A circular sector has radius R = 100 cm and central angle

A circular sector has radius $R = 100\ \text{cm}$ and central angle $\alpha = 60^{\circ}$. By how much does its area decrease if the central angle decreases by $30'$?

Step 1: The area of a sector

With the angle measured in radians, $S = \frac12 R^{2}\alpha$.

The radius is fixed, so $S$ depends on $\alpha$ alone — and it does so linearly.

Step 2: The differential

\[ \frac{dS}{d\alpha} = \frac12 R^{2} = \frac{100^{2}}{2} = 5000\ \text{cm}^{2}\text{ per radian}, \qquad \Delta S = \frac12R^{2}\,\Delta\alpha. \]

Step 3: Convert the angle change to radians

Half a degree, and negative because the angle decreases: \[\Delta\alpha = -30' = -\frac12\cdot\frac{\pi}{180} = -\frac{\pi}{360}\ \text{rad} \approx -0.0087266.\]

Step 4: Compute

\[ \Delta S = 5000\cdot\left(-\frac{\pi}{360}\right) = -\frac{5000\pi}{360} = -\frac{125\pi}{9} \approx -43.633\ \text{cm}^{2}. \]

Step 5: Verification

\[S(60^{\circ}) = \frac12\cdot10^{4}\cdot\frac{\pi}{3} = \frac{5000\pi}{3} \approx 5235.99\ \text{cm}^{2}\]

\[S(59.5^{\circ}) = \frac12\cdot10^{4}\cdot\frac{59.5\pi}{180} \approx 5192.35\ \text{cm}^{2}\]

\[5235.99-5192.35 = 43.63\ \text{cm}^{2} \;\;\checkmark \text{ — exact, not approximate}\]