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Thread: equztion of tangent

  1. December 15th 2008, 08:15 AM #1
    trythis
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    equztion of tangent

    (a) Find equation of the tangent line and the normal line to the curve
    y
    = (3x -2)e^-xat the point (0;-2);
    (b) Find     dy/dx given that cos(x + y) + y = 0.
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  2. December 15th 2008, 09:56 AM #2
    chabmgph
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    Quote Originally Posted by trythis View Post
    (a) Find equation of the tangent line and the normal line to the curve
    y
    = (3x -2)e^-xat the point (0;-2);
    (b) Find     dy/dx given that cos(x + y) + y = 0.
    Tangents and normals
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  3. December 15th 2008, 10:14 AM #3
    Greengoblin
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    1.)

    y=(3x-2)e^{-x}

    \frac{dy}{dx}=3e^{-x}-(3x-2)e^{-x}=(5-3x)e^{-x}

    y'(0)=5

    y-y_1=m(x-x_1)

    y+2=5(x-0)

    y=5x-2

    m_1=-\frac{1}{m}=-\frac{1}{5}

    y-y_1=m_1(x-x_1)

    y+2=-\frac{1}{5}(x-0)

    y=-\frac{x}{5}-2

    2.)

    \cos(x+y)+y=0

    y=-\cos(x+y)

    y'=\frac{d}{dx}(-\cos(x+y))

    y'=(1+y')\sin(x+y)=\sin(x+y)+y'\sin(x+y)

    y'-y'\sin(x+y)=\sin(x+y)

    y'[1-\sin(x+y)]=\sin(x+y)

    y'=\frac{\sin(x+y)}{1-\sin(x+y)}
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