Chain rule and rate of change base questions I hoped

[PA3040D]

Hi expert
I can answer following question with the help and support of ChatGPT and my friends. However, I'm unsure how to approach solving this type of question. What knowledge do I need to address this kind of question, and what topics should I study to think from this perspective? I'm having difficulty scoping out this type of question. Please advise."

Please Note
I searched on YouTube for answers to similar questions but couldn't find any. I would greatly appreciate it if you could provide the keywords to search for such answers on YouTube. or links of YouTube videos which I will be able to capture the method

 

[mario99]

You have posted multiple questions in the same post. We can only solve one question per thread, so which question of them do you want us to focus on?

 

[PA3040D]

Or sorry
Please help on this please

 

[mario99]

You wanna know what to study to be able to solve this kind of problems. Well it is obvious that you need to understand calculus and electric circuits. But both of these subjects are very deep, so at least understand the basics of them. Rate of change problems needs a little knowledge of how to solve differential equations. Electric circuits problems have a lot of formulas, so you will need to understand and memorize some of them if necessary. Chain rule and related problems are the easiest, a little of calculus and you can master them.

 

[PA3040D]

Thanks for the help
I believe that I have a solid understanding of the basics behind electrical and electronic engineering, as well as its equations as well mathematics also not too bad . The issue arises when trying to approach answers by understanding the content of the questions. It would be helpful if you could explain how to tackle questions using the first example. Also, I'm still confused about how to apply the chain rule for problems involving rates of change. I know how to use the chain rule to solve certain problems,

 

[mario99]

Let us focus on question number [imath]1[/imath] for this post. And please, move the other two questions separately in two threads if you are interested in solving them.

They want you to find the percentage error [imath]\displaystyle \frac{\Delta R_T}{R_T}[/imath] when two resistors are connected in parallel.

The formula for the percentage error for two resistors connected in parallel is:

[imath]\displaystyle \frac{\Delta R_T}{R_T} = R_T\left(\frac{1}{R_1}\frac{\Delta R_1}{R_1} + \frac{1}{R_2}\frac{\Delta R_2}{R_2}\right)[/imath]

To solve this problem, you just need to understand the formula above. Memorize it if necessary for future similar problems.


[imath]\displaystyle \frac{\Delta R_T}{R_T} = \ [/imath] percentage error in [imath]\displaystyle R_T \ \ \ [/imath] (What they want you to find.)


[imath]\displaystyle R_T = [/imath] the sum of the two resistors in parallel


[imath]\displaystyle R_1 = [/imath] the first resistor


[imath]\displaystyle R_2 = [/imath] the second resistor


[imath]\displaystyle \frac{\Delta R_1}{R_1} = [/imath] the percentage error in the first resistor


[imath]\displaystyle \frac{\Delta R_2}{R_2} = [/imath] the percentage error in the second resistor


Can you answer the problem now?

[imath]\displaystyle \frac{\Delta R_T}{R_T} = \ \ ?[/imath]



-----------------------------

Bonus question 1: How would you solve the problem if:
[imath]R_1 = (2 \pm 0.3) \ \Omega [/imath]
[imath]R_2 = (5 \pm 0.4) \ \Omega [/imath]

Bonus question 2: How would you solve the problem if:
[imath]R_1 = 2 \ \Omega \ \text{error} \ \pm 1\% [/imath]
[imath]R_2 = 5 \ \Omega \ \text{error} \ \pm 2\% [/imath]

 

[PA3040D]

Grate thanks or your continuous support

Just what I know is that when we apply the values to the equation, the answer comes out. However, I still struggle to understand the processes in between.

I'm unsure if the method I used is correct. I took partial derivatives for each resistor separately and obtained two answers: 9/25 and 4/25. Can you explain the significance of these two answers? I want to understand what happens when partial derivatives are taken.

 

[mario99]

I have already given you the formula for the percentage error for two resistors connected in parallel. You just need to plug the given values in it. If you want to derive this formula with calculus, I can show you.

