If two resistors with resistances R1 and R2 are connected in parallel, as in the figure below, then the total resistance R, measured in ohms (Ω), is given by

1/R= 1/R1+ 1/R2.
If R1 and R2 are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively, how fast is R changing when R1 = 80 Ω and R2 = 110 Ω?

Ohm my goodness, let's calculate this! To find how fast R is changing, we need to differentiate the formula 1/R = 1/R1 + 1/R2.

Now, let's differentiate both sides with respect to time (t):

d/dt(1/R) = d/dt(1/R1) + d/dt(1/R2)

Remember that R1 is increasing at a rate of 0.3 Ω/s and R2 at a rate of 0.2 Ω/s, which means:

d/dt(1/R1) = -0.3/R1^2 and d/dt(1/R2) = -0.2/R2^2

Now, let's substitute the values of R1 = 80 Ω and R2 = 110 Ω into the equation. Subtracting the rates, we get:

d/dt(1/R) = -0.3/80^2 + -0.2/110^2

Calculating this lovely mess will give us the rate of change of 1/R, so don't forget to take the negative sign into account.

Finally, to find how fast R is changing, we just need the reciprocal of the rate we found (since R is the reciprocal of 1/R).

Voilà! You've got your answer. I hope you didn't find this too resistible!

To find how fast the total resistance R is changing with respect to time, we can use the chain rule of differentiation.

Let's denote R as the total resistance, R1 as the resistance of the first resistor, and R2 as the resistance of the second resistor. We are given that R1 and R2 are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively.

First, we can differentiate the equation 1/R = 1/R1 + 1/R2 with respect to time t:

d/dt (1/R) = d/dt (1/R1) + d/dt (1/R2)

Next, we can simplify each term using the chain rule:

- (1/R^2) * dR/dt = (-1/R1^2) * dR1/dt + (-1/R2^2) * dR2/dt

Since we are interested in how fast R is changing, we can solve for dR/dt:

dR/dt = - (1/R^2) * (dR1/dt + dR2/dt) * R1^2 * R2^2

Given that R1 = 80 Ω and R2 = 110 Ω, we can substitute these values into the equation:

dR/dt = - (1/R^2) * (0.3 Ω/s + 0.2 Ω/s) * (80 Ω)^2 * (110 Ω)^2

Simplifying the equation further:

dR/dt = - (1/R^2) * (0.5 Ω/s) * (80 Ω)^2 * (110 Ω)^2

We can calculate the value of R using the equation 1/R = 1/R1 + 1/R2:

1/R = 1/80 Ω + 1/110 Ω
1/R = (110 + 80)/ (80 * 110) Ω
1/R = 190 / 8800 Ω
1/R = 19 / 880 Ω

Taking the reciprocal of both sides to find R:

R = 880 / 19 Ω

Substituting this value back into the equation for dR/dt:

dR/dt = - (1/(880/19)^2) * (0.5 Ω/s) * (80 Ω)^2 * (110 Ω)^2

Now we can calculate the numerical value.

To find how fast the total resistance R is changing, we can use the chain rule of differentiation. The equation given, 1/R = 1/R1 + 1/R2, can be written as R = (R1*R2)/(R1 + R2).

To find the rate of change of R with respect to time, we differentiate both sides of the equation:

dR/dt = (d/dt)((R1*R2)/(R1 + R2))

Now, we need to find the values of dR1/dt and dR2/dt at the given values of R1 and R2. It is given that dR1/dt = 0.3 Ω/s and dR2/dt = 0.2 Ω/s.

Substituting these values, we have:

dR/dt = (R2*(dR1/dt) - R1*(dR2/dt))/(R1 + R2)^2

Now we can plug in the given values of R1 = 80 Ω and R2 = 110 Ω:

dR/dt = (110*(0.3) - 80*(0.2))/(80 + 110)^2

Calculating the numerator:

(110*(0.3) - 80*(0.2)) = 33 - 16 = 17

Calculating the denominator:

(80 + 110)^2 = 190^2 = 36100

Now, substituting the numerator and denominator values into the equation:

dR/dt = 17/36100 Ω/s

Therefore, the rate at which R is changing when R1 = 80 Ω and R2 = 110 Ω is approximately 0.000471 ω/s.

this is very good. thanks for help

R^-1 = R1^-1 + R2^-1

-R^-2 dR/dt = -R1^-2 dR1/dt - R2^-2 dR2/dt

** 1/R^2 dR/dt = 1/R1^2 dR1/dt + 1/R2^2 dR2/dt **

when R1 = 80, R2 = 110
1/R = 1/80+1/110 = 19/880
R = 880/19

now you have the values in ** , and you can solve for dR/dt