Answer
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = \pi /3}} = 1$
Work Step by Step
We have $f\left( {x,y} \right) = {\tan ^{ - 1}}\frac{y}{x}$ and ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$.
Using the Derivative of Inverse Trigonometric Functions (Section 7.8 on page 371):
${f_x} = \frac{1}{{{{\left( {y/x} \right)}^2} + 1}}\left( { - \frac{y}{{{x^2}}}} \right) = - \frac{y}{{{x^2} + {y^2}}}$
${f_y} = \frac{1}{{{{\left( {y/x} \right)}^2} + 1}}\left( { \frac{1}{x}} \right) = \frac{x}{{{x^2} + {y^2}}}$
So,
$\nabla f = \left( { - \frac{y}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right)$
$\nabla {f_{{\bf{r}}\left( t \right)}} = \left( { - \sin t,\cos t} \right)$
${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t} \right)$
Using the Chain Rule for Paths (Theorem 2 of Section 15.5):
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \nabla {f_{{\bf{r}}\left( t \right)}}\cdot{\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t} \right)\cdot\left( { - \sin t,\cos t} \right) = 1$
Since $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = 1$ for all $t$, we have $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = \pi /3}} = 1$.
