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Polar cooordinates Watch

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    Not sure what my limits are for 7b? Ive worked out r^2 but not sure what limits to put in?
    Thanks
    http://m.imgur.com/5PZbn1R
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    -pi to pi, like it tells you at the start of the question.
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    (Original post by constellarknight)
    -pi to pi, like it tells you at the start of the question.
    http://m.imgur.com/DAECA2U

    Where am i going wrong? The answer is pi/2
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    I got it now, -pi to pi is the range of the entire polar curve, which is the loop shown on the diagram in all four quadrants. If you look at the diagram of just the one loop, theta has maximum value pi/4, at the maximum point of the top half of the loop, and minimum value -pi/4 at the minimum point of the bottom half of the loop. Thus you actually need to integrate between -pi/4 and pi/4, which does give pi/2.
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    (Original post by constellarknight)
    I got it now, -pi to pi is the range of the entire polar curve, which is the loop shown on the diagram in all four quadrants. If you look at the diagram of just the one loop, theta has maximum value pi/4, at the maximum point of the top half of the loop, and minimum value -pi/4 at the minimum point of the bottom half of the loop. Thus you actually need to integrate between -pi/4 and pi/4, which does give pi/2.
    Do you mind helping me with 8a. From post 1.
    Im not entirely sure what I have to do.
    Coshx= 1/2(e^x-e^-x)
    What I tried doing is just putting that into 2cosh^2x-1 which I dont think is right.

    Cheers
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    Yes, that's the right approach. If you get stuck, post your working.
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    Cos (2Th) must remain positive. Thus, you have to find the angle at which cos(2th) becomes negative.
    Cos(2th) = 0
    2th = pi/2
    th = pi/4.
    Beyond this angle, cos(2th) is negative and you get a negative amplitude which is wrong obviously, so thats your maximum angle.
    Then, you integrate from 0 to pi/4. Find the area, then multiply it by two to get the total area.
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    (Original post by constellarknight)
    I got it now, -pi to pi is the range of the entire polar curve, which is the loop shown on the diagram in all four quadrants. If you look at the diagram of just the one loop, theta has maximum value pi/4, at the maximum point of the top half of the loop, and minimum value -pi/4 at the minimum point of the bottom half of the loop. Thus you actually need to integrate between -pi/4 and pi/4, which does give pi/2.
    NEVER do that. Never combine areas under and above the horizontal, always split the integral. It does work in this because it's polar co-ordinates and the area is always positive, but generally it's bad practice and leads to errors.
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    (Original post by constellarknight)
    Yes, that's the right approach. If you get stuck, post your working.
    http://m.imgur.com/pRGz3bJ

    Thats what I have tried to do...
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    You forgot to multiply by the 2 at the start: you worked out cosh^2x-1 rather than 2cosh^2x-1.
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    (Original post by constellarknight)
    You forgot to multiply by the 2 at the start: you worked out cosh^2x-1 rather than 2cosh^2x-1.
    So you get 1/2e^(2x)+1/2e^(-2x)?
    Which is the same as cosh2x? Is that it
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    Yes.
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    (Original post by Super199)
    Do you mind helping me with 8a. From post 1.
    Im not entirely sure what I have to do.
    Coshx= 1/2(e^x-e^-x)
    What I tried doing is just putting that into 2cosh^2x-1 which I dont think is right.

    Cheers
    Do you know how to "convert" a trigonometric identity to a hyperbolic identity?

    From core, you should know that \displaystyle \cos2\theta \equiv 2\cos^{2}\theta - 1. If you've heard of Osborn's rule, it is easy to get from known trig identities to hyperbolic identities. Try converting it yourself from here.
    :yep:
    Spoiler:
    Show
    \displaystyle \cosh2x \equiv 2\cosh^{2}x -1
    By the way, for your integral I'd just use the symmetry of the shape rather than split it into two integrals by integrating it from 0 \ \text{to} \ \pi and then multiply by two to find the whole shape's area.

    \displaystyle 2\times \int_{0}^{\pi}\frac{1}{2}r^{2}\ \mathrm{d}\theta
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    Do you know that if  r \mapsto 0 as \theta \mapsto a then  \theta =a is a tangent to the curve?
    It can show the range of  \theta that you need to integrate between, but in this case they gave you it.
 
 
 
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