Maths Boffins Required!!!!

[Marky Mark88]

Hiya Folks, i have a question to do for an HNC Assignment which is proving to be a bit of a pain, any help would be greatly appreciated;

"The energy development in a resistive circuit is given by:

W = 6IR^2

Determine the approximate change in the energy developed if I is reduced by 2.5% and R is increased by 3.5% using the Binomial Theorem"

Using the Binomial Theorem, i have got a value for I of 0.975 and a value for R of 1.071225.

I then have no idea where to go from there!

Cheers,

Mark

[BoRED S2upid]

45.

My GF's 2 year old always says that when its a number question.

[bomb]

Using the Binomial Theorem, the answer is 'MARMALADE'.

[JJCW]

what were the initial values of I and R?

use the W = formula with initial values, then with new ones - sub one from the other?

[Marky Mark88]

No initial values have been given, i would assume they are just looking for a percentage change.

[s2art]

They probably want you to do something like (R + .035) squared. So you can ignore the contribution from the 0.035 squared bit.

[bluevelvet]

bluevelvet said:

If R = 1.071225,, doesn't that make W = 4,269.93 ????

Sorry, thought that was 61 not 6 I

on that basis is it 39.27, using the values of I & R given raised to the power of 2 ?

[johnfm]

W = 6 x I x R^2

I is now 0.975I
R is now 1.035R

W = 6 x (.975)I x (1.035R)^2

= 6 x 0.975 x 1.035^2 x IxR^2

= 6 x 1.044444 x IR^2

= 6.2666 IR^2

using original values for I and R

so, W has icreased by 4.444%

Binomial theroem?? WTF?

Edited by johnfm on Tuesday 9th December 15:54

[JJCW]

Marky Mark88 said:

Hiya Folks, i have a question to do for an HNC Assignment which is proving to be a bit of a pain, any help would be greatly appreciated;

"The energy development in a resistive circuit is given by:

W = 6IR^2

Determine the approximate change in the energy developed if I is reduced by 2.5% and R is increased by 3.5% using the Binomial Theorem"

Using the Binomial Theorem, i have got a value for I of 0.975 and a value for R of 1.071225.

I then have no idea where to go from there!

Cheers,

Mark

Hmm i'm rusty on this stuff..

If W = 6IR^2 is a standard formula, and no initial conditions are given, i would've that that finding values for I and R would be impossible; unless you take the initial conditions are I=1 and R=1? Therefore meaning your answer will be algebraic, or you solve through using the assumed initial conditions?

(i could be miiiiles off tho!!!!!)

Edited by JJCW on Tuesday 9th December 15:56

[Marky Mark88]

Heres what i did to get the values:

Seeming as I is an order of 1 (^1):

(1-0.025)^1 = 1 + 1(-0.025)
= 1 - 0.025
= 0.975

For R (^2):

(1+0.035)^2 = 1 + 2 x 0.035 + 2(2-1)/2! (0.035^2)
= 1 + 0.07 + (0.035^2)
= 1.071225

I beleive these are the right values for the percentage change, i just dont understand what the 'Ultimate' answer is!

[Marky Mark88]

johnfm said:

W = 6 x I x R^2

I is now 0.975I
R is now 1.035R

W = 6 x (.975)I x (1.035R)^2

= 6 x 0.975 x 1.035^2 x IxR

= 6 x 1.044444 x IR^2

= 6.2666 IR^2

using original values for I and R

so, W has icreased by 4.444%

Binomial theroem?? WTF?

I think i might just stick something like that down, it seems pretty reasonable to me.

[Geoff82]

Marky Mark88 said:

johnfm said:

W = 6 x I x R^2

I is now 0.975I
R is now 1.035R

W = 6 x (.975)I x (1.035R)^2

= 6 x 0.975 x 1.035^2 x IxR

= 6 x 1.044444 x IR^2

= 6.2666 IR^2

using original values for I and R

so, W has icreased by 4.444%

Binomial theroem?? WTF?

