{"id":4000,"date":"2022-04-12T18:12:06","date_gmt":"2022-04-12T18:12:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=4000"},"modified":"2022-11-01T04:41:05","modified_gmt":"2022-11-01T04:41:05","slug":"planar-transformations-and-jacobians","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/planar-transformations-and-jacobians\/","title":{"raw":"Planar Transformations and Jacobians","rendered":"Planar Transformations and Jacobians"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Determine the image of a region under a given transformation of variables.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Compute the Jacobian of a given transformation.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Planar Transformations<\/h2>\r\n<p id=\"fs-id1167794027118\">A\u00a0<span id=\"525bfbc5-13f3-461a-98dc-96c236b2031b_term226\" data-type=\"term\">planar transformation [latex]T[\/latex]\u00a0<\/span>is a function that transforms a region [latex]G[\/latex] in one plane into a region [latex]R[\/latex] in another plane by a change of variables. Both [latex]G[\/latex] and [latex]R[\/latex] are subsets of [latex]R^{2}[\/latex]. For example,\u00a0Figure 1\u00a0shows a region [latex]G[\/latex] in the [latex]uv[\/latex]-plane transformed into a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane by the change of variables [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex], or sometimes we write [latex]x=x(u, v)[\/latex] and [latex]y=y(u, v)[\/latex]. We shall typically assume that each of these functions has continuous first partial derivatives, which means [latex]g_u[\/latex], [latex]g_v[\/latex], [latex]h_u[\/latex], and [latex]h_v[\/latex] exist and are also continuous. The need for this requirement will become clear soon.<\/p>\r\n\r\n[caption id=\"attachment_1418\" align=\"aligncenter\" width=\"715\"]<img class=\"size-full wp-image-1418\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042718\/5-7-1.jpeg\" alt=\"On the left-hand side of this figure, there is a region G with point (u, v) given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v) and y = h(u, v). On the right-hand side of this figure there is a region R with point (x, y) given in the Cartesian xy- plane.\" width=\"715\" height=\"273\" \/> Figure 1.\u00a0The transformation of a region\u00a0[latex]G[\/latex] in the\u00a0[latex]uv[\/latex]-plane into a region\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nA transformation [latex]{T}{:} \\ {G}{\\rightarrow}{R}[\/latex], defined as [latex]T(u, v)=(x, y)[\/latex], is said to be a\u00a0<strong><span id=\"525bfbc5-13f3-461a-98dc-96c236b2031b_term227\" data-type=\"term\">one-to-one transformation<\/span><\/strong>\u00a0if no two points map to the same image point.\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794227719\">To show that [latex]T[\/latex] is a one-to-one transformation, we assume [latex]{T}{({{u}_{1}},{{v}_{1}})} = {T}{({{u}_{2}},{{v}_{2}})}[\/latex] and show that as a consequence we obtain [latex]{({{u}_{1}},{{v}_{1}})} = {({{u}_{2}},{{v}_{2}})}[\/latex]. If the transformation [latex]T[\/latex] is one-to-one in the domain [latex]G[\/latex], then the inverse [latex]{T}^{-1}[\/latex] exists with the domain [latex]R[\/latex] such that [latex]{T}^{-1}{\\circ}{T}[\/latex] and [latex]{T}{\\circ}{T}^{-1}[\/latex] are identity functions.<\/p>\r\n<p id=\"fs-id1167794023859\">Figure 1\u00a0shows the mapping [latex]{T}{({u},{v})} = {({x},{y})}[\/latex] where [latex]x[\/latex] and [latex]y[\/latex] are related to [latex]u[\/latex] and [latex]v[\/latex] by the equations [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex]. The region [latex]G[\/latex] is the domain of [latex]T[\/latex] and the region [latex]R[\/latex] is the range of [latex]T[\/latex], also known as the\u00a0<em data-effect=\"italics\">image<\/em>\u00a0of [latex]G[\/latex] under the transformation [latex]T[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1167794063017\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: determining how the transformation works<\/h3>\r\nSuppose a transformation [latex]T[\/latex] is defined as [latex]{T}{({r},{\\theta})} = {({x},{y})}[\/latex] where [latex]{x} = {r}{\\cos}{\\theta}, {y} = {r}{\\sin}{\\theta}[\/latex]. Find the image of the polar rectangle [latex]{G} = {\\left \\{{({r},{\\theta})}{\\mid}{0} \\leq {r} \\leq {{1},{0}} \\leq {\\theta} \\leq {{\\pi}\/{2}} \\right \\}}[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane to a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. Show that [latex]T[\/latex] is a one-to-one transformation in [latex]G[\/latex] and find [latex]{{T}^{-1}}{({x},{y})}[\/latex].\r\n\r\n[reveal-answer q=\"375348660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"375348660\"]\r\n\r\nSince [latex]r[\/latex] varies from 0 to 1 in the [latex]{r}{\\theta}[\/latex]-plane, we have a circular disc of radius 0 to 1 in the [latex]xy[\/latex]-plane. Because [latex]{\\theta}[\/latex] varies from 0 to [latex]{\\pi}\/{2}[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane, we end up getting a quarter circle of radius 1 in the first quadrant of the [latex]xy[\/latex]-plane (Figure 2). Hence [latex]R[\/latex] is a quarter circle bounded by [latex]x^{2}+y^{2}=1[\/latex] in the first quadrant.\r\n\r\n[caption id=\"attachment_1419\" align=\"aligncenter\" width=\"715\"]<img class=\"size-full wp-image-1419\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042750\/5-7-2.jpeg\" alt=\"On the left-hand side of this figure, there is a rectangle G with a marked subrectangle given in the first quadrant of the Cartesian r theta-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = r cos theta and y = r sin theta. On the right-hand side of this figure there is a quarter circle R with a marked subannulus (analogous to the rectangle in the other graph) given in the Cartesian x y-plane.