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    (Original post by duncanjgraham)
    Sub in really big postive X values eg X=100 - see where the curve approaches

    Sub in really negative X values eg X= -100- see where the curve approaches

    Sub in an X value extremely close to one side of a known asymptote , eg if an asymptote is X=1 sub in X=1.001 and see where the curve is going

    Sub in an X value extremely close to the OTHER side of the known asymptote , eg for the above you'd sub in X=0.009 and this gives you a very very very good idea of the nature of the curve either side of the asymptotes

    You then find points of intersections and any stationary points, and this is enough information to then intuitively join up the curve correctly
    Thank you so much. I will try to implement these steps from now.
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    (Original post by tangotangopapa2)
    June 2013, Q 8) A line (x + 2y = 2) cuts curve with polar equation (r=1+cos theta ) is two parts. Find the ratio of areas above and below the line enclosed by the curve.
    Mark Scheme does not seem to be helpful either. It simply states points of intersection as (0,1) and (2,0) out of nowhere and then proceeds in alien fashion. Could someone explain to me? I don't even know where to begin.
    The Cartesian equation of curve is x2 + y2 - x = sqrt( x2 + y2).
    I'm out atm, if it hasn't been answered by the time I'm back I'll try work through it


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    (Original post by tangotangopapa2)
    Thank you so much. I will try to implement these steps from now.
    I hope it helps, also if the graph has an unknown constant like y= kx/(X-1)(X-3)... When you're using similar steps to what I gave you, just kinda let k=1 so it doesn't interfere with you're graph drawing

    (You can then put k back into anything important like coordinates AFTER you've got the shape of the graph correct)

    This is what I do anyways
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    (Original post by tangotangopapa2)
    June 2013, Q 8) A line (x + 2y = 2) cuts curve with polar equation (r=1+cos theta ) is two parts. Find the ratio of areas above and below the line enclosed by the curve.
    Mark Scheme does not seem to be helpful either. It simply states points of intersection as (0,1) and (2,0) out of nowhere and then proceeds in alien fashion. Could someone explain to me? I don't even know where to begin.
    The Cartesian equation of curve is x2 + y2 - x = sqrt( x2 + y2).
    I was very confused on this question yesterday and I'm still not sure how the mark scheme has done it. The way I did it was convert x+2y=2 into polar coordinates and draw it with the other graph since it cuts it at theta= 0 and pi/2 when you draw it like that, then it becomes a lot clearer which areas you need to find!
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    (Original post by becksreills)
    I was very confused on this question yesterday and I'm still not sure how the mark scheme has done it. The way I did it was convert x+2y=2 into polar coordinates and draw it with the other graph since it cuts it at theta= 0 and pi/2 when you draw it like that, then it becomes a lot clearer which areas you need to find!
    Thank you, but how do you convert x+2y=2 to polar coordinates. (Also this line does not pass through the origin.)
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    (Original post by tangotangopapa2)
    Thank you, but how do you convert x+2y=2 to polar coordinates. (Also this line does not pass through the origin.)
    x=rcos(theta) and y=rsin(theta) so sub those in and rearrange to r=f(theta)
    It doesn't pass through the origin, but the value of r at theta=0 is the same as the value of r at theta=0 for the other graph so it cuts it there
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    How do I do 7iii. I know the solution but could somebody explain why the answer is what it is.. I have found the point in polar coordinates of P, but can't see how this helps me find the tangent...Name:  image.jpg
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    (Original post by becksreills)
    x=rcos(theta) and y=rsin(theta) so sub those in and rearrange to r=f(theta)
    It doesn't pass through the origin, but the value of r at theta=0 is the same as the value of r at theta=0 for the other graph so it cuts it there
    Thank you
    So, by substituting x=rcos(theta) and y=rsin(theta) and rearranging to r=f(theta), I got r = 2/(cos(theta) + 2 sin (theta)). Then equating 'r' with 'r' of r=cos(theta)sin2(theta). This equation is very hard to solve. (I cant ) but as you said, theta = 0 and pi/2 seem to satisfy the equation. Now how do we find the required areas.

