# How to do this related rates calculus problem about resistors connected in parallel

If two resistors with resistances $R_1$ and $R_2$ are connected in parallel, then the total resistance $R$, measured in ohms ($\Omega$), is given by

$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$$

If $R_1$ and $R_2$ are increasing at rates of $0.3~\Omega/s$ and $0.2~ \Omega/s$, respectively, how fast is $R$ changing when $R_1 = 70~\Omega$ and $R_2 = 80~\Omega$? (Round your answer to three decimal places.)

I honestly don't even know where to start with this other than taking the derivative of the given equation, but I can't even do that. When I tried I couldn't get an answer. I really need some help!

• "Low dee high..." The quotient rule coupled with the chain rule seems like a good start. – Sean Roberson Oct 6 '16 at 21:42

Differentiate both sides of the equation

$$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\tag{1}\label{eq1}$$

with respect to $t$ using the chain rule. This results in

$$-\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt}\tag{2}\label{eq2}$$

which can be rewritten

$$\frac{dR}{dt} = \left(\frac{R}{R_1}\right)^2 \frac{dR_1}{dt} + \left(\frac{R}{R_2}\right)^2 \frac{dR_2}{dt}$$

With $R_1 = 70$ Ω and $R_2 = 80$ Ω, you can find the corresponding $R$ using \eqref{eq1}. Then you'll have all the values needed to compute $\dfrac{dR}{dt}$, using \eqref{eq2}.

Hint:

from your equation you have: $$R=\frac{R_1R_2}{R_1+R_2}$$ and $R_1$ , $R_2$ are functions of time: $$R_1=70(1+0.3t)\qquad R_2=80(1+0.2t)$$

can you do from this?

• I'm still really confused and I don't understand what to do. Can you please help? – Maggie Oct 6 '16 at 21:57