\" width=\"715\" height=\"276\" \/> Figure 2.\u00a0A rectangle in the\u00a0[latex]r\\theta[\/latex]-plane is mapped into a quarter circle in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n<p id=\"fs-id1167793240536\">In order to show that [latex]T[\/latex] is a one-to-one transformation, assume [latex]{T}{({r_1},{{\\theta}_1})} = {T}{({r_2},{{\\theta}_2})}[\/latex] and show as a consequence that [latex]{({r_1},{{\\theta}_1})} = {({r_2},{{\\theta}_2})}[\/latex]. In this case, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{T}{({r_1},{{\\theta}_1})} &amp; = {T}{({r_2},{{\\theta}_2})} \\\\\r\n{({x_1},{y_1})} &amp; = {({x_2},{y_2})} \\\\\r\n{({r_1}{\\cos}{{\\theta}_{1}},{r_1}{\\sin}{{\\theta}_{1}})} &amp; = {({r_2}{\\cos}{{\\theta}_{2}},{r_2}{\\sin}{{\\theta}_{2}})} \\\\\r\n{r_1}{\\cos}{{\\theta}_{1}} &amp; = {r_2}{\\cos}{{\\theta}_{2}} \\\\\r\n{r_1}{\\sin}{{\\theta}_{1}} &amp; = {r_2}{\\sin}{{\\theta}_{2}}\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793456224\">Dividing, we obtain<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\frac{{r_1}{\\cos}{{\\theta}_{1}}}{{r_1}{\\sin}{{\\theta}_{1}}}} &amp; = {{\\frac{{r_2}{\\cos}{{\\theta}_{2}}}{{r_2}{\\sin}{{\\theta}_{2}}}}} \\\\\r\n{\\frac{{\\cos}{{\\theta}_{1}}}{{\\sin}{{\\theta}_{1}}}} &amp; = {\\frac{{\\cos}{{\\theta}_{2}}}{{\\sin}{{\\theta}_{2}}}} \\\\\r\n{\\tan}{{\\theta}_{1}} &amp; = {\\tan}{{\\theta}_{2}} \\\\\r\n{{\\theta}_{1}} &amp; = {{\\theta}_{2}}\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793953463\">since the tangent function is one-one function in the interval [latex]{0} \\leq {\\theta} \\leq {{\\pi}\/{2}}[\/latex]. Also, since [latex]{0} &lt; {r} \\leq {1}[\/latex], we have [latex]{{r}_{1}} = {{r}_{2}}, {{\\theta}_{1}} = {{\\theta}_{2}}[\/latex]. Therefore, [latex]{({{r}_{1}}, {{\\theta}_{1}})} = {({{r}_{2}}, {{\\theta}_{2}})}[\/latex] and [latex]T[\/latex] is a one-to-one transformation from [latex]G[\/latex] into\u00a0[latex]R[\/latex].<\/p>\r\n<p id=\"fs-id1167793936005\">To find [latex]{{T}^{-1}}{({x},{y})}[\/latex] solve for [latex]{r},{\\theta}[\/latex] in terms of [latex]x, y[\/latex]. We already know that [latex]{{r}^{2}} = {{x}^{2}} + {{y}^{2}}[\/latex] and [latex]{\\tan}{\\theta} = {\\frac{y}{x}}[\/latex]. Thus [latex]{{T}^{-1}}{({x},{y})} = {({r},{\\theta})}[\/latex] is defined as [latex]{r} = {\\sqrt{{x^{2}}+{y^{2}}}}[\/latex] and\u00a0[latex]{\\theta} = {{\\tan}^{-1}}{({\\frac{y}{x}})}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the image under T<\/h3>\r\nLet the transformation [latex]T[\/latex] be defined by [latex]T(u, v)=(x, y)[\/latex] where [latex]x=u^{2}+v^{2}[\/latex] and [latex]y=uv[\/latex]. Find the image of the triangle in the [latex]uv[\/latex]-plane with vertices [latex](0, 0)[\/latex],\u00a0[latex](0, 1)[\/latex], and\u00a0[latex](1, 1)[\/latex].\r\n\r\n[reveal-answer q=\"274529843\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"274529843\"]\r\n\r\nThe triangle and its image are shown in\u00a0Figure 3. To understand how the sides of the triangle transform, call the side that joins [latex](0, 0)[\/latex] and [latex](0, 1)[\/latex] side [latex]A[\/latex], the side that joins [latex](0, 0)[\/latex] and [latex](1, 1)[\/latex] side [latex]B[\/latex], and the side that joins [latex](1, 1)[\/latex] and\u00a0[latex](0, 1)[\/latex] side\u00a0[latex]C[\/latex].\r\n\r\n[caption id=\"attachment_1420\" align=\"aligncenter\" width=\"899\"]<img class=\"size-full wp-image-1420\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042830\/5-7-3.jpeg\" alt=\"On the left-hand side of this figure, there is a triangular region given in the Cartesian uv-plane with boundaries A, B, and C represented by the v axis, the line u = v, and the line v = 1, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u squared minus v squared and y = u v. On the right-hand side of this figure there is a complex region given in the Cartesian x y-plane with boundaries A\u2019, B\u2019, and C\u2019 given by the x axis, y axis, and a line curving from (negative 1, 0) through (0, 1), namely x = y squared minus 1, respectively.\" width=\"899\" height=\"348\" \/> Figure 3.\u00a0A triangular region in the\u00a0[latex]uv[\/latex]-plane is transformed into an image in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n<p id=\"fs-id1167793822404\">For the side [latex]A[\/latex]: [latex]u=0[\/latex],\u00a0[latex]{0} \\leq {v} \\leq {1}[\/latex] transforms to [latex]x=-v^{2}[\/latex], [latex]y=0[\/latex] so this is the side [latex]{A}^{\\prime}[\/latex] that joins [latex](-1, 0)[\/latex] and\u00a0[latex](0, 0)[\/latex].<\/p>\r\n<p id=\"fs-id1167793951154\">For the side [latex]B[\/latex]: [latex]u=v[\/latex], [latex]{0} \\leq {u} \\leq {1}[\/latex] transforms to [latex]x=0[\/latex], [latex]y=u^{2}[\/latex] so this is the side [latex]{B}^{\\prime}[\/latex] that joins [latex](0, 0)[\/latex] and\u00a0[latex](0, 1)[\/latex].<\/p>\r\n<p id=\"fs-id1167793929050\">For the side [latex]C[\/latex]: [latex]{0} \\leq {u} \\leq {1}[\/latex], [latex]v=1[\/latex] transforms to [latex]x=u^{2}-1[\/latex], [latex]y=u[\/latex] (hence [latex]x=y^{2}-1[\/latex]) so this is the side [latex]{C}{\\prime}[\/latex] that makes the upper half of the parabolic arc joining [latex](-1, 0)[\/latex] and\u00a0[latex](0, 1)[\/latex].<\/p>\r\n<p id=\"fs-id1167793844608\">All the points in the entire region of the triangle in the [latex]uv[\/latex]-plane are mapped inside the parabolic region in the [latex]xy[\/latex]-plane.