Mathematics

The key is to understand that projecting onto the $xy$-plane means dropping the $z$-coordinate entirely—you're looking straight down from above. So take your parametric curve and simply ignore the $\cos(t/2)$ term, leaving you with $\langle t + 1, 3t + 1 \rangle$. This projected curve is a straight line in the $xy$-plane, and you can find its equation by eliminating the parameter $t$ to relate $x$ and $y$ directly.

The key is to understand that projecting onto the $xy$-plane means dropping the $z$-coordinate entirely—you're looking straight down from above. So take your parametric curve and simply ignore the $\cos(t/2)$ term, leaving you with $\langle t + 1, 3t + 1 \rangle$. This projected curve is a straight line in the $xy$-plane, and you can find its equation by eliminating the parameter $t$ to relate $x$ and $y$ directly.