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Thread: Tangents to a curve

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    29

    Tangents to a curve

    This is a University level homework question that I am having some difficulty with....

    Find the equation of the line with the slope -1 that is the tangent to the curve
    $\displaystyle
    y=\frac{1}{x-1}
    $

    This is what I believe I should do:
    use the slope to get the equation
    $\displaystyle
    y=-1x+k
    $

    Then set the y-values to equal eachother
    $\displaystyle
    \frac{1}{x-1}=-1x+k
    $

    set it to equal 0:
    $\displaystyle
    -1x^{2}+kx-k=0
    $

    Am I correct so far and where do I go from here?

    Thank You
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  2. #2
    Member
    Joined
    Apr 2009
    Posts
    166
    You're working too hard. Find y'. That's the slope of any line tangent to y at that point.

    $\displaystyle y' = \frac{-1}{(x-1)^2}$

    Line has a slope of -1, set y' to -1 and solve for x

    $\displaystyle -1 = \frac{-1}{(x-1)^2}$

    Occurs @ x = 2

    Plug into original equation to get the point and you get (2, 1)

    Use $\displaystyle y_1 = -1x + b$, plug in your point

    $\displaystyle 1 = 1(2) + b$

    Equation is: $\displaystyle y_1 = -x - 1$
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  3. #3
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
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    If you want a tangent line that has slope of 1, then you must have some $\displaystyle x$ value where $\displaystyle \frac{dy}{dx}=-1$

    $\displaystyle \frac{dy}{dx}=-\frac{1}{(x-1)^{2}}=-1$

    Simply algebra will give you two values where that is true; $\displaystyle x=0,x=2$

    The equation of the line when $\displaystyle x=0$:

    $\displaystyle y-\frac{1}{0-1}=-1(x-0)$

    $\displaystyle y=-x-1$


    And a similar approach for $\displaystyle x=2$
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