Two parallel resistors have resistances R1 and R2. If the total resistance is R, then 1/R = 1/R1 + 1/R2. R1 is increasing at 0.04 ohm/sec,and R2 is decreasing at 0.03 ohm/sec. How fast is the total resistance changing at the
moment that R1 is 3 ohms and R2 is 4 ohms?
1/R = 1/R1 + 1/R2
-1/R^2 dR/dt = -1/R1^2 dR1/dt - 1/R2^2 dR2/dt
using the two given resistances,
1/R = 1/3 + 1/4 = 7/12
(7/12)^2 dR/dt = (1/3)^2(.04) + (1/4)^2(-.03)
dR/dt = 37/4900 = 0.0076
To find how fast the total resistance is changing at a given moment, we need to differentiate the equation 1/R = 1/R1 + 1/R2 with respect to time.
Let's assume that R1 and R2 are functions of time, where R1 = R1(t) and R2 = R2(t).
Differentiating both sides of the equation with respect to time, we get:
d(1/R)/dt = d(1/R1)/dt + d(1/R2)/dt
To simplify the equation, we can use the chain rule for differentiation. The chain rule states that if y = f(g(t)), then dy/dt = f'(g(t)) * g'(t), where f' is the derivative of f and g' is the derivative of g.
Applying the chain rule to our equation, we have:
-1/R^2 * dR/dt = -1/R1^2 * dR1/dt - 1/R2^2 * dR2/dt
Multiplying both sides of the equation by -R^2, we get:
dR/dt = R^2 * (dR1/dt)/R1^2 + R^2 * (dR2/dt)/R2^2
Now we can substitute the given values: R1 = 3 ohms, R2 = 4 ohms, dR1/dt = 0.04 ohm/sec, dR2/dt = -0.03 ohm/sec.
Plugging these values into the equation, we have:
dR/dt = R^2 * (0.04/3^2) + R^2 * (-0.03/4^2)
Simplifying further, we have:
dR/dt = (0.04/9)* R^2 - (0.03/16) * R^2
Combining the terms, we have:
dR/dt = (0.04/9 - 0.03/16) * R^2
Evaluating the expression within the parentheses, we get:
dR/dt = (0.0177) * R^2
Therefore, the total resistance is changing at a rate of 0.0177 R^2 ohms per second at the moment when R1 is 3 ohms and R2 is 4 ohms.