1. L'Hosptial's Rule Question

I was wondering if someone could help me solve this problem:

The figure shows a sector of a circle with central angle θ. Let A(θ) be the area of the segment between chord PR and the arc PR. Let B(θ) be the area of the triangle PQR. Find

Thanks a lot!

2. Originally Posted by ty2391
I was wondering if someone could help me solve this problem:

The figure shows a sector of a circle with central angle θ. Let A(θ) be the area of the segment between chord PR and the arc PR. Let B(θ) be the area of the triangle PQR. Find

Thanks a lot!
I'm not going to work through all of the details of the area calculations. It's just elementary trigonometry.

The area of POR is $\displaystyle \frac{r^2}{2}\sin\theta$. The area of the whole sector is $\displaystyle \pi r^2\cdot\frac{\theta}{2\pi}=\frac{r^2\theta}{2}$. So the area of $\displaystyle A(\theta)$ is $\displaystyle \frac{r^2}{2}(\theta-\sin\theta)$.

Furthermore, the area of POQ is $\displaystyle \frac{r^2}{2}\cos\theta\sin\theta=\frac{r^2}{2}\cd ot\frac{\sin2\theta}{2}$ so $\displaystyle B(\theta)=\frac{r^2}{2}(\sin\theta-\frac{\sin2\theta}{2})$

$\displaystyle \lim_{\theta\to0^+}\frac{A(\theta)}{B(\theta)}=\fr ac{\frac{r^2}{2}(\theta-\sin\theta)}{\frac{r^2}{2}(\sin\theta-\frac{\sin2\theta}{2})}=$ $\displaystyle \lim_{\theta\to0^+}\frac{\theta-\sin\theta}{\sin\theta-\frac{\sin2\theta}{2}}$

Use L'Hopital's Rule thrice to get that the limit equals $\displaystyle \frac{1}{3}$.