We make a graphic of this problem to define the angle.
The angle we can calculate through triangle relation, that is,
[tex]sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}[/tex]
With this function we should only calculate the derivate in function of c
[tex]\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}[/tex]
That is the rate of change of [tex]\theta[/tex].
b) At this point we need only make a substitution of 0 for c in the equation previously found.
[tex]\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}[/tex]
Hence we have finally the rate of change when c=0.