Answer
(a)
Average rate of change of $I$ with respect to $R$ for the interval from $R=8$ to $R=8.1$ is $\dfrac{-1}{5.4}$ amperes/ohm.
(b)
The rate of change of $I$ with respect to $R$ when $R = 8$ ohms is $\dfrac{-3}{16}$ amperes/ohm.
(c)
The rate of change of $R$ with respect to $I$ when $I = 1.5$ amperes is $\dfrac{-16}{3}$ ohms/ampere.
Work Step by Step
(a)
Firstly, calculate value of $I$ for $R=8$ ohms using $V=IR$.
Substitute $R=8$ ohms and $V=12$ volts in $V=IR$.
We get $12=I\times 8$.
This gives $I=\frac{12}{8}$ amperes.
Now, calculate value of $I$ for $R=8.1$ ohms using $V=IR$.
Substitute $R=8.1$ ohms and $V=12$ volts in $V=IR$.
We get $12=I\times 8.1$.
This gives $I=\frac{12}{8.1}$ amperes.
Now, calculate the average rate of change of $I$ with respect to $R$ as follows:
$\dfrac{\Delta I}{\Delta R}=\dfrac{\frac{12}{8.1}-\frac{12}{8}}{8.1-8}=\dfrac{\frac{96-97.2}{8.1\times8}}{0.1}=\dfrac{\frac{-1.2}{64.8}}{0.1}=\dfrac{-12}{64.8}=\dfrac{-1}{5.4}$ amperes/ohm.
(b)
Firstly, solve $V=IR$ for $I$.
We get $I=\dfrac{V}{R}=VR^{-1}$.
Now, calculate the rate of change of $I$ with respect to $R$ at $R=8$ ohms.
$\dfrac{dI}{dR}=\dfrac{d}{dR}(VR^{-1})=-VR^{-2}$.
Now substitute $V=12$ volts and $R=8$ ohms and then simplify.
$\dfrac{dI}{dR}=-12\times 8^{-2}=\frac{-12}{8^{2}}=\frac{-12}{64}=\frac{-3}{16}$ amperes/ohm.
(c)
Firstly, solve $V=IR$ for $R$.
We get $R=\dfrac{V}{I}=VI^{-1}$.
Now, calculate the rate of change of $R$ with respect to $I$ at $I=1.5$ amperes.
$\dfrac{dR}{dI}=\dfrac{d}{dI}(VI^{-1})=-VI^{-2}$.
Now substitute $V=12$ volts and $I=1.5$ amperes and then simplify.
$\dfrac{dR}{dI}=-12\times 1.5^{-2}=\frac{-12}{1.5^{2}}=\frac{-12}{2.25}=\frac{-1200}{225}=\frac{-16}{3}$ ohms/ampere.