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nLet a transformation [latex]T[\/latex]\u00a0 be defined as [latex]T(u, v)=(x, y)[\/latex] where [latex]x=u+v[\/latex], [latex]y=3v[\/latex]. Find the image of the rectangle [latex]{G} = {\\left \\{ {({u,v})}: {0} \\leq {u} \\leq {1,0} \\leq {v} \\leq {2}\\right \\}}[\/latex] from the [latex]uv[\/latex]-plane after the transformation into a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. Show that [latex]T[\/latex] is a one-to-one transformation and find [latex]{{T}^{-1}}{({x},{y})}[\/latex].\r\n\r\n[reveal-answer q=\"036725008\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"036725008\"]\r\n\r\n[latex]T^{-1}(x,y)=(u,v)[\/latex] where\u00a0[latex]u=\\frac{3x-y}3[\/latex] and\u00a0[latex]v=\\frac{y}3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Jacobians<\/h2>\r\n<\/div>\r\nRecall that we mentioned near the beginning of this section that each of the component functions must have continuous first partial derivatives, which means that [latex]g_u[\/latex], [latex]g_v[\/latex], [latex]h_u[\/latex], and [latex]h_v[\/latex] exist and are also continuous. A transformation that has this property is called a [latex]C^1[\/latex] transformation (here [latex]C[\/latex] denotes continuous). Let [latex]T(u, v)=(g(u, v), h(u, v))[\/latex], where [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex], be a one-to-one [latex]C^1[\/latex] transformation. We want to see how it transforms a small rectangular region [latex]S[\/latex], [latex]{\\Delta}{u}[\/latex] units by [latex]{\\Delta}{v}[\/latex] units, in the [latex]uv[\/latex]-plane (see the following figure).\r\n\r\n[caption id=\"attachment_1421\" align=\"aligncenter\" width=\"765\"]<img class=\"size-full wp-image-1421\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042913\/5-7-4.jpeg\" alt=\"On the left-hand side of this figure, there is a region S with lower right corner point (u sub 0, v sub 0), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with point (x sub 0, y sub 0) given in the Cartesian x y-plane with sides r(u, v sub 0) along the bottom and r(u sub 0, v) along the left.\" width=\"765\" height=\"251\" \/> Figure 4.\u00a0A small rectangle [latex]S[\/latex] in the\u00a0[latex]uv[\/latex]-plane is transformed into a region\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n<p id=\"fs-id1167793376348\">Since [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex], we have the position vector [latex]{\\bf{r}}{({u},{v})} = {g}{({u},{v})}{\\bf{i}} + {h}{({u},{v})}{\\bf{j}}[\/latex] of the image of the point [latex](u, v)[\/latex]. Suppose that [latex](u_0, v_0)[\/latex] is the coordinate of the point at the lower left corner that mapped to [latex](x_0, y_0)=T(u_0, v_0)[\/latex]. The line [latex]v=v_0[\/latex] maps to the image curve with vector function [latex]r(u, v_0)[\/latex], and the tangent vector at [latex](x_0, y_0)[\/latex] to the image curve is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{{\\bf{r}}_{u}} = {{g}_{u}}{({u_0},{v_0})}{\\bf{i}} + {h_{u}}{({u_0},{v_0})}{\\bf{j}} = {\\frac{{\\partial}{x}}{{\\partial}{u}}}{\\bf{i}} + {\\frac{{\\partial}{y}}{{\\partial}{u}}}{\\bf{j}}}[\/latex].<\/p>\r\nSimilarly, the line [latex]u=u_0[\/latex] maps to the image curve with vector function [latex]{\\bf{r}}(u_0, v)[\/latex], and the tangent vector at [latex](x_0, y_0)[\/latex] to the image curve is\r\n<p style=\"text-align: center;\">[latex]\\large{{{\\bf{r}}_{v}} = {{g}_{v}}{({u_0},{v_0})}{\\bf{i}} + {h_{v}}{({u_0},{v_0})}{\\bf{j}} = {\\frac{{\\partial}{x}}{{\\partial}{v}}}{\\bf{i}} + {\\frac{{\\partial}{y}}{{\\partial}{v}}}{\\bf{j}}}[\/latex].<\/p>\r\nNow, note that\r\n<p style=\"text-align: center;\">[latex]\\large{{{\\bf{r}}_{u}} = {\\displaystyle\\lim_{{\\Delta}{u}{\\rightarrow}{0}}}{\\frac{{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{u}}} \\ {\\text{so}} \\ {{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{\\approx}{\\Delta}{u}{\\bf{r}}_{u}}[\/latex].<\/p>\r\nSimilarly,\r\n<p style=\"text-align: center;\">[latex]\\large{{{\\bf{r}}_{v}} = {\\displaystyle\\lim_{{\\Delta}{v}{\\rightarrow}{0}}}{\\frac{{\\bf{r}}{({u_0},{v_0} + {{\\Delta}{v}}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{v}}} \\ {\\text{so}} \\ {\\bf{r}}{({u_0},{v_0} + {{\\Delta}{v}}) - {\\bf{r}}{({u_0},{v_0})}}{\\approx}{\\Delta}{v}{\\bf{r}}_{v}}[\/latex].<\/p>\r\nThis allows us to estimate the area [latex]{\\Delta}{A}[\/latex] of the image [latex]R[\/latex] by finding the area of the parallelogram formed by the sides [latex]{\\Delta}{v}{{\\bf{r}}_{v}}[\/latex] and [latex]{\\Delta}{u}{{\\bf{r}}_{u}}[\/latex]. By using the cross product of these two vectors by adding the\u00a0[latex]{\\bf{k}}[\/latex]th component as 0, the area [latex]{\\Delta}{A}[\/latex] of the image [latex]R[\/latex] (refer to\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-the-cross-product\/\" data-page-slug=\"2-4-the-cross-product\" data-page-uuid=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a\" data-page-fragment=\"page_d53b2426-3a76-4d95-af4f-6532ad8e4e9a\">The Cross Product<\/a>) is approximately [latex]{\\mid}{\\Delta}{u}{{\\bf{r}}_{u}} \\ {\\times} \\ {\\Delta}{v}{{\\bf{r}}_{v}}{\\mid} = {\\mid}{{\\bf{r}}_{u}} \\ {\\times} \\ {{\\bf{r}}_{v}}{\\mid} \\ {\\Delta}{u}{\\Delta}{v}[\/latex]. In determinant form, the cross product is\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1em;\">[latex]\\large{{\\bf{r}}_u\\times{\\bf{r}}_v=\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\ \\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{u}}&amp;0 \\\\ \\frac{\\partial{x}}{\\partial{v}}&amp;\\frac{\\partial{y}}{\\partial{v}}&amp;0\\end{vmatrix}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{u}} \\\\ \\frac{\\partial{x}}{\\partial{v}}&amp;\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}{\\bf{k}}=\\left(\\frac{\\partial{x}}{\\partial{u}}\\frac{\\partial{y}}{\\partial{v}}-\\frac{\\partial{x}}{\\partial{v}}\\frac{\\partial{y}}{\\partial{u}}\\right){\\bf{k}}}[\/latex].