Calculus: The Language of Change

Exercise Set 15.1
Cartesian Coordinates

solutions by Randall Childs and David Theobald

Exercise 1

(a) Give a general formula for the distance between a point on the line with number coordinate 'a' and a point with coordinate 'b'.  Explain how your formula covers the various cases of positive and negative signs for 'a' and 'b'.

(b) Give a general formula for the directed distance from a point with coordinate 'a' to a point with the coordinate 'b'.  Explain how your formula covers the various cases of positive and negative signs for 'a' and 'b'.
(HINT: How much is: a-b?, b-a?, |b-a|?)

Answers:
	(a) 	| b - a |	(Note:	| b - a |  =  | a - b |.)
	(b)	b - a

Explanation

The answers may be explained correctly in many different ways.  Compare your explanation with those of other students.  The main questions that your explanation should answer are:

(1)	for part (a)	What's the connection between absolute value and distance (as opposed to directed distance)?
(There's a certain piece of information that neither cares about.)

(2)	for part (b)	Why 'b - a' instead of 'a - b'?

(3)	for both	Why subtraction?
(This is the part where you break it down into the different cases of a and b being positive or negative.)

Exercise 2

2-D

Locate the points X(-3,4), Y(3,-4), and Z(3,4).

Enter the input cell below each point's name in order to see a two-dimensional graph that locates that point.

Point X

\(xpt\  = \ Graphics[\ {\ PointSize[0.03], \ Point[{\(-3\), 4}]\ }\ ]; 

    Show[{xpt}, \ Axes\  -> \ True, 
      PlotRange\  -> \ {{\(-5\), 5}, {\(-5\), 5}}]; \)

Point Y

\(ypt\  = \ Graphics[\ {\ PointSize[0.03], \ Point[{3, \(-4\)}]\ }\ ]; 

    Show[{ypt}, \ Axes\  -> \ True, 
      PlotRange\  -> \ {{\(-5\), 5}, {\(-5\), 5}}]; \)

Point Z

\(zpt\  = \ Graphics[\ {\ PointSize[0.03], \ Point[{3, 4}]\ }\ ]; 

    Show[{zpt}, \ Axes\  -> \ True, 
      PlotRange\  -> \ {{\(-5\), 5}, {\(-5\), 5}}]; \)

Why do these cells work?

Each of the above cells has two lines.  The first line takes a set of two-dimensional graphical information and assigns it a name.  The second line tells Mathematica to actually draw or "Show" that graphical information with a set of guidelines telling the computer how to draw the graph.

The result is that Mathematica draws the x and y axes, showing all points for which -5≤x≤+5 and -5≤y≤+5.  Then it draws a "big" point at the given set of coordinates.

3-D

Locate the points X(-3,2,4), Y(3,2,-4), Z(3,2,4).

Enter the input cell below each point's name in order to see a three-dimensional graph that locates that point.  You may wish to make the graph larger.  Mathematica doesn't draw the (Subscript[x, 1]), (Subscript[x, 2]), and (Subscript[x, 3]) axes in its 3-D plots, but the graph has been oriented so that the positive and negative directions of (Subscript[x, 1]), (Subscript[x, 2]), and (Subscript[x, 3]) in these plots correspond to the same directions in the 3-D diagrams in the text.

So "positive (Subscript[x, 1])" is "coming at ya" and to the left.  "Positive (Subscript[x, 2])" is "coming at ya" and to the right.  "Positive (Subscript[x, 3])" goes straight up.

Point X

\(x1\  = \ 
      Graphics3D[
        \ {\ GrayLevel[0.5], \ Line[{{0, 0, 0}, \ {\(-3\), 0, 0}}]\ }\ ]; 

    x2\  = \ Graphics3D[
        \ {\ GrayLevel[0.5], \ Line[{{\(-3\), 0, 0}, \ {\(-3\), 2, 0}}]\ }
          \ ]; 
x3\  = \ 
      Graphics3D[
        \ {\ GrayLevel[0.5], \ Line[{{\(-3\), 2, 0}, \ {\(-3\), 2, 4}}]\ }
          \ ]; 
ptx\  = \ 
      Graphics3D[\ {\ PointSize[0.03], \ Point[{\(-3\), 2, 4}]\ }\ ]; 

    Show[{x1, x2, x3, ptx}, \ Axes\  -> \ True, \ 
      PlotRange\  -> \ {{\(-5\), 5}, {\(-5\), 5}, {\(-5\), 5}}, \ 
      ViewPoint\  -> \ {1.5, 1, 1}]; \)

Point Y

\(y1\  = \ 
      Graphics3D[\ {\ GrayLevel[0.5], \ Line[{{0, 0, 0}, \ {3, 0, 0}}]\ }\ ]; 
    
y2\  = \ 
      Graphics3D[\ {\ GrayLevel[0.5], \ Line[{{3, 0, 0}, \ {3, 2, 0}}]\ }\ ]; 
    
y3\  = \ 
      Graphics3D[
        \ {\ GrayLevel[0.5], \ Line[{{3, 2, 0}, \ {3, 2, \(-4\)}}]\ }\ ]; 

    pty\  = \ Graphics3D[\ {\ PointSize[0.03], \ Point[{3, 2, \(-4\)}]\ }\ ]; 
    
Show[{y1, y2, y3, pty}, \ Axes\  -> \ True, \ 
      PlotRange\  -> \ {{\(-5\), 5}, {\(-5\), 5}, {\(-5\), 5}}, \ 
      ViewPoint\  -> \ {1.5, 1, 0.9}]; \)

Point Z

\(z1\  = \ 
      Graphics3D[\ {\ GrayLevel[0.5], \ Line[{{0, 0, 0}, \ {3, 0, 0}}]\ }\ ]; 
    
z2\  = \ 
      Graphics3D[\ {\ GrayLevel[0.5], \ Line[{{3, 0, 0}, \ {3, 2, 0}}]\ }\ ]; 
    
z3\  = \ 
      Graphics3D[\ {\ GrayLevel[0.5], \ Line[{{3, 2, 0}, \ {3, 2, 4}}]\ }\ ]; 
    
ptz\  = \ Graphics3D[\ {\ PointSize[0.03], \ Point[{3, 2, 4}]\ }\ ]; 

    Show[{z1, z2, z3, ptz}, \ Axes\  -> \ True, \ 
      PlotRange\  -> \ {{\(-5\), 5}, {\(-5\), 5}, {\(-5\), 5}}, \ 
      ViewPoint\  -> \ {1.5, 1.1, 1}]; \)

Why do these cells work?

Each of the above cells has five lines.  The first four take sets of three-dimensional graphical information and assigns a name to each set.  The fifth one tells Mathematica to actually draw or "Show" those four sets of graphical information with a set of guidelines telling the computer how to draw the graph.

The result is that Mathematica draws a box showing all points in three-dimensional space for which -5≤x1≤+5, -5≤x2≤+5, and -5≤x3≤+5.  Then it draws gray line segments which for a given point (x1,x2,x3) draw three line segments: from (0,0,0) to (x1,0,0); from (x1,0,0) to (x1,x2,0); and from (x1,x2,0) to (x1,x2,x3).  These gray lines suggest to the eye the location within three-dimensional space of the given set of coordinates.  Finally, it draws a "big" point at the given set of coordinates.