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Data-Roxas

quick maths question

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by , 3rd December 2012 at 02:05 AM (338 Views)
alright, so I was doing my Calc 2 homework, and this problem came up:

Instructions: find the area of the region that lies inside both (polar) curves.

I'm given r^2=sin(2 * 'theta') and r^2=cos(2 * 'theta')

according to the back of the text book, I'm supposed to get (pi/8) - .25 at the end, but i'm not seeing how that answer is possible. I checked Yahoo answers, and found this-http://answers.yahoo.com/question/index;_ylt=Av1k8SdaO1Bjo9PtNMh8BogjzKIX;_ylv=3?qid=20120814143312AAHpW0X

but this doesn't get the answer in the book. I'm currently going to move on to another problem, and check this in a minute or so. In the meantime, all youse fellow math peeps should try to help me out...

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  1. Froakie's Avatar
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    Not gonna look at the link or I'll get put off

    let theta = a so it's easier
    sin2a = cos2a
    sin2a/cos2a = 1
    tan2a = 1
    a = 1/2arctan(1)
    a = 1/2*π/4
    a = π/8
    Therefore they intersect when a = π/8

    Find the first area between 0 and π/8
    1/2∫[0,π/8] cos(2a) da
    = 1/2[1/2sin2a]|0,π/8
    = 1/4[sin(π/4)-sin(0)]
    = 1/4[1/√2]
    = 1/4√2

    Do you know what the graph of cos2a and sin2a look like? They're like two loops about the origin but slightly rotated from each other. What I found above was a small quarter of the area you want, you can just times it by four and you get 1/√2 or √2/2 which is what that yahoo answer apparently got.

    I can't possibly see how it could be π/8 - 1/4 since what you are integrating doesn't have any constants, and therefore would not give you any theta to sub in values of pi from. Using the double angle formulae is unnecessary since integrating cos^2(a) or sin^2(a) (which would give you the constant that you need) is impossible without converting it back to cos2a/sin2a.

    Sorry I couldn't help more, but for all purposes since this is a simple problem, getting what the text book says is pretty impossible, and unless I'm not seeing it either, the book is maybe wrong. x)

  2. Karamazov's Avatar
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    I just saw math and backed into a corner, sobbing.

    But of course Croag would know what to do. :P
    Yoshi-san likes this.
  3. Data-Roxas's Avatar
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    Quote Originally Posted by Croag
    Not gonna look at the link or I'll get put off

    let theta = a so it's easier
    sin2a = cos2a
    sin2a/cos2a = 1
    tan2a = 1
    a = 1/2arctan(1)
    a = 1/2*π/4
    a = π/8
    Therefore they intersect when a = π/8

    Find the first area between 0 and π/8
    1/2∫[0,π/8] cos(2a) da
    = 1/2[1/2sin2a]|0,π/8
    = 1/4[sin(π/4)-sin(0)]
    = 1/4[1/√2]
    = 1/4√2

    Do you know what the graph of cos2a and sin2a look like? They're like two loops about the origin but slightly rotated from each other. What I found above was a small quarter of the area you want, you can just times it by four and you get 1/√2 or √2/2 which is what that yahoo answer apparently got.

    I can't possibly see how it could be π/8 - 1/4 since what you are integrating doesn't have any constants, and therefore would not give you any theta to sub in values of pi from. Using the double angle formulae is unnecessary since integrating cos^2(a) or sin^2(a) (which would give you the constant that you need) is impossible without converting it back to cos2a/sin2a.

    Sorry I couldn't help more, but for all purposes since this is a simple problem, getting what the text book says is pretty impossible, and unless I'm not seeing it either, the book is maybe wrong. x)

    thanks, That's what I was thinking too, but I needed an outside source, fresh on the problem to confirm it. Where's the +1 Karma button on BMGf, again?

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