Results 1 to 5 of 5

Math Help - Parametric equations, circle, center, radius

  1. #1
    Member nautica17's Avatar
    Joined
    Aug 2009
    Posts
    121

    Parametric equations, circle, center, radius

    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. :|
    Last edited by nautica17; February 8th 2011 at 06:16 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member nautica17's Avatar
    Joined
    Aug 2009
    Posts
    121
    Center = origin
    Radius = 2

    I think that's right?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by nautica17 View Post
    Center = origin
    Radius = 2

    I think that's right?

    Yes

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,739
    Thanks
    645
    Hello, nautica17!

    I don't agree with their answer . . .


    \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.

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470
    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.
    Last edited by HallsofIvy; February 10th 2011 at 08:01 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: March 15th 2010, 04:10 PM
  2. Center of a circle and the radius.
    Posted in the Geometry Forum
    Replies: 5
    Last Post: January 11th 2010, 03:45 AM
  3. center and radius of a circle?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 29th 2009, 09:46 PM
  4. Center (h,k) and the radius r of each circle.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 11th 2008, 11:29 PM
  5. Center and radius of a circle?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 28th 2007, 12:26 PM

Search Tags


/mathhelpforum @mathhelpforum