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Math Help - equation of the tangent line?

  1. #1
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    equation of the tangent line?

    the question is
    find the equation of the tangent line to the curve 2x^3 + 2y^3 = 9xy when x=1, y=2


    i tried
    d2x^3/dx + d2y^3/dx = d9xy/dx

    d2x^3/dx = 6x^2
    d2y^3/dx = 6y^2
    d9xy/dx = ?

    i don't know how to do
    how do i do this problem?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by haebinpark View Post
    the question is
    find the equation of the tangent line to the curve 2x^3 + 2y^3 = 9xy when x=1, y=2


    i tried
    d2x^3/dx + d2y^3/dx = d9xy/dx

    d2x^3/dx = 6x^2
    d2y^3/dx = 6y^2
    d9xy/dx = ?

    i don't know how to do
    how do i do this problem?
    Use implicit differentiation. The RHS will require the product rule.

    \frac{d}{dx}(2x^3+2y^3) = \frac{d}{dx}(9xy)

    Looking at the RHS:

    u = 9x \: \rightarrow \: u' = 9

    v = y  \: \rightarrow \: v' = \frac{dy}{dx}

    \frac{d}{dx}(9xy) = 9y + 9x \frac{dy}{dx}

    The LHS is simpler:

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

    Set both sides equal to each other and isolate dy/dx. You can then sub in the value you know to find the gradient of the tangent

    Spoiler:
    6x^2 + 6y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}

    9x \frac{dy}{dx} - 6y^2 \frac{dy}{dx} = 9y - 6x^2

    \frac{dy}{dx}(9x-6y^2) = 9y - 6x^2

    A factor of 3 will also cancel out: \frac{dy}{dx} = \frac{3y-2x^2}{3x-2y^2}
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  3. #3
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    but
    you have dy/dx on both side?
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  4. #4
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    anyone...?
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    Use implicit differentiation. The RHS will require the product rule.

    \frac{d}{dx}(2x^3+2y^3) = \frac{d}{dx}(9xy)

    Looking at the RHS:

    u = 9x \: \rightarrow \: u' = 9

    v = y  \: \rightarrow \: v' = \frac{dy}{dx}

    \frac{d}{dx}(9xy) = 9y + 9x \frac{dy}{dx}

    The LHS is simpler:

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

    Set both sides equal to each other and isolate dy/dx. You can then sub in the value you know to find the gradient of the tangent

    Spoiler:
    6x^2 + 6y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}

    9x \frac{dy}{dx} - 6y^2 \frac{dy}{dx} = 9y - 6x^2

    \frac{dy}{dx}(9x-6y^2) = 9y - 6x^2

    A factor of 3 will also cancel out: \frac{dy}{dx} = \frac{3y-2x^2}{3x-2y^2}






    can i go

    from 6x^2 + 6y^2 dy/dx = 9y+9x dy/dx

    to 6y^2 dy/dx - 9x dy/dx = 9y - 6x^2 ?


    and after finding dy/dx
    how do i find the equation?
    because that was the question... hmm
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