    Clearly, finding total area enclosed by curve and area between theta = 0 and pi/2 won't give the required areas, as area of sector (i.e enclosed by curve and half lines theta = 0 and pi/2) is not what we are interested in. We are interested in areas cut by lines. I couldn't find a way to proceed.
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    (Original post by Mathematicus65)
    How do I do 7iii. I know the solution but could somebody explain why the answer is what it is.. I have found the point in polar coordinates of P, but can't see how this helps me find the tangent...Name:  image.jpg
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    The line of symmetry is y=x. At P, (intuitively) the gradient of the tangent should be perpendicular to the line of symmetry. So the gradient at this point is -1. Now the equation of line passing through P and having gradient -1 is the required equation of line.
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    IN JUNE 2012 YOU ARE MADE TO FACTORISE A CUBIC EQUATION IN TWO DIFFERENT QUESTIONS - wtf? I don't even know how to do that **** methodically?

    *edit* it's a paper with low grade boundaries and you don't lose too many marks for not being able to factorise the cubic , still annoying tho: P
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    (Original post by tangotangopapa2)
    The line of symmetry is y=x. At P, (intuitively) the gradient of the tangent should be perpendicular to the line of symmetry. So the gradient at this point is -1. Now the equation of line passing through P and having gradient -1 is the required equation of line.
    Thank you!
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    (Original post by tangotangopapa2)
    Thank you
    So, by substituting x=rcos(theta) and y=rsin(theta) and rearranging to r=f(theta), I got r = 2/(cos(theta) + 2 sin (theta)). Then equating 'r' with 'r' of r=cos(theta)sin2(theta). This equation is very hard to solve. (I cant ) but as you said, theta = 0 and pi/2 seem to satisfy the equation. Now how do we find the required areas.

    Clearly, finding total area enclosed by curve and area between theta = 0 and pi/2 won't give the required areas, as area of sector (i.e enclosed by curve and half lines theta = 0 and pi/2) is not what we are interested in. We are interested in areas cut by lines. I couldn't find a way to proceed.
    You don't need to convert x+2y=2 in to polar form. Draw the polar function onto an x-y grid and you can see where it intersects.

    Here's my diagram:


    And here's my working if you're interested
    Spoiler:
    Show
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    (Original post by duncanjgraham)
    IN JUNE 2012 YOU ARE MADE TO FACTORISE A CUBIC EQUATION IN TWO DIFFERENT QUESTIONS - wtf? I don't even know how to do that **** methodically?

    *edit* it's a paper with low grade boundaries and you don't lose too many marks for not being able to factorise the cubic , still annoying tho: P
    I don't remember these questions? Which ones are you talking about?
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    (Original post by Parallex)
    You don't need to convert x+2y=2 in to polar form. Draw the polar function onto an x-y grid and you can see where it intersects.

    Here's my diagram:


    And here's my working if you're interested
    Spoiler:
    Show
    Thank you so, so much. Brilliant solution.
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    (Original post by duncanjgraham)
    IN JUNE 2012 YOU ARE MADE TO FACTORISE A CUBIC EQUATION IN TWO DIFFERENT QUESTIONS - wtf? I don't even know how to do that **** methodically?

    *edit* it's a paper with low grade boundaries and you don't lose too many marks for not being able to factorise the cubic , still annoying tho: P
    Which question?


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    (Original post by marioman)
    I don't remember these questions? Which ones are you talking about?
    june 2012 q 3ii and 8iii)
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    (Original post by drandy76)
    Which question?


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    june 2012 3ii and 8iii
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    (Original post by tangotangopapa2)
    Thank you so, so much. Brilliant solution.
    No problem. :P
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    (Original post by duncanjgraham)
    june 2012 3ii and 8iii
    Doesn't 3 ii reduce to a quadratic? At least that's what it looks like from a glance


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    (Original post by duncanjgraham)
    june 2012 3ii and 8iii
    Thats really annoying but everytime you have cubic equation then try putting x=1,x=2 and x=-1. This should give you one of the factors. (They should not give cubics with hard to guess factors)
 
 
 
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