Differentiation: rates of change

DevonZA
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Homework Statement

Homework Equations

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?[/B]
 

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DevonZA said:

Homework Statement

Homework Equations

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?[/B]

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##.
 
Ray Vickson said:
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##.

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
 
DevonZA said:
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?
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.
DevonZA said:
##\frac{dR1}{dt}## = +1ohm and ##\frac{dR2}{dt}## = +1.5ohm
 
Mark44 said:
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.

I don't follow. Could you show me the partial for R1?
 
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?
DevonZA said:
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?
I don't follow. Could you show me the partial for R1?
Finding the partial with respect to R1, you treat R2 as if it were a constant. How would you find this ordinary derivative?

$$\frac{d}{dx} (\frac{kx}{x+k})$$
 
DevonZA said:
I don't follow. Could you show me the partial for R1?

Are you saying you do not know how to find the derivative
\frac{d}{dx} \left( \frac{75 x}{75+x} \right) \; ?
 
Final answer attached. Thanks to all who helped.
 

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