Calculating Rate of Change in Parallel Resistors

[tangents]

Hey Guys, I learning about Related rates and although I understand the basic concepts, I'm stuggling with this problem

- When two resistors r1 and R2 are connected in parallel, the total resistance R is given by the equation 1/R=1/R1+1/R2. If R1 and R2 are increasing at rates of .01 ohm/sec and .02 ohm/sec respectively, at what rate is R changing at the instant R1=30 ohms and R2= 90 ohms?

My Work:

dr1/dt= .01
dr2/dt= .02
dr/dt= Trying to find this when r1=30 and r2=90
First what I did was plug in R1 and R2 to get total resistance which i got to be 1/22.5. Next I took the derivative of that equation: -1/r1²(dr/dt)=-1/r²(dr1/dt)-1/r²2(dr/dt). I plugged in R1 and R2 along with dr1/dt and dr2/dt along with the total resistance and tried to solve for dr/dt (total resistance change). But I end up with .018 ohms/sec whereas the answer the book shows is 11/1600 or .006 ohm/sec.

Where did i go wrong? Could someone please show me a easy to follow step by step in getting the answer ^_^, i would greatly appreciate it!.

 

[LeonhardEuler]

-1/r1²(dr/dt)=-1/r²(dr1/dt)-1/r²2(dr/dt).

Don't you mean:
$$\frac{-1}{R^2}\frac{dR}{dt}=\frac{-1}{R_1^2}\frac{dR_1}{dt} - \frac{1}{R_2^2}\frac{dR_2}{dt}$$
Then its just a matter of solving for dr/dt:
$$\frac{dR}{dt}=-R^2(\frac{-1}{R_1^2}\frac{dR_1}{dt} - \frac{1}{R_2^2}\frac{dR_2}{dt})$$
You can get the value of R from the first equation. The answer comes out right.

 

[LeonhardEuler]

First what I did was plug in R1 and R2 to get total resistance which i got to be 1/22.5.

I didn't see this before. This is a mistake. The formula gives you 1/R, so R is actually 22.5, not 1/22.5.