Death and Decay question (1 Viewer)

Grey Council

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Growth and Decay question

Question:
In work on electricity, the formula
L (di/dt) + Ri = E
occurs, where i, t are variables and L, R, E are constants.

Show that i = E(1 - e^(-Rt/L)/R satisfies the equation and find limiting value of i as t--> infinite.

My problem:
I've got
i = E/R + A*e^(-Rt/L)
How is A = E/R?

Worded very trickily, isn't it? It's a question that involves:
dN/dt = k(N - n1) ____ where k and n1 are constants.

The formula:
N - B*e^(kt)
etc, don't get scared by the looks. Give it a try.
 
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Grey Council

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i = E(1 - e^(-Rt/L)/R
is

i = E/R * (1 - e^(-Rt/L))

hopefully that makes it easier

A is supposed to be a constant. I can't get it to equal to E/R. :S
 

BlackJack

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Isn't it easier just to show that the given equation satisfies by differentiating it and obtaining the same conservation of energy?

That is a good enough test. :p
 

Grey Council

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hrm, not sure if we are allowed to do that blackjack

abdoo, this is part of "applications of calculus to the physical world" so maybe you haven't done it. hehe, don't worry about it. I thought as you'd finished the 4u course you'd have done the 3u one. heheh never mind. i hafta go. Can someone just do the question?
 

evilc

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i messed about with it using differential equations, i will have a proper go soon..are there any initial conditions in the question GuardiaN? such as i = 0 etc
 

BlackJack

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Okay, 'ere's my integration way:

Denote S as integration symbol.
We'll start with
L (di/dt) + Ri = E.
--
di/dt = (E-Ri)/L
Therefore, dt/di = L/(E-Ri).

We integrate w.r.t. i. This is a natural log with chain rules.
Hence,
S dt = S L/(E-Ri) di
t + C = -L/R * S -R/(E-Ri) di
= -L/R * ln(E-Ri)
rearrange,
ln(E-Ri) = -R(t+C)/L
E-Ri = e<sup>-R(t+C)/L</sup> = Ae<sup>-Rt/L #</sup>

initial conditions will be i=0, t=0, as we just opened the switch in the circuit in this LR circuit. (physics stuff, I assume this is what the question entails.) <sup>+</sup>
Hence,
E - 0 = A(1)
A=E (this A is sl. different from the A in Guardian's post)

E - Ri = Ee<sup>-Rt/L</sup>
i = E(1-e<sup>-Rt/L</sup>) / R as required.

<sup>#</sup> - consider this:
e<sup>x+c</sup> = e<sup>x</sup>*e<sup>c</sup>, and the latter part will equal a constant at all times. Similarly if we multiply by -R/L it will still be a constant.
<sup>+</sup> - you can find out the initial conditions by subbing t=0 in the equation you want to show i.e. the one written in the question, E(1-e<sup>-Rt/L</sup>) / R, and you'll get a corresponding i value. btw, GuardiaN, your A should be equal to -E/R once done.

Now, as t increases indefinitely, e<sup>-Rt/L</sup> decreases towards zero.
.'. i -> E( 1- (0) ) / R -> E/R
 
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BlackJack

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With all honesty, my opinion is that you are allowed the differentiation method (a.k.a. assume/guess equation given is true, deduce that it is true) in high school.

They can't really argue with it if you show by differentiation that the equations are the same.

btw, this is the differentiating solution:

We will show that our equation is a satisfying equation because these calculus maths are reversible. :D

i = E(1 - e<sup>-Rt/L</sup>)/R
= E/R - E/R * e<sup>-Rt/L</sup>

di/dt = -E/R * -R/L * e<sup>-Rt/L</sup>
= E/L * e<sup>-Rt/L</sup>

Also, rearranging i = E(1 - e<sup>-Rt/L</sup>)/R gives:
e<sup>-Rt/L</sup> = 1 - Ri/E
Substitute into main equation:

.'. di/dt = E/L * (1 - Ri/E)
L(di/dt) = E - Ri
L(di/dt) + Ri = E

Since that is the satisfied equation... etc.

Continue as in last post for infinity-ness.
 
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Grey Council

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initial conditions will be i=0, t=0, as we just opened the switch in the circuit in this LR circuit. (physics stuff, I assume this is what the question entails.) +
you can find out the initial conditions by subbing t=0 in the equation you want to show i.e. the one written in the question, E(1-e-Rt/L) / R, and you'll get a corresponding i value. btw, GuardiaN, your A should be equal to -E/R once done.
LOL! Thats what i was asking! How do i get A to equal to -E/R! heheh, thanks, but i didn't know that we have just opened a switch in this LR circuit (whatever the heck that means ;) )

hehe, its clear now. As soon as you said i=0 when t=0, it was solved. :D

btw, That was the full question. THere is no extra information. What Blackjack did the first time is the correct way to do it. Not his second method, heck, they tell us to prove something true and you use the answer and work backwards? Sorry, you live life forwards, not backwards. :p hehe, thanks blackjack.
 

sylvia87

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hey ! i am doing extension maths.. but i am just doing alrgiht! and i mean JUST! but like i want to keep it becoz i dont want to have exactly 10 units u noe!

but as i said i am not doing as good! as other ppl!
will it affect my 2unit maths mark becoz i did extension?.. when in the uai it will count all my other subjects.
 

Grey Council

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whaa? Is this the place to post that?

no, your extension maths mark does not in anyway affect your two unit maths mark. However, if you put the effort in (not all that much) you can easily do well in maths. If you do your homework (only homework, no more, and then revise before exams) you should get a fairly decent mark.
 

Grey Council

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lol, go victorling. :D

the undead guy uses growth and decay. lol, believe me when i say I made the title of the post accidently. ^_^
 
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CM_Tutor

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Originally posted by BlackJack
Isn't it easier just to show that the given equation satisfies by differentiating it and obtaining the same conservation of energy?

That is a good enough test. :p
Originally posted by Grey Council
hrm, not sure if we are allowed to do that blackjack
Sorry Grey Council, but BlackJack is definitely right on this one. The question says:
Originally posted by Grey Council
In work on electricity, the formula
L (di/dt) + Ri = E
occurs, where i, t are variables and L, R, E are constants.

Show that i = E(1 - e^(-Rt/L)/R satisfies the equation and find limiting value of i as t--> infinite.
In other words, you must show that i = E(1 - e<sup>-Rt/L</sup>) / R satisfies the differential equation L * di/dt + Ri = E. This can only be done by a differential method. The other solutions offered here actually solve the differential equation, and establish that not only is i = E(1 - e<sup>-Rt/L</sup>) / R a solution, it is the only solution. This is acceptable as you have effectively answered a harder question.
 

Grey Council

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o

its been around 50 days since i did that question. lol, can't remember at all what it was about. heh, i guess i should revise, huh?

but I see what you mean. Although the reason I thought that is the book was kinda teaching the method that Blackjack used originally. i think.

btw, do you have harder growth and decay questions? ^_^ it'll be good revision for me, haven't done that type of stuff for ages. mostly forgotten it. hrm
 

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