C4 (Not MEI) - Thursday June 21 2012, PM

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Original post by Killjoy-

No, I did FP3 and that was bad (for me anyway).
Need 95%+ in S2 if I want an A* in FM.

FP2 was worse, I'd put money on it :tongue:

:tongue:

Beresford George

What is the hardest C4 paper to date, based on grade boundaries for an A.

Or what is your hardest C4 paper generally?

anyone has the jan 2012 c4 paper and markscheme?

Beresford George

Original post by jamieward

anyone has the jan 2012 c4 paper and markscheme?

Just type it in to google and it comes up.

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Original post by Luppy021

Expand, and you should get 1+2sinx+sin^2x, use the half angle formula for sin^2x and proceed as normal, sorry if this was a bit brief

yeah i realised i should have used that after i posted, cheers tho

cant find it :frown:

:frown:

Original post by jamieward

cant find it :frown:

:frown:

someone posted it on another thread, here's the link :smile:

:smile:

http://www.thestudentroom.co.uk/attachment.php?attachmentid=149944&d=1337609482

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thanks, do u have the question paper to go with it?

Original post by jamieward

thanks, do u have the question paper to go with it?

Ill repost them both here for convenience :smile:

:smile:

The thread with lots of OCR Jan 2012 papers on (including S2!!)
http://www.thestudentroom.co.uk/showthread.php?t=2003667

4724-01Jan12.pdf275.0KB

4724_MS_Jan12.pdf268.3KB

Just need help to figure out question 8, it gets a little bit too messy for me :tongue:

:tongue:

uploadfromtaptalk1340054809347.jpg68.0KB

Hi,
really struggling with the differential equations with the natural occurace, like the ones with temperature and things. I dont really know how to solve them and they're always big markers, any help?

Original post by tashbolton

Hi,
really struggling with the differential equations with the natural occurace, like the ones with temperature and things. I dont really know how to solve them and they're always big markers, any help?

An example question would help

Original post by Doctor.

An example question would help

Newton’s law of cooling states that the rate at which the temperature of an object is falling at any
instant is proportional to the difference between the temperature of the object and the temperature of
its surroundings at that instant. A container of hot liquid is placed in a room which has a constant
temperature of 20 ◦C. At time t minutes later, the temperature of the liquid is θ ◦C.
(i) Explain how the information above leads to the differential equation
dθ
dt
= −k(θ − 20),
where k is a positive constant. [2]
(ii) The liquid is initially at a temperature of 100 ◦C. It takes 5 minutes for the liquid to cool from
100 ◦C to 68◦C. Show that
θ = 20 + 80e−(1
5 ln 5
3)t. [8]

Original post by Doctor.

Just need help to figure out question 8, it gets a little bit too messy for me :tongue:

:tongue:

I don't know which page in the book that is from, could you tell me the answer so I can see whether it is correct before posting my working?

(edited 11 years ago)

Original post by wibletg

FP2 was worse, I'd put money on it :tongue:

:tongue:

Grade boundaries will adjust accordingly I suppose.
I just hope S2 is decent. :s-smilie:

:s-smilie:

Original post by Killjoy-

I don't know which page in teh book that is from, could you tell me the answer so I can see whether it is correct before posting my working?

Page 238, 0.5(tan^-1(0.5e^x))+k

Original post by Doctor.

Page 238, 0.5(tan^-1(0.5e^x))+k

Great

:smile:

I'll post my working soon.

Original post by Killjoy-

Great

:smile:

I'll post my working soon.

Aw thank you :smile:

:smile:

it looked so long winded lol :tongue:

:tongue:

Just had to give up in the end :facepalm:

:facepalm:

Original post by Doctor.

Aw thank you :smile:

:smile:

it looked so long winded lol :tongue:

:tongue:

Just had to give up in the end :facepalm:

:facepalm:

It's the right way round now :P I forgot k BTW.

(edited 11 years ago)

WP_000074.jpg494.7KB

Original post by tashbolton

Newton’s law of cooling states that the rate at which the temperature of an object is falling at any
instant is proportional to the difference between the temperature of the object and the temperature of
its surroundings at that instant. A container of hot liquid is placed in a room which has a constant
temperature of 20 ◦C. At time t minutes later, the temperature of the liquid is θ ◦C.
(i) Explain how the information above leads to the differential equation
dθ
dt
= −k(θ − 20),
where k is a positive constant. [2]
(ii) The liquid is initially at a temperature of 100 ◦C. It takes 5 minutes for the liquid to cool from
100 ◦C to 68◦C. Show that
θ = 20 + 80e−(1
5 ln 5
3)t. [8]

Gradient of a curve of x against t would give the rate of change of the quantity x with respect to time. Now the symbol for this would be dx/dt, and you know that.

There is a reason for the negative sign, the question says the rate at which the temperature falls is prop. to the temperature difference at that instant, but

dtheta/dt would be the rate of change of temp of the object.

So we may write

-\dfrac{d \theta}{dt} \alpha (\theta -20)

or

\dfrac{d \theta}{dt} \alpha -(\theta -20)

So

\dfrac{d \theta}{dt} -k(\theta -20)

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