Answer
(i) false
(ii) true
(iii) false
(iv) true
Work Step by Step
Let $g\left( t \right) = f\left( {{\bf{r}}\left( t \right)} \right)$. The Chain Rule for Paths (Theorem 2 of Section 15.5) gives the derivative:
$g'\left( t \right) = \frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \nabla {f_{{\bf{r}}\left( t \right)}}\cdot{\bf{r}}'\left( t \right)$
(i) At $t=1$, we have $g'\left( 1 \right) = \nabla {f_{{\bf{r}}\left( 1 \right)}}\cdot{\bf{r}}'\left( 1 \right)$.
From Figure 4, we see that ${\bf{r}}'\left( 1 \right)$ is pointing to the right since the particle travels along the path counterclockwise. However, the gradient $\nabla {f_{{\bf{r}}\left( 1 \right)}}$ points in the direction of maximum rate of increase of $f$. So, it is pointing to the left (from level curve $0$ to $2$). Thus, the angle is obtuse between ${\bf{r}}'\left( 1 \right)$ and $\nabla {f_{{\bf{r}}\left( 1 \right)}}$. It follows that $\nabla {f_{{\bf{r}}\left( 1 \right)}}\cdot{\bf{r}}'\left( 1 \right) < 0$. So, $g'\left( 1 \right) < 0$. Therefore, statement (i) is false.
(ii) As the particle travels in the range $1 \le t \le 2$, it moves from level curve $0$ to $-2$, to $-4$, to $-6$ and then moves up again from $-6$ to $-4$, to $-2$, and back again to $0$. So, it hits the bottom at $-6$. We conclude that $g\left( t \right)$ has a local minimum for some $1 \le t \le 2$. Therefore, statement (ii) is true.
(iii) At $t=2$, we have $g'\left( 2 \right) = \nabla {f_{{\bf{r}}\left( 2 \right)}}\cdot{\bf{r}}'\left( 2 \right)$.
From Figure 4, we see that the tangent vector ${\bf{r}}'\left( 2 \right)$ points in the positive direction. Since the level curve moves from $0$ to $2$, $\nabla {f_{{\bf{r}}\left( 2 \right)}} > 0$. So, $g'\left( 2 \right) > 0$. Therefore, statement (iii) is false.
(iv) At $t=3$, we have $g'\left( 3 \right) = \nabla {f_{{\bf{r}}\left( 3 \right)}}\cdot{\bf{r}}'\left( 3 \right)$.
From Figure 4, we see that at $t=3$ the velocity vector ${\bf{r}}'\left( 3 \right)$ is tangent to the level curve $2$. Since $\nabla {f_{{\bf{r}}\left( 3 \right)}}$ is perpendicular to the level curve $2$, $\nabla {f_{{\bf{r}}\left( 3 \right)}}\cdot{\bf{r}}'\left( 3 \right) = 0$. So, $g'\left( 3 \right) = 0$. Therefore, statement (iv) is true.
