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 = 90 Ω? (Round your answer to three decimal places.)
Well, calculating the rate of change can sometimes be a little resist-istance, but fear not! I'm here to help!
We'll start by differentiating the given equation,
1/R = 1/R1 + 1/R2.
So, d(1/R) = d(1/R1) + d(1/R2).
Next, let's substitute R1 = 80 Ω and R2 = 90 Ω, and their respective rates of change, dR1/dt = 0.3 Ω/s and dR2/dt = 0.2 Ω/s.
Now, let's solve for dR/dt, which represents the rate of change of total resistance R.
d(1/R)/dt = d(1/R1)/dt + d(1/R2)/dt.
To make things easier, we'll divide through by dt:
d(1/R)/dt = (d(1/R1)/dt) + (d(1/R2)/dt).
Now, let's plug in the values:
d(1/R)/dt = (d(1/80)/dt) + (d(1/90)/dt).
d(1/R)/dt = (-1/80^2)(dR1/dt) + (-1/90^2)(dR2/dt).
d(1/R)/dt = (-1/6400)(0.3) + (-1/8100)(0.2).
Simplifying,
d(1/R)/dt = -0.0000375 - 0.00002469.
d(1/R)/dt ≈ -0.00006219 Ω/s.
Finally, to find dR/dt, the rate of change of total resistance R, we want the negative reciprocal of d(1/R)/dt:
dR/dt ≈ -1/(-0.00006219).
dR/dt ≈ 16.072 Ω/s.
So, when R1 = 80 Ω and R2 = 90 Ω, the total resistance R is changing at a rate of approximately 16.072 Ω/s. Keep sparking those calculations!
To find how fast R is changing, we need to use the chain rule from calculus. Let's start by differentiating the equation with respect to time t:
d/dt (1/R) = d/dt (1/R1) + d/dt (1/R2)
Now, let's differentiate each term separately:
d/dt (1/R) = -1/R^2 * dR/dt
d/dt (1/R1) = -1/R1^2 * dR1 / dt
d/dt (1/R2) = -1/R2^2 * dR2 / dt
Now, let's substitute the given values for R1, R2, and dR1/dt, and dR2/dt:
R1 = 80 Ω
R2 = 90 Ω
dR1/dt = 0.3 Ω/s
dR2/dt = 0.2 Ω/s
Now, plug in the values:
-1/R^2 * dR/dt = -1/(80^2) * 0.3 - 1/(90^2) * 0.2
Simplifying further:
-1/R^2 * dR/dt = -0.000480384 - 0.000246914
Adding the two values:
-1/R^2 * dR/dt = -0.000727298
Now, let's solve for dR/dt:
-0.000727298 = -1/R^2 * dR/dt
Rearranging the equation:
dR/dt = 1/R^2 * 0.000727298
Substituting the given values for R1 and R2:
dR/dt = 1/(170^2) * 0.000727298
Calculating the final answer:
dR/dt ≈ 0.0000190226 Ω/s
Therefore, R is changing at a rate of approximately 0.0000190226 Ω/s.
To find how fast R is changing, we need to differentiate the equation that relates R1, R2, and R:
1/R = 1/R1 + 1/R2
Applying the power rule of differentiation, we get:
d(1/R)/dt = d(1/R1)/dt + d(1/R2)/dt
To find d(1/R)/dt, we differentiate both sides with respect to time (t):
d(1/R)/dt = -1/R^2 * dR/dt
Similarly, d(1/R1)/dt = -1/R1^2 * dR1/dt and d(1/R2)/dt = -1/R2^2 * dR2/dt.
Now, substituting the given values:
d(1/R)/dt = -1/R^2 * dR/dt
d(1/R1)/dt = -1/R1^2 * dR1/dt = -1/(80^2) * 0.3
d(1/R2)/dt = -1/R2^2 * dR2/dt = -1/(90^2) * 0.2
We can rearrange the equation 1/R = 1/R1 + 1/R2 to solve for R:
1/R = 1/R1 + 1/R2
1/R = (R1 + R2)/(R1 * R2)
R = R1 * R2 / (R1 + R2)
Now, substitute R1 = 80 Ω and R2 = 90 Ω:
R = (80 * 90) / (80 + 90) = 40 Ω
Now, substitute all the values into the equation for d(1/R)/dt:
d(1/R)/dt = -1/R^2 * dR/dt
dR/dt = -R^2 * d(1/R)/dt
dR/dt = - (40^2) * (-(1/(80^2))*0.3 - (1/(90^2))*0.2)
Simplifying the expression:
dR/dt = -1600 * (-(1/6400)*0.3 - (1/8100)*0.2)
Calculating the value:
dR/dt ≈ -1600 * (-0.000046875 - 0.0000246914) ≈ -1600 * (-0.0000715664)
dR/dt ≈ 0.1145
Therefore, the value of dR/dt (how fast R is changing) when R1 = 80 Ω and R2 = 90 Ω is approximately 0.1145 Ω/s.