1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parametric equations

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider the curve of intersection of the cylinders [x^2+y^2=4] and [z+x^2=4]. Find parametric equations for this curve and use them to write a position vector.

    2. Relevant equations
    Thats what im looking for. What to set t equal to.

    3. The attempt at a solution
    I set t=x and got a square root for y. So if i set t=x^2, i get rid of the square root for y, but im not sure if this is correct. i really dont know the rules for parametric equations. Any input is appreciated.

  2. jcsd
  3. Feb 13, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    An easy choice for t is the angular coordinate in the x,y plane. So x=2*cos(t), y=2*sin(t). Given this, can you figure out what z is in terms of t? Unfortunately, I don't think there are any general rules for doing this. You just have to make a picture in your mind of what the curve looks like and then look for a choice for t.
  4. Feb 13, 2008 #3
    I see. Yes i could find z with those coordinates. But is there a reason you chose those? Are those cylindrical coordinates? (im a little rusty).
    Is there anything wrong with setting t=x^2?
  5. Feb 13, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper

    The x^2+y^2=4 cylinder intersects the x,y plane in a circle. It's easy to parametrize that in polar coordinates. If you take t=x^2 then you still have a square root for y (contrary to what you said), so you'd have to define the curve as a union of pieces.
  6. Feb 13, 2008 #5
    All yes your right.
    So my position vector should be: r = < 4cos^2(t), 4sin^2(t), 4-4cos^2(t) >
    Is this correct?
  7. Feb 13, 2008 #6


    User Avatar
    Science Advisor
    Homework Helper

    Noooo. x=2*cos(t), not 4cos^2(t)!!!
  8. Feb 13, 2008 #7
    yea i just realized i did that.
    So, r = < 2cos(t), 2sin(t), 4-4cos^2(t) >

    This look good?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Parametric equations
  1. Parametric equations (Replies: 1)

  2. Parametric Equations (Replies: 6)

  3. Parametric Equation (Replies: 3)

  4. Parametric equation (Replies: 10)