We then looked in some detail at circles in space [Transparancy 2], coming up with a general formula for an arbitrarily oriented circle in 3-space.
An example was worked out, in detail, of how to find the vector equation of a circle cutting the axes at x=1, y=1, and z=1. [Transparancies 3 to 5.] Use was made of one of the results worked out in Homework Assignment #1, which enabled us to find the centre of the circle and its corresponding radius.
Next, we considered the vector equation of an arbitrarily oriented helix in 3-space [Transparancy 6]. The previous example, of a circle cutting the axes at x=1, y=1, z=1, was extended into a helix of the same radius oriented along the same axis. The equation was plotted in Maple, and here's what we obtained:
> x:=1/3+sqrt(3)/3*cos(t)+1/3*sin(t)+t/10/Pi;
> y:=1/3-sqrt(3)/3+1/3*sin(t)+t/10/Pi;
> z:=1/3-2/3*sin(t)+t/10/Pi;
> with(plots):
> spacecurve([x,y,z],t=0..12*Pi,axes=NORMAL,scaling=CONSTRAINED,numpoints=1000,orientation=[-25,70]);
This graph agrees with what we expected.
By interpreting the paramater "t" as time, one can consider the velocity vector of a spacecurve [Transparancy 7]. This gives us an expression for the Unit Tangent Vector to the spacecurve.
Finally, by approximating a spacecurve by a large number of short segments, and taking the limit of an infinite number of infinitesimally small segments, one finds an integral for the arc length along the spacecurve [Transparancies 8 and 9]. The space curve can, in principle, be re-paramaterized in terms of arc length rather than "t", but this is usually too complicated to do.