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Parametric and cartesian equations? HELP

  1. Jul 25, 2007 #1
    parametric and cartesian equations?? HELP!!!

    1. x = 3t, y = 9t^2, negative infinity<t<positive infinity



    2. a) What are the initial and terminal points, if any? Find a Cartesian equation for a curve that contains the parametrized curve. What portion of the grap of the Cartesian equation is traced by the parametrized curve?:bugeye:



    3. There isn't any attempt. I have no idea. Different language here.:confused:
     
  2. jcsd
  3. Jul 25, 2007 #2

    cristo

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    You need to show some work before we can help you. What do you think the question means by initial and terminal points? Do you think this graph has any?

    To find the cartesian form for the equation, why not use one equation to eliminate t from the other?
     
  4. Jul 25, 2007 #3
    okay, i'm guessing the initial and terminal points mean where the graph starts and ends, and is that what a cartesian equation is? My book doesn't explain and the examples it gives don't help me grasp what the cartesian equation actually is.
     
  5. Jul 25, 2007 #4

    cristo

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    Yea, so to find the starting point (x,y) evaluate (3t,9t^2) as t tends to negative infinity. To find the ending point, do the same, but evaluate it as t tends to infinity. (You should note the part of question that says find the points if any!)
    Well, I presume it just means an equation of the form y=f(x) where f is some function of x. So, you can find this by eliminating t as I said above.
     
  6. Jul 25, 2007 #5
    to get this straight:

    to get the starting point i'd take negative infinity = 3t, 9t^2 and those would me my initial coordinates. I think my previous teacher said something about the number being so large that it doesn't exist so is that when i put none? and is that the same for the positive infinity when i set it to 3t and 9t^2? There aren't any points, are there?

    oh and thanks for the cartesian equation now that i see the book it makes sense.
     
  7. Jul 25, 2007 #6

    cristo

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    Well, yes there are no start or end points since the limits are not defined.
    In fact, if you now know what the equation is, then you will see there are no start or end points.
     
  8. Jul 25, 2007 #7
    wait, how do i know that there are no endpoints just by putting 3t, 9t^2 into the cartesian equation?
     
  9. Jul 25, 2007 #8

    cristo

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    Have you found the equation in cartesian form yet? It should be quite a familiar graph.
     
  10. Jul 25, 2007 #9
    so i set 9t^2 in for y and 3t in for x and then the equation looks like 9t^2=3t(x) and then solve for x so x=3t, but that doesn't leave me with a parametric curve. And then how is that supposed to help me find the starting and end points? sorry i haven't taken math in half a year and i feel super dumb right now.
     
  11. Jul 25, 2007 #10

    cristo

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    Ok, so you have two equations: x=3t; y=9t^2 and you want to make this into an equation of the form y=f(x). So, the first equation can be rearranged to give t=x/3, which can then be substituted into the second equation, yielding y=9(x/3)^2. Can you simplify this?
     
  12. Jul 25, 2007 #11
    okay so now that i've simplified down to the cartesian equation (y=x^2), then i substitute x^2=f(x). then how do i find out that there aren't any existing start and end points?

    sorry it takes so long to reply. at the beach with limited connectivity. i've moved so it should be faster now.
     
  13. Jul 25, 2007 #12

    cristo

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    No, now you've got your equation. (Forget about the f(x) part if it's confusing you; it was just my notation). So, the equation is y=x^2.

    If you plot this (or simply recall what it looks like) you should be able to see that it has no start or end point. If you let x tend to infinity, the function tends to infinity, and if you let x tend to negative infinity, the function tends to negative infinity. So, you see, there are no possible start or end points.

    Ok, now you've just made me incredibly jealous, as I'm sat here in rainy England!
     
  14. Jul 25, 2007 #13
    okay so now that i've got the cartesian equation for my graph, what is the problem asking when it says what portion of the graph is being traced?

    as for being in england, i'm no better off. probably worse as it's gorgeous and sunny and i have to do this homework inside all day before it's due on friday :P
     
  15. Jul 25, 2007 #14

    cristo

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    Well, I may be overlooking something, but I'm sure the parametrised equation covers the whole of the graph y=x^2.
     
  16. Jul 25, 2007 #15
    okay now i've got the answer but how did you get there? how do you know that it covers the whole thing?

    And is the parametrized equation the same as teh cartesian equation?
     
    Last edited: Jul 25, 2007
  17. Jul 25, 2007 #16

    cristo

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    Sorry, I know it's both against forum rules, and not of much use to you to just say something like that, but I'm having some difficulty in how to actually explain this.

    Ok, so the parametric equation holds for [itex]-\infty<t<\infty[/itex]. Therefore this is valid for [itex]-\infty<x<\infty[/itex] and [itex]0<y<\infty[/itex]. But this is the entire domain and range of y=x^2, and so the parametric equation covers the entire cartesian graph.
     
  18. Jul 25, 2007 #17
    well cristo you did a mighty fine job of explaining that for having difficulty. on my next problem i might have to ask for help understanding how to get the portion of the graph that it covers.

    as for the questions: The initial and terminal points are nonexistant. The cartesian equation (aka parametric equation??) is y=x^2, and the whole graph is traced by the parametrized curve. is that correct?
     
  19. Jul 25, 2007 #18

    cristo

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    Yup, they would be my answers. Apart from one small point. The parametric equation and the cartesian equations are not the same, but are two different forms of the same equation. The cartesian form is the standard one, whereas the parametric form is a set of two equations for x and y, each in terms of a parameter, t (Hence the name!).

    Enjoy the sun :smile:
     
    Last edited: Jul 25, 2007
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