<\/span><\/p>\r\n<p id=\"fs-id1167793371835\">Since [latex]{\\mid}{\\bf{k}}{\\mid} = {1}[\/latex], we have\u00a0[latex]{\\Delta}{A} \\ {\\approx} \\ {\\mid}{{\\bf{r}}_{u}} \\ {\\times} \\ {{\\bf{r}}_{v}}{\\mid} \\ {\\Delta}{u}{\\Delta}{v} = {\\left ({\\frac{{\\partial}{x}}{{\\partial}{u}}} \\ {\\frac{{\\partial}{y}}{{\\partial}{v}}} - {\\frac{{\\partial}{x}}{{\\partial}{v}}} \\ {\\frac{{\\partial}{y}}{{\\partial}{u}}} \\right )}{\\Delta}{u}{\\Delta}{v}[\/latex].<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nThe\u00a0<strong><span id=\"525bfbc5-13f3-461a-98dc-96c236b2031b_term228\" data-type=\"term\">Jacobian<\/span><\/strong>\u00a0of the [latex]C^1[\/latex] transformation [latex]T(u, v)=(g(u, v), h(u, v))[\/latex] is denoted by [latex]J(u, v)[\/latex] and is defined by the [latex]2\\times 2[\/latex] determinant\r\n<p style=\"text-align: center;\">[latex]\\large{J(u,v)=\\Bigg|\\frac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{u}} \\\\ \\frac{\\partial{x}}{\\partial{v}}&amp;\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\left(\\frac{\\partial{x}}{\\partial{u}}\\frac{\\partial{y}}{\\partial{v}}-\\frac{\\partial{x}}{\\partial{v}}\\frac{\\partial{y}}{\\partial{u}}\\right)}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793421425\">Using the definition, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\Delta}{A}{\\approx}{J}{({u},{v})}{\\Delta}{u}{\\Delta}{v} = {\\bigg\\vert}{\\frac{{\\partial}{(x,y)}}{{\\partial}{(u,v)}}}{\\bigg\\vert}{\\Delta}{u}{\\Delta}{v}}.[\/latex]<\/p>\r\n<p id=\"fs-id1167794171496\">Note that the Jacobian is frequently denoted simply by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{J}{({u},{v})} = {\\frac{{\\partial}{(x,y)}}{{\\partial}{(u,v)}}}}.[\/latex]<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">Note also that<\/span>\r\n<p style=\"text-align: center;\">[latex]\\large{\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{u}} \\\\ \\frac{\\partial{x}}{\\partial{v}}&amp;\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\left(\\frac{\\partial{x}}{\\partial{u}}\\frac{\\partial{y}}{\\partial{v}}-\\frac{\\partial{x}}{\\partial{v}}\\frac{\\partial{y}}{\\partial{u}}\\right)=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}}[\/latex]<\/p>\r\nHence the notation [latex]{J}{({u},{v})} = {\\frac{{\\partial}{(x,y)}}{{\\partial}{(u,v)}}}[\/latex] suggests that we can write the Jacobian determinant with partials of [latex]x[\/latex] in the first row and partials of [latex]y[\/latex] in the second row.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the jacobian<\/h3>\r\nFind the Jacobian of the transformation given in\u00a0Example \"Determining How the Transformation Works\".\r\n\r\n[reveal-answer q=\"602743587\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"602743587\"]\r\n<p id=\"fs-id1167793338722\">The transformation in the example is [latex]{T} = {({r},{\\theta})} = {({r}{\\cos}{\\theta},{r}{\\sin}{\\theta})}[\/latex] where [latex]{x} = {r}{\\cos}{\\theta}[\/latex] and [latex]{y} = {r}{\\sin}{\\theta}[\/latex]. Thus the Jacobian is<\/p>\r\n[latex]\\hspace{4cm}\\large{\\begin{align}\r\n\r\nJ(r,\\theta)&amp;=\\frac{\\partial(x,y)}{\\partial(r,\\theta)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{r}}&amp;\\frac{\\partial{x}}{\\partial{\\theta}} \\\\ \\frac{\\partial{y}}{\\partial{r}}&amp;\\frac{\\partial{y}}{\\partial{\\theta}}\\end{vmatrix}=\\begin{vmatrix}\\cos\\theta&amp;-r\\sin\\theta \\\\ \\sin\\theta&amp;r\\cos\\theta \\end{vmatrix} \\\\\r\n\r\n&amp;=r\\cos^2\\theta+r\\sin^2\\theta=r(\\cos^2\\theta+\\sin^2\\theta)=r \\end{align}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the jacobian<\/h3>\r\nFind the Jacobian of the transformation given in\u00a0Example \"Finding the Image under [latex]T[\/latex]\".\r\n\r\n[reveal-answer q=\"435278520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"435278520\"]\r\n<p id=\"fs-id1167793267327\">The transformation in the example is [latex]T(u, v)=(u^{2}-v^{2}, uv)[\/latex] where [latex]x=u^{2}-v^{2}[\/latex] and [latex]y=uv[\/latex]. Thus the Jacobian is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\begin{vmatrix}2u&amp;v \\\\ -2v&amp;u\\end{vmatrix}=2u^2+2v^2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the Jacobian of the transformation given in the previous Try It: [latex]T(u, v)=(u+v, 3v)[\/latex].\r\n\r\n[reveal-answer q=\"203975843\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"203975843\"]\r\n\r\n[latex]\\large{J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\begin{vmatrix}1&amp;0 \\\\ 1&amp;3\\end{vmatrix}=3}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197114&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=51E1qqcwWzY&amp;video_target=tpm-plugin-jejxrzic-51E1qqcwWzY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.44_transcript.html\">transcript for \u201cCP 5.44\u201d here (opens in new window).<\/a><\/center>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Determine the image of a region under a given transformation of variables.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Compute the Jacobian of a given transformation.