation, de?nition 5.1 says that the total derivative at a point (p; q) of a coordinate transformation T (u; v) is a matrix J (u; v) evaluated at (p; q) : In a manner analogous to that in section 2-5, it can be shown that this matrix is given by J (u; v) = xu xv yu yv (see exercise 46). The total derivative is also known as the Jacobian Matrix of the transformation T (u; v) : EXAMPLE 1 What is the Jacobian matrix for the polar coordinate transformation? Solution: Since x = r cos () and y = r sin () ; the Jacobian matrix is J (r; ) = xr x yr y = cos () ??r sin () sin () r cos () If u (t) = hu (t) ; v (t)i is a curve in the uv-plane, then x (t) = T (u (t) ; v (t)) is the image of u (t) in the xy-plane. Moreover, dx dt = dx dt dy dt = xu du dt + xv dv dt yu du dt + yv dv dt = xu xv yu yv du dt dv dt 1 The last vector is du=dt: Thus, we have shown that if x (t) = T (u (t)) ; then dx dt = J (u) du dt That is, the Jacobian maps tangent vectors to curves in the uv-plane to tangent vectors to curves in the xy-plane. In general, the Jacobian maps any tangent vector to a curve at a given point to a tangent vector to the image of the curve at the image of the point. EXAMPLE 2 Let T (u; v) = u2 ?? v2; 2uv a) Find the velocity of u (t) = t; t2 when t = 1: b) Find the Jacobian and apply it to the vector in a) c) Find x (t) = T (u (t)) in the xy-plane and then ?nd its velocity vector at t = 1: Compare to the result in (b). Solution: a) Since u0 (t) = h1; 2ti ; the velocity at t = 1 is u0 (1) = h1; 2i : b) Since x (u; v) = u2??v2 and y (u; v) = 2uv; the Jacobian of T (u; v) is J (u; v) = xu xv yu yv = 2u ??2v 2v 2u Since u0 = h1; 2ti ; we have J (u; v) u0 = 2u ??2v 2v 2u 1 2t = 2u (1) ?? 2v (2t) 2v (1) + 2u (2t) = 2u ?? 4tv 2v + 4tu 2 Substituting hu; vi = t; t2 yields x0 = J (u; v) u0 = 2t ?? 4t ?? t2 2t2 + 4t (t) = 2t ?? 4t3 6t2 In vector form, x0 (t) = 2t ?? 4t3; 6t2 ; so that x0 (1) = h??2; 6i : c) Substituting u = t; v = t2 into T (u; v) = u2 ?? v2; 2uv results in x (t) = ?? t2 ?? t4; 2t3 which has a velocity of x0 (t) = 2t ?? 4t3; 6t2 . Moreover, x0 (1) = h??2; 6i : Check your Reading: At what point in the xy-plane is x0 (1) tangent to the curve? The Jacobian Determinant The determinant of the Jacobian matrix of a transformation is given by det (J) = xu xv yu yv = @x @u @y @v ?? @x @v @y @u 3 However, we often use a notation for det (J) that is more suggestive of how the determinant is calculated. @ (x; y) @ (u; v) = @x @u @y @v ?? @x @v @y @u The remainder of this section explores the Jacobian determinant and some of its more important properties. EXAMPLE 3 Calculate the Jacobian Determinant of T (u; v) = u2 ?? v; u2 + v Solution: If we identify x = u2 ?? v and y = u2 + v; then @ (x; y) @ (u; v) = @x @u @y @v ?? @x @v @y @u = (2u) (1) ?? (??1) (2u) = 4u Before we consider applications of the Jacobian determinant, let?s develop some of the its properties. To begin with, if x (u; v) and y (u; v) are di¤erentiable functions, then @ (y; x) @ (u; v) = @y @u @x @v ?? @y @v @x @u = ?? @x @u @y @v ?? @x @v @y @u = ?? @ (x; y) @ (u; v) from which it follows immediately that @ (x; x) @ (u; v) = @ (y; y) @ (u; v) = 0 Similarly, if f (u; v) ; g (u; v) ; and h (u; v) are di¤erentiable, then @ (f + g; h) @ (u; v) = @ (f + g) @u @h @v ?? @ (f + g) @v @h @u = @f @u @h @v + @g @u @h @v ?? @f @v @h @u ?? @g @v @h @u = @f @u @h @v ?? @f @v @h @u + @g @u @h @v ?? @g @v @h @u = @ (f; h) @ (u; v) + @ (g; h) @ (u; v) 4

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