Answer
Graph VI.
Work Step by Step
Because $\sin^{2}A+\cos^{2}A=1$,
the x and y coordinates satisfy
$x^{2}+y^{2}=1$
So, as z changes, this circle translates into a cylinder propagating around the z-axis. The curve will lie on this cylinder.
The expression for z relies on $t^{2}$ so z has only positive values.
So, we look for a cylinder above the xy plane.
Two graphs offer the above, IV and VI.
Looking at z, the values for z are 1/(positive number greater than 1),
so they are betwen 0 and 1, and
as t increases, z will rapidly approach zero, while x and y rotate about their circle of radius 1.
When t=0, the point on the curve is (1,0,1),
when t=$\pi$/2, the point is above (0,1,0), and z is less than 1/2.
This tells us the difference between IV and VI. Thus, the graph is VI.