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Planar Transformations<\/h2>\n<p id=\"fs-id1167794027118\">A\u00a0<span id=\"525bfbc5-13f3-461a-98dc-96c236b2031b_term226\" data-type=\"term\">planar transformation [latex]T[\/latex]\u00a0<\/span>is a function that transforms a region [latex]G[\/latex] in one plane into a region [latex]R[\/latex] in another plane by a change of variables. Both [latex]G[\/latex] and [latex]R[\/latex] are subsets of [latex]R^{2}[\/latex]. For example,\u00a0Figure 1\u00a0shows a region [latex]G[\/latex] in the [latex]uv[\/latex]-plane transformed into a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane by the change of variables [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex], or sometimes we write [latex]x=x(u, v)[\/latex] and [latex]y=y(u, v)[\/latex]. We shall typically assume that each of these functions has continuous first partial derivatives, which means [latex]g_u[\/latex], [latex]g_v[\/latex], [latex]h_u[\/latex], and [latex]h_v[\/latex] exist and are also continuous. The need for this requirement will become clear soon.<\/p>\n<div id=\"attachment_1418\" style=\"width: 725px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1418\" class=\"size-full wp-image-1418\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042718\/5-7-1.jpeg\" alt=\"On the left-hand side of this figure, there is a region G with point (u, v) given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v) and y = h(u, v). On the right-hand side of this figure there is a region R with point (x, y) given in the Cartesian xy- plane.\" width=\"715\" height=\"273\" \/><\/p>\n<p id=\"caption-attachment-1418\" class=\"wp-caption-text\">Figure 1.\u00a0The transformation of a region\u00a0[latex]G[\/latex] in the\u00a0[latex]uv[\/latex]-plane into a region\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p>A transformation [latex]{T}{:} \\ {G}{\\rightarrow}{R}[\/latex], defined as [latex]T(u, v)=(x, y)[\/latex], is said to be a\u00a0<strong><span id=\"525bfbc5-13f3-461a-98dc-96c236b2031b_term227\" data-type=\"term\">one-to-one transformation<\/span><\/strong>\u00a0if no two points map to the same image point.<\/p>\n<\/div>\n<p id=\"fs-id1167794227719\">To show that [latex]T[\/latex] is a one-to-one transformation, we assume [latex]{T}{({{u}_{1}},{{v}_{1}})} = {T}{({{u}_{2}},{{v}_{2}})}[\/latex] and show that as a consequence we obtain [latex]{({{u}_{1}},{{v}_{1}})} = {({{u}_{2}},{{v}_{2}})}[\/latex]. If the transformation [latex]T[\/latex] is one-to-one in the domain [latex]G[\/latex], then the inverse [latex]{T}^{-1}[\/latex] exists with the domain [latex]R[\/latex] such that [latex]{T}^{-1}{\\circ}{T}[\/latex] and [latex]{T}{\\circ}{T}^{-1}[\/latex] are identity functions.<\/p>\n<p id=\"fs-id1167794023859\">Figure 1\u00a0shows the mapping [latex]{T}{({u},{v})} = {({x},{y})}[\/latex] where [latex]x[\/latex] and [latex]y[\/latex] are related to [latex]u[\/latex] and [latex]v[\/latex] by the equations [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex]. The region [latex]G[\/latex] is the domain of [latex]T[\/latex] and the region [latex]R[\/latex] is the range of [latex]T[\/latex], also known as the\u00a0<em data-effect=\"italics\">image<\/em>\u00a0of [latex]G[\/latex] under the transformation [latex]T[\/latex].<\/p>\n<div id=\"fs-id1167794063017\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: determining how the transformation works<\/h3>\n<p>Suppose a transformation [latex]T[\/latex] is defined as [latex]{T}{({r},{\\theta})} = {({x},{y})}[\/latex] where [latex]{x} = {r}{\\cos}{\\theta}, {y} = {r}{\\sin}{\\theta}[\/latex]. Find the image of the polar rectangle [latex]{G} = {\\left \\{{({r},{\\theta})}{\\mid}{0} \\leq {r} \\leq {{1},{0}} \\leq {\\theta} \\leq {{\\pi}\/{2}} \\right \\}}[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane to a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. Show that [latex]T[\/latex] is a one-to-one transformation in [latex]G[\/latex] and find [latex]{{T}^{-1}}{({x},{y})}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q375348660\">Show Solution<\/span><\/p>\n<div id=\"q375348660\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since [latex]r[\/latex] varies from 0 to 1 in the [latex]{r}{\\theta}[\/latex]-plane, we have a circular disc of radius 0 to 1 in the [latex]xy[\/latex]-plane. Because [latex]{\\theta}[\/latex] varies from 0 to [latex]{\\pi}\/{2}[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane, we end up getting a quarter circle of radius 1 in the first quadrant of the [latex]xy[\/latex]-plane (Figure 2). Hence [latex]R[\/latex] is a quarter circle bounded by [latex]x^{2}+y^{2}=1[\/latex] in the first quadrant.<\/p>\n<div id=\"attachment_1419\" style=\"width: 725px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1419\" class=\"size-full wp-image-1419\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042750\/5-7-2.jpeg\" alt=\"On the left-hand side of this figure, there is a rectangle G with a marked subrectangle given in the first quadrant of the Cartesian r theta-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = r cos theta and y = r sin theta. On the right-hand side of this figure there is a quarter circle R with a marked subannulus (analogous to the rectangle in the other graph) given in the Cartesian x y-plane.\" width=\"715\" height=\"276\" \/><\/p>\n<p id=\"caption-attachment-1419\" class=\"wp-caption-text\">Figure 2.