Let $\textbf r(t)=\langle\sin(t),\cos(t),9\sin(t)+5\cos(2t)\rangle$.
Find the projection of $\textbf r(t)$ onto the $xz$-plane for $−1\le x\le1$.
I have no clue how to start.
2026-04-13 00:49:11.1776041351
How do I find the projection of a vector in three dimensions?
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For all $t \in \mathbb{R}$ $$ x = \sin(t)\\ z = 9 \sin(t) + 5 \cos(2t) $$ that means $$ t=\sin^{-1}(x) $$ and $$ z(x) = 9 \sin(\sin^{-1}(x)) + 5 \cos(2\sin^{-1}(x)) \\ = 9 x + 5 \cos(2\sin^{-1}(x)) $$