Results 1 to 4 of 4

Math Help - Find all points where the tangent line is horizontal.

  1. #1
    Junior Member moonman's Avatar
    Joined
    Jan 2009
    From
    Earth, the best place in the solar system.
    Posts
    66

    Find all points where the tangent line is horizontal.

    Find all points where the tangent line is horizontal.
    The equation is x^(2) + xy + y^(2) = 1

    Okay so from a previous thread I learned that the horizontal tangent line is when a single point touches the equation and in this case the cirlce/oval.

    I took the derivative and i got:
    2x + xy' + y + 2y^(2) y' = 0

    I'm looking at the graph of this and I think there are only two horizontal tangents but I dont know how to compute them. Please show me how. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2006
    From
    San Francisco
    Posts
    145
    2x + xy' + y + 2y^(2) y' = 0

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

    y'= -(2x+y)/(x+2y^2)

    so where y'=0 there's a horizontal tangent so:

    -(2x+y)=0
    y+2x=0 divided by -1 but 0/-1 =0 so


    where y=-2x there will be a tangent line i think you can sub this back into your original so

    x^(2) + x*(-2x)+ (-2x)^(2) = 1

    so x^2-2x^2+4x^2=1

    so 3x^2=1

    x= (1/3)^(1/2)


    i did this on the computer while typing it and my maths a bit rusty but there


    ***********mistake it should be x= + or - (1/3)^(1/2)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by moonman View Post
    Find all points where the tangent line is horizontal.
    The equation is x^(2) + xy + y^(2) = 1

    Okay so from a previous thread I learned that the horizontal tangent line is when a single point touches the equation and in this case the cirlce/oval.

    I took the derivative and i got:
    2x + xy' + y + 2y^(2) y' = 0 Mr F says: Wrong. It's 2x + xy' + y + 2y y' = 0. Now solve this equation for y'.

    I'm looking at the graph of this and I think there are only two horizontal tangents but I dont know how to compute them. Please show me how. Thanks.
    \frac{dy}{dx} = - \frac{2x + y}{x + 2y}.

    \frac{dy}{dx} = 0 \Rightarrow 2x + y = 0.

    So you have the following equations:

    x^2 + xy + y^2 = 1 .... (1)

    2x + y = 0 .... (2)

    Solve equations (1) and (2) simultaneously: x \pm \frac{1}{\sqrt{3}}. Therefore y = \, .....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2006
    From
    San Francisco
    Posts
    145
    Okay i didn't bother checking your other math, nice catch Mr. Fantastic,

    my works is still fine, just change my denominator even though I get the same answer

    Oh yeah forgot to find the way but as mr. fantastic said "..." -you can do that yourself
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Stationary points - find horizontal line
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 8th 2012, 06:29 AM
  2. Replies: 5
    Last Post: January 11th 2009, 11:19 PM
  3. Replies: 2
    Last Post: December 9th 2008, 03:44 AM
  4. Replies: 2
    Last Post: October 26th 2008, 05:28 AM
  5. Replies: 0
    Last Post: October 24th 2007, 06:23 PM

Search Tags


/mathhelpforum @mathhelpforum