\u00a0A rectangle in the\u00a0[latex]r\\theta[\/latex]-plane is mapped into a quarter circle in the\u00a0[latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<p id=\"fs-id1167793240536\">In order to show that [latex]T[\/latex] is a one-to-one transformation, assume [latex]{T}{({r_1},{{\\theta}_1})} = {T}{({r_2},{{\\theta}_2})}[\/latex] and show as a consequence that [latex]{({r_1},{{\\theta}_1})} = {({r_2},{{\\theta}_2})}[\/latex]. In this case, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {T}{({r_1},{{\\theta}_1})} & = {T}{({r_2},{{\\theta}_2})} \\\\  {({x_1},{y_1})} & = {({x_2},{y_2})} \\\\  {({r_1}{\\cos}{{\\theta}_{1}},{r_1}{\\sin}{{\\theta}_{1}})} & = {({r_2}{\\cos}{{\\theta}_{2}},{r_2}{\\sin}{{\\theta}_{2}})} \\\\  {r_1}{\\cos}{{\\theta}_{1}} & = {r_2}{\\cos}{{\\theta}_{2}} \\\\  {r_1}{\\sin}{{\\theta}_{1}} & = {r_2}{\\sin}{{\\theta}_{2}}  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793456224\">Dividing, we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\frac{{r_1}{\\cos}{{\\theta}_{1}}}{{r_1}{\\sin}{{\\theta}_{1}}}} & = {{\\frac{{r_2}{\\cos}{{\\theta}_{2}}}{{r_2}{\\sin}{{\\theta}_{2}}}}} \\\\  {\\frac{{\\cos}{{\\theta}_{1}}}{{\\sin}{{\\theta}_{1}}}} & = {\\frac{{\\cos}{{\\theta}_{2}}}{{\\sin}{{\\theta}_{2}}}} \\\\  {\\tan}{{\\theta}_{1}} & = {\\tan}{{\\theta}_{2}} \\\\  {{\\theta}_{1}} & = {{\\theta}_{2}}  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793953463\">since the tangent function is one-one function in the interval [latex]{0} \\leq {\\theta} \\leq {{\\pi}\/{2}}[\/latex]. Also, since [latex]{0} < {r} \\leq {1}[\/latex], we have [latex]{{r}_{1}} = {{r}_{2}}, {{\\theta}_{1}} = {{\\theta}_{2}}[\/latex]. Therefore, [latex]{({{r}_{1}}, {{\\theta}_{1}})} = {({{r}_{2}}, {{\\theta}_{2}})}[\/latex] and [latex]T[\/latex] is a one-to-one transformation from [latex]G[\/latex] into\u00a0[latex]R[\/latex].<\/p>\n<p id=\"fs-id1167793936005\">To find [latex]{{T}^{-1}}{({x},{y})}[\/latex] solve for [latex]{r},{\\theta}[\/latex] in terms of [latex]x, y[\/latex]. We already know that [latex]{{r}^{2}} = {{x}^{2}} + {{y}^{2}}[\/latex] and [latex]{\\tan}{\\theta} = {\\frac{y}{x}}[\/latex]. Thus [latex]{{T}^{-1}}{({x},{y})} = {({r},{\\theta})}[\/latex] is defined as [latex]{r} = {\\sqrt{{x^{2}}+{y^{2}}}}[\/latex] and\u00a0[latex]{\\theta} = {{\\tan}^{-1}}{({\\frac{y}{x}})}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding the image under T<\/h3>\n<p>Let the transformation [latex]T[\/latex] be defined by [latex]T(u, v)=(x, y)[\/latex] where [latex]x=u^{2}+v^{2}[\/latex] and [latex]y=uv[\/latex]. Find the image of the triangle in the [latex]uv[\/latex]-plane with vertices [latex](0, 0)[\/latex],\u00a0[latex](0, 1)[\/latex], and\u00a0[latex](1, 1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q274529843\">Show Solution<\/span><\/p>\n<div id=\"q274529843\" class=\"hidden-answer\" style=\"display: none\">\n<p>The triangle and its image are shown in\u00a0Figure 3. To understand how the sides of the triangle transform, call the side that joins [latex](0, 0)[\/latex] and [latex](0, 1)[\/latex] side [latex]A[\/latex], the side that joins [latex](0, 0)[\/latex] and [latex](1, 1)[\/latex] side [latex]B[\/latex], and the side that joins [latex](1, 1)[\/latex] and\u00a0[latex](0, 1)[\/latex] side\u00a0[latex]C[\/latex].<\/p>\n<div id=\"attachment_1420\" style=\"width: 909px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1420\" class=\"size-full wp-image-1420\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042830\/5-7-3.jpeg\" alt=\"On the left-hand side of this figure, there is a triangular region given in the Cartesian uv-plane with boundaries A, B, and C represented by the v axis, the line u = v, and the line v = 1, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u squared minus v squared and y = u v. On the right-hand side of this figure there is a complex region given in the Cartesian x y-plane with boundaries A\u2019, B\u2019, and C\u2019 given by the x axis, y axis, and a line curving from (negative 1, 0) through (0, 1), namely x = y squared minus 1, respectively.\" width=\"899\" height=\"348\" \/><\/p>\n<p id=\"caption-attachment-1420\" class=\"wp-caption-text\">Figure 3.\u00a0A triangular region in the\u00a0[latex]uv[\/latex]-plane is transformed into an image in the\u00a0[latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<p id=\"fs-id1167793822404\">For the side [latex]A[\/latex]: [latex]u=0[\/latex],\u00a0[latex]{0} \\leq {v} \\leq {1}[\/latex] transforms to [latex]x=-v^{2}[\/latex], [latex]y=0[\/latex] so this is the side [latex]{A}^{\\prime}[\/latex] that joins [latex](-1, 0)[\/latex] and\u00a0[latex](0, 0)[\/latex].<\/p>\n<p id=\"fs-id1167793951154\">For the side [latex]B[\/latex]: [latex]u=v[\/latex], [latex]{0} \\leq {u} \\leq {1}[\/latex] transforms to [latex]x=0[\/latex], [latex]y=u^{2}[\/latex] so this is the side [latex]{B}^{\\prime}[\/latex] that joins [latex](0, 0)[\/latex] and\u00a0[latex](0, 1)[\/latex].<\/p>\n<p id=\"fs-id1167793929050\">For the side [latex]C[\/latex]: [latex]{0} \\leq {u} \\leq {1}[\/latex], [latex]v=1[\/latex] transforms to [latex]x=u^{2}-1[\/latex], [latex]y=u[\/latex] (hence [latex]x=y^{2}-1[\/latex]) so this is the side [latex]{C}{\\prime}[\/latex] that makes the upper half of the parabolic arc joining [latex](-1, 0)[\/latex] and\u00a0[latex](0, 1)[\/latex].