Planar Transformations

A planar transformation [latex]T[/latex] 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, Figure 1 shows 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.

Figure 1. The transformation of a region [latex]G[/latex] in the [latex]uv[/latex]-plane into a region [latex]R[/latex] in the [latex]xy[/latex]-plane.

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.

Figure 1 shows 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 image of [latex]G[/latex] under the transformation [latex]T[/latex].

Example: determining how the transformation works

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].

Show Solution

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.

Figure 2. A rectangle in the [latex]r\theta[/latex]-plane is mapped into a quarter circle in the [latex]xy[/latex]-plane.

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

[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]

Dividing, we obtain

[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]

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 [latex]R[/latex].

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 [latex]{\theta} = {{\tan}^{-1}}{({\frac{y}{x}})}[/latex].

Example: finding the image under T

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], [latex](0, 1)[/latex], and [latex](1, 1)[/latex].

Show Solution

The triangle and its image are shown in Figure 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 [latex](0, 1)[/latex] side [latex]C[/latex].

Figure 3. A triangular region in the [latex]uv[/latex]-plane is transformed into an image in the [latex]xy[/latex]-plane.

For the side [latex]A[/latex]: [latex]u=0[/latex], [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 [latex](0, 0)[/latex].

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 [latex](0, 1)[/latex].

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 [latex](0, 1)[/latex].

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.

try it

Let a transformation [latex]T[/latex]  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].

Show Solution

[latex]T^{-1}(x,y)=(u,v)[/latex] where [latex]u=\frac{3x-y}3[/latex] and [latex]v=\frac{y}3[/latex].

Jacobians

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).

Figure 4. A small rectangle [latex]S[/latex] in the [latex]uv[/latex]-plane is transformed into a region [latex]R[/latex] in the [latex]xy[/latex]-plane.

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

[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].

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

[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].

Now, note that

[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].

Similarly,

[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].

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 [latex]{\bf{k}}[/latex]th component as 0, the area [latex]{\Delta}{A}[/latex] of the image [latex]R[/latex] (refer to The Cross Product) 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

[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].

Since [latex]{\mid}{\bf{k}}{\mid} = {1}[/latex], we have [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].

Using the definition, we have

[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]

Note that the Jacobian is frequently denoted simply by

[latex]\large{{J}{({u},{v})} = {\frac{{\partial}{(x,y)}}{{\partial}{(u,v)}}}}.[/latex]

Note also that

[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]

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.

Watch the following video to see the worked solution to the above Try It