The study of transformations finds many applications in math, physics, chemistry, biology and engineering. Indeed, the much studied subject of linear algebra may be considered an investigation of “linear” transformations. The problems in these worksheets focus on the transformations of rotation, reflection, and translation, and their action on curves. The solutions stress general aspects of the action of transformations on curves.

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- Consider a line segment anchored at the origin of a Cartesian coordinate system and ending at the point \((1,2)\). Reflect it about the x-axis. Can you get to the transformed segment by instead rotating about the origin? If yes, by how much?
- Place \(\textbf{L}\) with its vertex at the origin and its arms along the two axes. Reflect it about the y-axis. Can you get to the same configuration by rotating instead?
- Consider the curve \(y=x\) and rotate it by \(\theta\) radians. What equation describes the transformed curve?
- Consider the curve \(y = m x\) (m is a real number) and reflect the x-axis about it. What curve is the x-axis transformed into?
- Consider the curve \(2 x^2 + y^2 = 1\) and rotate it by \(\theta\) degrees. What equation describes the transformed curve?
- Consider the line \(x=0\). Translate it by the vector \((2,3)\). Then rotate about the origin by \(\theta\) radians. What equation describes the transformed curve? Is the result the same if the transformations are performed in the reverse order?
- Consider two intersecting lines in the 2-D plane. What effect does a rotation about some point P by \(\theta\) radians have on the angle between the lines?

- Under a reflection a line is transformed into another line (see the solution to problem 4 for an explicit derivation). The origin remains fixed and \((1,2) \rightarrow (1,-2)\) upon a reflection about the x-axis. Thus the slope of the resulting segment is \(\frac{-2}{1}\) and it subtends an angle of \(tan^{-1}(-2) ~\sim~ -1.1\) radians to the positive x-axis. The original line segment subtended an angle of \(tan^{-1}(2) ~\sim~ 1.1\) radians, thus you can obtain the result of the reflection by instead rotating clockwise by \(\sim~ 2.2\) radians or counterclockwise by \(\sim~ 2\pi ~-~ 2.2\) radians.
**No, and this serves to illustrate that reflections and rotations are genuinely distinct transformations.**The effect of reflecting \({|{\underline{\hspace{0.3cm}}}}\) about the y-axis is \(\underline{\hspace{0.3cm}}|\). The results of rotating \(|\underline{\hspace{0.3cm}}\) by 90, 180, and 270 degrees are \(\underline{\hspace{0.5cm}}\rule{0.4pt}{1ex} ~,~ ^{\rule{1ex}{0.5pt}}_{\hspace{0.12cm}\big{|}} ~,~ {\rule{0.4pt}{1ex}}^{\overline{\hspace{0.6cm}}}\) respectively. We see that identical arms never overlap and the effect of a reflection about the y-axis cannot be reproduced by means of a rotation. The same holds for reflection about any line \(y=mx\).The action of a general transformation may be specified by stating that, under it, the point \((x,y)\) moves to the point \((X(x,y), Y(x,y))\). Now, consider the set of points which make up the curve in question, that is satisfy the equation \(y=x\); denote these points by \((x_c, y_c)\). They will be moved to the points \((x^T_c,y^T_c) ~=~ (X(x_c,y_c), Y(x_c,y_c))\) and these points will make up the transformed curve. Let’s assume the transformation is invertible. The inverse transformation, \(\left(~X^{-1} (x,y), Y^{-1}(x,y)~\right)\), is defined by,

\[\boxed{\left(~X^{-1}(X(x,y), Y(x,y))~,~Y^{-1}(X(x,y), Y(x,y))~\right) ~=~ \left(x,~y\right).}\]Consequently, points on the transformed curve will obey the equation: \(Y^{-1}(x,y) = X^-1(x,y)\) because points on the original curve obey the equation \(y=x\).

In the present case, the transformation in question is a rotation by \(\theta\) radians: \[\boxed{(X(x,y),~Y(x,y)) = \left(~cos(\theta) x ~+~ sin(\theta)~ y~,~-sin(\theta)~ x ~+~ cos(\theta)~y~\right).}\]

The inverse transformation is simply a rotation by \(-\theta\) radians:

\[\boxed{\left(X^{-1}(x,y),~Y^{-1}(x,y)\right) ~=~ \left(cos(\theta) x ~-~ sin(\theta)~ y ~,~ sin(\theta)~ x ~+~ cos(\theta)~y\right).}\]Notice that both these transformations are linear (if you’ve studied matrices, what is the matrix representation of this transformation?).

