# Reciprocal Differentiation Help!

1. May 23, 2005

### Orion1

If two resistors with resistances R1 and R2 are connected in parallel, then the total resistance Rt, measured in ohms, is:
$$\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}$$

If R1 and R2 are increasing at rates:
$$\frac{d \Omega_1}{dt} = 0.3 \; \; \frac{d \Omega_2}{dt} = 0.2 \; \; R_1 = 80 \; \Omega \; \; R_2 = 100 \; \Omega$$

How fast is Rt changing?

$$\frac{d \Omega_t}{dt} = \frac{d}{dt} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)^{-1}$$

Is this the correct initial setup to differentiate this problem?

I am uncertain of the initial differential setup, due to the reciprocals...

This was my initial setup, however does not appear any simpler...

Any suggestions?

Last edited: May 23, 2005
2. May 23, 2005

### TenaliRaman

well i am assuming that the resistances here are variable resistances (ofcourse otherwise the problem would make little sense, but there was no mention of this in the problem itself).

Anyways,
Rearrange to get Rt as,
Rt = R1R2/(R1+R2)
now differentiate this w.r.t to t.

This now corresponds to ur,
$$\frac{d \Omega_t}{dt} = \frac{d}{dt} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)^{-1}$$

Well this should be relatively simple,
first differentiate R1R2/(R1+R2) as D(u/v) form.
That should give,
(vdu - udv)/v^2

Now du is nothing but D(R1R2) which can be differentiated as D(uv) form.
dv is nothing but D(R1+R2) which can be differentiated as D(u+v) form.

The final expression might be a bit "inelegant" but it shouldnt be a problem.

-- AI

3. May 23, 2005

### dextercioby

Here's the elegant way to do it

$$-\frac{1}{R_{t}^{2}}\frac{dR_{t}}{dt}=-\frac{1}{R_{1}^{2}}\frac{dR_{1}}{dt}-\frac{1}{R_{2}^{2}}\frac{dR_{2}}{dt}$$

Multiply by $-R_{t}^{2}$ and substitute the numerical values.

Daniel.