Answer
(a) Since $f\left( {{\bf{r}}\left( t \right)} \right)$ is constant, we conclude that $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = 0$.
(b) Using the Chain Rule for Paths:
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = 0$.
The result agrees with part (a).
Work Step by Step
(a) Since $f\left( {x,y} \right) = {x^2} + {y^2}$ and ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$. So, $f\left( {{\bf{r}}\left( t \right)} \right) = {\cos ^2}t + {\sin ^2}t = 1$.
Since $f\left( {{\bf{r}}\left( t \right)} \right)$ is constant, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = 0$.
(b) We are given $f\left( {x,y} \right) = {x^2} + {y^2}$ and ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$. So,
${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t} \right)$
The gradient of $f$ is
$\nabla f = \left( {2x,2y} \right)$, ${\ \ \ }$ $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {2\cos t,2\sin t} \right)$
Using the Chain Rule for Paths we have
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \nabla {f_{{\bf{r}}\left( t \right)}}\cdot{\bf{r}}'\left( t \right)$
$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \left( {2\cos t,2\sin t} \right)\cdot\left( { - \sin t,\cos t} \right)$
$ = - 2\cos t\sin t + 2\sin t\cos t$
So, $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = 0$.
The result agrees with part (a).
