Answer:
[tex]\frac{dR(t)}{dt}=0.06\Omega[/tex]
Explanation:
Since [tex]R(t)=\frac{V}{I(t)}[/tex], we calculate the resistance rate by deriving this formula with respect to time:
[tex]\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})[/tex]
Deriving what is left (remember that [tex](\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)[/tex]):
[tex]\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}[/tex]
So we have:
[tex]\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}[/tex]
Which for our values is (the rate of I(t) is decreasing so we put a negative sign):
[tex]\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega[/tex]
