Each student will get individual data for this assignment.

Here is some help for the first *Maple* assignment. The
assignment requests several pictures.
Pictures are
very important. Not many people can get much information from vast
tables of numbers, but humans possess a large capacity to receive and
organize visual information. **You are allowed, and indeed
encouraged, to copy and modify the commands discussed here.**
Remember that all commands should be entered into Maple in one
line - this page has some commands broken across two lines because of
space limitiations.

Maple has thousands of commands, but many of them are kept in storage
in "packages" to save memory. Once Maple is started, you can call up
packages via the *with* command. One package, called *plots*,
contains several three dimensional plotting commands. Enter *with(plots);* in *Maple* to see a large number of commands appear which are now easily
accessible. One of them is *spacecurve* which will plot three-dimensional parametric curves. To see more about this
command you can type *help(spacecurve);* to bring up the help page.

The command

* spacecurve([4*t-11,3*t+7,-5*t+2], t=0..1, axes=normal, color=black, thickness=2, labels=[x,y,z]);
*

draws the parametric function

*
x = 4t - 11
y = 3t + 7
z = -5t + 2
*
for the range 0 ≤ t ≤ 1. These equations should give you a
line segment from the point (-11,7,2) where t=0, to the point
(-7,10,-3) where t=1.

The picture *Maple* displays (shown above)
is deceptive. The line segment is *not*
vertical although it is shown that way. In this case this problem
arises because the three coordinate axes are not drawn to the same scale.

Add the "scaling=constrained" option to change the command to

*spacecurve([4*t-11,3*t+7,-5*t+2], t=0..1, axes=normal, color=black, thickness=2, labels=[x,y,z], scaling=constrained);
*

The result is shown below. The *scaling=constrained* makes all the axes have
the same scale, and so the line segment is no longer displayed vertically.

The command

*spacecurve([3*cos(t)+10,3*sin(t)+4,-3], t=0..Pi, axes=normal, color=black, thickness=2, labels=[x,y,z]);*

draws the parametric function

*
x = 3 cos(t) + 10
y = 3 sin(t) + 4
z = -4
*
for the range 0 ≤ t ≤ π. (Notice that to get the constant π
you need to use the capitalization

Change the command to

*spacecurve([3*cos(t)+10,3*sin(t)+4,-3], t=0..Pi, axes=normal, color=black, thickness=2, labels=[x,y,z], scaling=constrained);*

The result is shown below. The curve looks much more like a semicircle.

The image can be manipulated in various ways before "exporting" or printing it. It can be rotated, scaled, etc. with some mouse clicks on the picture.

Often one will want to draw multiple graphs on the same axes for easier comparison. In *Maple* this is achieved via the *display* command found in the *plots* package. Store each graph to its own variable, and then display the set of those variables. For example,

*
A:=spacecurve([4*t-11,3*t+7,-5*t+2], t=0..1, axes=normal, color=black, thickness=2, labels=[x,y,z], scaling=constrained):
B:=spacecurve([3*cos(t)+10,3*sin(t)+4,-3],t=0..Pi, axes=normal, color=black, thickness=2, labels=[x,y,z], scaling=constrained):
display({A, B});*

The first two commands assign values to the variables

We will now show several views of the same pair of curves given above.
We show **very bad** versions of
the pictures to emphasize that poor graphs can actually
decrease the effectiveness of technical communication instead of
helping.

This is the plot with *scaling=unconstrained*.

This is the result with

Here is a a *very bad* version of the picture. It is
taken from "the side", with the x-axis coming straight out of the
image. This picture seems to show two line segments, when one of
the segments is actually the semicircle viewed edge-on.

Here's another *very bad* version of the picture. The
axes have been taken away, and the viewing angle makes the line
segment seem to cross what could be a parabolic arc.

Now we analyze more vectors and show more pictures.

Suppose p is the point (3,10,-7), q is the point (9,8,3), and r
is the point (6,5,7). Then we can get the vector from p to q via:

[9,8,3] - [3,10,-7]; [6, -2, 10]

This is the vector from p to q. Notice that square brackets [] are
used to represent points/vectors instead of parentheses. A picture
of the vector can be created with the *spacecurve* command.
We will store that plot to the variable *PQplot*.

*PQplot:=spacecurve([6*t+3,-2*t+10,10*t-7], t=0..1, axes=normal, color=black, thickness=2, labels=[x,y,z], scaling=constrained);*

A similar computation gets the vector from p to r and the corresponding picture:

[6,5,7] - [3,10,-7]; [3, -5, 14]

What about the cross product of the two vectors? The *VectorCalculus* package has a *CrossProduct* command with a short version *&x*. When using the *VectorCalculus* package, denote vectors via angle brackets < and >.

What are these ewith(VectorCalculus): <6,-2,10> &x <3,-5,14 > ; 22 e_{x}- 54 e_{y}- 24 e_{z}

We then can draw the cross product vector as a line segment "based" at p:

*CPplot:=spacecurve([22*t+3,-54*t+10,-24*t-7], t=0..1, axes=normal, color=black, thickness=2, labels=[x,y,z], scaling=constrained);*

and all three graphs can be displayed with

*display3d({PQplot,PRplot,CPplot});*

The above picture was rotated so that the cross product vector is more clearly perpendicular to the other two. Other views, such as the one below, are not nearly as clear and should be avoided.

Here is an even worse view: one vector is entirely hidden behind one of the others, making it look like there are only two vectors drawn.

Below is a picture of the triangle in R^{3} whose
vertices are the points p and q and r. This picture is very easy to
create using a command in the *plots* package. Find this
command yourself: look at the list of commands in *plots*,
guess, and then use *help*. A hint: The command you're looking for can do more than triangles. It can also plot quadrilaterals, pentagons, hexagons, and many other two-dimensional shapes in three dimensions.

By the way, by rotating the viewpoint, you can make any of the angles of this triangle seem to be a right angle! Perspective can be misleading and irritating.

Finally, here are the triangle, two of the vectors along the
triangle's edges, and the cross product of these two edge
vectors. You would also generate this picture via the *display* command.
The view has been chosen so that the cross product
appears to be perpendicular to the triangle, which it is. Twice the area
of the triangle is equal to the length of the cross product.

**
Maintained by
greenfie@math.rutgers.edu and last modified 6/11/2008 by Andrew Baxter.
**