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Math Help - implicit differentiation help

  1. #1
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    implicit differentiation help

    A curve is defined implicitly by the equation

    x^3 + y^3 = 6xy

    a)Using Implicit Differentiation
    b)Determine the location of each point on the curve whose tangent line has a slope of -1.

    what i got for a) is

    y' = \frac{6y-3x^2}{3y^2-6x}

    i just want get clarified if thats correct or not,

    and also, how do i get the answer to b? i don't know where to start.

    can anyone help?
    thanks!
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  2. #2
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    your differentiation is correct

    Now set y ' = -1

    6y - 3x^2 =-3y^2 +6x

    3x^2 + 6x + 3y^2 - 6y = 0

    3(x^2 +2x) + 3(y^2 -2y) = 0

    Complete the squares

    3(x+1)^2 + 3(y-1)^2 = 6

    (x+1)^2 + (y-1)^2 =2

    so at all pts on this circle the tangent line to the curve defined by

    has a slope of -1
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  3. #3
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    Quote Originally Posted by break View Post
    A curve is defined implicitly by the equation

    x^3 + y^3 = 6xy

    a)Using Implicit Differentiation
    b)Determine the location of each point on the curve whose tangent line has a slope of -1.

    what i got for a) is

    y' = \frac{6y-3x^2}{3y^2-6x}

    i just want get clarified if thats correct or not,

    and also, how do i get the answer to b? i don't know where to start.

    can anyone help?
    thanks!
    x^3 + y^3 = 6xy

    \frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(6xy)

    3x^2 + \frac{d}{dx}(y^3) = 6\left[x\frac{d}{dx}(y) + y\frac{d}{dx}(x)\right]

    3x^2 + \frac{d}{dy}(y^3)\,\frac{dy}{dx} = 6x\,\frac{dy}{dx} + 6y

    3x^2 + 3y^2\,\frac{dy}{dx} = 6x\,\frac{dy}{dx} + 6y

    3y^2\,\frac{dy}{dx} - 6x\,\frac{dy}{dx} = 6y - 3x^2

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

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


    You need to determine where the tangent line has slope = -1.

    So -1 = \frac{2y - x^2}{y^2 - 2x}

    2x - y^2 = 2y - x^2

    y^2 + 2y = x^2 + 2x

    y^2 + 2y + 1^2 = x^2 + 2x + 1^2

    (y + 1)^2 = (x + 1)^2

    y + 1 = |x + 1|

    y = |x + 1| - 1
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  4. #4
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    i'm sort of lost, am i not supposed to find points?

    points as in the example below,

    ex:
    (3,6)
    (-3,-6)
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  5. #5
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    A function is still a set of points...

    If you have y = |x + 1| - 1, it really means

    y = x + 1 - 1 if x + 1 \geq 0, i.e. x \geq -1

    y = x if x \geq -1.


    It also means

    y = -(x + 1) - 1 if x + 1 < 0, i.e. x < -1

    y = -x - 2 if x < -1.


    So all the points which satisfy these two linear equations will be your solution.
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