<\/p>\n<p id=\"fs-id1167793844608\">All the points in the entire region of the triangle in the [latex]uv[\/latex]-plane are mapped inside the parabolic region in the [latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Let a transformation [latex]T[\/latex]\u00a0 be defined as [latex]T(u, v)=(x, y)[\/latex] where [latex]x=u+v[\/latex], [latex]y=3v[\/latex]. Find the image of the rectangle [latex]{G} = {\\left \\{ {({u,v})}: {0} \\leq {u} \\leq {1,0} \\leq {v} \\leq {2}\\right \\}}[\/latex] from the [latex]uv[\/latex]-plane after the transformation into a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. Show that [latex]T[\/latex] is a one-to-one transformation and find [latex]{{T}^{-1}}{({x},{y})}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q036725008\">Show Solution<\/span><\/p>\n<div id=\"q036725008\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]T^{-1}(x,y)=(u,v)[\/latex] where\u00a0[latex]u=\\frac{3x-y}3[\/latex] and\u00a0[latex]v=\\frac{y}3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Jacobians<\/h2>\n<\/div>\n<p>Recall that we mentioned near the beginning of this section that each of the component functions must have continuous first partial derivatives, which means that [latex]g_u[\/latex], [latex]g_v[\/latex], [latex]h_u[\/latex], and [latex]h_v[\/latex] exist and are also continuous. A transformation that has this property is called a [latex]C^1[\/latex] transformation (here [latex]C[\/latex] denotes continuous). Let [latex]T(u, v)=(g(u, v), h(u, v))[\/latex], where [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex], be a one-to-one [latex]C^1[\/latex] transformation. We want to see how it transforms a small rectangular region [latex]S[\/latex], [latex]{\\Delta}{u}[\/latex] units by [latex]{\\Delta}{v}[\/latex] units, in the [latex]uv[\/latex]-plane (see the following figure).<\/p>\n<div id=\"attachment_1421\" style=\"width: 775px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1421\" class=\"size-full wp-image-1421\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042913\/5-7-4.jpeg\" alt=\"On the left-hand side of this figure, there is a region S with lower right corner point (u sub 0, v sub 0), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with point (x sub 0, y sub 0) given in the Cartesian x y-plane with sides r(u, v sub 0) along the bottom and r(u sub 0, v) along the left.\" width=\"765\" height=\"251\" \/><\/p>\n<p id=\"caption-attachment-1421\" class=\"wp-caption-text\">Figure 4.\u00a0A small rectangle [latex]S[\/latex] in the\u00a0[latex]uv[\/latex]-plane is transformed into a region\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<p id=\"fs-id1167793376348\">Since [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex], we have the position vector [latex]{\\bf{r}}{({u},{v})} = {g}{({u},{v})}{\\bf{i}} + {h}{({u},{v})}{\\bf{j}}[\/latex] of the image of the point [latex](u, v)[\/latex]. Suppose that [latex](u_0, v_0)[\/latex] is the coordinate of the point at the lower left corner that mapped to [latex](x_0, y_0)=T(u_0, v_0)[\/latex]. The line [latex]v=v_0[\/latex] maps to the image curve with vector function [latex]r(u, v_0)[\/latex], and the tangent vector at [latex](x_0, y_0)[\/latex] to the image curve is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{\\bf{r}}_{u}} = {{g}_{u}}{({u_0},{v_0})}{\\bf{i}} + {h_{u}}{({u_0},{v_0})}{\\bf{j}} = {\\frac{{\\partial}{x}}{{\\partial}{u}}}{\\bf{i}} + {\\frac{{\\partial}{y}}{{\\partial}{u}}}{\\bf{j}}}[\/latex].<\/p>\n<p>Similarly, the line [latex]u=u_0[\/latex] maps to the image curve with vector function [latex]{\\bf{r}}(u_0, v)[\/latex], and the tangent vector at [latex](x_0, y_0)[\/latex] to the image curve is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{\\bf{r}}_{v}} = {{g}_{v}}{({u_0},{v_0})}{\\bf{i}} + {h_{v}}{({u_0},{v_0})}{\\bf{j}} = {\\frac{{\\partial}{x}}{{\\partial}{v}}}{\\bf{i}} + {\\frac{{\\partial}{y}}{{\\partial}{v}}}{\\bf{j}}}[\/latex].<\/p>\n<p>Now, note that<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{\\bf{r}}_{u}} = {\\displaystyle\\lim_{{\\Delta}{u}{\\rightarrow}{0}}}{\\frac{{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{u}}} \\ {\\text{so}} \\ {{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{\\approx}{\\Delta}{u}{\\bf{r}}_{u}}[\/latex].<\/p>\n<p>Similarly,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{{\\bf{r}}_{v}} = {\\displaystyle\\lim_{{\\Delta}{v}{\\rightarrow}{0}}}{\\frac{{\\bf{r}}{({u_0},{v_0} + {{\\Delta}{v}}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{v}}} \\ {\\text{so}} \\ {\\bf{r}}{({u_0},{v_0} + {{\\Delta}{v}}) - {\\bf{r}}{({u_0},{v_0})}}{\\approx}{\\Delta}{v}{\\bf{r}}_{v}}[\/latex].<\/p>\n<p>This allows us to estimate the area [latex]{\\Delta}{A}[\/latex] of the image [latex]R[\/latex] by finding the area of the parallelogram formed by the sides [latex]{\\Delta}{v}{{\\bf{r}}_{v}}[\/latex] and [latex]{\\Delta}{u}{{\\bf{r}}_{u}}[\/latex]. By using the cross product of these two vectors by adding the\u00a0[latex]{\\bf{k}}[\/latex]th component as 0, the area [latex]{\\Delta}{A}[\/latex] of the image [latex]R[\/latex] (refer to\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-the-cross-product\/\" data-page-slug=\"2-4-the-cross-product\" data-page-uuid=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a\" data-page-fragment=\"page_d53b2426-3a76-4d95-af4f-6532ad8e4e9a\">The Cross Product<\/a>) is approximately [latex]{\\mid}{\\Delta}{u}{{\\bf{r}}_{u}} \\ {\\times} \\ {\\Delta}{v}{{\\bf{r}}_{v}}{\\mid} = {\\mid}{{\\bf{r}}_{u}} \\ {\\times} \\ {{\\bf{r}}_{v}}{\\mid} \\ {\\Delta}{u}{\\Delta}{v}[\/latex]. In determinant form, the cross product is<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1em;\">[latex]\\large{{\\bf{r}}_u\\times{\\bf{r}}_v=\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ \\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{u}}&0 \\\\ \\frac{\\partial{x}}{\\partial{v}}&\\frac{\\partial{y}}{\\partial{v}}&0\\end{vmatrix}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{u}} \\\\ \\frac{\\partial{x}}{\\partial{v}}&\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}{\\bf{k}}=\\left(\\frac{\\partial{x}}{\\partial{u}}\\frac{\\partial{y}}{\\partial{v}}-\\frac{\\partial{x}}{\\partial{v}}\\frac{\\partial{y}}{\\partial{u}}\\right){\\bf{k}}}[\/latex].