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Math Help - Implicit Line Finding?

  1. #1
    No one in Particular VonNemo19's Avatar
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    Implicit Line Finding?

    Find the equation of the line(s) tangent to the curve x^3+2xy-y^2=-2 at the point(s) where x=1.

    My answer is

    y=-\frac{3}{4}x+\frac{15}{4}

    and

    y=-\frac{5}{4}x+\frac{1}{4}.

    Is this correct?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by VonNemo19 View Post
    Find the equation of the line(s) tangent to the curve x^3+2xy-y^2=-2 at the point(s) where x=1.

    My answer is

    y=-\frac{3}{4}x+\frac{15}{4}

    and

    y=-\frac{5}{4}x+\frac{1}{4}.

    Is this correct?
    It would be better to post your work rather than just your answers.
    But since you do: No, those are not correct at all. Post your work and we might be able to point out where you went wrong.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    It would be better to post your work rather than just your answers.
    But since you do: No, those are not correct at all. Post your work and we might be able to point out where you went wrong.
    x^3+2xy-y^2=-2

    Differentiating with respect to x...

    3x^2+2x*\frac{dy}{dx}+y*2-2y*\frac{dy}{dx}=0

    (2x-2y)\frac{dy}{dx}=-3x^2-2y

    \frac{dy}{dx}=\frac{-3x^2-2y}{2x-2y}

    Now, the points of tangecy are as follows

    (1)^3+2(1)y-y^2=-2
    y^2-2y-3=0
    (y-3)(y+1)=0\Rightarrow{y=3,-1}

    Therefore

    \frac{dy}{dx}\text{ at x=1 is }

    \frac{-3(1)^2-2(3)}{2(1)-2(3)}=\frac{9}{4}

    and

    \frac{-3(1)^2-2(-1)}{2(1)-2(-1)}=\frac{-1}{4}

    So finding the equation...

    Is this right so far? is there an easier way to do this?
    Last edited by VonNemo19; June 14th 2009 at 12:54 PM.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    x^3+2xy-y^2=-2

    Differentiating with respect to x...

    3x^2+2x*\frac{dy}{dx}+y*2-2y*\frac{dy}{dx}=0

    (2x-2y)\frac{dy}{dx}=-3x^2-2y

    \frac{dy}{dx}=\frac{-3x^2-2y}{2x-2y}

    Now, the points of tangecy are as follows

    (1)^3+2(1)y-y^2=-2
    y^2-2y-3=0
    (y-3)(y+1)=0\Rightarrow{y=3,-1}

    Therefore

    \frac{dy}{dx}\text{ at x=1 is }

    \frac{-3(1)^2-2(3)}{2(1)-2(3)}=\frac{9}{4}

    and

    \frac{-3(1)^2-2(-1)}{2(1)-2(-1)}=\frac{-1}{4}

    So finding the equation...

    Is this right so far? is there an easier way to do this?
    yes, you are correct so far. now what are the tangent lines for each point? (state the point as well)
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