We know that the formula for calculating the equivalent resistance connected in parallel is:

[imath]\displaystyle \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}[/imath]

And we know that the percentage error in [imath]\displaystyle R_T[/imath] is [imath]\displaystyle \frac{\mathrm{d}R_T}{R_T}[/imath].

We want to find a way to get this percentage error from the given equivalent resistance formula.

Let [imath]\displaystyle \mathrm{z} = \frac{1}{R_T}[/imath]

And

[imath]\displaystyle x = \frac{1}{R_1}[/imath]

And

[imath]\displaystyle y = \frac{1}{R_2}[/imath]

Then, we have:

[imath]\displaystyle \mathrm{z} = x + y[/imath]

From calculus, we know that the total differential [imath]\displaystyle \mathrm{dz} [/imath] is:

[imath]\displaystyle \mathrm{dz} = \frac{\partial \mathrm{z}}{\partial x}\mathrm{d}x + \frac{\partial \mathrm{z}}{\partial \mathrm y}\mathrm{d}y[/imath]


[imath]\displaystyle \frac{\partial \mathrm{z}}{\partial x} = 1[/imath] And [imath]\displaystyle \ \frac{\partial \mathrm{z}}{\partial \mathrm y} = 1[/imath]

Then we have:

[imath]\displaystyle \mathrm{dz} = \mathrm{d}x + \mathrm{d}y[/imath]

We will find the derivatives of the formulas at the beginning above.

[imath]\displaystyle \frac{\mathrm{dz}}{\mathrm{d}R_T} = -\frac{1}{R^2_T} \ \ \ \ \ [/imath] then, [imath]\displaystyle \ \ \ \ \ \mathrm{dz} = -\frac{\mathrm{d}R_T}{R^2_T} [/imath]


[imath]\displaystyle \frac{\mathrm{d}x}{\mathrm{d}R_1} = -\frac{1}{R^2_1} \ \ \ \ \ [/imath] then, [imath]\displaystyle \ \ \ \ \ \mathrm{d}x = -\frac{\mathrm{d}R_1}{R^2_1} [/imath]


[imath]\displaystyle \frac{\mathrm{d}y}{\mathrm{d}R_2} = -\frac{1}{R^2_2} \ \ \ \ \ [/imath] then, [imath]\displaystyle \ \ \ \ \ \mathrm{d}y = -\frac{\mathrm{d}R_2}{R^2_2} [/imath]

Then, we have:

[imath]\displaystyle -\frac{\mathrm{d}R_T}{R^2_T} = -\frac{\mathrm{d}R_1}{R^2_1} - \frac{\mathrm{d}R_2}{R^2_2}[/imath]

Or

[imath]\displaystyle \frac{\mathrm{d}R_T}{R^2_T} = \frac{\mathrm{d}R_1}{R^2_1} + \frac{\mathrm{d}R_2}{R^2_2}[/imath]

Or

[imath]\displaystyle \frac{\mathrm{d}R_T}{R_T} = R_T\left(\frac{\mathrm{d}R_1}{R^2_1} + \frac{\mathrm{d}R_2}{R^2_2}\right) = R_T\left(\frac{1}{R_1}\frac{\mathrm{d}R_1}{R_1} + \frac{1}{R_2}\frac{\mathrm{d}R_2}{R_2}\right)[/imath]

It does not matter if you will use in the formula [imath]\displaystyle \mathrm{d} R_T [/imath] or [imath]\displaystyle \Delta R_T[/imath]. The main point in the formula is that [imath]\mathrm{d} R_T[/imath] (or [imath]\Delta R_T[/imath]) is the error and [imath]\displaystyle \frac{\mathrm{d} R_T}{R_T} \left(\text{or} \ \displaystyle \frac{\Delta R_T}{R_T}\right)[/imath] is the percentage error. The same thing applies to each resistor, for example, [imath]\mathrm{d} R_1[/imath] is the error in the resistor [imath]R_1[/imath] and [imath]\displaystyle \frac{\mathrm{d} R_1}{R_1}[/imath] is the percentage error in [imath]R_1[/imath].

