If tan^2x=1 the expression is undefined, this occurs when x=(pi/4)+kpi/2 for k=0,1,2...n a positive integer. Similar to if x=0, lnx is undefined and so there is a asymptote there.

Reply 8

roshanhero

So basically does it means that when "r" tends to infinity,there is asymptote.How can we picturise it graphically.Can you post some graph of it as well,pls.

Reply 9

qgujxj39

roshanhero

So basically does it means that when "r" tends to infinity,there is asymptote.How can we picturise it graphically.Can you post some graph of it as well,pls.

Well r represents the distance of the point of the curve from the pole, so if r tends to infinity, it gets greater and greater and closer and closer to the line

\theta = \alpha

(where

\alpha

is the angle of the point for which r is undefined), and any line which a curve grows arbitrarily close to without ever touching it is, by definition, an asymptote.

Reply 10

Ze_Rebel_Mole

I have just quickly composed a graph using a plotter, which can be found in the attachment (sorry for its very large size!).

An explanation

Untitled-1.jpg286.5KB

Reply 11

roshanhero

Thanks a lot for all your help,now I get it.Here is the second question.
find the asymptote of
r^2cos2(theta)=a^2
If I plug in cos2(theta) equals zero,I would get r=infinity,so does it means that I would get asymptote at theta=45 degrees.

Reply 12

roshanhero

Can we have some posts here.

hey all

Yes

What about when

\tan{2\theta} < 0

Reply 16

roshanhero

n1r4v

Yes

But the answer is different.

What's the answer.

We have been teached to find the asymptotes of polar curve by using formula-
Rsin(q -a )=1/f'(a )

Where a denotes the roots of f (q )=0

I appiled it but failed.

Can we show me how to do it by using this formula.