Differentiation: rates of change

In summary, the homework statement states that the combined electrical resistance R of two resistors connected in parallel is ##R = \frac{R1R2}{R1+R2}## and that you need to calculate the partials. The partial with respect to R1 is ##\frac{dR1}{dt}## and the partial with respect to R2 is ##\frac{dR2}{dt}.##
  • #1
DevonZA
181
6

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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  • #2
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
[tex] \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) [/tex]
as well as ##d R_1 / dt## and ##d R_2 / dt##.
 
  • #3
Ray Vickson said:
You need to calculate the partials
[tex] \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) [/tex]
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
 
  • #4
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
 
  • #5
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?
 
  • #6
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?

$$\frac{d}{dx} (\frac{kx}{x+k})$$
 
  • #7
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
[tex] \frac{d}{dx} \left( \frac{75 x}{75+x} \right) \; ? [/tex]
 
  • #8
Final answer attached. Thanks to all who helped.
 

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Related to Differentiation: rates of change

1. What is differentiation?

Differentiation is a mathematical process that measures the rate at which one quantity changes with respect to another. It is used to find the slope of a curve at any given point, and is an essential tool in calculus.

2. Why is differentiation important?

Differentiation allows us to analyze and understand the behavior of complex functions. It is used to solve a variety of real-world problems, from finding maximum and minimum values to calculating rates of change in physics and engineering.

3. What is the difference between average rate of change and instantaneous rate of change?

The average rate of change is the total change in a quantity over a specific interval, while the instantaneous rate of change is the rate of change at a specific moment in time. In other words, the average rate of change gives an overall picture of how a quantity is changing, while the instantaneous rate of change gives a more precise understanding of the rate of change at a particular point.

4. How is differentiation related to derivatives?

Derivatives are a type of mathematical function that represents the rate of change of a quantity. Differentiation is the process used to calculate derivatives, and it is often referred to as the inverse of integration. In other words, differentiation allows us to find the derivative of a function, while integration allows us to find the original function from its derivative.

5. What are some practical applications of differentiation?

Differentiation has numerous practical applications in various fields such as physics, engineering, economics, and biology. For example, it is used to calculate velocity and acceleration in physics, to optimize production in economics, and to model population growth in biology. It is also used in fields such as finance, medicine, and computer science.

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