Homework Help: Differentiation: rates of change

1. Aug 13, 2015

DevonZA

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. The combined electrical resistance R of two resistors connected in parallel is $R = \frac{R1R2}{R1+R2}$, where R, R1 and R2 are measured in ohm. R1 and R2 are increasing at rates of 1 and 1,5 ohm per second respectively. Find the rate of change of R when R1 = 50 ohm and R2 = 75 ohm.

2. Differentiation: rates of change.

3. See attached..

So I'm not sure what the derivative of R1 and R2 would be unless I put in their values of 50 and 75 ohm but then the derivative would be 0? Is it possible to rearrange the given equation?

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2. Aug 13, 2015

Ray Vickson

You need to calculate the partials
$$\frac{\partial}{\partial R_1} \left( \frac{R_1 R_2}{R_1 + R_2}\right) \; \;\text{and} \;\; \frac{\partial}{\partial R_2}\left( \frac{R_1 R_2}{R_1 + R_2}\right)$$
as well as $d R_1 / dt$ and $d R_2 / dt$.

3. Aug 13, 2015

DevonZA

How do I calculate the partials? Do I put in the value of R1 when doing $\frac{\partial}{\partial R1}$ $(\frac{R1R2}{R1+R2})$ and same for R2?
$\frac{dR1}{dt}$ = +1ohm and $\frac{dR2}{dt}$ = +1.5ohm

4. Aug 13, 2015

Staff: Mentor

No. R1 and R2 are variables (functions of t). It is only at a particular moment that they have the values below. When you take the partial with respect to one of the variables, you treat the other variable as if it were a constant.

5. Aug 14, 2015

DevonZA

I don't follow. Could you show me the partial for R1?

6. Aug 14, 2015

Staff: Mentor

How do I calculate the partials? Do I put in the value of R1 when doing $\frac{\partial}{\partial R1}$ $(\frac{R1R2}{R1+R2})$ and same for R2?
Finding the partial with respect to R1, you treat R2 as if it were a constant. How would you find this ordinary derivative?

$$\frac{d}{dx} (\frac{kx}{x+k})$$

7. Aug 14, 2015

Ray Vickson

Are you saying you do not know how to find the derivative
$$\frac{d}{dx} \left( \frac{75 x}{75+x} \right) \; ?$$

8. Aug 16, 2015

DevonZA

Final answer attached. Thanks to all who helped.

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