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Math Help - Horizontal Tangent - Implicit Differentiation

  1. #1
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    Horizontal Tangent - Implicit Differentiation

    a curve in the plane is given by the following equation
    x^4+2x^2y+3y^2=6

    A) Find an expression for dy/dx using implicit differentiation
    B) Find the four points on the curve where the tangent line is horizontal

    Ok I know how to solve for the derivative. I got
    dy/dx = (4x^3-4xy)/(2x^2+6y)

    However Im not quite sure how to solve when the tangent is horizontal. I that the tangent is horizontal when the derivative equals 0. However I don't know how solve for this when it is an implicit derivative.
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    Quote Originally Posted by rawkstar View Post
    a curve in the plane is given by the following equation
    x^4+2x^2y+3y^2=6

    A) Find an expression for dy/dx using implicit differentiation
    B) Find the four points on the curve where the tangent line is horizontal

    Ok I know how to solve for the derivative. I got
    dy/dx = (4x^3-4xy)/(2x^2+6y)

    However Im not quite sure how to solve when the tangent is horizontal. I that the tangent is horizontal when the derivative equals 0. However I don't know how solve for this when it is an implicit derivative.
    I am assuming your derivative is correct ...

    \frac{dy}{dx} = 0 if 4x^3 - 4xy = 0

    4x(x^2-y) = 0

    x = 0 , y = x^2

    sub x = 0 and find the corresponding y-value(s)

    sub y = x^2 and find the corresponding x-values
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    sub it into the original equation?
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    Quote Originally Posted by rawkstar View Post
    sub it into the original equation?
    yes ... you are trying to find points on the curve.
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    ok so i subbed those into the orignial equation and got two x values and two y values
    do i then plug those into the original to get the corresponding opposite value
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    Quote Originally Posted by rawkstar View Post
    ok so i subbed those into the orignial equation and got two x values and two y values
    do i then plug those into the original to get the corresponding opposite value
    when you sub in x = 0 , you get two values for y ... the two points will be (0,\sqrt{2}) and (0,-\sqrt{2})<br />

    when you sub in x^2 for y, you get two values for y ... square those to get the corresponding x-values.
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