Ahmed's Calculus Parallel Resistors Q: Rate of Change Help

[MarkFL]

Here is the question:

Calculus Parallel Resistors Question Help?

Need some detailed help with the following problem please!

An electrical circuit consists of two parallel resistors, with resistances R1 and R2 respectively. The total resistance R of the circuit (measured in Ohms) is specified by (1/R = (1/R1) + (1/R2).

The resistors are heating up, so their resistances are increasing over time. Suppose that R1 is increasing at a rate of .3 Ohms/s and R2 is increasing at a rate of .2 ohms/s.

When R1 = 80 Ohms and R2 = 100 Ohms, how fast is the total resistance increasing?

Here is a link to the question:

Calculus Parallel Resistors Question Help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Ahmed,

We are given the following information:

The relationship of the total resistance to the individual resistances of the two resistors:

(1) \(\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\)

The time rate of change of the resistance of the two resistors in Ohms/second:

(2) \(\displaystyle \frac{dR_1}{dt}=0.3\)

(3) \(\displaystyle \frac{dR_2}{dt}=0.2\)

We are asked to find \(\displaystyle \frac{dR}{dt}\). If we implicitly differentiate (1) with respect to time $t$, we find:

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

Multiplying through by $-R^2$, we have:

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

We have everything we need except $R$, and so solving (1) for $R$, we obtain:

\(\displaystyle R=\frac{R_1R_2}{R_1+R_2}\) and so we have:

\(\displaystyle \frac{dR}{dt}=\left(\frac{R_1R_2}{R_1+R_2} \right)^2\left(\frac{1}{R_1^2}\cdot\frac{dR_1}{dt}+\frac{1}{R_2^2}\cdot\frac{dR_2}{dt} \right)\)

Now, plugging in the given values, we find:

\(\displaystyle \left.\frac{dR}{dt} \right|_{(R_1,R_2)=(80,100)}=\left(\frac{80\cdot100}{80+100} \right)^2\left(\frac{1}{80^2}\cdot\frac{3}{10}+ \frac{1}{100^2}\cdot\frac{1}{5} \right)=\frac{107}{810}\)

To Ahmed and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.

 

[akbarali]

Fabulous explanation! Thank you so much!