Expert: Alon Mandes - 8/24/2008

Question
We have been teached to find the asymptotes of polar curve by using formula-
Rsin(Theta -alpha )=1/f'(alpha )
Where alpha denotes the roots of f (theta)=0
How can we find the asymptote of the polar curve
2r^2=tan2(theta) by using this formula.If there is any links for solving this pls post it as well.


Answer
The formula of the asymptotic line is :

    g'(Ø)
r = ----------
      sin(Ø - Øo)

Where g(Ø)=1/f(Ø) & Øo is the root or solution of g(Ø)=0 or
1/sqrt[0.5tan(2Ø)]=0. Let's solve it :
1/sqrt[0.5tan(2Ø)]=0 means sqrt[0.5tan(2Ø)]=∞ (infinity)
The same as 0.5tan(2Ø)=∞, which also the same as tan(2Ø)=∞.
Now, when does the function tan(2Ø) goes to infinity ? the answer
is when 2Ø=π/2, meaning Ø=π/4, & this is the solution  Øo.
Now let's calculate g'(Ø) : g'(Ø)= [1/sqrt[0.5tan(2Ø)]]'=

0.5*2*1/cos^2(2Ø)          1              
------------------  *   --------           
2*sqrt[0.5tan(2Ø)]       0.5tan(2Ø)             


                           0.5*2*1/cos^2(2*π/4)
So, now we find g'(Øo) =    --------------------     =  ∞
                           2*sqrt[0.5tan(2*π/4)]

Hence, our asymptotic line is Ø= π/4, & r=∞ , which is a vertical
line .

Alon.