一个扇形的半径为 $r$ cm，角为 $\theta$ 弧度。该扇形的面积为 $A\text{ cm}^2$，周长为固定值 $k$ cm。

1. Formulate equations for the perimeter and area of the sector.

The fixed perimeter $k$ of the sector is the sum of its two radii and the arc length ($r\theta$):
$k = 2r + r\theta$

The area $A$ of the sector is given by:
$A = \frac{1}{2}r^2\theta$

2. Express the area $A$ purely in terms of $r$ and the constant $k$.

From the perimeter equation, we can isolate the arc length $r\theta$:
$r\theta = k - 2r$

Substitute this expression into the area equation to eliminate $\theta$:
$A = \frac{1}{2}r(r\theta)$
$A = \frac{1}{2}r(k - 2r)$
$A = \frac{k}{2}r - r^2$

3. Differentiate $A$ with respect to $r$ to show the given result.

Differentiating both sides with respect to $r$ (treating $k$ as a constant):
$\frac{\text{d}A}{\text{d}r} = \frac{\text{d}}{\text{d}r} \left( \frac{k}{2}r - r^2 \right)$
$\frac{\text{d}A}{\text{d}r} = \frac{k}{2} - 2r$

1. Find the value of $r$ in terms of $k$ when the arc length equals the radius.

Let the arc length be $s$.
We are given that $s = r$.
The perimeter of the sector is $k = 2r + s$.
Substituting $s = r$:
$k = 2r + r = 3r$
$r = \frac{k}{3}$

2. Evaluate $\frac{\text{d}A}{\text{d}r}$ at this specific value of $r$.

From part (i), we have $\frac{\text{d}A}{\text{d}r} = \frac{k}{2} - 2r$.
Substitute $r = \frac{k}{3}$:
$\frac{\text{d}A}{\text{d}r} = \frac{k}{2} - 2\left(\frac{k}{3}\right)$
$\frac{\text{d}A}{\text{d}r} = \frac{3k}{6} - \frac{4k}{6} = -\frac{k}{6}$

3. Apply the chain rule to find the rate of change of the area, $\frac{\text{d}A}{\text{d}t}$.

Using the chain rule for connected rates of change:
$\frac{\text{d}A}{\text{d}t} = \frac{\text{d}A}{\text{d}r} \times \frac{\text{d}r}{\text{d}t}$
We are given that $\frac{\text{d}r}{\text{d}t} = \frac{k}{10}$.
$\frac{\text{d}A}{\text{d}t} = \left(-\frac{k}{6}\right) \times \left(\frac{k}{10}\right) = -\frac{k^2}{60}$