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    Hi

    Really stuck on 6b of the Jan 09 AQA C4 paper. (I've linked mark scheme and question below). 6a is fine, following this I've rearranged to find dy/dx in terms of x and y, then equated to zero. Tried to find a substitution as similar to y=(x), with no luck.

    Would anybody mind pointing me in the right direction, and explaining how to approach this Q?

    http://www.mrbartonmaths.com/resourc...AQAJan09MS.pdf - mark scheme

    http://www.mrbartonmaths.com/resourc...AQAJan09QP.pdf - question is at the bottom of page 3
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    (Original post by Serine)
    Hi

    Really stuck on 6b of the Jan 09 AQA C4 paper. (I've linked mark scheme and question below). 6a is fine, following this I've rearranged to find dy/dx in terms of x and y, then equated to zero. Tried to find a substitution as similar to y=(x), with no luck.

    Would anybody mind pointing me in the right direction, and explaining how to approach this Q?

    http://www.mrbartonmaths.com/resourc...AQAJan09MS.pdf - mark scheme

    http://www.mrbartonmaths.com/resourc...AQAJan09QP.pdf - question is at the bottom of page 3
    What do you get for dy/dx?
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    (Original post by Zacken)
    What do you get for dy/dx?
    dy/dx=2(1-xy) / x^2 +3y^2
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    (Original post by Serine)
    dy/dx=2(1-xy) / x^2 +3y^2
    Yes, that's good! What do you get when you set that equal to zero?
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    Set dy/dx = 0, rearrange the equation to find y in terms of x then substitute that back into the original equation to eliminate the y.
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    (Original post by Zacken)
    Yes, that's good! What do you get when you set that equal to zero?
    0=2(1-xy) - this sounds really silly but am I correct in rejecting the denominator of the expression as a potential source of solutions? :/
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    (Original post by Serine)
    0=2(1-xy) - this sounds really silly but am I correct in rejecting the denominator of the expression as a potential source of solutions? :/
    Definitely!! That's perfect. Good work. You can simplify and re-arrange that to 2(1-xy) = 0 \Rightarrow 1-xy = 0 \Rightarrow 1 = xy, agreeing with me so far?

    So now you have two equations that x and y satisfy: xy = 1 and x^2y + y^3 = 2x  + 1.

    What if you re-arrange for y in the first equation and plug it into the second?
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    (Original post by Serine)
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    Ta-da! Well done.
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    I think I can see it now - in essence this is just like finding a corresponding value of y for a value of x given by f'(x) at a stationary point, because this function is defined implicitly dy/dx can give the value of x or y at the stationary point in terms of the other variable. Hence when substituting this into the original expression you get an expression of the x co-rd!
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    (Original post by Zacken)
    Ta-da! Well done.
    Thank you so much!!!
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    (Original post by Serine)
    I think I can see it now - in essence this is just like finding a corresponding value of y for a value of x given by f'(x) at a stationary point, because this function is defined implicitly dy/dx can give the value of x or y at the stationary point in terms of the other variable. Hence when substituting this into the original expression you get an expression of the x co-rd!
    Basically. The question gives you two bits of information - the equation of the curve, so x,y satisfies this and then asks you to find the stationary point, so you know that \frac{\mathrm{d}y}{\mathrm{d}x} = 0 this gives you two equations and the rest is just simultaneous equations except leaving your answer as an equation in x instead of solving for it.

    P.S: The reason why you can ignore the denominator is because if you have something of the form \frac{a}{b} = 0, multiply both sides by b to get \frac{a}{b} \times b = 0\times b , simplifying: a = 0.
 
 
 
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