<\/span><\/p>\n<p id=\"fs-id1167793371835\">Since [latex]{\\mid}{\\bf{k}}{\\mid} = {1}[\/latex], we have\u00a0[latex]{\\Delta}{A} \\ {\\approx} \\ {\\mid}{{\\bf{r}}_{u}} \\ {\\times} \\ {{\\bf{r}}_{v}}{\\mid} \\ {\\Delta}{u}{\\Delta}{v} = {\\left ({\\frac{{\\partial}{x}}{{\\partial}{u}}} \\ {\\frac{{\\partial}{y}}{{\\partial}{v}}} - {\\frac{{\\partial}{x}}{{\\partial}{v}}} \\ {\\frac{{\\partial}{y}}{{\\partial}{u}}} \\right )}{\\Delta}{u}{\\Delta}{v}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p>The\u00a0<strong><span id=\"525bfbc5-13f3-461a-98dc-96c236b2031b_term228\" data-type=\"term\">Jacobian<\/span><\/strong>\u00a0of the [latex]C^1[\/latex] transformation [latex]T(u, v)=(g(u, v), h(u, v))[\/latex] is denoted by [latex]J(u, v)[\/latex] and is defined by the [latex]2\\times 2[\/latex] determinant<\/p>\n<p style=\"text-align: center;\">[latex]\\large{J(u,v)=\\Bigg|\\frac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{u}} \\\\ \\frac{\\partial{x}}{\\partial{v}}&\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\left(\\frac{\\partial{x}}{\\partial{u}}\\frac{\\partial{y}}{\\partial{v}}-\\frac{\\partial{x}}{\\partial{v}}\\frac{\\partial{y}}{\\partial{u}}\\right)}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167793421425\">Using the definition, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\Delta}{A}{\\approx}{J}{({u},{v})}{\\Delta}{u}{\\Delta}{v} = {\\bigg\\vert}{\\frac{{\\partial}{(x,y)}}{{\\partial}{(u,v)}}}{\\bigg\\vert}{\\Delta}{u}{\\Delta}{v}}.[\/latex]<\/p>\n<p id=\"fs-id1167794171496\">Note that the Jacobian is frequently denoted simply by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{J}{({u},{v})} = {\\frac{{\\partial}{(x,y)}}{{\\partial}{(u,v)}}}}.[\/latex]<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">Note also that<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{u}} \\\\ \\frac{\\partial{x}}{\\partial{v}}&\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\left(\\frac{\\partial{x}}{\\partial{u}}\\frac{\\partial{y}}{\\partial{v}}-\\frac{\\partial{x}}{\\partial{v}}\\frac{\\partial{y}}{\\partial{u}}\\right)=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}}[\/latex]<\/p>\n<p>Hence the notation [latex]{J}{({u},{v})} = {\\frac{{\\partial}{(x,y)}}{{\\partial}{(u,v)}}}[\/latex] suggests that we can write the Jacobian determinant with partials of [latex]x[\/latex] in the first row and partials of [latex]y[\/latex] in the second row.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding the jacobian<\/h3>\n<p>Find the Jacobian of the transformation given in\u00a0Example &#8220;Determining How the Transformation Works&#8221;.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q602743587\">Show Solution<\/span><\/p>\n<div id=\"q602743587\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793338722\">The transformation in the example is [latex]{T} = {({r},{\\theta})} = {({r}{\\cos}{\\theta},{r}{\\sin}{\\theta})}[\/latex] where [latex]{x} = {r}{\\cos}{\\theta}[\/latex] and [latex]{y} = {r}{\\sin}{\\theta}[\/latex]. Thus the Jacobian is<\/p>\n<p>[latex]\\hspace{4cm}\\large{\\begin{align}    J(r,\\theta)&=\\frac{\\partial(x,y)}{\\partial(r,\\theta)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{r}}&\\frac{\\partial{x}}{\\partial{\\theta}} \\\\ \\frac{\\partial{y}}{\\partial{r}}&\\frac{\\partial{y}}{\\partial{\\theta}}\\end{vmatrix}=\\begin{vmatrix}\\cos\\theta&-r\\sin\\theta \\\\ \\sin\\theta&r\\cos\\theta \\end{vmatrix} \\\\    &=r\\cos^2\\theta+r\\sin^2\\theta=r(\\cos^2\\theta+\\sin^2\\theta)=r \\end{align}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding the jacobian<\/h3>\n<p>Find the Jacobian of the transformation given in\u00a0Example &#8220;Finding the Image under [latex]T[\/latex]&#8220;.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q435278520\">Show Solution<\/span><\/p>\n<div id=\"q435278520\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793267327\">The transformation in the example is [latex]T(u, v)=(u^{2}-v^{2}, uv)[\/latex] where [latex]x=u^{2}-v^{2}[\/latex] and [latex]y=uv[\/latex]. Thus the Jacobian is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\begin{vmatrix}2u&v \\\\ -2v&u\\end{vmatrix}=2u^2+2v^2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the Jacobian of the transformation given in the previous Try It: [latex]T(u, v)=(u+v, 3v)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q203975843\">Show Solution<\/span><\/p>\n<div id=\"q203975843\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\large{J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\begin{vmatrix}1&0 \\\\ 1&3\\end{vmatrix}=3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197114&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=51E1qqcwWzY&amp;video_target=tpm-plugin-jejxrzic-51E1qqcwWzY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.44_transcript.html\">transcript for \u201cCP 5.44\u201d here (opens in new window).<\/a><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-4000\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 5.44. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":29,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 5.44\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4000","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4000","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4000\/revisions"}],"predecessor-version":[{"id":6383,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4000\/revisions\/6383"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4000\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4000"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4000"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4000"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4000"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}