I'm stuck on this one... I've worked through the algebra multiple times, and am like positive the issue isn't there, which leads me to believe that I'm setting the problem up wrong. Would someone work through this with me, and help me understand what I'm doing wrong? I would appreciate it so much! The problem:

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, is given by: "1/R = 1/R1 + 1/R2". If R1 and R2 are increasing at rates of .3ohms/sec & .2ohms/sec. respectively, how fast is R changing when R1=80ohms and R2=100ohms?

Question

I'm stuck on this one... I've worked through the algebra multiple times, and am like positive the issue isn't there, which leads me to believe that I'm setting the problem up wrong. Would someone work through this with me, and help me understand what I'm doing wrong? I would appreciate it so much! The problem:

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, is given by: "1/R = 1/R1 + 1/R2". If R1 and R2 are increasing at rates of .3ohms/sec & .2ohms/sec. respectively, how fast is R changing when R1=80ohms and R2=100ohms?

amonoconnor · Answer

The figure provided in the book:
(I didn't find this very helpful, but you may have insight I do not)
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amonoconnor · Answer

Here's my work:
$$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2}$$
$$\frac{1}{R} = \frac{1}{80}+\frac{1}{100}$$
$$\frac{1}{R} = .0125 + .01$$
$$R = \frac{1}{.0225}$$
$$R = 44.4(repeating)$$

amonoconnor · Answer

$$R^{-1} = R1^{-1} + R2^{-1}$$
$$[-1*R^{-2}*\frac{dR}{dt}] = [-1*R1^{-2}*\frac{dR1}{dt}] +[-1*R2^{-2}*\frac{dR2}{dt}]$$

displayerror · Answer

In your last post, you're forgetting the exponents on your resistances.
You differentiated $R^{-1}=R_1^{-1} + R_2^{-2}$, so all of your derivatives should be in terms of resistance raised to the exponent of -1

displayerror · Answer

Ignore the exponent on the $R_2$, it should be a $-1$.
You don't know the rate at which $R_1^{-1}$ and $R_2^{-1}$ are changing, so I would instead use a different form of the equation that is provided.
If you simplify the sum of parallel resistances equation that's provided to you, you'll find that $$R_T=\frac{R_1R_2}{R_1+R_2}$$
Using this equation makes things a lot easier since the derivatives of the resistances will yield a rate of change of the actual resistance rather than that of the inverse resistance.

amonoconnor · Answer

Are not dR1/dt and dR2/dt the .3 and .2? Why does just plugging stuff in not work?

displayerror · Answer

$\dfrac{dR_1}{dt}=0.3 \dfrac{\Omega}{s}$ $\dfrac{dR_2}{dt}=0.2 \dfrac{\Omega}{s}$ 
Notice how in the two derivatives above, the functions that I'm taking the derivative of are $R_1$ and $R_2$, not $R_1^{-1}$ and $R_2^{-1}$ 
We don't know what $\dfrac{dR_1^{-1}}{dt}$ and $\dfrac{dR_2^{-1}}{dt}$ are --- that information isn't provided, so we can't just plug the given derivatives into the derivative that you got. That and the derivative you got is in terms of $R_1$ and $R_2$ which is not correct as I mentioned in the previous post.