1. Not finding help here? Sign up for a free 30min 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!

Is this a valid parametrisation of a curve?

  1. Nov 27, 2016 #1
    • Member warned that the homework template is required
    Hi, so I am learning how to parametrise curves.

    For the curve y^2=16x, I have said let x=t^2. Then, we can say y^2=16t^2, so that we can take the root of this and get y=4t. What I wanted to ask was do we have to say "plus or minus" in front of the 4t, or do we just leave it as positive to get a valid parametrisation?

    Thanks.
     
  2. jcsd
  3. Nov 27, 2016 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The question to ask is what values can ##t## take?
     
  4. Nov 27, 2016 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Is ##y## allowed to be ##<0## along the curve? Is ##t < 0## allowed?
     
  5. Nov 27, 2016 #4
    I suppose so? The question doesn't specify so I am not sure.
     
  6. Nov 27, 2016 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    How would you draw a graph of the relation ##x = \frac{1}{16} y^2##?
     
  7. Nov 27, 2016 #6
    You would have a sideways parabola where y can be any real number, but x cannot be below 0 because you are squaring.
     
  8. Nov 27, 2016 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    OK, so now what does that tell you about ##t## in your parametrization?
     
  9. Nov 27, 2016 #8
    From my parametrisations, t can be both positive and negative. Would you say that is correct?
     
  10. Nov 27, 2016 #9

    Mark44

    Staff: Mentor

    No, that isn't right. If ##y^2 = 16t^2##, then ##y = \pm 4t##.
     
  11. Nov 27, 2016 #10
    Thank you. That is what I was asking in the sentence after that.
     
  12. Nov 27, 2016 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    But @rickonstark999, you do understand that your first post above is correct and your parameterization ##x=t^2,~y=4t## with ##t## both positive and negative, ie ##-\infty < t < \infty## is a correct parameterization of your curve, right? You don't need any ##\pm## sign.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Is this a valid parametrisation of a curve?
  1. Parametrising a curve (Replies: 1)

  2. Valid proof? (Replies: 3)

  3. Valid mapping? (Replies: 5)

Loading...