Solutions To Differential Geometry Problems (Part 2)

 1)The student will need to understand that the group of translations required will be for a type of transformation that takes each point in the graph (in this case y = x +1) and slides it the same distance in the same direction, in this case to y = x + 3. Considering a flat 2-d plane he will want to translate  points (x,y) to:   (x+a,y+b)(x+a,  y+ b),  where a = -2 and b = 0.

Or in matrix form:

By taking assorted separate (x,y) mappings he will see, for example:

(3,4)  ->   (1, 4)

(2, 3) ->  (0, 3)

(1,2) -> (-1, 2)

(0, 1) ->  (-2, 1)

(-1, 0) -> (-3, 0)

(-2, -1)  -> (-4, -1)

We see in each case b = 0, a =  -2  But on doing even a few such mappings it should occur to the savvy reader that it isn’t limited to integer values. For example, let x = 2.3111 then x + a =  0.3111, or let x  =  ½,  x + a =  ½ +(-2) = -3/2 and so on.  In effect, one has an infinite set.

Meanwhile, y ->  (y + b) remains the same for any (x+ a), (y + b)

To obtain his desired parallelogram the student needs to see that there must be a boundary or termination point to the mapping to avoid an infinite set, i.e.  extending beyond the given domain and range. Then he needs to set out limits for x->x+ a:

  {x,y : x->x+ a :  x< 2, x +a > -6}  

2) Rem. Coordinates have been transformed to different origin, viz.

So displaced origin is now at:  (x’, y’, z’)  = (x+0,  y+6, z+0), =  (0, 6, 0)

In original system Q is at:  

x= a= 4,  y= 0,  z = c= 2, 

So from (0,0, 0) Q position was:  (4, 0, 2) and length of  Q =

 Ö (4)  ²  +  (2) ²  ) =   Ö (20)  =  4.47

But in new system, coordinates are:  (x’ = 0, y’ = 6,  z’ = 0)

And length of Q =

 Ö (4)  ²  +  (2) ²  + (6) ²     =   Ö (56)  =  7.48