I have a homework question that I am just stuck on, I have been doing it for a while so I will show you what I have. Let $r(t) = (\sin 2t, \cos 2t, \sin2t\cos4t)$ . Find the point where $r(t) $intersects the $xy$-plane on the interval $1/2 π < t < 3/4 π.$
What I did was:
I set: $\cos4t=0$ to get $\pi/8 +4n\pi, 3\pi/8 +4n\pi.$ I set: $\sin2t=0$ to get $\pi/2 +2n\pi.$
I know that the curve intersects the $xy$-plane at the points where z = 0. But what do I do next? Appreciate any help, I just want to know what I did wrong and where to proceed from this, I would like to understand how to get the answer. Thanks! :)
Your answers could be written a bit more compact (and correct): $$\cos4t = 0 \implies 4t = \pi/2+n\pi \implies t = \pi/8 + n\pi/4.$$ Similarly: $$\sin2t = 0 \implies t = n\pi/2.$$
You now need to find the value(s) of the integer $n$ that make $t$ fall into the given range. Once you have these values for $t$, plug it back in $r(t)$ to get the points (...,...,0) that belongs to $r(t)$ and lie on the $xy$-plane.