Calculus 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 the resistance at a certain moment, we need to use the chain rule of differentiation.
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)
We know that V = IR, where V is the voltage, I is the current, and R is the resistance.
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = d(I * R)/dt
Using the product rule, this becomes:
dV/dt = I * dR/dt + R * dI/dt
Plugging in the given values, we have:
4 = (0.5) * dR/dt + (250) * (0.01)
Simplifying the equation, we have:
4 = 0.5 * dR/dt + 2.5
Subtracting 2.5 from both sides of the equation, we get:
1.5 = 0.5 * dR/dt
Now, we can solve for the rate of change of the resistance:
dR/dt = 1.5 / 0.5
dR/dt = 3 ohms/min
Therefore, the rate of change of the resistance at that moment is 3 ohms per minute.