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    The curve C has equation 16y^3 +9x^2 y - 54x=0

    I need to find the coords of the points on C where dy÷dx= 0

    Using implicit differentiation i worked out dy/dx to be = (54-18xy)/ (48y^2 +9x^2)

    How do you work out the coords for this if we have two unknown variables in the equation? Do you take simultaneous equations? I even tried mmaking x and y 0 to work out the other but that doesnt work. Can you explain please?
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    (Original post by Jyashi)
    The curve C has equation 16y^3 +9x^2 y - 54x=0

    I need to find the coords of the points on C where dy÷dx= 0

    Using implicit differentiation i worked out dy/dx to be = (54-18xy)/ (48y^2 +9x^2)

    How do you work out the coords for this if we have two unknown variables in the equation? Do you take simultaneous equations? I even tried mmaking x and y 0 to work out the other but that doesnt work. Can you explain please?
    If you look at your equation for dy/dx, what can you say about when dy/dx = 0?

    (Hint: consider the denominator and the numerator and what 'trick' you can use).
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    (Original post by Jyashi)
    The curve C has equation 16y^3 +9x^2 y - 54x=0

    I need to find the coords of the points on C where dy÷dx= 0

    Using implicit differentiation i worked out dy/dx to be = (54-18xy)/ (48y^2 +9x^2)

    How do you work out the coords for this if we have two unknown variables in the equation? Do you take simultaneous equations? I even tried mmaking x and y 0 to work out the other but that doesnt work. Can you explain please?
    Remember that there's always 16y^3 + 9x^2 y - 54 x = 0 lying around and begging to be used. (that wasn't meant to sound dirty).
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    (Original post by Zacken)
    Remember that there's always 16y^3 + 9x^2 y - 54 x = 0 lying around and begging to be used. (that wasn't meant to sound dirty).
    Oh zacken! You sound so naughty when you talk dirty like that 🙊
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    (Original post by Zacken)
    Remember that there's always 16y^3 + 9x^2 y - 54 x = 0 lying around and begging to be used. (that wasn't meant to sound dirty).
    Well i tried taking all the y figures from the equation that is beggingto be used and divided it by 0 to cancel it out. That gave me a quadratic equation which gives me the answer x=5.68 & x=0.31

    Am i on the right track?
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    (Original post by Jyashi)
    Well i tried taking all the y figures from the equation that is beggingto be used and divided it by 0 to cancel it out. That gave me a quadratic equation which gives me the answer x=5.68 & x=0.31

    Am i on the right track?
    :erm:

    Not at all, you know that \frac{dy}{dx} = 0 \iff 54 - 18xy = 0

    So now you have two simultaneous equations, 54-18xy = 0 and 16y^3 + 9x^2y - 54x = 0
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    (Original post by Jyashi)
    Well i tried taking all the y figures from the equation that is beggingto be used and divided it by 0 to cancel it out. That gave me a quadratic equation which gives me the answer x=5.68 & x=0.31

    Am i on the right track?
    What you've done doesn't seem to get you anyway (dividing by 0 is nasty business )

    What do you think about the previous hint that I've given?
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    (Original post by Zacken)
    :erm:

    Not at all, you know that \frac{dy}{dx} = 0 \iff 54 - 18xy = 0

    So now you have two simultaneous equations, 54-18xy = 0 and 16y^3 + 9x^2y - 54x = 0
    Thanks Zacken saved my ass again. Ill try to work this out abd keep it in my head. I knew when you have two unkown variables you can use simultaneous equations but didnt know exactly how to set it up.
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    (Original post by SeanFM)
    What you've done doesn't seem to get you anyway (dividing by 0 is nasty business )

    What do you think about the previous hint that I've given?
    Hey Sean Fm your hint got me as far as i got with that girl at the pub after i bought her a pint of beer and 2 min later she decided to leave with her friends.

    But thats only cause im really stupid and didnt know what you meant by the denominator "trick"
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    (Original post by Jyashi)
    Hey Sean Fm your hint got me as far as i got with that girl at the pub after i bought her a pint of beer and 2 min later she decided to leave with her friends.

    But thats only cause im really stupid and didnt know what you meant by the denominator "trick"
    :lol: nice analogy.

    If \frac{a}{b} = 0 then the only way for that to be 0 is for a=0. There is no value of b for which a/b = 0 .
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    (Original post by Jyashi)
    Thanks Zacken saved my ass again. Ill try to work this out abd keep it in my head. I knew when you have two unkown variables you can use simultaneous equations but didnt know exactly how to set it up.
    Glad it helped.

    (Original post by SeanFM)
    :lol: nice analogy.

    If \frac{a}{b} = 0 then the only way for that to be 0 is for a=0. There is no value of b for which a/b = 0 .
    Sorry for butting in, by the way.
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    (Original post by Zacken)
    Glad it helped.



    Sorry for butting in, by the way.
    Sorry Zacken i feel really stupid. Cant seem to work out the simultaneous equation either. Ive done it on a simple level before but nothing with x^2 y and y^3 and. I need a breakdown about this if possible. Come to the rescue one more time will you?
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    (Original post by Jyashi)
    Sorry Zacken i feel really stupid. Cant seem to work out the simultaneous equation either. Ive done it on a simple level before but nothing with x^2 y and y^3 and. I need a breakdown about this if possible. Come to the rescue one more time will you?
    54 - 18xy = 0 \Rightarrow x = \frac{54}{18y} - simplify that and then substitute it into the other equation to find y and then plug it back into the above to find x.
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    (Original post by Zacken)
    54 - 18xy = 0 \Rightarrow x = \frac{54}{18y} - simplify that and then substitute it into the other equation to find y and then plug it back into the above to find x.
    You know that saying "hiding in plain sight"? I finally know what it means now...
 
 
 
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