A turning point in cartesian coordinates is the solution to
We cannot form this expression in polar coordinates directly. If in polar coordinates we have
as a function of
then
and
so now
and
are expressed in terms of the angle
and we can use the parametric formula for the gradient
to find the gradient at any point.
If we want a turning point then we solve
From this we obtain value(s) for
and can then use
to obtain values for![]()
Example
Find the turning point.

![]()
Clear all the factions by multiplying by
and cancel the non zero constant
obtaining
![]()
Divide throughout by
to obtain![]()
Now use the double angle formula
to express everything in terms of![]()
![]()
Clear all the fractions to obtain![]()