Answer
(a)$$A=f( \theta )=50(\tan \theta -\theta ), \qquad D_A=\mathbb{R}- \{ \theta \mid \theta = (2n+1)\frac{ \pi }{2}, \, n \in \mathbb{Z} \}. $$
(b)$$f(0.3) \approx 0.467, \quad f(0.6) \approx 4.207, \quad f(0.9) \approx 18.008, \quad f(1.2) \approx 68.608, \quad f(1.5) \approx 630.071$$
(c)$$\lim_{\theta \to \frac{\pi }{2}}f(\theta )= + \infty $$
Work Step by Step
(a)
To find the area of the shaded region, we need to subtract the area of the sector from the area of the right triangle.
Knowing that the height of the right triangle can be obtained from $\tan \theta$, we can find the area of the right triangle:$$h=r \tan \theta \quad \Rightarrow \quad S_1= \frac{(r)(r \tan \theta )}{2}=50 \tan \theta .$$We can find the area of the sector as follows.$$\frac{S_2}{\pi r^2}=\frac{\theta }{2 \pi} \quad \Rightarrow \quad S_2=\frac{1}{2}r^2 \theta =50 \theta$$So, the area of the shaded region equals$$A=f( \theta )=S_1-S_2=50 (\tan \theta - \theta ).$$The domain of $A=f ( \theta )$ is the set of all real numbers except those vanishing the denominator of $\tan \theta = \frac{ \sin \theta }{\cos \theta }$; that is,$$D_A=\mathbb{R}- \{ \theta \mid \cos \theta=0\} = \mathbb{R}- \{ \theta \mid \theta = (2n+1)\frac{ \pi }{2}, \, n \in \mathbb{Z} \}. $$
(b)
$$f(0.3) \approx 0.467, \quad f(0.6) \approx 4.207, \quad f(0.9) \approx 18.008, \quad f(1.2) \approx 68.608, \quad f(1.5) \approx 630.071$$
(c)
As $\theta$ approaches to $\frac{ \pi }{2}$ from the left, $\sin \theta$ approaches $1$, $\cos \theta$ approaches $0$, and $\theta$ approaches $\frac{\pi }{2}$. Thus, we conclude that$$\lim_{\theta \to \frac{\pi }{2}}f(\theta )= + \infty .$$
