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Cartesian equation

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data
    eliminate the parameter to find cartesian equation of the curve


    2. Relevant equations



    3. The attempt at a solution
    @ means delta

    x = 4cos@, y=5sin@, -pi/2 <= @ <= pi/2

    i have no idea how to do it.. i read the whole chapter and it doesnt make any sense..
    so i make x=y in some way.. or what?
     
  2. jcsd
  3. Nov 28, 2009 #2

    rock.freak667

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    do you know any trigonometric identities involving cosδ and sinδ?
     
  4. Nov 28, 2009 #3
    yeah.. obv.. i just didnt know how to put this delta symbol on here
     
  5. Nov 28, 2009 #4

    lanedance

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    so which trig identity would you use here?
     
  6. Nov 29, 2009 #5
    cos2x + sin2x = 1
    but then what do i do about the 4 and 5 that are in there?
     
  7. Nov 29, 2009 #6

    rock.freak667

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    Right so

    cos2δ+sin2δ=1

    4 and 5 are constants, so you can just divide by them to get sinδ and cosδ
     
  8. Nov 29, 2009 #7
    so that would mean, cos2δ = x2 = (4cosδ)2?

    x2+y2 = (4cosδ)2 + (5sinδ)2 = 1
    is this correct?
     
  9. Nov 29, 2009 #8

    lanedance

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    no, that doesn't make sense, the identity is:
    cos2δ + sin2δ = 1

    start with that form and think about what you have to divide x & y by to substitute into it
     
  10. Nov 29, 2009 #9

    HallsofIvy

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    [itex](4sin(\delta))^2+ (5cos(\delta))^2= 16sin^2(\delta)+ 25cos^2(\delta)[/itex] is NOT equal to 1. [itex]sin^2(\delta)+ cos^2(\delta)[/itex] is equal to 1.

    If [itex]x= 4 cos(\delta)[/itex] then [itex]cos(\delta)= ?[/itex]
     
  11. Nov 29, 2009 #10
    oh so [itex]x/4= cos(\delta)[/itex]
    and then same thing for [itex]x/5= sin(\delta)[/itex]

    then plug it in the equation..
    [itex](x/5)^2+ (x/4)^2[/itex] = 1

    and solve for x or how do i find cartesian equation of the curve?
     
  12. Nov 29, 2009 #11

    lanedance

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    [itex]x/4= cos(\delta)[/itex] is true

    [itex]x/5= sin(\delta)[/itex] is definitely not, where did you get that?
     
  13. Nov 29, 2009 #12
    that was my mistake... i meant [itex]y/5= sin(\delta)[/itex]

    so from there [itex](y/5)^2+ (x/4)^2[/itex] = 1
    ==> [itex](y^2/25)+ (x^2/16)[/itex] = 1

    what i did next on my paper is, solved for y, and this cartesian equation of the curve..
    is that right?
     
  14. Nov 30, 2009 #13

    lanedance

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    be careful though, there is two solutions for y... the one you should use is based on the orginal domainof [itex] \delta [/itex]
     
  15. Dec 1, 2009 #14
    how do you know what the graph gonna look like though?
    i mean, what if you have a question like this:
    x = 3cos6t
    y = 4sin2t

    I can do the same thing like x/3=cos6t.. cuz then what about this 6 in there?
     
  16. Dec 2, 2009 #15

    lanedance

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    yeah, that is a bit more complicated, the first one simplfies a lot because it is an ellipse

    what do you think will happen? think about cycles
     
  17. Dec 2, 2009 #16
    well cos6t just means it gonna be horizontally compressed (by 1/6 from normal)
    same for sin2t just by 1/2

    the only thing i can come up with is: cos2x = 2sinxcosx
    so would cos6x = 6sinxcosx ? my guess is no..
    then i have no idea what to do :/
     
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