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Polar Slope

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data
    What is the slope of the line tangent to the polar curve r=2theta at the point theta= pie/2


    2. Relevant equations
    r = xcos theta
    r= ysin theta


    3. The attempt at a solution
    I kept getting zero but the answer is -2/pie.
    Can anyone give me a hint? :) Thanks!
     
  2. jcsd
  3. Apr 7, 2009 #2

    Mark44

    Staff: Mentor

    Your "relevant equations" are irrelevant and incorrect. They should be
    x = r cos([itex]\theta[/itex])
    y = r sin([itex]\theta[/itex])
    pie is something to eat. [itex]\pi[/itex], the Greek letter pi, is a number.

    For the slope, you want dy/dx. From the equations for x and y in terms of r and [itex]\theta[/itex], you can get y/x = tan([itex]\theta[/itex]), and from this, you can get
    [itex]\theta = tan^{-1}(y/x)[/itex].
    Use these to convert your polar equation into Cartesian form, and then calculate the derivative dy/dx, and evaluate this derivative at the point where [itex]\theta[/itex] is [itex]\pi/2[/itex].
     
  4. Apr 7, 2009 #3
    I apologize for those mistakes.
    However, I am still confused in what you are saying to do.

    Tan (pi/2) is undefined.
    I'm not sure how to get the slope.
    I know the slope is the derivative or dy/dtheta/dx/dtheta but I don't know how that helps me.

    Thanks!
     
  5. Apr 7, 2009 #4

    Mark44

    Staff: Mentor

    The polar curve r = 2[itex]\theta[/itex] is a spiral in the counterclockwise direction. If we're talking about the slope of the tangent line at (pi, pi/2) (polar coordinates), we have to be talking about dy/dx, because the other derivative, dr/d[itex]\theta[/itex] is constant and equal to 2. I drew a quick sketch of this curve and convince myself that dy/dx at the point in question was negative, which agrees with the answer you gave, at least in sign.

    You need to convert your polar equation into Cartesian coordinates, and then take the derivative dy/dx.

    Since [itex]\theta[/itex] = tan-1(y/x), the polar equation becomes
    [tex]\sqrt{x^2 + y^2} = tan^{-1}(y/x)[/tex]
    Rather than trying to solve for y in that equation, I think it would be easier to differentiate implicitly, and then solve for dy/dx in the resulting equation. If you get that far, you want to evaluate the derivative at x = 0, since your polar point (pi, pi/2) has an x-coordinate of 0.
     
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