If V volts are measured across a resistance of R ohms in a circuit carrying a current of I amperes, V=IR. At a certain instant, the voltage is 250 and is increasing at a rate of 4 volts/min, while the current is 0.5 AMP and is increasing at a rate of 0.01 amp/min. Find the rate of change of the resistance at that moment.

To find the rate of change of resistance, we need to apply the chain rule of differentiation as the resistance depends on both voltage and current.

Given:
V = 250 volts (voltage)
dV/dt = 4 volts/min (rate of change of voltage)
I = 0.5 amperes (current)
dI/dt = 0.01 amp/min (rate of change of current)

The relationship between voltage (V), current (I), and resistance (R) is given by Ohm's Law: V = IR.

Differentiating both sides with respect to time (t), we get:
dV/dt = d(I*R)/dt

Using the product rule, the derivative of I*R with respect to time is:
d(I*R)/dt = dI/dt * R + I * dR/dt

We can rearrange the equation to solve for dR/dt, which represents the rate of change of resistance:
dR/dt = (dV/dt - dI/dt * R) / I

Substituting the given values, we get:
dR/dt = (4 - 0.01 * R) / 0.5

So, at that moment, the rate of change of resistance is given by (4 - 0.01 * R) / 0.5.