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Math Help - Implicit Differentiation, Tangent Lines

  1. #1
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    Question Implicit Differentiation, Tangent Lines

    Given the curve x+xy+2y^2=6.
    a)Find an expression for the slope of the curve at any point (x,y) on the curve.
    b)Write an expression for the line tangent to the curve at the point (2,1).
    c)Find the coordinates of all other points on this curve with slope equal to the slope at (2,1).
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    a.)  x+xy+2y^2 = 6

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

    Note that due to the chain rule  \frac{d}{dx} (f(y)) = f'(y) \frac{dy}{dx}

     1 + \left( x\frac{dy}{dx} + y \right) +4y\frac{dy}{dx} = 0

     x\frac{dy}{dx} + 4y\frac{dy}{dx}=-1-y

     \frac{dy}{dx}(x+4y) = -(y+1)

     \frac{dy}{dx} = -\frac{y+1}{x+4y}
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    b.)  y-y_0 = m(x-x_0) where  (x_0,y_0) = (2,1)

    We have  m from part a.

     m = -\frac{y_0+1}{x_0+4y_0}

    now just plug in the values for  x_0 and  y_0 and solve for  y and you're done.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    c.) from part b we have that  m = -\frac{1}{3} , so just set  \frac{dy}{dx} = -\frac{1}{3}

     -\frac{y+1}{x+4y} = -\frac{1}{3}

     3y+3 = x+4y

     y = 3-x

    So the slope of the curves tangent line is the same when [tex] y = 3-x [tex] and  x+xy+2y^2 = 6 .

    Solving this system of equations we get  x=6, \; y=-3
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