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p4 polars watch

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    arg
    from the heinmann book, exercise 7a q6

    6. The polar equation of a curve C is r^2 = 2sin2$
    $ = theta

    Show that C could be represented parametrically by the equations

    x= 2t/(1+t^4)
    y=2t^3/(1+t^4)

    where t is a parameter


    the cartesian equation is (x^2 + y^2)^2 = 4xy btw

    danke!
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    damn... i would try to help ya... but i got phy6 today... if its not done by the time i get back from phy6... ill give it a shot for ya
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    (Original post by kikzen)
    arg
    from the heinmann book, exercise 7a q6

    6. The polar equation of a curve C is r^2 = 2sin2$
    $ = theta

    Show that C could be represented parametrically by the equations

    x = 2t/(1+t^4)
    y =2t^3/(1+t^4)

    where t is a parameter


    the cartesian equation is (x^2 + y^2)^2 = 4xy btw

    danke!
    x = 2t/(1+t^4)
    y = 2t³/(1+t^4)

    also,

    x = rsinø
    y = rcosø

    dividing y by x gives

    y/x = t²
    t = √(y/x)

    now,

    y - x = (2t³ - 2t)/(1+t^4)
    y - x = 2t(t²-1)/(1+t^4)

    substituting for t and t²,

    y - x = 2√(y/x)(y/x - 1)/(1 + (y/x)²)

    multiplying top and bottom of rhs by x²,

    y - x = 2√(yx)(y - x)/(x² + y²)
    (x² + y²) = 2√(yx)

    now substitute for x = rsinø and y = rcosø,

    r²(sin²ø + cos²ø) = 2r√(cosø.sinø)
    r = 2√(½sin(2ø))
    r² = 2sin(ø)
    ========

    bitte!
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    (Original post by Fermat)
    ...
    danke(this is the only german i know)

    BUT...

    i get how you need to find t in terms of y and x

    but then why do you do y - x = ...
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    (Original post by kikzen)
    danke(this is the only german i know)

    BUT...

    i get how you need to find t in terms of y and x

    but then why do you do y - x = ...
    I want to find a (mathematical) relationship between r and ø.
    I started off getting an expression for t and t² in terms of x and y.
    Then I simply noticed that subtracting x and y would also give me an expression in t and t² (actually t^4).
    So I just did that and substituted, for t and t², giving an eqn in x and y only. Then it was just a matter of using the x = rsinø and y = rcosø eqns
    • Thread Starter
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    (Original post by Fermat)
    I want to find a (mathematical) relationship between r and ø.
    I started off getting an expression for t and t² in terms of x and y.
    Then I simply noticed that subtracting x and y would also give me an expression in t and t² (actually t^4).
    So I just did that and substituted, for t and t², giving an eqn in x and y only. Then it was just a matter of using the x = rsinø and y = rcosø eqns
    hey hey stay there ive got another

    well, its more the integration i cant do

    curve C has polar eq r=asintsin2t between 0 and pi/2
    find the area of C

    guh i used to be good at integration! see what happens when you dont do p3 !
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    I gotta go just now, BUT...
    You have to do an integration involving

    sin²t.sin²(2t)

    which becomes,

    sin(^4)t.cos(^2)t
    sin(^4)t - sin(^6)t

    convert the sin(^4)t using sin²t = ½(1-cos2t), expand the result and use cos²t = ½(1+cos2t)
    Do similar for the sin(^6) term.
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    (Original post by Fermat)
    I gotta go just now, BUT...
    You have to do an integration involving

    sin²t.sin²(2t)

    which becomes,

    sin(^4)t.cos(^2)t
    sin(^4)t - sin(^6)t

    convert the sin(^4)t using sin²t = ½(1-cos2t), expand the result and use cos²t = ½(1+cos2t)
    Do similar for the sin(^6) term.
    thanks ill try again
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    I've just been trying it again. I got an answer of the area of,

    A = (aπ)/256

    But I'm not at all happy with that. I can't find any fault in my working, but when I drew the graph in Graphmatica, I roughly measured the area and got (approx)

    A = 0.28π (using a=1)

    What did you get ?
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    (Original post by Fermat)
    I've just been trying it again. I got an answer of the area of,

    A = (aπ)/256

    But I'm not at all happy with that. I can't find any fault in my working, but when I drew the graph in Graphmatica, I roughly measured the area and got (approx)

    A = 0.28π (using a=1)

    What did you get ?
    ive been doing some other things but im having a go now (just a few lines in i can tell this will take a bit of time )
    but the book got (pi*a^2) / 16
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    (Original post by kikzen)
    ive been doing some other things but im having a go now (just a few lines in i can tell this will take a bit of time :p)
    but the book got (pi*a^2) / 16
    Yeah, I shoulda had a² instead of just a.
    But I found my mistake :)
    I had divided by 4 instead of multiplying by 4. So I was out by a factor of 16. Which is right. That gives me,

    A = (a²π)/256 * 16
    A = πa²/16
    =======

    And, like you said, there is a bit of work involved!
 
 
 
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