Bonus answer: If the percentage error of [imath]R_1 = [/imath] the percentage error of [imath]R_2[/imath] for two resistors connected in parallel, then the percentage error of [imath]R_T [/imath] must be the same.

 

[mario99]

Grate thanks or your continuous support

Just what I know is that when we apply the values to the equation, the answer comes out. However, I still struggle to understand the processes in between.

I'm unsure if the method I used is correct. I took partial derivatives for each resistor separately and obtained two answers: 9/25 and 4/25. Can you explain the significance of these two answers? I want to understand what happens when partial derivatives are taken.

View attachment 37947

For the main problem in post #1, we don't need partial derivatives. But they can be used in other problems if you want to compare the change of error between the equivalence resistance and the resistor [imath]R_1[/imath] or [imath]R_2[/imath].

For examples [imath]\displaystyle \frac{\partial R_T}{\partial R_2} = \frac{4}{25}[/imath]

is equal to

[imath]\displaystyle \partial R_T = \frac{4}{25}\partial R_2[/imath]

The expression above means that if we have a small change of error equal to [imath]\partial R_2[/imath] in the resistor [imath]R_2[/imath], the error of equivalence resistance will also change by a small amount equal to [imath]\displaystyle \frac{4}{25}\partial R_2[/imath].

 

[PA3040D]

Now I can understand around 90% of the concept base on your continuous support
Thanks once again

Can you please help find the wrong with the final answer of the following

Quection


Answer

 

[mario99]

You are not listening to me

Why is it difficult for you to post a new question in a new thread?

 

[PA3040D]

Of Couse yes sorry . I wiil

 

[mario99]

For the main problem in post #1, we don't need partial derivatives.

I am sorry when I said we don't need partial derivatives in post #1. I was wrong. I was confused by your reply, but now I understand why you said that.

When [imath]R_T[/imath] is written like this:

[imath]\displaystyle R_T = \frac{R_1R_2}{R_1+ R_2}[/imath]

We can use the the partial derivatives [imath]\displaystyle \frac{\partial R_T}{\partial R_1}[/imath] and [imath]\displaystyle \frac{\partial R_T}{\partial R_2}[/imath] to derive the equation of the percentage error [imath]\displaystyle \frac{dR_T}{R_T}[/imath].

The total differential [imath]\displaystyle dR_T[/imath] is:

[imath]\displaystyle dR_T = \frac{\partial R_T}{\partial R_1}dR_1 + \frac{\partial R_T}{\partial R_2}dR_2[/imath]


[imath]\displaystyle = \left(\frac{R_2^2}{(R_1 + R_2)^2}\right)dR_1 + \left(\frac{R_1^2}{(R_1 + R_2)^2}\right)dR_2[/imath]


[imath]\displaystyle = \frac{R_1^2}{R_1^2}\left(\frac{R_2^2}{(R_1 + R_2)^2}\right)dR_1 + \frac{R_2^2}{R_2^2}\left(\frac{R_1^2}{(R_1 + R_2)^2}\right)dR_2[/imath]


[imath]\displaystyle = \left(\frac{R_1^2 R_2^2}{(R_1 + R_2)^2}\right)\left(\frac{1}{R_1^2}dR_1 + \frac{1}{R_2^2}dR_2\right)[/imath]


[imath]\displaystyle = R_T^2\left(\frac{1}{R_1^2}dR_1 + \frac{1}{R_2^2}dR_2\right)[/imath]

Then we have:

[imath]\displaystyle dR_T = R_T^2\left(\frac{1}{R_1^2}dR_1 + \frac{1}{R_2^2}dR_2\right)[/imath]


[imath]\displaystyle \frac{dR_T}{R_T} = R_T\left(\frac{1}{R_1^2}dR_1 + \frac{1}{R_2^2}dR_2\right) = R_T\left(\frac{1}{R_1}\frac{dR_1}{R_1} + \frac{1}{R_2}\frac{dR_2}{R_2}\right)[/imath]

Which is another way to derive the percentage error of the equivalence resistance.