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Parametric Expression

  1. Feb 25, 2006 #1
    The lemniscate of Bernoulli is the curve that is the locus of points the product of whose distances from two fixed centres (called the foci) a distance of 2c apart is the cosntant [tex][c^2[/tex]. If the foci have Cartesian coordinates [tex](\pmc, 0)[/tex] the Cartesian equation of the lemniscate is

    [tex]([x-c]^2 + y^2)([x + c]^2 + y^2) = c^4[/tex]​

    or

    [tex](x^2 + y^2)^2 = 2c^2(x^2 - y^2).[/tex]​

    a) Show that the lemniscate of Bernoulli may be expressed parametrically by

    [tex] x(t) = \sqrt{2c}\frac{\cost}{1 + \sin^2t}[/tex], [tex] y(t) = \sqrt{2c}\frac{costsint}{1 + sin^2t}[/tex]​

    where [tex] t \epsilon[-\pi, \pi). [/tex] For t out of this interval the curve repeats on itself.)
     
  2. jcsd
  3. Feb 25, 2006 #2

    benorin

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    Should read "If the foci have Cartesian coordinates [tex](\pm c, 0)[/tex]..."
     
  4. Feb 25, 2006 #3

    Fermat

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    and x(t) is missing a cos(t) on the numerator
     
  5. Feb 25, 2006 #4

    benorin

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    that explains a lot
     
  6. Feb 25, 2006 #5

    benorin

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    I'll fix the question

    Modified quote:

    note that the c is no longer under the square root
     
    Last edited: Feb 26, 2006
  7. Feb 25, 2006 #6

    benorin

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    a) Just plug-in x=x(t) and y=y(t) into either of the given equations and show that it reduces to an identity and note that x(t) and y(t) are periodic functions, having as a funamental period [tex] t \in [-\pi, \pi). [/tex]. BTW, use \in rather than \epsilon for "element of".
     
  8. Feb 26, 2006 #7
    Thanks very much for fixing my post and for your help! So i just plug

    [tex] x = \sqrt{2c}\frac{cost}{1 + \sin^2t} and y = \sqrt{2c}\frac{costsint}{1 + sin^2t} [/tex] into the 2 equations?
     
    Last edited: Feb 26, 2006
  9. Feb 26, 2006 #8

    Fermat

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    That's right, as benorin posted. Also remember about the rt(2c). It should be rt(2)*c.

    BTW, if you'ld like a challenge, try and find the slope of the curve at the origin :smile:
     
  10. Feb 26, 2006 #9

    0rthodontist

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    Just substituting in doesn't prove it though. It shows that any point of the form (x(t), y(t)) is on the lemniscate--it doesn't show that every point on the lemniscate is of the form (x(t), y(t)).

    I'm not sure what the simplest way to show the latter is. One way I can think of is expanding out your second formula for the lemniscate, which will give you a quadratic equation in x^2. Solve that equation for x^2 and from that find how many solutions there are for a given y (which is the number of values of x that are possible for a given y). Then show that for a given y(t), the parametric equation yields the same number of x(t) as there are solutions x for that y.
     
    Last edited: Feb 26, 2006
  11. Feb 27, 2006 #10
    Ok i'm still stuck on this and get the feeling im doing something silly and overcomplicated. I tried fitting those x and y values to the first equation and got:

    [tex] ([\sqrt{2}c\frac{cost}{1 + sin^2t} - c]^2 + \sqrt{2}c\frac{costsint}{1 + sin^2t}) \times ([\sqrt{2}c\frac{cost}{1 + sin^2t} + c]^2 + \sqrt{2}c\frac{costsint}{1 + sin^2t}) [/tex]

    So thismultiplies to:

    [tex] (2c^2\frac{\cos^2t}{1 + \sin^2t} - c^2 + 2c^2\frac{\cos^2t\sin^2t}{(1 + sin^2t)^2}) \times (2c^2\frac{\cos^2t}{(1 + \sin^2t)^2} + c^2 + 2c^2\frac{\cos^2t\sin^2t}{(1 + sin^2t)^2}) [/tex]

    I have to run, ill continue posting my calculations later, tho im sure its wrong already.
     
    Last edited: Feb 27, 2006
  12. Feb 27, 2006 #11
    Argh why is my latex code not working. There should be cos's and sin's on the top line of the first part
     
  13. Feb 27, 2006 #12

    benorin

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    If you use \cos or \sin, either leave a space after them and before the arguement or use {} around the arguement, for example: \cos t = [tex]\cos t[/tex], but \cost = [tex]\cost[/tex]
     
  14. Feb 27, 2006 #13

    benorin

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    Sometimes I get just a blank space where an unrecognized function is like \log is OK, but \Log is blank.
     
  15. Feb 27, 2006 #14

    benorin

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    And, by the way, using the second form of the equation given is far easier...
     
  16. Feb 27, 2006 #15

    benorin

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    Try the easier equation...

    Let us substitute

    [tex] x(t) = \sqrt{2}c \frac{\cos t}{1 + \sin^2t}[/tex]

    [tex] y(t) = \sqrt{2}c\frac{\cos t \sin t}{1 + sin^2t}[/tex]

    into

    [tex](x^2 + y^2)^2 = 2c^2(x^2 - y^2)[/tex]

    to get


    [tex]\left[\left( \sqrt{2}c \frac{\cos t}{1 + \sin^2t}\right) ^2 + \left( \sqrt{2}c\frac{\cos t \sin t}{1 + sin^2t}\right) ^2 \right] ^2 = 2c^2\left[\left( \sqrt{2}c \frac{\cos t}{1 + \sin^2t}\right) ^2 - \left( \sqrt{2}c\frac{\cos t \sin t}{1 + sin^2t}\right) ^2 \right][/tex]

    now simplify to get something like 0=0 :smile:
     
    Last edited: Feb 27, 2006
  17. Feb 28, 2006 #16
    Im being really dumb here and just cannot simplify this. So far i get:

    [tex] 4c^4\frac{\cos^4 t}{(1 + \sin^2 t)^4} + 4c^4\frac{\cos^4 t\sin^2 t}{(1 + sin^2 t)^4} + 4c^4\frac{\cos^4 t\sin^4 t}{(1 + sin^2 t)^4} = 4c^4(\frac{cos^2 t}{(1 + \sin^2 t)^2} - \frac{\cos^4 t\sin^2 t}{(1 +sin^2 t)^2}) [/tex]

    This somehow equates

    [tex] \frac{\cos^4 t}{(1 + \sin^2 t)^4} + \frac{\cos^4 t\sin^2 t}{(1 + sin^2 t)^4} + \frac{\cos^4 t\sin^4 t}{(1 + sin^2 t)^4} = \frac{cos^2 t}{(1 + \sin^2 t)^2} - \frac{\cos^4 t\sin^2 t}{(1 +sin^2 t)^2} ??? [/tex]
     
    Last edited: Feb 28, 2006
  18. Feb 28, 2006 #17
    ok i finally got this, il code it later when i get a chance, but thanks for your help
     
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