Thus, the transformed curve is: \[\boxed{sin(\theta)~x ~+~ cos(\theta)~y ~=~ cos(\theta)~x ~-~ sin(\theta)~y.}\]

- Let’s tackle this problem in a general way. Before proceeding, take a look at the previous solution for a discussion of the action of a transformation on a curve. Under a reflection about the line \(y=mx\), the point \((x,y)\) moves to the point

\[\boxed{(X,Y) ~=~ \left(\frac{1-m^2}{1+m^2}~x ~+~ \frac{2m}{1+m^2}~y ~,~ \frac{2m}{1+m^2}~x ~-~ \frac{1-m^2}{1+m^2}~y\right).}\]

(See if you can work this out and check if it makes sense for reflections about the x and y axes).Next, note that a reflection is idempotent; reflecting twice is the same as doing nothing. Thus, the inverse of this transformation is identical

\[\boxed{\left(X^{-1}(x,y),~Y^{-1}(x,y)\right) ~=~ \left(\frac{1-m^2}{1+m^2}~x ~+~ \frac{2m}{1+m^2}~y ~,~ \frac{2m}{1+m^2}~x ~-~ \frac{1-m^2}{1+m^2}~y\right).}\]Thus, following the logic of the previous solution, the curve \(y = 0\) (the x-axis) is transformed to:

\[\boxed{Y^{-1}(x,y) ~=~ \frac{2m}{1+m^2}~x ~-~ \frac{1-m^2}{1+m^2}~y ~=~ 0.}\] - Under a rotation, the point \((x,y)\) moves to the point

\[\boxed{(X(x,y), Y(x,y)) ~=~ \left(cos(\theta) x ~+~ sin(\theta)~ y~,~ -sin(\theta)~ x ~+~ cos(\theta)~y)\right).}\]

Thus the transformed curve is \(2 (X^{-1})^2 + (Y^{-1})^2 = 1\), that is,

\[\boxed{2(cos(\theta) x ~-~ sin(\theta)~ y)^2 + (sin(\theta)~ x ~+~ cos(\theta)~y)^2 ~=~ 1.}\] (See the solution to problem 3 if this is not clear). Under a translation by the vector \((2,3)\), the point \((x,y)\) moves to the point \((X,Y) ~=~ (x+2, y+3)\). The inverse transformation is \((X^{-1}(x,y),Y^{-1}(x,y)) `~=~ (x-2, y-3)\) and the line \(x=0\) is transformed to \(x-2 = 0\). Under a further rotation, this curve is transformed to (see solution 3, if this is not clear):

\[\boxed{cos(\theta) x ~-~ sin(\theta)~ y ~-~ 2 ~=~ 0.}\]Performing the transformations in the reverse order, the line \(x=0\) is first transformed to the line \(cos(\theta) x ~-~ sin(\theta)~ y ~=~ 0\) and then, under the translation, to

\[\boxed{cos(\theta) (x-2) ~-~ sin(\theta)~ (y-3) ~=~ cos(\theta) x ~-~ sin(\theta)~ y ~-~ 2 cos(\theta) ~+~ 3 sin(\theta) ~=~ 0.}\]

Thus, performing the transformations in the reverse order results in a different line.- A rotation will not change the angle between the two lines, regardless of the point about which it is performed. We can clarify this as follows: A rotation with the point P as “origin” will manifestly keep the distances between P and all other points unchanged, along with the angle subtended at P by any pair of points. Pick two points, one on each line, and consider the triangle formed by these two points and the intersection point. Under a rotation, the lengths of the sides of this triangle will remain unchanged since the distance of each vertex from P along with the angle subtended by each side at P is unchanged. Consequently all angles of the triangle, including the intersection angle of the original lines, will be unchanged. (It may help to draw this out. A solution with a diagram included is forthcoming).

I have taught physics at levels ranging from introductory classical mechanics to advanced graduate quantum mechanics, along with calculus and linear algebra.