Results 1 to 4 of 4

Math Help - something to do with derivatives?

  1. #1
    Junior Member
    Joined
    Oct 2007
    Posts
    36
    Awards
    1

    something to do with derivatives?

    Can someone please help me out with this problem? I don't even know how to approach it. Thankyou


    The graph of the tilted ellipse x^2 - x*y + y^2 = 3 is shown above. What are the dimensions and the location of the box containing the ellipse?
    Note: the sides of the box are vertical and horizontal and also are tangent to the ellipse.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by coe236 View Post
    Can someone please help me out with this problem? I don't even know how to approach it. Thankyou


    The graph of the tilted ellipse x^2 - x*y + y^2 = 3 is shown above. What are the dimensions and the location of the box containing the ellipse?
    Note: the sides of the box are vertical and horizontal and also are tangent to the ellipse.
    the note says it all. the line you see are the vertical and horizontal tangent lines of the figure. so find the derivative and set it equal to zero to find the horizontal tangent lines, then make its denominator zero to find the vertical ones.

    you do know how to find the derivative here, right? differentiate implicitly
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,807
    Thanks
    116
    Quote Originally Posted by coe236 View Post
    Can someone please help me out with this problem? I don't even know how to approach it. Thankyou


    The graph of the tilted ellipse x^2 - x*y + y^2 = 3 is shown above. What are the dimensions and the location of the box containing the ellipse?
    Note: the sides of the box are vertical and horizontal and also are tangent to the ellipse.
    Hello,

    consider a horizontal tangent to the ellipse. This tangent has the equation

    t: y = k (k is a constant)

    Calculate the intersection between the ellipse and the tangent: You must get only one single point:

    x^2-k\cdot x + k^2 - 3=0~\implies~x=\frac12 \cdot k \pm \sqrt{\frac14 \cdot k^2 - k^2 + 3}

    You get exactly one point if the radical is zero:

    \frac14 \cdot k^2 - k^2 + 3~\implies~k=2\ \vee\ k= -2

    The upper and lower side of the square are placed on the lines with the equations: y = 2 or y = -2

    To get the vertical lines you use x = r. All calculations are nearly the same as above. You'll get x = 2 or x = -2
    Attached Thumbnails Attached Thumbnails something to do with derivatives?-ellip_kasten.gif  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2007
    Posts
    36
    Awards
    1
    Thankyou, I appreciate it
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives and Anti-Derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 6th 2011, 06:21 AM
  2. Replies: 1
    Last Post: July 19th 2010, 04:09 PM
  3. Derivatives with both a and y
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2009, 09:17 AM
  4. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  5. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum