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Thread: Finding the equation of a tangent line for an implicit function

  1. #1
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    Finding the equation of a tangent line for an implicit function

    Given:

    $\displaystyle
    xy^{2} + 5x^{2} = 4y
    $

    Find the equation of the tangent line at x = 1

    Say what?

    Deriving the implicit function of y with respect to x yields:
    $\displaystyle
    \frac{dy}{dx} = \frac{5}{y-2}
    $
    but I'm completely stuck on how to find the equation of the tangent line from this (assuming I did this correctly).

    This equation doesn't seem to be a function. I get multiple values of y for x equal 1.

    Was I tricked?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by mbacarella View Post
    Given:

    $\displaystyle
    xy^{2} + 5x^{2} = 4y
    $

    Find the equation of the tangent line at x = 1

    Say what?

    Deriving the implicit function in terms of x yields
    $\displaystyle
    \frac{dy}{dx} = \frac{5}{y-2}
    $
    but I'm completely stuck on how to find the equation of the tangent line from this (assuming I did this correctly).
    Find $\displaystyle f(1)$ (note this is the original function!) to give you the corresponding $\displaystyle y$ coordinate.

    You can then sub the ordered pair $\displaystyle (1, f(1))$ into $\displaystyle \frac{dy}{dx}$
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  3. #3
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    My algebra is broken

    Thanks.

    btw, that implicit function should have read:

    $\displaystyle
    xy^{2} - 5x^{2} = 4y
    $

    Anyway, here's what I got:
    $\displaystyle
    (1, f(1)) = (1, (1)y^{2} - 5(1)^{2} = 4y) = (1, -1)
    $

    Re-evaluating the derivative:
    $\displaystyle
    \frac{d}{dx}( xy^{2} -5x^{2} = 4y ) = \frac{dx}{dx}2y^{1}\frac{dy}{dx} - 10x\frac{dx}{dx} =4y^{0}\frac{dy}{dx} = (1)2y\frac{dy}{dx} - 10x(1) = 4(1)\frac{dy}{dx}
    $

    Now substituting y and x
    $\displaystyle
    = (1)2(-1)\frac{dy}{dx} - 10(1) = 4(1)\frac{dy}{dx} = \frac{dy}{dx} = \frac{10}{2(-1)-4} = -\frac{5}{3}
    $

    AKA
    $\displaystyle
    f^{\prime}(1) = m_{tan} = -\frac{5}{3}
    $

    Using the point-slope form
    $\displaystyle
    y - y_{1} = m_{tan}(x - x_1) = y - (-1) = -\frac{5}{3}(x - 1) = y = -\frac{5}{3}x + \frac{5}{3} - 1
    $

    Or finally?

    $\displaystyle
    y = -\frac{5}{3}x + \frac{2}{3}
    $

    I'd appreciate any comments or criticism on my thought processes or approach, even if the answer isn't wrong.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    For



    Use the product rule on the first term

    y^2 - 2xydy/dx -10 x = 4dy/dx


    dy/dx(-2xy-4) = 10x - y^2

    dy/dx = [ y^2 -10 x]/[2xy+4]

    When x =1 y^2 -4y -5 = 0 (y-5)(y +1) = 0

    There are 2 possibilities (1,5) and (1,-1)

    Now finish
    Last edited by Calculus26; Nov 19th 2009 at 06:22 AM.
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  5. #5
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    Oops

    Thanks for the correction. How did I miss that?

    From the implicit derivation:
    $\displaystyle
    \frac{d}{dx}\left[ xy^{2} -5x^{2} = 4y \right]
    $

    $\displaystyle
    \frac{dx}{dx}y^{2} + x2y^{1}\frac{dy}{dx} - 10x\frac{dx}{dx} =4y^{0}\frac{dy}{dx}
    $

    $\displaystyle
    (1)y^2 + 2xy\frac{dy}{dx} - 10x(1) = 4(1)\frac{dy}{dx}
    $

    $\displaystyle
    y^2 + 2xy\frac{dy}{dx} - 10x = 4\frac{dy}{dx}
    $

    $\displaystyle
    2xy\frac{dy}{dx} - 4\frac{dy}{dx} = 10x - y^{2}
    $

    $\displaystyle
    \frac{dy}{dx}(2xy - 4) = 10x-y^2
    $

    We seem to disagree on this point:
    $\displaystyle
    \frac{dy}{dx} = \frac{10x-y^{2}}{2xy-4}
    $
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  6. #6
    MHF Contributor Calculus26's Avatar
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    yes--- I had a sign mistake and have changed my last post to reflect this.
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  7. #7
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    Ah, thank you.

    Substituting for x and y:
    $\displaystyle
    \frac{dy}{dx} = m_{tan} = \frac{10(1) - (-1)^{2}}{2(1)(-1)-4} = -\frac{9}{6} = -\frac{3}{2}
    $

    Equation of a tangent line in point-slope form
    $\displaystyle
    y - y_{1} = m_{tan}(x - x_1)
    $

    Performing the substitutions:
    $\displaystyle
    y - (-1) = -\frac{3}{2}(x - 1)
    $
    $\displaystyle
    y = -\frac{3}{2}x + \frac{3}{2} - 1
    $

    Or finally?
    $\displaystyle
    y = -\frac{3}{2}x + \frac{1}{2}
    $

    I'm curious why you said there were two possibilities (1, -1) and (1, 5). Wouldn't that make the dependent value y not a function?
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  8. #8
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    n/m

    For completeness sake:

    I don't know why I expected to be unable to find tangent lines of relations. Please disregard my blathering on "not a function".
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