# Parametric equations, circle, center, radius

• Feb 8th 2011, 06:01 PM
nautica17
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. :|
• Feb 8th 2011, 06:27 PM
nautica17
Center = origin

I think that's right?
• Feb 8th 2011, 06:42 PM
tonio
Quote:

Originally Posted by nautica17
Center = origin

I think that's right?

Yes

Tonio
• Feb 8th 2011, 07:51 PM
Soroban
Hello, nautica17!

I don't agree with their answer . . .

Quote:

$\text{Show that the graph of: }\:{\bf r} \:=\:\sin(t){\bf i} + 2\cos(t){\bf j} + \sqrt{3}\sin(t){\bf k}$

$\text{is a }circle\text{ and find its radius and center.}$
. . . . .?

$\text{We have: }\;\begin{Bmatrix}x &=& \sin t & [1] \\ y &=& 2\cos t & [2] \\ z &=& \sqrt{3}\sin t & [3] \end{Bmatrix}$

$\text{Square [1], [2], [3]: }\:\begin{Bmatrix}x^2 &=& \sin^2\!t \\ y^2 &=& 4\cos^2\!t \\ z^2 &=& 3\sin^2\!t \end{Bmatrix}$

$\text{Add: }\:x^2 + y^2 + z^2 \;=\;\sin^2\!t + 4\cos^2\!t + 3\sin^2\!t$

. . . . . . . . . . . . . . $=\;4\sin^2\!t + 4\cos^2\!t$

. . . . . . . . . . . . . . $=\;4\underbrace{(\sin^2\!t + \cos^2\!t)}_{\text{This is 1}}$

Therefore, we have: . $x^2 + y^2 + z^2 \:=\:4$

This is a sphere with center (0,0,0) and radius 2.

• Feb 9th 2011, 04:42 AM
HallsofIvy
No, Soroban, you are wrong. (Quick, someone go check if hell has frozen over!)

Showing that $x^2+ y^2+ z^2= 4$ 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 $x= sin(t)$ and $z= \sqrt{3}sin(t)$, the graph is, specifically, the great circle on that sphere where the plane $z= \sqrt{3}x$ intersects it.