Calculus Problem!!!?

Question

A voltage V across a resistance R generates a current C = V/R. Suppose a constant voltage of 12 volts is put across a resistance that is decreasing at a rate of 0.2 ohms per second when the resistance is 6 ohms.
At what rate is the current changing?

Answer 1

voltage = V
current = I
resistance = R

I = V/R
I = VR^(-1)
dI / dR = (-1)VR^(-2)
dR/dt = - 0.2

dI/dt = (-1)VR^(-2) (- 0 .2)
dI/dt = (0.2) V / R²
dI/dt = (0.2) (12) / 36
dI/dt = 0.2 / 3
dI/dt = 2 / 30
dI/dt = 1 / 15 amps / sec

Answer 2

This is a rate of change problem.

You need to compute C when R1 = 6 ohms and V = 12 volts AND R2 = 6.2 and V = 12 volts.

C2 - C1 = 12 / 6.2 - 12/6 = 1.935 - 2 = -0.0645 v / ohm

Check that as resistance increases by 0.2 ohm/sec, the current decreases by 0.0656 per second.

Answer 3

C = V/R

Differentiate keeping V as constant with respect to time

dC/dt = - (V/ R^2) (dR/dt)

substituting V = 12 , R = 6 and dR/dt = 0.2

dC/dt = -(12/6^2)* (0.2)

negative sign indicates resistance is decreasing

dC/dt = (12/36)*0.2

dC/dt = 0.2/3 = 0.067

so current is changing at the rate of 0.067 amperes /s

Answer 4

constant voltage means the dV/dt = 0
resistance decreasing at 0.2, dR/dt= -0.2 ohms/sec
With voltage 12 and resistance 6, then C = V/R = 12/6 = 2
You are trying to find the rate of change in current, dC/dt=?

Now, take the derivative of the equation with respect to time
C= V/R
dC/dt = [(dV/dt)R- V(dR/dt)]/R^2

plugging in the above values
dC/dt = [(0)(6) - 12(-2)]/6^2
dC/dt = 24/36
dC/dt = 2/3 amps/sec (I think amps)