Chapter 4.5, Problem 68E

To determine

To find:

The value of $\underset{R\to r^{+}}{\lim }v\text{and}\underset{r\to 0^{+}}{\lim }v$ where $\theta$ is the central angle.

Answer to Problem 68E

  $\underset{R\to r^{+}}{\lim }v=0\text{and}\underset{r\to 0^{+}}{\lim }v=0$

Explanation of Solution

Given:

The velocity $v$ of an electrical in the cable is

  $v=-c{\left(\frac{r}{R}\right)}^2\ln \left(\frac{r}{R}\right)$

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:

  $\underset{R\to r^{+}}{\lim }v=\underset{R\to r^{+}}{\lim }\left[-c{\left(\frac{r}{R}\right)}^2\ln \left(\frac{r}{R}\right)\right]\mathrm{.....................}\left(1\right)$

Applying the limit

  $\begin{array}{l}\underset{R\to r^{+}}{\lim }v=\underset{R\to r^{+}}{\lim }\left[-c{\left(\frac{r}{R}\right)}^2\ln \left(\frac{r}{R}\right)\right]\\ \underset{R\to r^{+}}{\lim }v=-c\left(1\right)\ln \left(1\right)\\ \underset{R\to r^{+}}{\lim }v=-c\left(1\right)\left(0\right)\\ \underset{R\to r^{+}}{\lim }v=0\end{array}$

  $\begin{array}{l}\underset{r\to 0^{+}}{\lim }v=\underset{r\to 0^{+}}{\lim }\left[-c{\left(\frac{r}{R}\right)}^2\ln \left(\frac{r}{R}\right)\right]\mathrm{....................}\left(2\right)\\ \underset{r\to 0^{+}}{\lim }v=\underset{r\to 0^{+}}{\lim }\left[-c{\left(\frac{r}{R}\right)}^2\ln \left(\frac{r}{R}\right)\right]\\ \underset{r\to 0^{+}}{\lim }v=-c\left(1\right)\ln \left(1\right)\\ \underset{r\to 0^{+}}{\lim }v=-c\left(1\right)\left(0\right)\\ \underset{r\to 0^{+}}{\lim }v=0\end{array}$