I think i might just stick something like that down, it seems pretty reasonable to me.

While that is the correct answer it is not a binomial expansion of the original equation, so for full marks you would need to show that

((1+0.035)R)^2 when binomially expanded = (1)^2 + (2x1x0.035) + (0.035)^2

ETA then plug that in to get the numbers that johnfm gave you for the correct figure of 4.4% as well as showing binomial expasnion for I.

Edited by Geoff82 on Tuesday 9th December 16:03

[Marky Mark88]

Geoff82 said:

Marky Mark88 said:

johnfm said:

W = 6 x I x R^2

I is now 0.975I
R is now 1.035R

W = 6 x (.975)I x (1.035R)^2

= 6 x 0.975 x 1.035^2 x IxR

= 6 x 1.044444 x IR^2

= 6.2666 IR^2

using original values for I and R

so, W has icreased by 4.444%

Binomial theroem?? WTF?

I think i might just stick something like that down, it seems pretty reasonable to me.

While that is the correct answer it is not a binomial expansion of the original equation, so for full marks you would need to show that

((1+0.035)R)^2 when binomially expanded = (1)^2 + (2x1x0.035) + (0.035)^2

ETA then plug that in to get the numbers that johnfm gave you for the correct figure of 4.4% as well as showing binomial expasnion for I.

Edited by Geoff82 on Tuesday 9th December 16:03

How do you get the value of 4.4% when the numbers that come out of the 2 expansions are 0.975 and 1.071225 for I and R respectively?

[Xaero]

Marky Mark88 said:

Hiya Folks, i have a question to do for an HNC Assignment which is proving to be a bit of a pain, any help would be greatly appreciated;

"The energy development in a resistive circuit is given by:

W = 6IR^2

Determine the approximate change in the energy developed if I is reduced by 2.5% and R is increased by 3.5% using the Binomial Theorem"

Using the Binomial Theorem, i have got a value for I of 0.975 and a value for R of 1.071225.

I then have no idea where to go from there!

Cheers,

Mark

From what I can make out, if you are saying I= 0.975 and R= 1.041225 (how did you get these numbers? - I don't know that theorem)

Then substitute those numbers in the equation and add the question into the equation:

W = 6 x 0.975 x 1.041225^2 = 5.85 x 1.084149500625 = 6.092745785625

changedW = 6 x (0.975 x 0.975) x (1.041225^2 x 1.035) = 6 x 0.950625 x 1.22094733146875 = 6.40014783418648828125

So the difference is 6.40014783418648828125 - 6.092745785625 = an increase of 0.30740204856148828125

But I'm probably completely wrong

[johnfm]

0.975 x 1.071225 = 1.044444375

This is 1 + 0.044444375

.044444375 x 100 = 4.44%

So, the new value of W is 4.44% larger than the old value.

[Geoff82]

johnfm said:

0.975 x 1.071225 = 1.044444375

This is 1 + 0.044444375

.044444375 x 100 = 4.44%

So, the new value of W is 4.44% larger than the old value.

Exactly!

[Marky Mark88]

Thank you very much!

Now all i have to do is write that down and make it look like i know what im doing and im quids in!

Cheers guys!

[Geoff82]

No worries, I also do exams at £40 an hour.

[turbobloke]

From the wording, is this a question relating to an AC circuit? If the load is purely resistive i.e. no reactance then calculations resemble an equivalent DC circuit.

W = I t V (presumanly W = work, equivalent to Energy, but the symbol is confusing as W is also a unit of power, the watt)

but V = I R

so W = I t I R = t I^2 R

Looks like they're giving you t = 6 for some reason.

Is the initial equation really I R^2 rather than I^2 R?

Just checking, as I'm under the Night Nurse just now

[Spiritual_Beggar]

BoRED S2upid said:

45.

My GF's 2 year old always says that when its a number question.

No no....that's the meaning of life!! Or was it 47? I cant remember now