Chapter 4.5, Problem 70E

To determine

To find:

The value of $\underset{\theta \to 0^{+}}{\lim }\frac{A\left(\theta \right)}{B\left(\theta \right)}$ where $\theta$ is the central angle.

Answer to Problem 70E

  $\underset{\theta \to 0^{+}}{\lim }\frac{A\left(\theta \right)}{B\left(\theta \right)}=\frac{1}{3}$

Explanation of Solution

Given:

The given figure represents a sector with central angle $\theta$ and a chord $PR$ .

The sector $PR$ is named as $A\left(\theta \right)$ and the area of the right triangle $OPR$ .

Concept used:

The left derivative and right derivative of a function $f$ at a point $x=c$ are , equal

  $\begin{array}{l}{f}_{-}\text{'}\left(c\right)=\underset{h\to 0^{-}}{\lim }\frac{f\left(c-h\right)-f\left(c\right)}{-h}\\ f_{+}\text{'}\left(c\right)=\underset{h\to 0^{+}}{\lim }\frac{f\left(c+h\right)-f\left(c\right)}{h}\end{array}$

Calculation:

Let the radius of the circle be $r$ .

  $OP=OR=r$

The value of $A\left(\theta \right)$ is the difference between the area of the sector $PR$ and the area of the triangle $OPR$ .

Area of the sector $PR$ is

  $A=\left(\frac{\theta }{2\pi }\cdot \pi {\left(r\right)}^2\right)$

Where $\theta$ is the central angle.

area of the triangle $OPR$

  $=\frac{1}{2}PQ\cdot OR$

the height of the triangle $OPR$

  $\begin{array}{l}PQ=OP\sin \theta \\ PQ=r\cdot \sin \theta \end{array}$

Hence area of the triangle $OPR$

A $\left(OPR\right)$

  $\begin{array}{l}=\frac{1}{2}PQ\cdot OR\\ =\frac{1}{2}r\cdot r\sin \theta \end{array}$

  $\begin{array}{l}A\left(\theta \right)=\left(\frac{\theta }{2\pi }\cdot \pi {\left(r\right)}^2\right)-\frac{1}{2}r\cdot r\sin \theta \\ A\left(\theta \right)=\left(\frac{\theta }{2}{\left(r\right)}^2\right)-\frac{r^2}{2}\sin \theta \\ A\left(\theta \right)=\frac{r^2}{2}\left(\theta -\sin \theta \right)\end{array}$

The value of $B\left(\theta \right)$

  $\begin{array}{l}=\frac{1}{2}PQ\cdot QR\\ =\frac{1}{2}\left(r-r\cos \theta \right)\left(r\sin \theta \right)\\ =\frac{r^2}{2}\left(\sin \theta -\sin \theta \cos \theta \right)\end{array}$

The value of $\underset{\theta \to 0^{+}}{\lim }\frac{A\left(\theta \right)}{B\left(\theta \right)}$

  $\begin{array}{l}\underset{\theta \to 0^{+}}{\lim }\frac{A\left(\theta \right)}{B\left(\theta \right)}=\underset{\theta \to 0^{+}}{\lim }\frac{\frac{r^2}{2}\left(\theta -\sin \theta \right)}{\frac{r^2}{2}\left(\sin \theta -\sin \theta \cos \theta \right)}\\ \underset{\theta \to 0^{+}}{\lim }\frac{A\left(\theta \right)}{B\left(\theta \right)}=\underset{\theta \to 0^{+}}{\lim }\frac{\left(\theta -\sin \theta \right)}{\left(\sin \theta -\sin \theta \cos \theta \right)}\end{array}$

The value of the function as $\theta$ approaches $0^{+}$

  $\underset{\theta \to 0^{+}}{\lim }\frac{A\left(\theta \right)}{B\left(\theta \right)}=\frac{1}{3}$