Chapter 5.6, Problem 4E

a.

To determine

To relate: $\frac{dp}{dt}$ to $\frac{dr}{dt}$ and $\frac{ds}{dt}$ .

Answer to Problem 4E

The value of $\frac{dp}{dt}$ is $2\text{RI}\frac{d\text{I}}{dt}+{\text{I}}^2\frac{d\text{R}}{dt}$

Explanation of Solution

Given information:

All variables are differentiable function of $t$ .

  $\text{P}={\text{RI}}_{}^2$ ; P = power, R = resistance, I = current

Calculation :

We have to relate $\frac{dp}{dt}$ to $\frac{dr}{dt}$ and $\frac{ds}{dt}$ .

Since all variables are differentiable function of $t$

Therefore,

  $\begin{array}{l}\text{P}={\text{RI}}_{}^2\\ \frac{dp}{dt}=\text{R}\left(2\text{I}\frac{d\text{I}}{dt}\right)+{\text{I}}^2\frac{d\text{R}}{dt}\left[\text{Use product rule}\text{.}\right]\\ \frac{dp}{dt}=2\text{RI}\frac{d\text{I}}{dt}+{\text{I}}^2\frac{d\text{R}}{dt}\end{array}$

Hence, the value $\frac{dp}{dt}$ is $2\text{RI}\frac{d\text{I}}{dt}+{\text{I}}^2\frac{d\text{R}}{dt}$ .

b.

To determine

To relate: $\frac{d\text{R}}{dt}$ to $\frac{d\text{I}}{dt}$ if $p$ is constant.

Answer to Problem 4E

The equation that relates $\frac{d\text{R}}{dt}$ to $\frac{d\text{I}}{dt}$ is $\frac{d\text{R}}{dt}=-\frac{2\text{RI}}{\text{I}}\frac{d\text{I}}{dt}$ .

Explanation of Solution

Given information:

All variables are differentiable function of $t$ .

  $\text{P}={\text{RI}}_{}^2$ ; P = power, R = resistance, I = current

Calculation :

We have to relate $\frac{d\text{R}}{dt}$ to $\frac{d\text{I}}{dt}$ if $p$ is constant.

Since all variables are differentiable function of $t$

Therefore,

  $\begin{array}{l}\text{P}={\text{RI}}_{}^2\\ \frac{dp}{dt}=2\text{RI}\frac{d\text{I}}{dt}+{\text{I}}^2\frac{d\text{R}}{dt}\left[\text{Use product rule}\text{.}\right]\\ 0=2\text{RI}\frac{d\text{I}}{dt}+{\text{I}}^2\frac{d\text{R}}{dt}\left[\frac{dp}{dt}=0\text{ since }p\text{ is constant}\text{.}\right]\\ -{\text{I}}^2\frac{d\text{R}}{dt}=2\text{RI}\frac{d\text{I}}{dt}\\ \frac{d\text{R}}{dt}=-\frac{2\text{RI}}{{\text{I}}^2}\frac{d\text{I}}{dt}\\ \frac{d\text{R}}{dt}=-\frac{2\text{RI}}{\text{I}}\frac{d\text{I}}{dt}\end{array}$

Hence, the equation that relates $\frac{d\text{R}}{dt}$ to $\frac{d\text{I}}{dt}$ is $\frac{d\text{R}}{dt}=-\frac{2\text{RI}}{\text{I}}\frac{d\text{I}}{dt}$ .