Answer
(a) The radius of the ring, $r = \frac{3}{\pi}$.
(b) The circumference, $C(r)$ varies from 5.5 to 6.5. So, the radius,$r$ vary from $\frac{5.5}{2\pi}$ to $\frac{6.5}{2\pi}$.
(c) If $\epsilon = 0.5$ and $\delta = \frac{0.5}{2\pi}$, we get
$|C(r) - 6| < \epsilon$ whenever $|r - \frac{3}{2\pi}|< \delta$.
Work Step by Step
The inner circumference of ring, $C(r) = 2\pi r$ for radius, $r$.
(a) When $C(r) = 6$, we get $2\pi r = 6 \Rightarrow r = \frac{3}{\pi}$.
(b) For $C(r) = 5.5$, we get $2\pi r = 5.5 \Rightarrow r = \frac{5.5}{2\pi}$.
For $C(r) = 6.5$, we get $2\pi r = 6.5 \Rightarrow r = \frac{6.5}{2\pi}$.
When inner circumference,$C(r)$ varies from $5.5$ to $6.5$, then, inner radius,$r$ varies from $\frac{5.5}{2\pi}$ to $\frac{6.5}{2\pi}$.
(c) From previous part, we have $5.5 < C(r) < 6.5$ whenever $\frac{5.5}{2\pi} < r < \frac{6.5}{2\pi}$.
Now, $5.5 < C(r) < 6.5$ gives
$6 - 0.5 < C(r) < 6 + 0.5 \Rightarrow -0.5 < C(r) - 6 < 0.5$, or, $|C(r)- 6|< 0.5$.
Also, $\frac{5.5}{2\pi} < r < \frac{6.5}{2\pi}$ gives
$\frac{6-0.5}{2\pi} < r < \frac{6+0.5}{2\pi} $
$\Rightarrow \frac{6}{2\pi} - \frac{0.5}{2\pi} < r < \frac{6}{2\pi} + \frac{0.5}{2\pi}$
$\Rightarrow - \frac{0.5}{2\pi} < r - \frac{6}{2\pi} < \frac{0.5}{2\pi}$
$\Rightarrow |r - \frac{6}{2\pi} |< \frac{0.5}{2\pi}$.
Thus, if $\epsilon = 0.5$ and $\delta = \frac{0.5}{2\pi}$, we get
$|C(r) - 6 | < \epsilon$ whenever $|r - \frac{6}{2\pi}|< \frac{0.5}{2\pi}$.
