# Thread: Basic Related rates question

1. ## Basic Related rates question

Hi guys,

I have a basic Cal I related rates question and am unsure how to correctly solve. Can you check my solution? The part I'm a little confused about is where I differentiated both sides. Can you do as I did?
THanks for any help,
james

If two resistors of $\displaystyle R_1$ and $\displaystyle R_2$ are connected in parallel in an electric circuit to make an $\displaystyle R$-ohm resister, the value of $\displaystyle R$ can be found from the equation:
$\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$
If $\displaystyle R_1$ is decreasing at the rate of $\displaystyle 3$ ohms/sec and $\displaystyle R_2$ is increasing at a rate of $\displaystyle 5$ ohms/sec at what rate is $\displaystyle R$ changing when $\displaystyle R_1 = 60$ ohms and $\displaystyle R_2 = 50$ ohms

Substituing $\displaystyle R_1 = 60$ and $\displaystyle R_2 = 50$ gives that $\displaystyle R = \frac{11}{300}$

Differentiating both sides with respect to time gives:

$\displaystyle -\frac{1}{R^2}\left(\frac{dR}{dt}\right) = -\frac{1}{R_1^2}\left(\frac{dR_1}{dt}\right) -\frac{1}{R_2^2}\left(\frac{dR_2}{dt}\right)$

Substituting all known values into the above equation give:
$\displaystyle -\frac{300}{11}\left(\frac{dR}{dt}\right) = -\frac{1}{60^2}(-3) - \frac{1}{50^2}(5)$

Then we solve for $\displaystyle \frac{dR}{dt}$ to obtain the answer.

2. finish. solve for $\displaystyle \frac{dR}{dt}$ ...

$\displaystyle \frac{dR}{dt} = \frac{11}{300}\left(-\frac{3}{60^2} + \frac{5}{50^2}\right)$