Center = origin
Radius = 2
I think that's right?
This is a question from my calculus 3 class.
"Show that the graph of r = sin(t)i + 2cos(t)j + sqrt(3) * sin(t)k is a circle and find its radius and center."
In my head, I know that in 3-space r will be a circle that is sort of at an angle due to the k component oscillating. But how do I prove this (like where do I start since this seams so trivial)? And for whatever reason I can't figure out the radius and center. 3-d and parametric equations don't seem to sit in my head and my textbook isn't helping either.
Edit: I'm assuming the center is the origin, but not a hundred percent sure.
Edit: Would taking the norm or r give me the radius if I plug in some value t?
Edit: I think I got it. :|
No, Soroban, you are wrong. (Quick, someone go check if hell has frozen over!)
Showing that shows that the path lies ON a sphere of radius 2. But since x, y, and z depend on a single real parameter, the graph is a curve in that sphere, not the entire surface.
Since we have and , the graph is, specifically, the great circle on that sphere where the plane intersects it.