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Thread: Find Point where tangent is Horizontal

  1. #1
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    Find Point where tangent is Horizontal

    Find a point on the curve x3 + y3 = 5xy
    other than the origin at which the tangent line is horizontal.


    Here's what i got so far...

    dy/xy = (5y-3x^2)/(3y^2-5x)

    solved for y... y=3x^2/5


    I have to plug y back into the initial equation right? I keep getting a wrong value .. I'm getting (250/27)^1/3. Can someone tell me what I did wrong and how i can get the right answer?!
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  2. #2
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    Re: Find Point where tangent is Horizontal

    Looks like you got to the right place.

    Plugging $y=\dfrac{3x^2}{5}$ in and solving for $x$ you get 1 real solution

    $x = \dfrac{5}{3}2^{1/3},~y = \dfrac{5}{3} 2^{2/3}$

    In other words you have the right answer.
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  3. #3
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    Re: Find Point where tangent is Horizontal

    Thanks! That's good to know, but i was wondering how you solved for x algebraically? I couldn't for the life of me get x to simplify to your answer :/
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    Re: Find Point where tangent is Horizontal

    Quote Originally Posted by lc99 View Post
    Find a point on the curve x3 + y3 = 5xy
    other than the origin at which the tangent line is horizontal.


    Here's what i got so far...

    dy/xy = (5y-3x^2)/(3y^2-5x)

    solved for y... y=3x^2/5


    I have to plug y back into the initial equation right? I keep getting a wrong value .. I'm getting (250/27)^1/3. Can someone tell me what I did wrong and how i can get the right answer?!
    Also note that the derivative of y does not actually exist at the origin. The graph crosses itself at the origin with two different slopes.

    Find Point where tangent is Horizontal-graph.jpg

    -Dan
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    Re: Find Point where tangent is Horizontal

    Quote Originally Posted by romsek View Post
    Looks like you got to the right place.

    Plugging $y=\dfrac{3x^2}{5}$ in and solving for $x$ you get 1 real solution

    $x = \dfrac{5}{3}2^{1/3},~y = \dfrac{5}{3} 2^{2/3}$

    In other words you have the right answer.
    How do you simplify the (250/27) ^1/3 into that answer is what i mean? Ahh nvmd. I figured that it was like simplifying radicals except that it's cubed root! THanks anyways!
    Last edited by lc99; Oct 8th 2017 at 06:28 PM.
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    Re: Find Point where tangent is Horizontal

    Quote Originally Posted by lc99 View Post
    How do you simplify the (250/27) ^1/3 into that answer is what i mean?
    $250=5^3\cdot 2$ so that $\sqrt[3]{\left(\dfrac{250}{27}\right)}=\dfrac{5\sqrt[3]{2}}{3}$
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