Answer
$\bf{Graph{(IV)}}$
Work Step by Step
The parametric equations of a circle having radius $r$ are; $x=r \cos t ; y =r \sin t$
Here, we have $x= \cos t , y=\sin t, z=\cos 2t$
This shows that the projection of the curve on the xy plane will be a circle of radius $1$.
The term $z=e^{0.8} t$ tells us that this is greater than $0$ for all $t$.
This matches with $\bf{Graph{(IV)}}$.
