Recall that converting from Cartesian to polar coordinates involves the identities

As a function in polar coordinates,

depends on

, so you can write

.
Differentiating the identities with respect to

gives

The slope of the tangent line to

is given by

Given

, you have

. So the tangent line to

has a slope of

When

, the tangent line has slope

This line is tangent to the point

which in Cartesian coordinates is equivalent to

, so the equation of the tangent line is

In polar coordinates, this line has equation


The tangent line passes through the y-axis when

, so the y-intercept is

.
The vector from this point to the point of tangency on

is given by the difference of the vector from the origin to the y-intercept (which I'll denote

) and the vector from the origin to the point of tangency (denoted by

). In the attached graphic, this corresponds to the green arrow.

The angle between this vector and the vector pointing to the point of tangency is what you're looking for. This is given by



The second problem is just a matter of computing the second derivative of

with respect to

and plugging in

.


