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    Find the coordinates of the turning points on the curve with equation

    y^3 + 3x(y^2) - x^3 = 3

    So I differentiated implicitly to get

    dy/dx = (3x^2 - 3y^2)/(3y^2 + 6xy)

    Put it equal to 0

    3x^2 - 3y^2 = 0

    so x = y ... doesnt really help

    I tried rearranging the original equation to get y^3 in terms of x but there is still y's in there and they dont cancel, so Im still left with an equation that has both x's and y's in. Any ideas on how to do this?

    Quick answers would be appreciated, Im at school :/
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    firstly, x = +/- y

    secondly, you stick x=y into the original eqn, y^3 + 3x(y^2) - x^3 = 3, to get

    x³ + 3x³ - x² = 3

    x³ = 1 => x = 1 => (1,1) is a turning point.

    do the same with x=-y

    -x³ + 3x³ - x³ = 3

    x³ = 3 => x = cbrt(3) => (cbrt(3),-cbrt(3)) is a turning point.
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    (Original post by imasillynarb)
    Find the coordinates of the turning points on the curve with equation

    y^3 + 3x(y^2) - x^3 = 3
    OK, differentiating implicitly gives:

    3y^2 dy/dx + 3y^2 + 6xy dy/dx - 3x^2 = 0.

    remove common factors, and rearrange to get:

    dy/dx (y^2 + 2xy) = x^2 - y^2

    or dy/dx = (x - y) (x + y)/ (y)(y + 2x), ie there is a turning point where
    x = y or x = -y.

    so you just substitute x = y, or x = -y into your curve equation, to solve, and get coordinates...

    eg x^3 +3x^3 - x^3 = 3
    and -x^3 + 3x^3 - x^3 = 3.

    so either x = 1, or x = cube root 3.

    have i gone wrong anywhere?
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    (Original post by elpaw)
    x³ = 3 => x = cbrt(3) => (cbrt(3),-cbrt(3)) is a turning point.
    bah i always post 10 seconds too late... :confused:
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    (Original post by 4Ed)
    have i gone wrong anywhere?
    if you have, then so have i.
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    They reckon the answers are (1,1) and (10/3,10.3)
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    (Original post by imasillynarb)
    They reckon the answers are (1,1) and (10/3,10.3)
    you do realise that the book is often wrong!

    i think both Elpaw and I have the correct answers.... what do you think yourself, based on your working?
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    --------------------------------------------------------------------------------

    Hey to anyone out there struggling with differentiation or finding it hard tocheck answers correctly then i found a site which has various programs including one on differentiation that will differentiate for you, using many different ways. Its not on the site as i emailed the siteowner directly and he sent me the program. however if u want it jus do the same.....

    The site is www.geocities.com/softpro101
 
 
 

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