Physics 374
Fall 2005
Prof. E. F. Redish

Solution, Problem MP14

a.     One fairly straightforward way to do this is to consider the polar coordinates of the object.  Originally, the object is at
					x = r cos ϕ
					y = r sin ϕ
And after the object is rotated through an angle θ, its coordinates must be
					\(TraditionalForm\`\(x\^'\)\) = r cos (ϕ+θ)
					\(TraditionalForm\`\(y\^'\)\) = r sin (ϕ+θ)
And using the familiar trig identities to expand the cosine and sine of the sums gives:
	\(TraditionalForm\`\(x\^'\)\) = r cos (ϕ+θ) = r cos ϕ cos θ - r sin ϕ sin θ = x cos θ - y sin θ
	\(TraditionalForm\`\(y\^'\)\) = r sin (ϕ+θ) = r cos ϕ sin θ - r sin ϕ cos θ = x sin θ + y cos θ
And these two equations can be written in matrix form:
				(({{\(x\^'\)}, {\(y\^'\)}})) = (({{cos θ, -sin θ}, {sin θ, cos θ}}))(({{x}, {y}}))
Which agrees with what is stated in the problem.

b.     There are several ways to find these mirror-x and mirror-y matrices.   To flip a point about the x-axis, you'll need to notice that doing that, keeps the same x coordinate but replaces the y-coordinate by its negative.  To flip a point about the y-axis, the y coordinate stays the same, but the x-coordinate reverses sign.  This gives us
				x-mirror  = (({{1, 0}, {0, -1}}))
				
				y-mirror = (({{-1, 0}, {0, 1}}))

c.     First off, we should think about what it means to "imbed our 2-D vector in a 3-D space whose last component is always one".  Perhaps an analogy to a 3 dimensional x,y,z system will be useful.  Whenever we plot things in just the x-y plane, we are tacitly assuming that the z coordinate is always the same...often zero.  We may write the point as (2,3) but what we may really mean is (2,3,0).  Here, we're just going to be working in the x and y directions within the z = 1 plane instead of the z = 0 plane.  It'll turn out that this extra dimension turns out to be quite useful.
      Now we can better define what our translation operator has to do.  It must take the vector (x,y,1) and change it to (x+a,y+b,1).  The operator will be a 3x3 matrix:
      					(({{x + a}, {y + b

                     1}})) = (({{    ?
?}, {   ?}})({{    ?
?}, {   ?}})({{    ?
?}, {    ?}}))  (({{x}, {y

                 1}}))
First, we note that if we want the matrix multiplication to give the 1 in the bottom row, then the bottom row of the translation matrix is set.
					(({{x + a}, {y + b

                     1}})) = (({{    ?
?}, {   0}})({{    ?
?}, {   0}})({{    ?
?}, {    1}}))  (({{x}, {y

                 1}}))
Since we see that the top row of the translated vector involves an x, but no y, and the middle row of the translated vector involves a y, but no x, putting a mini-identity matrix in the upper left would probably be a good idea.
					(({{x + a}, {y + b

                     1}})) = (({{  1

                  0}, {   0}})({{    0

                  1}, {   0}})({{    ?
?}, {    1}}))  (({{x}, {y

                 1}}))
Where now it's easy to see what the final two question marks have to be
					(({{x + a}, {y + b

                     1}})) = (({{  1

                  0}, {   0}})({{    0

                  1}, {   0}})({{    a

                     b}, {    1}}))  (({{x}, {y

                 1}}))
...so there's our translation matrix.

d.     Finding the 3-D forms of the rotation and reflection matrices will be easier since there's no "+ a" or "+ b" to worry about...the new coordinates \(TraditionalForm\`\(x\^'\)\) and \(TraditionalForm\`\(y\^'\)\) in these cases are always strictly of the form
					(x^(' )) =  α x + β y
and this straight multiply-and-add form is exactly what matrix multiplication is all about.  We can just imbed the 2-D matrices we found in parts "a" and "b" up in the 2x2 corner and let the 3rd component "1" just come along for the ride:
 
	Rotation:                    		(({{\(x\^'\)}, {\(\(y\^'\)

                \ 1\)}})) = (({{  cos θ

                  sin θ}, {   0}})({{    -sin θ

                     cos θ}, {    0}})({{    0

                    0}, {    1}}))  (({{x}, {y

                 1}}))
	
	Mirror-x 			(({{\(x\^'\)}, {\(\(y\^'\)

                \ 1\)}})) = (({{ 1

                 0}, {  0}})({{    0

               -1}, {    0}})({{    0

                    0}, {    1}}))  (({{x}, {y

                 1}}))
	
	Mirror-y			(({{\(x\^'\)}, {\(\(y\^'\)

                \ 1\)}})) = (({{        -1

                           0}, {        0}})({{     0

                   1}, {    0}})({{    0

                    0}, {    1}}))  (({{x}, {y

                 1}}))

(Solution by T .  Bing)