Question

derive:dR/dT = d/dT ((R1*R2)/(R2+R1))

Answer 1

\[\frac{ dR }{ dT } = \frac{ d }{ dT } \left( \frac{ R_1.R_2 }{ R_2 + R_1 } \right)\]

Answer 2

do i derive the num and denom separately?

Answer 3

You don't "Derive"; You "Differentiate". Now moving on...

Answer 4

@Rowa So what are you asked to do in this problem?

Answer 5

it's alot to explain, but it comes down to the equation i gave

Answer 6

I know but what are we supposed to do with the equation? @Rowa

Answer 7

just says, calculate dR/dT

Answer 8

and R=R_1*R_2/R_2+R_1

Answer 9

Oh ok. So Are \(\bf R_1 \) and \(\bf R_2\) functions or constants? @rowa

Answer 10

functions

Answer 11

What are they functions of? Like are they both functions of 'x' or what? @Rowa

Answer 12

of T

Answer 13

OOOOOOOOOOOH. So basically, this is what it is:\[\bf R(T)=\frac{ R_1(T) \frac{}{} R_2(T) }{ R_1(T)+R_2(T) }\]
@Rowa Correct?

Answer 14

guess so

Answer 15

Ok so then we basically use the Division rule to differentiate:\[\bf \frac{ dR }{ dT }=\frac{ (R_1 \frac{}{} R_2)'(R_1(T)+R_2(T))-(R_1(T)+R_2(T))'(R_1 \frac{}{} R_2)}{ (R_1(T)+R_2(T))^2 }\]

Answer 16

done, i was doing that while waiting

Answer 17

and why is it differentiating?

Answer 18

ya it takes a while to type out lol

Answer 19

@Rowa When we take the "Derivative of something like a function", we are differentiating the function not deriving it. This is a common misconception.

Answer 20

alright, it's just a way of saying it

Answer 21

@Rowa Have you simplified the top and bottom yet?

Answer 22

Leave the denominator for last, finish up with the top first.

Answer 23

i applied the product rule and addition rule

Answer 24

Ya exactly. The product rule and the addition rule and then you're done. You can simplify further after that but I believe there won't be much simplification in a situation like this. So after applying that, you should be fine, assuming you applied it correctly. @Rowa

Answer 25

maybe i should type the complete question, because i'm not understanding some steps written by my professor

Answer 26

somehow, R_1 dR_2/dT is zero..

Answer 27

I'm pretty sure that this is not good for me because I haven't seen the whole question so I can't really say anything. The best thing to do is to take a picture of the question and then attach it and post.

Answer 28

problem is, it's in a different language

Answer 29

i'll have to translate it

Answer 30

Where are you from lol

Answer 31

europe

Answer 32

.......Europe isn't the greatest answer..........

Answer 33

western europe?

Answer 34

isn't there a country you live in?

Answer 35

well yea, but is that important?

Answer 36

shall i translate?

Answer 37

ok just post the whole question translated. that would be best. and post the original question alongside as well.

Answer 38

2 resistors R_1 and R_2 are parallel to each other. The total R is 1/R = 1/R_1 + 1/R_2
R_1 is heated slowly, this causes an increase of the resistor given by : R_1(T) = R_0 (1+T-T_0/300) T is temperature, R_0 and T_0 are constants that givethe resistance and temperature of R_1 at time 0. The question: Give the instantaneous increase in R (in function of T), so : dR/dT

Answer 39

R2 is a constant then, not a function. Makes this problem simpler.

Answer 40

how is R_2 a constant

Answer 41

Only R1 is heated. R2's resistance is constant.

Answer 42

@Rowa Aren't you given what R2 is? Or is R2 defined as a function of T as well?

Answer 43

don't know what R2 is, i've typed everything it says here

Answer 44

@agent0smith R2 isn't the same. \[\bf \frac{ 1 }{ R }=\frac{ 1 }{ R_1 }+\frac{ 1 }{ R_2 }\]Since R1 is a function then 1/R2 varies as R1 changes hence R2 is not constant.

Answer 45

sure? in the steps, it says the derivative of R2 is zero

Answer 46

Then R2 is a constant!

Answer 47

R2 is a constant... A resistors resistance can't change all on it's own.

Answer 48

hmm, i see

Answer 49

@Rowa You gotta tell the whole story or we won't know otherwise lol.

