A turning point in cartesian coordinates is the solution toWe 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 solveFrom this we obtain value(s) for
and can then use
to obtain values for
ExampleFind the turning point.
Clear all the factions by multiplying byand cancel the non zero constant
obtaining
Divide throughout byto obtain
Now use the double angle formulato express everything in terms of
Clear all the fractions to obtain