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Math Help - Implicit Diff and an Ellipse

  1. #1
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    Implicit Diff and an Ellipse

    Consider the graph of x^2+xy+y^2=1. Use implicit differentiation to find all points on the graph where the tangent line is parallel to the line y=-x.

    Where do I go after I implicitly differentiate this thing?

    2x+y'+2y*y'=0
    y'+2y*y'=-2x
    y'(1+2y)=-2x
    y'=-(2x)/(1+2y)

    Did I do that right?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ebonyscythe View Post
    Consider the graph of x^2+xy+y^2=1. Use implicit differentiation to find all points on the graph where the tangent line is parallel to the line y=-x.

    Where do I go after I implicitly differentiate this thing?

    2x+y'+2y*y'=0
    y'+2y*y'=-2x
    y'(1+2y)=-2x
    y'=-(2x)/(1+2y)

    Did I do that right?
    you did not do this correctly. note that you need the product rule to differentiate the xy

    when you get the correct derivative, solve for y' and set it equal to -1 (the slope of y = -x)
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  3. #3
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    Hello, ebonyscythe!

    Your differentiation is off . . .


    Consider the graph of x^2+xy+y^2\:=\:1
    Use implicit differentiation to find all points on the graph
    where the tangent line is parallel to the line y =-x

    \text{We have: }\;x^2 + \underbrace{xy}_{\text{product}} + y^2\:=\:1\;\;{\color{blue}[1]}
    . . \text{then: }\;2x + \overbrace{xy' + y} + 2yy' \:=\:0

    Then we have: . xy' + 2yy' \:=\:-2x - y

    Factor: . (x + 2y)y' \:=\:-(2x + y)

    . . Hence: . y' \:=\:-\frac{2x + y}{x + 2y}


    \text{The slope of }\, y \,= \,\text{-}x\:\text{  is }-1.

    Set: . -\frac{2x + y}{x + 2y} \:=\:-1\quad\Rightarrow\quad y \,=\,x

    Substitute into [1]: . x^2 + x\!\cdot\!x + x^2\:=\:1\quad\Rightarrow\quad3x^2\:=\:1\quad\Righ  tarrow\quad x \:=\:\pm\frac{\sqrt{3}}{3}
    . . Then: . y \:=\:\pm\frac{\sqrt{3}}{3}

    Answers: . \left(\frac{\sqrt{3}}{3},\:\frac{\sqrt{3}}{3}\righ  t),\;\left(\text{-}\,\frac{\sqrt{3}}{3},\:\text{-}\,\frac{\sqrt{3}}{3}\right)

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