Answer 50

@genius12 1/R varies as 1/R1 varies. 1/R2 is just a constant.

Answer 51

@agent0smith ya ya I realise lol.

Answer 52

Total resistance R is NOT constant.

Answer 53

alright, but the steps i took to differentiate are correct?

Answer 54

comparing them to the steps my prof. took, they look correct

Answer 55

Ok, now going back to what you said: \[\bf R=\frac{ R_1 \frac{}{} R_2 }{ R_1 + R_2 }\]Then the derivative is evaluated the same way as we did with the division rule except now wherever you see R1, replace it with the function R1 given in full form and consider R2 to be a constant and simplify from there on.

@Rowa

Answer 56

\[\frac{ R_2+R_1(R_1+R_2\frac{ dR_1 }{ dT }) -(R_1R_2) + \frac{ dR_1 }{ dT } }{ (R_2+R_1)^2 }\]

Answer 57

alright

Answer 58

@Rowa Scroll up to the division rule derivative that I gave you. Just replace R1 with the full form of the function given and consider R2, R0, T0 as constants. The only variable will be T. And simplify from there.

Answer 59

is what i wrote incorrect?

Answer 60

@Rowa Yes.

Answer 61

forgot to remove the R_1 there

Answer 62

\[\frac{ R_2+R_1(R_2\frac{ dR_1 }{ dT }) -(R_1R_2) + \frac{ dR_1 }{ dT } }{ (R_2+R_1)^2 }\]

Answer 63

@Rowa Now that we know that R2 is a constant, it should be:\[\bf \frac{ dR }{ dt }=\frac{ \left( R_2\frac{ dR_1 }{ dT }\right)(R_1(T)+R_2)-\left( \frac{ dR_1 }{ dT } \right)(R_1(T) \frac{}{} R_2) }{ (R_1(T)+R_2)^2 }\]

Answer 64

put a plus there by accident, but other than that..my equation is good..?

Answer 65

\[\frac{ R_2+R_1(R_2\frac{ dR_1 }{ dT }) -(R_1R_2) \frac{ dR_1 }{ dT } }{ (R_2+R_1)^2 }\]

Answer 66

Yup that's good.

Answer 67

alright

Answer 68

now to replace R1 with R_0 (1+T-T_0/3000)

Answer 69

Now, is R1 this:\[\bf R_1(T)=R_0\left( 1+T-\frac{ T_0 }{ 300 } \right)\]?
@Rowa

Answer 70

\[\bf R_1(T)=R_0\left( 1+\frac{ T-T_0 }{ 3000 } \right)\]

Answer 71

Then:\[\bf \frac{ dR_1 }{ dT }=\frac{ R_0 }{ 3000 }\]Since \(\bf R_0, T_0\) are constants. @Rowa

Answer 72

So now you can replace dR/dt with R0/3000 and replace R1 with the full form. If simplification is possible, simplify and you're done.

Answer 73

i see, so: \[\frac{ dR_0 }{ dT } +\frac{ d }{ dT }(\frac{ R_0T-T_0 }{ 3000})\]

Answer 74

@Rowa I just realised the R2 + R1 should be in brackets in the numerator of your derivative.

Answer 75

dR_0 is zero, T_0 is zero, leaves me with R_0T/3000 = R_0/3000

Answer 76

oh yes, forgot

Answer 77

So dR_1/dT = R_0/3000

Answer 78

just writing the steps i take, to see if its good

Answer 79

Can you carry on from here or do you still need me?

Answer 80

i think i can, thanks for helping!

Answer 81

the answer should be \[\frac{ R_2^2R_0 }{ 3000(R_1+R_2)^2 }\]

Answer 82

but i get \[\frac{ R_2^2\frac{ R_0 }{ 3000 } }{ (R_1+R_2)^2 }\]

Answer 83

@genius12

Answer 84

@genius12 could you verify?

Answer 85

@rowa
Both answers are the same thing. You see this when you write your answer like this:\[\bf \frac{ R_2^2 \frac{ R_0 }{ 3000 } }{ (R_1+R_2)^2 }=\frac{ 1 }{ (R_!+R_2)^2 } \frac{}{} R_2^2\frac{ R_0 }{ 3000 }=\frac{ R_2^2R_0 }{ 3000(R_1+R_2)^2 }\]So you see, they're basically the same answer written in differently. Hope that helps.