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Slopes of tangent lines of parametric curves.

  1. May 5, 2014 #1
    1. The problem statement,ll variables and given/known data
    I have the first and second derivatives of a parametric function and the book is asking for when the slope of the tangent is vertical and horizontal. I get that horizontal is when dy/dx is 0. But what about vertical, is that dy/dx is 1?



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 5, 2014 #2

    LCKurtz

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    If the tangent line has slope 1, wouldn't it be at a 45 degree angle? Think about dx/dy at a vertical tangent line point.
     
  4. May 5, 2014 #3

    Mentallic

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    When it's vertical, dy/dx is undefined. This is an abuse of notation, but it may help you to think of an undefined value as being [itex]1/0 = \pm \infty[/itex] So if you have

    [tex]\frac{dy}{dx}=\frac{f(x)}{g(x)}[/tex]

    Then to find where the tangent is horizontal, you need to evaluate f(x)=0 and to find where it is vertical you calculate g(x)=0. But keep in mind that when solving either one, the x value you find cannot also be a zero of the other. So for example,

    [tex]y=log(x-1)[/tex] x>1

    [tex]\frac{dy}{dx}=\frac{1}{x-1}=\frac{x}{x(x-1)}[/tex] I just added a factor of x into the numerator and denominator for illustrative purposes. It doesn't change the function over real values of x.

    There is a vertical tangent to the function y at x=1, but at x=0 there is neither a vertical or horizontal even though x=0 gives us 0 in the numerator and in the denominator.
     
  5. May 5, 2014 #4
    Thanks so very very much for this clears up a lot.
     
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