>	with(plottools):   # if you type : in the end # then there will be no/less text output

>	with(plots):

Warning, the name changecoords has been redefined

Warning, the previous binding of the name arrow has been removed and it now has an assigned value

>	cylinderplot(1,theta=0..2*Pi,z=-1..1);

>

# OK, let's do the math first, # if x^2 + y^2 = 1+z^2 # and for a given z we have an x/y plane # where the ray with angle theta and length r # projects onto the x and y axis with x and y, respectively, # by Pythagoras, we have x^2 + y^2 = r^2, i.e. # r = sqrt(x^2+y^2) = sqrt(1 + z^2) by the above. # hence cylinderplot(sqrt(1+z^2),theta=0..2*Pi,z=-1..1);

>	# we can give the plot a name, i.e. assign it to # a variable:

>	H := cylinderplot(sqrt(1+z^2),theta=0..2*Pi,z=-2..2);

>	display(H); # here it is:

>	# let's intersect the Hyperboloid with a plane # at first, a plane:

>	plot3d(1,x=-2..2,y=-2..2);

>	P := plot3d(1,x=-2..2,y=-2..2);

>	display(H,P);

>	# we need labels for the corrdinate axes, # let's replot with labels:

>	H := cylinderplot(sqrt(1+z^2),theta=0..2*Pi,z=-2..2,axes=BOXED,labels=[x,y,z]);

>	display(H);

>	display(H,P);

>	# we need to rotate the plane around the y axis

>	# by 90 degrees = 9*Pi/18 (or x-axis) # the rotate command can do that # it is called # rotate(object to rotate, alpha, beta, gamma) # where alpha, beta, gamma are the angles # of rotation around the x, y, z axes, resp.

>	R := rotate(P,0,9*Pi/18,9*Pi/18):  # beware of the output !

>	display(H,R);

>	display(H,R);

>

# we can see the plane at distance 1 # from the z-axis. It intersects the Hyperboloid # in two lines. # Let's see if we can get hold of those two lines: # the right thing to do is to define a spacecurve, # which is just a parametrized curve in 3 space. # The function is # spacecurve([x(t),y(t),z(t)],t=-...);

>	# lets do the math first. # the curve is in the plane y=1, i.e. # we substitute y=1 into the equation x^2+y^2-z^2=1 # an get # x^2=z^2 # i.e. # x=+z and x=-z # so, we may parametrize x(t)=t,y(t)=1,z(t)=+-t

>	C1 := spacecurve([t,1,t],t=-2..2,color=red,thickness=3): # beware of too much output!

>	C2 := spacecurve([t,1,-t],t=-2..2,color=red,thickness=3):

>	display(H,R,C1,C2);

>	# let's see the hyperbolas. # no we cut with a plane x=0 to get from # the equation x^2+y^2-z^2=1 # y^2=z^2+1 # i.e. y = +- sqrt(1+z^2) # i.e. x(t)=0,y(t) =sqrt(1+z^2),z(t) = t

>	C3 := spacecurve([0,sqrt(1+t^2),t],t=-2..2,color=red,thickness=3):

>	display(H,C3);

>	# let's draw 4 hyperbolas, one in each direction. # we do that using a list (recall how we # handled lists in the first lecture?)

>	L := []: for i from 0 to 4 do  L := [ op(L), rotate(C3,0,0,i*9*Pi/18)]: end:

>	display(H,L);

>

>	# irem(a, b) gives the remainder r of a upon division by b, # i.e. the r in a = q * b + r with 0 <= r < b. # For example,

>	irem(17,5);

>	irem(18,5);

>	irem(19,5);

>	irem(20,5);

>

#however:

>	irem(-1,5);

>	# so for negative inputs we get a negative remainder # we need to be aware of that fact.

>

>	# this is how we make an if construct:

>	if 3 > 2 then print("hello"); end;

>	if 2 > 3 then print("hello"); end;

>	if 2 > 3 then print("hello"); else print("good-bye"); end;

>	# the igcd function computes the greates common divisor (gcd) for integers

>	igcd(6,30);

>	igcd(176654,675364);

>	igcd(8!,9!+7!);

>	# is that correct? # let us check it: # 9! + 7! = 7! * (1 + 9 * 8) # 8! = 8 * 7!

>	1 + 9 * 8;

>

7!;

>	# since 8 and 73 are relatively prime we have that the gcd is 7!= 5040

>

>

>

>

>	HOMEWORK PROBLEMS:

>	# hw 1 # the hyperboloid of one sheet admits two sets of # disjoint lines on it (called rulings). # They can be obtained from the two lines we # have drawn above by rotation. # Plot these lines, but don't plot the # hyperboloid any more. # use a list of objects to draw # as in the last example.

>

>	# hw 2 # Plot the Hyperbolic paraboloid x^2-y^2+z = 0 # Using the command spacecurve, # show that one parabola and one hyperbola # on the surface. #

>

>	# hw 3 # a) Make a 10 by 10 matrix and fill the entry in row i column j #    with the gcd of i and j. # b) Do the same, but this time let the entry (i,j) be the gcd of #    i! and j!

>

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