Answer
(a) ${\bf{r}}\left( t \right)$ parametrizes a helix of radius $R$ and height $h$ making $N$ complete turns.
(b) The radius of the spring in Figure 5 (A) is larger, we guess that it uses more wire than the other one.
(c) The length of the spring in Figure 5 (A) is $\simeq 132.01$.
The length of the spring in Figure 5 (B) is $\simeq 125.7$.
we conclude that the spring in Figure 5 (A) uses more wire than the one in Figure 5 (B).
Work Step by Step
(a) We have
${\bf{r}}\left( t \right) = \left( {R\cos \left( {\frac{{2\pi Nt}}{h}} \right),R\sin \left( {\frac{{2\pi Nt}}{h}} \right),t} \right)$
for $0 \le t \le h$
The component functions are
$x\left( t \right) = R\cos \left( {\frac{{2\pi Nt}}{h}} \right)$,
$y\left( t \right) = R\sin \left( {\frac{{2\pi Nt}}{h}} \right)$,
$z\left( t \right) = t$
So,
$x{\left( t \right)^2} + y{\left( t \right)^2} = {R^2}{\cos ^2}\left( {\frac{{2\pi Nt}}{h}} \right) + {R^2}{\sin ^2}\left( {\frac{{2\pi Nt}}{h}} \right)$
$x{\left( t \right)^2} + y{\left( t \right)^2} = {R^2}$
Thus, the curve is a circle of radius $R$ on the $xy$-plane.
The height of the curve is represented by $z\left( t \right) = t$. At $t=h$, the height is $h$. For the interval $0 \le t \le h$, the circle on the $xy$-plane has made rotation of angle $\frac{{2\pi Nh}}{h} = 2\pi N$, which corresponds to $N$ complete turns. Hence, ${\bf{r}}\left( t \right)$ parametrizes a helix of radius $R$ and height $h$ making $N$ complete turns.
(b) Comparing the two figures in Figure 5; and since the radius of the spring in Figure 5 (A) is larger, we guess that it uses more wire than the other one.
(c) Recall the parametrization in part (a):
${\bf{r}}\left( t \right) = \left( {R\cos \left( {\frac{{2\pi Nt}}{h}} \right),R\sin \left( {\frac{{2\pi Nt}}{h}} \right),t} \right)$
for $0 \le t \le h$
The spring in Figure 5 (A) has $R=7$, $N=3$ and $h=4$, so
${{\bf{r}}_A}\left( t \right) = \left( {7\cos \left( {\frac{{2\pi \cdot 3\cdot t}}{4}} \right),7\sin \left( {\frac{{2\pi \cdot 3\cdot t}}{4}} \right),t} \right)$
${{\bf{r}}_A}\left( t \right) = \left( {7\cos \frac{{3\pi }}{2}t,7\sin \frac{{3\pi }}{2}t,t} \right)$
for $0 \le t \le 4$
The derivative is
${{\bf{r}}_A}'\left( t \right) = \left( { - \frac{{21\pi }}{2}\sin \frac{{3\pi }}{2}t,\frac{{21\pi }}{2}\cos \frac{{3\pi }}{2}t,1} \right)$
The spring in Figure 5 (B) has $R=4$, $N=5$ and $h=3$, so
${{\bf{r}}_B}\left( t \right) = \left( {4\cos \left( {\frac{{2\pi \cdot 5\cdot t}}{3}} \right),4\sin \left( {\frac{{2\pi \cdot 5\cdot t}}{3}} \right),t} \right)$
${{\bf{r}}_B}\left( t \right) = \left( {4\cos \frac{{10\pi }}{3}t,4\sin \frac{{10\pi }}{3}t,t} \right)$
for $0 \le t \le 3$
The derivative is
${{\bf{r}}_B}'\left( t \right) = \left( { - \frac{{40\pi }}{3}\sin \frac{{10\pi }}{3}t,\frac{{40\pi }}{3}\cos \frac{{10\pi }}{3}t,1} \right)$
By Theorem 1, the length of the spring in Figure 5 (A) is
${s_A} = \mathop \smallint \limits_0^4 \sqrt {{{\left( { - \frac{{21\pi }}{2}\sin \frac{{3\pi }}{2}t} \right)}^2} + {{\left( {\frac{{21\pi }}{2}\cos \frac{{3\pi }}{2}t} \right)}^2} + 1} {\rm{d}}t$
${s_A} = 4\sqrt {{{\left( {\frac{{21\pi }}{2}} \right)}^2} + 1} \simeq 132.01$
Similarly, the length of the spring in Figure 5 (B) is
${s_B} = \mathop \smallint \limits_0^3 \sqrt {{{\left( { - \frac{{40\pi }}{3}\sin \frac{{10\pi }}{3}t} \right)}^2} + {{\left( {\frac{{40\pi }}{3}\cos \frac{{10\pi }}{3}t} \right)}^2} + 1} {\rm{d}}t$
${s_B} = 3\sqrt {{{\left( {\frac{{40\pi }}{3}} \right)}^2} + 1} \simeq 125.7$
Comparing ${s_A}$ and ${s_B}$, we conclude that the spring in Figure 5 (A) uses more wire than the one in Figure 5 (B). This verifies our guessing in part (b).
