The voltage V (volts), current I (amperes), and resistence R (ohms) of an electric circuit are related by the equation V=IR.

Suppose that V is increasing at a rate of 3 volt/sec while I is decreasing at a rate of 1/8 amp/sec. Let t denote time in seconds.

Determine the rate at which R is changing when V = 9 volts and I = 2 amperes.

Answer = _________ ohms/sec

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To determine the rate at which R is changing when V = 9 volts and I = 2 amperes, we need to find the derivative of the equation V = IR with respect to time (t). The derivative will give us the rate of change of voltage (dV/dt) and the rate of change of current (dI/dt).

The given rates inform us that dV/dt = 3 volts/sec and dI/dt = -1/8 amp/sec (decreasing at a rate of 1/8 amp/sec).

Using the equation V = IR, we can express the derivative of V with respect to t as follows:

dV/dt = d/dt (IR) = R(dI/dt) + I(dR/dt)

Now, since we want to find the rate at which R is changing (dR/dt), we can solve for it:

dR/dt = (dV/dt - I(dR/dt)) / R

Plugging in the provided values V = 9 volts and I = 2 amperes, along with the known values of dV/dt and dI/dt, we have:

3 = 9(dR/dt) / R - 2(dR/dt)

To simplify the equation, we can multiply through by R:

3R = 9(dR/dt) - 2R(dR/dt)

Rearranging the equation:

3R = (9 - 2R)(dR/dt)

Dividing both sides by (9 - 2R):

3R / (9 - 2R) = dR/dt

Now, we can substitute the given values V = 9 volts and I = 2 amperes:

dR/dt = 2 / (9 - 2R)

Calculating the rate at which R is changing when V = 9 volts and I = 2 amperes:

dR/dt = 2 / (9 - 2(2))
dR/dt = 2 / (9 - 4)
dR/dt = 2 / 5

Therefore, the rate at which R is changing when V = 9 volts and I = 2 amperes is 2/5 ohms/sec.