Answer
${\frac{dV}{dt}}=1 {\frac{volt}{sec}}$
${\frac{dI}{dt}}={\frac{-1}{3}} {\frac{ampere}{sec}}$
${\frac{dR}{dt}}={\frac{1}{I}({\frac{dV}{dt}-{\frac{V}{I}{\frac{dI}{dt}}}})}$
${\frac{dR}{dt}}={\frac{3}{2}{\frac{ohm}{sec}}}$
Work Step by Step
a) Rate of increase in voltage is 1 volt per sec
${\frac{dV}{dt}}=1 {\frac{volt}{sec}}$
b) Rate of decreasing in current is
${\frac{dI}{dt}}={\frac{-1}{3}} {\frac{ampere}{sec}}$
c) To have${\frac{dR}{dt}}$ we have to solve $V=IR$
$R={\frac{V}{I}}$
${\frac{dR}{dt}}={\frac{I\frac{dV}{dt}-{V\frac{dI}{dt}}}{I^2}}$
${\frac{dR}{dt}}={\frac{1}{I}({\frac{dV}{dt}-{\frac{V}{I}{\frac{dI}{dt}}}})}$
d) From above equations a, b, c:
${\frac{dR}{dt}}={\frac{1}{I}({\frac{dV}{dt}-{\frac{V}{I}{\frac{dI}{dt}}}})}$
${\frac{dR}{dt}}={\frac{1}{2}{(1-{\frac{12}{2}{\frac{-1}{3}}})}}$ where V=12 volt and I =2 amp
${\frac{dR}{dt}}={\frac{3}{2}{\frac{ohm}{sec}}}$
