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Polar tangent lines

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    r=2-3cosθ Find the tangent line at any point, and at the point (2,∏) Find the tangent line(s) at the pole


    2. Relevant equations

    Do I have to use x=rcosθ and y=rsinθ to convert it to rectangular to find slopes?


    3. The attempt at a solution

    Is the point 2∏ even a point in the graph. It is a limicon (sp?) graph?
     
  2. jcsd
  3. Mar 28, 2012 #2
    You can find the tangent in any coordinate system. You can also convert into rectangular coordinates but it will probably require a lot more work.
     
  4. Mar 28, 2012 #3

    HallsofIvy

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    "The point [itex]2\pi[/itex]" doesn't make any sense. Was that a typo for [itex](2, \pi)[/itex], the given point? When [itex]\theta= \pi[/itex], [/itex]r= 2- 3cos(\pi)= 5[/itex] so, no, that point is not on the graph. However, [itex]r= 2- 3cos(\pi/2)= 2[/itex] so perhaps that was what was meant. Alternatively, perhaps the problem intended [itex]r= 2- 3sin(\theta)[/itex].

    The slope of the tangent line at any point is, by definition, dy/dx. Since, in polar coordinates, [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex], we have [itex]dx= dr cos(\theta)- r sin(\theta)d\theta[/itex] and [itex]dy= dr sin(\theta)d\theta+ r cos(\theta) d\theta[/itex] so that
    [tex]\frac{dy}{dx}= \frac{cos(\theta) dr- r sin(\theta)d\theta}{sin(\theta)dr+ r cos(\theta)d\theta}= \frac{cos(\theta)\frac{dr}{d\theta}- r sin(\theta)}{sin(\theta)\frac{dr}{d\theta}+ r cos(\theta)}[/tex]
    That can be simplified.
     
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